Pharmaceutical Calculations: A Conceptual Approach 3030203344, 9783030203344, 3030203352, 9783030203351

Table of contents :
Preface......Page 5
Dedication......Page 9
How to Use This Book......Page 10
Contents......Page 11
Chapter 1: Measurement......Page 16
1.1 Accuracy and Precision of Measurements......Page 17
1.2 Principles of Weighing Balances and Scales......Page 19
1.3 Sensitivity Requirement (SR) of a Balance, Error, and Percentage Error......Page 21
1.4 Formulas......Page 22
1.5 Practice Examples......Page 24
Exercises......Page 27
Additional Exercises......Page 30
Chapter 2: Error Assessment of Drug Dose in Pharmaceutical Mixtures......Page 34
2.1 A Method to Determine Uncertainties of Drug Concentration in Pharmaceutical Mixtures (Savva 2006, 2016)......Page 36
2.2 Applications of the Method......Page 38
2.3 The Aliquot Method of Weighing (Savva 2015)......Page 44
2.4 Calculating the Error Associated with the Drug Dose When Two-Component Formulations Are Measured on Balances of Different .........Page 56
2.5 The Method of Geometric Dilution......Page 60
2.6 Concluding Remarks......Page 61
Exercises......Page 62
Additional Exercises......Page 65
References......Page 68
Chapter 3: Density and Specific Gravity......Page 69
3.2 Specific Gravity of Liquids and Solids......Page 70
3.3 Application of Density and Specific Gravity......Page 71
Exercises......Page 73
Chapter 4: Units of Concentration and the Salt Factor......Page 75
4.1.2 Solvent......Page 76
4.2.2 Percent by Weight (% w/w)......Page 77
4.2.5 Molarity (M)......Page 78
4.2.7 Normality (N)......Page 79
4.2.8 Ratio Strength......Page 80
4.2.11 Mole Fraction (x)......Page 81
4.3 Practice Examples......Page 82
4.4 Salt Factor, S......Page 93
Exercises......Page 95
Additional Exercises......Page 100
Chapter 5: Dilution and Concentration of Pharmaceutical Solutions and Other Physical Mixtures......Page 107
5.1.1 Mass Balance Equation......Page 108
5.1.2 Multiple Serial Dilutions and the Dilution Factor (DF)......Page 113
5.2 Concentrating Solutions......Page 118
Exercises......Page 122
Additional Exercises......Page 126
Chapter 6: Mixture Problems in Pharmaceutical Sciences......Page 127
6.1 General Mixture Problems......Page 129
6.2 Mixture Composition Problems: Extended Form of the Mass Balance Equation......Page 135
6.3 A Conceptual Problem-Solving Approach for Higher Than Two-Component Mixture Composition Problems......Page 138
6.3.1 Three-Component Mixtures......Page 139
6.3.1.1 Extracting the Whole Set of Ordered Triple Solutions......Page 143
6.3.2.1 Four-Component Mixtures in Which the Concentration of the Three Principal Solutions Lie Below (or Above) the Desired C.........Page 150
6.3.2.2 Four-Component Mixtures in Which Two of the Principal Solutions Have a Concentration Below (or Above) the Desired Conc.........Page 154
6.4.1 Alligation Medial......Page 158
6.4.2 Alligation Alternate......Page 159
6.4.2.1 Alligation Alternate for Higher Than Two-Component Mixtures......Page 161
Exercises......Page 164
Additional Exercises......Page 167
References......Page 169
Chapter 7: Isotonic Solutions......Page 170
7.1 The Importance of Isotonic Solutions in Pharmacotherapy......Page 171
7.2.1 Osmolarity and Van´t Hoff Dissociation Factor......Page 173
7.3.1 The Sodium Chloride Equivalent Method......Page 178
7.3.2 The Freezing Point Depression Method......Page 182
7.4 Adjusting the Osmotic Pressure of Hypertonic Solutions......Page 183
Exercises......Page 185
Additional Exercises......Page 191
Chapter 8: Diffusion......Page 194
8.1 Drug Dissolution......Page 195
8.2 Transmembrane Diffusion......Page 198
8.3 A Closer Look at the Osmotic Pressure......Page 207
Exercises......Page 211
Reference......Page 221
Chapter 9: Use of Prefabricated Dosage Forms in Extemporaneous Compounding......Page 222
9.1 Tablets and Capsules......Page 223
Exercises......Page 228
Chapter 10: Fluids and Electrolytes......Page 232
10.2 Composition of Body Fluids and Fluid Regulation in Body Compartments......Page 234
10.3 Conditions and Diseases That Cause Abnormal Fluid Shifts......Page 236
10.4 Rules of Physics That Could Prove Useful in Clinical Intervention......Page 237
10.5 Maintenance and Replacement Fluids......Page 243
10.6 Osmotic Diuretics......Page 245
10.7 The Gram-Equivalent Weight (EW)......Page 248
10.7.1 Converting Mass Expressed in Grams (g) into Equivalents (Eq) and Vice Versa......Page 249
10.7.2 Calculating Equivalent of Ions in Solutions That Are Made by Multiple Common Ion Salts......Page 253
Exercises......Page 255
Additional Exercises......Page 262
Chapter 11: Essential Mathematics for Pharmacokinetics......Page 263
11.2 Multiplication and Division of Fractions......Page 264
11.3 Exponents......Page 265
11.4 Algebraic Expressions: Solving Equations......Page 266
11.6 Properties of Exponents......Page 267
11.7 Solving Systems of Equations......Page 268
11.8 Graphing Equations of the Form y = mx + b......Page 269
11.9 Logarithms......Page 272
11.11 Solving Exponential and Logarithmic Equations......Page 273
11.12 Graphing Special Exponential Functions......Page 274
Exercises......Page 278
Chapter 12: Rates and Orders of Reactions......Page 281
12.2 Rate Law, Rate Constant, and Order of Reaction......Page 282
12.3 Zero-Order Reactions......Page 284
12.3.1 Calculating the Half-Life in Zero-Order Excretion Processes......Page 285
12.4 First-Order Reactions......Page 287
12.4.1 Calculating the Half-Life in First-Order Kinetic Processes......Page 289
12.5 Shelf-Life of Drug Products (t0.9)......Page 291
12.6 The Importance of the Leading Coefficients of Special Exponential Functions and the Meaning of the Zero- and First-Order .........Page 292
12.7 Horizontal and Vertical Shifts of Exponential Functions......Page 296
Exercises......Page 303
Chapter 13: Fundamental Concepts of Dosage Calculations......Page 312
13.1 Understanding the Drug-Mediated Therapeutic Effect......Page 313
13.2.2 Drug Distribution......Page 316
13.2.3 Drug Metabolism......Page 317
13.3 The Concepts of Bioavailability, Effective Dose, and Dose Conversions Between Dosage Forms......Page 318
13.4 One-Compartment Pharmacokinetic Model......Page 320
13.5 Drug Apparent Volume of Distribution (Vd)......Page 322
13.6 Drug Clearance (CL)......Page 325
13.7 Accumulation and Fluctuation of Drug in the Plasma......Page 327
13.8 Loading (DL) and Maintenance Dose (DM)......Page 329
Exercises......Page 332
Chapter 14: Dosage Calculations Based on Body Surface Area (BSA)......Page 339
14.1 Fundamental Equation for Surface Area Determination of Cylindrical Objects......Page 341
14.2 Formulas for Body Surface Area......Page 343
14.3 Derivation of BSA Formulas......Page 344
14.4 BSA or BW?......Page 345
14.6 Advantages and Disadvantages of BSA Approaches in Dosage Calculations......Page 347
Exercises......Page 350
References......Page 352
Chapter 15: Intravenous Infusion and Flow Rate......Page 353
15.1 IV Infusion Accessories......Page 354
Exercises......Page 359
Appendix A. Interpretation of Prescriptions and Medication Orders......Page 363
Appendix B. Miscellaneous Abbreviations Commonly Used in Pharmacy......Page 365
Appendix C. Units of Measurement and Equivalent Conversions......Page 366
Appendix E. Specific Gravity of Miscellaneous Liquids......Page 367
Appendix F. Table of Atomic Masses......Page 368
Chapter 1......Page 370
Chapter 8......Page 371
Chapter 13......Page 372
Chapter 14......Page 373
Index......Page 374

Citation preview

Michalakis Savva

Pharmaceutical Calculations A Conceptual Approach

Pharmaceutical Calculations

Michalakis Savva

Pharmaceutical Calculations A Conceptual Approach

Michalakis Savva School of Pharmacy South University Savannah, GA, USA

ISBN 978-3-030-20334-4 ISBN 978-3-030-20335-1 https://doi.org/10.1007/978-3-030-20335-1

(eBook)

© Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Cover Caption: The journey of a drug after its synthesis ‘till its biological activity. Chemical structures (left inset) dictate compounding calculations (middle inset), compounded drugs dictate calculations related to dosage regimen design (inset to the right), and drug released in plasma (blue spheres) from dosage forms bind selectively to receptors (drawn on cell membrane) to elicit specific pharmacological responses. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Pharmacists counsel patients on medication use, side effects and drug interactions, design dosage regimens for specific drugs and extemporaneously compound medications to meet unique needs of individual patients. All of those responsibilities require knowledge of Chemistry/Physical Chemistry, Biology, and Mathematics.

Hierarchy of skills in pharmacy education

As per the sketch above, pharmacists should be able to read chemical structures and classify drugs in different therapeutic classes that have the same mechanism of action, similar side effects, and similar drug interactions. They should also be able to look at a chemical structure of a drug and tell many of its physicochemical properties that can be utilized in compounding. Some of these properties will be used for excipient selection, compounding calculations (pharmaceutics), and dosage regimen design (pharmacokinetics). Biology, Chemistry, and Mathematics are all very important subjects in Pharmacy education. v

vi

Preface

The whole journey of a drug after its synthesis ‘till its biological activity is depicted in the front cover of the book. The sketches in the three bubbles emphasize, starting from the left, that chemical structures will dictate calculations (middle bubble) related to drug compounding into dosage forms. Compounded drugs will in turn dictate calculations related to dosage regimen design (bubble to the right). Administration of the finished products according to the calculated dose and dosing interval will result in drug release in the plasma (blue spheres), its diffusion into the cells of the targeted tissues and its selective binding to receptors (three different receptors are drawn on cell membrane) to elicit the so desired specific pharmacological response. Each receptor will generate a different biological signal. The book is written with a purpose to describe pharmaceutical calculations conceptually as they apply to the different parts of the drug’s journey after its chemical synthesis, into our body. Descriptions of key processes like drug solubility, dissolution, and diffusion at the molecular level, as they are related to the drug’s chemical structure, will help you understand and carry out calculations involved in these and subsequent processes like drug absorption and tissue distribution. At the time this book was written, deficiencies in pharmacy students’ ability to perform pharmaceutical calculations was high, worldwide, commonly blamed on student’s poor algebra skills. This view is far from being accurate. Undeniably most of the pharmacy students don’t have advanced mathematical education, but all of them can carry out basic algebra operations like addition and multiplication. In my experience, there seems to be a very small percentage that have forgotten how to perform division of fractions and a small percentage of them that cannot reorganize equations and solve for a variable. These simple issues can be addressed effectively after few days of practice, but cannot be overlooked. A relatively large portion of pharmacy students cannot handle exponential/logarithmic transformations involved in pharmacokinetics later in their studies. It is recommended to spend a lecture or two describing and practicing logarithmic properties so that they can at least handle transformations related to the formalized simple first-order exponential equations (see Chap. 11). I have tried it and it makes a huge difference. A large number of the pharmacy students avoided practicing mathematics during their high school and undergraduate studies because they found it difficult to relate to, as mathematics is a very abstract subject. That should no longer be an issue in pharmaceutical calculations if the subject is taught conceptually. Teaching excessive number of topics with very low complexity problems that only emphasize repetition should be avoided, as that would leave students with shallow knowledge and understanding. Pharmacy students need to learn how to do things rather than “learning” for exam purposes. The topic of pharmaceutical calculations must be taught conceptually because all the problems require making connections to Chemistry, Biology, Medicine, and Physics. For example, some students have trouble converting % w/w into % w/v. When was the last time that the difference between drug density as compared to drug concentration was taught? Have we ever told them that the density given is the solution’s density and not the drug’s density, and therefore they have to apply it to the solution which is the denominator? Is the chapter of dilutions taught using the concept of mass balance preservation? Do they

Preface

vii

know that one can perform dilutions by keeping the mass of the drug constant (increasing the volume of the solution) or by reducing the mass of the drug (keeping the volume constant)? If these concepts are not taught, how do you expect them to carry out calculations when drug is added to or removed from a solution or a solid physical mixture? When the Aliquot method of weighing is taught, do they learn to calculate the error in the final dosage form or we just demonstrate the method for a single drug measurement? When the method of alligation alternate is described, how did you rationalize subtracting concentrations and setting them equal to parts or volume of mixtures? When you teach isotonic solutions, do you explain the difference between osmotic pressure, osmolarity, and tonicity? How do we expect the students to understand the concepts of absorption and distribution as they relate to the concentration gradient on the two sides of a membrane, as the driving force, if we have never covered the chapter of diffusion? How do we expect the students to calculate dosages using the BSA if they don’t understand how BSA equations were generated, what their limitations are, how is BSA different from body weight method (BW) and how pharmacokinetics, as the only scientific method that calculates dosages, is related to these methods? Some pharmacy schools, in their attempt to develop competence in performing pharmaceutical calculations, have integrated pharmaceutical calculation exercises into other subjects across the curriculum. Although, frequent review of the topics is necessary, it may not be very efficient, unless pharmaceutical calculations is recognized as a separate subject that requires educated instructors to teach these important principles and concepts conceptually. You need to bring and study the concepts of all other subjects into the course of Pharmaceutical Calculations, in the beginning, from the calculations point of view, not the other way around. This book is written for the purpose of providing you with the necessary skills to carry out accurate compounding and dosage regimen calculations. Descriptions and activities related to the most important foundational pharmaceutical concepts are included so that you become a better scientist early on in your studies. You cannot conceptually calculate the error associated with a dose measurement unless you first understand the concept of accuracy versus precision in a measurement. Similarly, you cannot fully understand the concept of drug absorption and distribution to tissues unless you have a solid understanding of the process of transmembrane diffusion. Early understanding of these concepts will allow reinforcement and deeper understanding of other related concepts taught in other courses. The book places more weight on the qualitative understanding of concepts before considering other quantitative problems. It demonstrates how to approach and solve pharmaceutical and clinical problems using a conceptual approach. You have to be able to diagnose a problem correctly before attempting to solve it. To this end, every chapter contains conceptual questions. The origin and physical significance of all final forms of critical equations is always described in detail, thus, allowing you to recognize the real application and limitations of an equation. Unless you know how a particular equation is derived and what are the assumptions and approximations made for its derivation, you can’t really be aware of its limitations and full potential. Specific strategies are explained step by step in more than 100 practice examples.

viii

Preface

Practice exercises are interesting, challenging and they aim to probe and improve students’ understanding in many different ways. Many of them are written as challenging multiple-choice questions with successful distractors aiming to increase content knowledge and understanding or to assess your ability to analyze and evaluate relevant information. A connection between basic sciences, pharmaceutical sciences, and pharmacy is frequently made in order to motivate you to study the book. Examples and exercises are taken from the fields of compounding pharmacy, pharmaceutics, pharmacokinetics, pharmacology, and medicine, and they are written for your learning and your understanding. In every chapter, the necessary background information with enough details is included so that it can be used as a tutorial book. You may read it all by yourself. Solutions to selected exercises can be found at www.springer.com. We would appreciate any feedback on general content errors or omissions ([email protected]) Savannah, GA, USA

Michalakis Savva

Dedication

Dedicated to my wife Maria Kovani and my two daughters Irene and Helen.

ix

How to Use This Book

If you are a PharmD and you are interested only in compounding calculations, then you may only study Chap. 1, Sect. 2.3; Chaps. 3–5, Sects. 6.2 and 6.4 and exercises 6.6–6.18; Chap. 7; Chaps. 9 and 10; and Chaps. 14 and 15. You don’t have to solve the additional exercises. Chapter 11 is mandatory only if you choose to study Chaps. 12 and 13 which are related to pharmacokinetics/clinical pharmacokinetics. For all other scientists, I encourage them to study all the chapters and to solve all the exercises, including the additional ones. The chapters were kept small where the theory is described within less than 30 pages. Practice examples serve the purpose of demonstration, but they should also be used for practice. You should hide the solution and try to solve it yourself. If you can’t solve it, go back and read the theory again. Proceed to the next section only if you have solved all the practice examples. The number of essential problems that you have to solve at the end of each chapter was also kept to a minimum so that you don’t get overwhelmed. Spending a couple of days in each chapter should be enough to finish it, receive your mental reward, feel good about your accomplishment, and move to the next chapter.

xi

Contents

1

2

3

Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Accuracy and Precision of Measurements . . . . . . . . . . . . . . . 1.2 Principles of Weighing Balances and Scales . . . . . . . . . . . . . 1.3 Sensitivity Requirement (SR) of a Balance, Error, and Percentage Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Practice Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . .

1 2 4

. . . . .

6 7 9 12 15

Error Assessment of Drug Dose in Pharmaceutical Mixtures . . . . 2.1 A Method to Determine Uncertainties of Drug Concentration in Pharmaceutical Mixtures (Savva 2006, 2016) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Applications of the Method . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Aliquot Method of Weighing (Savva 2015) . . . . . . . . . . 2.4 Calculating the Error Associated with the Drug Dose When Two-Component Formulations Are Measured on Balances of Different Sensitivity . . . . . . . . 2.5 The Method of Geometric Dilution . . . . . . . . . . . . . . . . . . . . 2.6 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

19

. . .

21 23 29

. . . . . .

41 45 46 47 50 53

Density and Specific Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Mathematical Definitions and Equations . . . . . . . . . . . . . . . . 3.2 Specific Gravity of Liquids and Solids . . . . . . . . . . . . . . . . . 3.3 Application of Density and Specific Gravity . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

55 56 56 57 59

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Contents

4

Units of Concentration and the Salt Factor . . . . . . . . . . . . . . . . . . 4.1 Solute, Solvent, and Solution . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Solute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Solvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Concentration Units in Pharmaceutical Sciences . . . . . . . . . . 4.2.1 Mass per Unit Volume . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Percent by Weight (% w/w) . . . . . . . . . . . . . . . . . . 4.2.3 Percent Weight per Volume (% w/v) . . . . . . . . . . . . 4.2.4 Percent by Volume (% v/v) . . . . . . . . . . . . . . . . . . . 4.2.5 Molarity (M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.6 Molality (m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.7 Normality (N) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.8 Ratio Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.9 Parts per Million (ppm) by Mass . . . . . . . . . . . . . . . 4.2.10 Parts per Billion (ppb) by Mass . . . . . . . . . . . . . . . . 4.2.11 Mole Fraction (x) . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.12 Activity Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Practice Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Salt Factor, S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

Dilution and Concentration of Pharmaceutical Solutions and Other Physical Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Dilution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Mass Balance Equation . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Multiple Serial Dilutions and the Dilution Factor (DF) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Concentrating Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

. . . . . . . . . . . . . . . . . . . . . .

61 62 62 62 63 63 63 63 64 64 64 65 65 66 67 67 67 68 68 79 81 86

. . .

93 94 94

. 99 . 104 . 108 . 112

Mixture Problems in Pharmaceutical Sciences . . . . . . . . . . . . . . . . 6.1 General Mixture Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Mixture Composition Problems: Extended Form of the Mass Balance Equation . . . . . . . . . . . . . . . . . . . . . . . . 6.3 A Conceptual Problem-Solving Approach for Higher Than Two-Component Mixture Composition Problems . . . . . . 6.3.1 Three-Component Mixtures . . . . . . . . . . . . . . . . . . . . 6.3.2 Four-Component Mixtures . . . . . . . . . . . . . . . . . . . . 6.4 Alligation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Alligation Medial . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Alligation Alternate . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

113 115 121 124 125 136 144 144 145 150

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xv

Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 7

Isotonic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The Importance of Isotonic Solutions in Pharmacotherapy . . . . 7.2 Preparing Isotonic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Osmolarity and Van’t Hoff Dissociation Factor . . . . . 7.3 Adjusting the Osmotic Pressure of Hypotonic Solutions . . . . . . 7.3.1 The Sodium Chloride Equivalent Method . . . . . . . . . . 7.3.2 The Freezing Point Depression Method . . . . . . . . . . . 7.4 Adjusting the Osmotic Pressure of Hypertonic Solutions . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

157 158 160 160 165 165 169 170 172 178

8

Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Drug Dissolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Transmembrane Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 A Closer Look at the Osmotic Pressure . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

181 182 185 194 198 208

9

Use of Prefabricated Dosage Forms in Extemporaneous Compounding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 9.1 Tablets and Capsules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

10

Fluids and Electrolytes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Body Fluid Compartments . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Composition of Body Fluids and Fluid Regulation in Body Compartments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Conditions and Diseases That Cause Abnormal Fluid Shifts . . 10.4 Rules of Physics That Could Prove Useful in Clinical Intervention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Maintenance and Replacement Fluids . . . . . . . . . . . . . . . . . . 10.6 Osmotic Diuretics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 The Gram-Equivalent Weight (EW) . . . . . . . . . . . . . . . . . . . 10.7.1 Converting Mass Expressed in Grams (g) into Equivalents (Eq) and Vice Versa . . . . . . . . . . . . . . . 10.7.2 Calculating Equivalent of Ions in Solutions That Are Made by Multiple Common Ion Salts . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

. . . . . .

. 219 . 221 . 221 . 223 . . . .

224 230 232 235

. 236 . 240 . 242 . 249

Essential Mathematics for Pharmacokinetics . . . . . . . . . . . . . . . . . . 11.1 Rounding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Multiplication and Division of Fractions . . . . . . . . . . . . . . . . . 11.3 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

251 252 252 253

xvi

12

13

Contents

11.4 Algebraic Expressions: Solving Equations . . . . . . . . . . . . . . . 11.5 Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Properties of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . 11.8 Graphing Equations of the Form y ¼ m∙x + b . . . . . . . . . . . . . 11.9 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.10 Properties of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.11 Solving Exponential and Logarithmic Equations . . . . . . . . . . . 11.12 Graphing Special Exponential Functions . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

254 255 255 256 257 260 261 261 262 266

Rates and Orders of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Rate of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Rate Law, Rate Constant, and Order of Reaction . . . . . . . . . . . 12.3 Zero-Order Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Calculating the Half-Life in Zero-Order Excretion Processes . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 First-Order Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4.1 Calculating the Half-Life in First-Order Kinetic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Shelf-Life of Drug Products (t0.9) . . . . . . . . . . . . . . . . . . . . . . 12.6 The Importance of the Leading Coefficients of Special Exponential Functions and the Meaning of the Zeroand First-Order Rate Constants . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Horizontal and Vertical Shifts of Exponential Functions . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

269 270 270 272

Fundamental Concepts of Dosage Calculations . . . . . . . . . . . . . . . 13.1 Understanding the Drug-Mediated Therapeutic Effect . . . . . . 13.2 The Process of Drug Absorption, Distribution, Metabolism, and Excretion (ADME) . . . . . . . . . . . . . . . . . . 13.2.1 Drug Absorption . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 Drug Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.3 Drug Metabolism . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.4 Drug Excretion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 The Concepts of Bioavailability, Effective Dose, and Dose Conversions Between Dosage Forms . . . . . . . . . . . 13.4 One-Compartment Pharmacokinetic Model . . . . . . . . . . . . . . 13.5 Drug Apparent Volume of Distribution (Vd) . . . . . . . . . . . . . 13.6 Drug Clearance (CL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 Accumulation and Fluctuation of Drug in the Plasma . . . . . . . 13.8 Loading (DL) and Maintenance Dose (DM) . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

273 275 277 279

280 284 291

. 301 . 302 . . . . .

305 305 305 306 307

. . . . . . .

307 309 311 314 316 318 321

Contents

14

15

Dosage Calculations Based on Body Surface Area (BSA) . . . . . . . . 14.1 Fundamental Equation for Surface Area Determination of Cylindrical Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Formulas for Body Surface Area . . . . . . . . . . . . . . . . . . . . . . 14.3 Derivation of BSA Formulas . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 BSA or BW? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Using BSA Formulas for Dosage Adjustment . . . . . . . . . . . . . 14.6 Advantages and Disadvantages of BSA Approaches in Dosage Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xvii

329 331 333 334 335 337 337 340 342

Intravenous Infusion and Flow Rate . . . . . . . . . . . . . . . . . . . . . . . . 343 15.1 IV Infusion Accessories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A. Interpretation of Prescriptions and Medication Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix B. Miscellaneous Abbreviations Commonly Used in Pharmacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix C. Units of Measurement and Equivalent Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix D. Selected Roman Numerals . . . . . . . . . . . . . . . . . . . . . . . Appendix E. Specific Gravity of Miscellaneous Liquids . . . . . . . . . . . . Appendix F. Table of Atomic Masses . . . . . . . . . . . . . . . . . . . . . . . . . Answers to Multiple-Choice Questions . . . . . . . . . . . . . . . . . . . . . . . .

353 353 355 356 357 357 358 360

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

Chapter 1

Measurement

Abstract Emphasis is placed on understanding the measurements, the operation of prescription balances, and the responsibility of the operator to carry out measurements as accurate as possible in order to avoid causing human casualties. The concept of measurement is explained first by an extensive description of accuracy versus precision of measured values. You can use the exact quantitative definition described in this chapter to classify measurements as accurate and/or precise, and you can further perfect your skills by solving the carefully selected multiple-choice, true or false, fill in the gaps, or computational problems. A connection to the pharmaceutical field is made through various examples that demonstrate that, as with all other applied sciences, appropriate units should accompany a meaningful measurement. It is also explained that measurements are obtained with laboratory instruments, and as such, they are always accompanied by an error. The magnitude of the error or uncertainty in a measurement is dependent not only on the analytical instrument but also on the skills and knowledge of the operator. The impact of accurate dose measurements in the livelihood of patients is also demonstrated, thus explicitly denoting that it is the responsibility of every scientist to be aware of measurement uncertainties. Keywords Measurement · Desired value · Actual value · Accuracy · Precision · Prescription balance · Sensitivity requirement · Error · Percentage error · Maximum allowable error · Least allowable weight

Learning Objectives After reading this chapter, you should be able to: • Understand the difference between a number from a measurement • Define the concepts of accuracy and precision and describe their differences • Remember the main operational principles of a prescription balance Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_1) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_1

1

2

1 Measurement

• Describe the concept of a sensitivity requirement (SR) of a balance • Calculate the error and % error associated with a measurement • Calculate the least allowable weight (LAW) of a drug that is measured at a maximum allowable error (MAE) and vice versa

1.1

Accuracy and Precision of Measurements

Measurement can be defined as the determination of properties of matter using scientific instruments that are configured to perform measurement in standard units. Due to the solid and liquid nature of most drugs and finished dosage forms, the two most common laboratory measurements are those of the mass and the volume. Electronic and mechanical balances routinely measure the mass or weight of a substance while the volume is measured by graduated cylinders, pipets, and syringes. For a measured quantity to be meaningful, the number must always be accompanied by a unit. For example, the description of a dose written as “Take atenolol, 50” is meaningless. It becomes a bit more informative when it is written as “Take atenolol, 50 mg.” What we also need to realize is that measurements are not exact. They are characterized by an uncertainty because the instruments have a limited sensitivity. Errors affect the reliability of the process, that is, the accuracy and precision of measurement. Accuracy refers to how close is the average value of a set of measurements or how close is a measurement to the true or desired value of the quantity that is measured. Precision refers to how close measurements (actual values) agree with each other (Fig. 1.1). We can designate a measurement as being accurate if its value is close to the true value
the balance sensitivity (SR). Similarly, we can designate a set of measurements as being accurate if their average or mean value is close to the true value of the sample. On the other hand, we can designate a measurement as being precise if its value is close to the mean value
the balance sensitivity (SR) or if the absolute value of the difference of the mean value of all measurements minus the measured value is less than or equal to the balance sensitivity, that is, |mean value  measured value|  SR, regardless of how close is the mean value to the true value. We can designate a set of measurements as being precise, regardless of how close is the average value of the measurements to the true value, if the difference of the higher from the lower value of the measurements is less than or equal () to twice the sensitivity of the instrument, that is, maximum measured value  minimum measured value  2 SR. Notice that accuracy is related to the true value of a measurement, whereas the concept of precision is related to the mean value (not to the true value) of all samples. Conclusively, a measurement is composed of a number, a unit, and an uncertainty. In this book, the term uncertainty is used equivalently to the probable error or just error. The inevitable uncertainties that accompany each measurement result in an average value that lies within a confidence interval. An experiment is described in Fig. 1.1 where four groups of students (II, IP, AI, and AP) were each given five tablets. They were asked to weigh them individually on a prescription balance (SR ¼ 6 mg), sum the masses, and divide by 5 to calculate the average tablet weight.

1.1 Accuracy and Precision of Measurements

3

Fig. 1.1 Inaccuracy is associated with the difference between the average value of a set of data from the true value, whereas imprecision is associated with the difference of each data set from each other. In this figure, the true mass of each sample is 100 mg. II: Inaccurate and imprecise measurements since the average value (dashed line) is very different (SR ¼ 6 mg) from each other. IP: Inaccurate ( 12 mg). AP: Accurate and precise measurements ((maximum value  minimum value) < 2∙SR ¼ 12 mg)

The true or real value of each tablet is 100 mg. Group II and group IP did badly since the average value was way off the true value of the tablets. Group II is both inaccurate and imprecise since the average value was determined to be 80 mg while the individual measurements were very different from each other. They definitely have to improve their weighing technique. The measurements conducted by group IP were precise since they were all close to each other but the average was 79 mg instead of 100 mg. These results suggest that their weighing skills are good, but perhaps the balance they used was not calibrated properly. These kind of errors are called systematic because they are always biased in one particular direction. The best measurement was carried out by the AP group. They found an average value equal to the true value of the tablet with a standard deviation of
4 mg from the average value. We say that a tablet contains 100 mg
4 mg of drug, meaning that the average drug content in each tablet is 100 mg with 4 mg being the maximum deviation of the measurement from the mean. Also, the difference between the highest from the lowest value is 104 – 96 ¼ 8 mg, which is less than twice the sensitivity requirement of the prescription balance (2SR ¼ 26 ¼ 12 mg). Essentially the amount of drug in all the tablets varies between 96 and 104 mg. One could easily discern why precision without accuracy is not acceptable. However, accuracy without precision is also not acceptable (see Example 1.3),

4

1 Measurement

regardless of whether the measurements are for preparative or analytical purpose. Measurements with errors much higher than the sensitivity requirement (SR) of the balance are not allowed. If the purpose was analytical, for example, verifying the true weight of a drug product in the market, just because it happened this time and the average fell close to the true value, it does not mean that you will get lucky all the time. Furthermore, drug products having a big variation in their weight are not acceptable because of the potential some of the dosage forms to contain drug way higher or way lower than the regular dose. Therefore, a measurement is considered acceptable if it is both accurate and precise. In general, any weighing measurement that deviates much higher than the SR of the balance signifies an improper technique or a faulty balance, and steps have to be taken to set them right. Errors in the measurements are a very big problem in all sciences. Advances made in basic sciences such as physics, chemistry, and biology are intimately related to the level of accuracy and precision with which experiments are carried out. Inaccuracies are often the cause of confusion and inability to draw conclusions from experiments. In applied science fields, they are more closely associated with unhappiness, pain, and in extreme cases human casualties. For example, inaccuracies in the design of car engines produce unreliable cars making a lot of drivers unhappy. Inaccuracies in the construction of roads, bridges, and buildings designed by civil engineers would also make a lot of drivers and homeowners unhappy. If the inaccuracies are very big, they could even cause loss of human lives. It is therefore of utmost importance to always be informed and try to carry out your work with excellence, so that you can achieve the smallest possible errors in the products and services you offer. Dispensing medication is among the primary responsibilities of pharmacists. To prevent accidents, you must calculate and measure the dose for a patient as accurately as possible. The accuracy in the calculation and measurement of drug dose eliminates the possibility of toxic effects due to overdose and ineffective treatment due to lower dose, respectively. To this end, I have included an extensive selection of solved examples and practice exercises, so that you can perfect your mathematical skills to accurately prepare compounded prescriptions.

1.2

Principles of Weighing Balances and Scales

The difference between a balance and a scale is that a balance has two pans, whereas a scale has only one pan. Typically, you place the unknown sample on one of the pans and you measure its weight (or mass) by adding known weights on the other pan until the two pans remain steady at equal height (lever is balanced). A scale, on the other hand, measures the weight of an object by measuring and restoring the deformation of a spring or a wire that was caused in proportion to the load placed on the pan. Drug molecules can exist in one of the three states, namely solid, liquid, or gas. Most commonly, they are supplied as solids, and in some cases, they are sold as liquid formulations. Accurate determination of the amounts of solids and the volume of liquids is required for precise execution of medication orders. Electronic scales and mechanical prescription balances are used for weighing drugs in powder form.

1.2 Principles of Weighing Balances and Scales

5

Fig. 1.2 Two different types of spring balances

To be more precise, balances and scales measure weight of matter and not mass. Mass refers to the amount of matter in a body, whereas weight is the force that gravity exerts on that body (F ¼ m  g). In this book, we consider the weight of a body to be its mass in units of milligrams, grams, or kilograms, although in physics they are strictly two different quantities. It is very important to know the sensitivity or the detection limit of a balance and the factors that govern accurate measurement of liquids. In order to become more familiar with the operating principles of balances, let us take a closer look of a spring balance (Fig. 1.2). The sensitivity of a spring balance is dependent on the elasticity of the spring accessory. Weights that fail to stretch (Fig. 1.2, left) or compress (Fig. 1.2, right) the spring significantly cannot be measured. On the other hand, weights large enough to cause permanent deformation or distortion of the spring also cannot be measured. In the first case, the weight is smaller than the lower weighing limit, whereas in the second case, the weight exceeds the upper limit of measurement of the device. Pharmacists in the United States of America have been traditionally using dualarm equal beam mechanical balances (prescription balances) although use of other kind of more sensitive electronic balances is increasing in recent days. Standard electronic balances have sensitivity requirement (SR) close to 5 mg while analytical balances have SR as small as 0.1 μg. The balances must meet all specific standards as set by the National Bureau of Standards at all times. The most common class A and class III mechanical prescription balances have a capacity of 60–120 g and a sensitivity requirement of 5–10 mg. In other words, weights smaller than 5–10 mg and weights greater than 60–120 g cannot be measured on these balances. A sketch showing a simple beam balance is drawn in Fig. 1.3, in order to better understand the principles of operation of prescription balances. Certainly, current

6

1 Measurement

(c)

Fig. 1.3 Simple beam balance: (a) beam, (b) pivot, (c) weights

(a)

(b)

prescription balances are not as simple as the one drawn above, but the main principle of weighing using this kind of balance is still the same. Assuming that there is no friction between the different parts of the balance, the sensitivity of a dualarm equal beam balance depends on the length of the beam and the total mass of the various parts of the scale. Since balances measure in essence weight and not mass, the importance of leveling the balance prior to any weighing cannot be overstated. This is true for electronic balances, as well. The angle between the pivot (the axis of the pin) and the beam should be 90 . Also the weights on the beam should be placed at equal distances from the pivot. This limitation is eliminated from the current prescription balances by employing a dual arm beam in a parallel assembly. Nonetheless, positioning the weights in the center of the weighing pans is always a good practice. The prescription balances class A and class III are also called torsion balances because the read out of an applied load depends on the twisting of a wire or a thread.

1.3

Sensitivity Requirement (SR) of a Balance, Error, and Percentage Error

Prior to start using a balance, make sure that it is clean and that it lies horizontal on your bench at an angle of 180 with the pointer of the balance readout at zero position and the pans resting exactly at equal heights. In general, sensitivity is defined as the ratio of the change in the instrument response induced by a change in the stimulus. With respect to a mechanical equal-arm beam balance, the sensitivity requirement is defined as the smallest weight required to cause a shift of the balance marker, one division from the rest point (zero position of the index; Fig. 1.4). Obviously, the smaller the SR, the more sensitive is the balance. In a mechanical prescription balance, weights are always placed in the right pan because the balance by default adds weight—up to 1 g—to the right pan by turning the readout knob clockwise. In other words, you do not need to balance the weight of the sample compound in the left pan manually by adding weights in the right pan. If your sample is less than 1 g, you can counterbalance its weight by turning the readout knob of the balance until the beam is horizontal and the two pans are leveled

1.4 Formulas

7 Weights pan

Sample pan

Balance pointer Equal-arm beam

0

Balance marking

Fig. 1.4 Schematic of the readout of a mechanical balance

at exactly the same height. The weight written in the balance marking is the weight of your sample. But, how does the SR affect the measurement? As mentioned previously, weights smaller than the SR of the balance are not perceptible. What about measuring weights bigger than the SR of a balance? What would the measured value of a known mass, for example, 100 mg, read on a balance with a SR equal to 6 mg? Well, it could be anywhere between 100
6 mg. The
sign is due to the fact that the dualarm beam balance and all other balances measure weights by balancing. In other words, a 100 mg weight placed on the right pan of Fig. 1.4 could balance with a weight on the left pan equal to 94 mg or 106 mg (100
6). We can, therefore, conclude that the SR of a balance is always a plus/minus value, and the maximum potential error of a measurement is equal to the SR of the balance, assuming that the operators are well trained and flawless in their job. As a result, the uncertainty in the true value of a measurement spans a range twice as large as the SR, that is, 12 mg in the case above. Another direct consequence is that the error and % error associated with a weight measurement could be positive or negative.

1.4

Formulas

The equations listed below are valid for all analytical instruments, not only for balances and other liquid handling equipment. Error ¼ measured or actual value  desired or true value

ð1:1Þ

8

1 Measurement

SR ¼ measured value  desired value

ð1:2Þ

, Error ¼ SR

ð1:3Þ

) %Error ¼

measured value  desired value  100 desired value

ð1:4Þ

SR  100 desired value

ð1:5Þ

) %Error ¼

Two points relevant to the equations above must be discussed. First, for mechanical balances, the maximum potential error is equal to the SR of the balance as given by the manufacturer only when errors are solely due to the instrument and not due to the operator. It is necessary that the SR of a mechanical balance be determined by the particular operator and that the newly estimated value should be used in all relevant calculations. For electronic scales, we recommend using the SR as given by the manufacturer. Second, it is not always easy to distinguish in a given practice problem between the desired, theoretical, ideal, true, or real value and the measured, experimental, or the actual value (see practice Example 1.2). Unfortunately, there is no easy way to overcome this problem because it depends on how the question is phrased, but as always, practice makes perfect. Sometimes, the true or the desired value is called the least allowable weight of the drug (LAW). This terminology is used when we need to calculate the smallest amount of drug that can be weighed on a given balance without exceeding the maximum allowable error (MAE) in the measurement. For example, the minimum allowable weight that can be weighed on a balance of sensitivity SR ¼ 6 mg with a maximum possible error associated with the measurement being equal to 5% is 120 mg. %Error ¼

SR SR  100 ) true value ¼  100 true value %Error

) LAW ¼

SR 6 mg  100 ¼  100 ¼ 120 mg %error 5

The smaller we want the error to be in a measurement, the greater should be the amount of the drug we weigh. For example, for MAE  1%, the smaller amount of the drug (LAW) which can be weighed without exceeding the limit of the maximum allowable error is 600 mg (solve Eq. 1.5 for the desired value).

1.5 Practice Examples

1.5

9

Practice Examples

Most of the problems are solved using Eq. (1.5) in conjunction with Eq. (1.1). With Eq. (1.5), we can calculate three parameters: the maximum allowable error or % error, the true or ideal value of the drug and the sensitivity requirement of the prescription balance. It is understood that to be able to compute one of the above three parameters with only one equation, the other two have to be given. For example, to calculate the lowest allowable weight of the drug (LAW), the maximum allowable error associated with the measurement and the sensitivity of the balance have to be given. Example 1.1 You wish to dissolve 6 g of drug X in 100 mL of water. You measured 100 mL of water in a graduated cylinder, but the actual amount of water transferred was only 98 mL. The mistake in the volume occurred because you did not bring the graduated cylinder at eye level to mark the level of the water in the cylinder correctly (parallax error). (a) Calculate the error and % error in the measurement. (b) What is the drug concentration ordered and what is the actual concentration you prepared? (c) What is the % error in drug concentration? (d) How many milligrams of drug are there in one dose assuming that the dose is ½ tsp? Solution Using Eq. (1.1) (a) Error ¼ actual value  desired value ¼ 98 mL  100 mL ¼ 2 mL Using Eq. (1.4), measured value  desired value error  100 ¼  100 desired value desired value 2 mL ¼  100 ¼ 2 100 mL

%Error ¼

Contrary to the % error which is unitless, the error has units. (b) Desired concentration ¼ 6 g/100 mL ¼ 6% w/v Actual concentration ¼ 6 g/98 mL ¼ 6.12 g/100 mL ¼ 6.12% w/v Note: Units of concentration are extensively discussed in Chap. 4. (c) Substituting desired and actual concentration in Eq. (1.4) (continued)

10

1 Measurement

Example 1.1 (continued) 6:12  6  100 ¼ 2 6 Recognize that although the error in the volume is negative, the corresponding error in the dose is positive. This is expected since the drug is dissolved in volume less than 100 mL, and thus, the actual drug solution is more concentrated than the desired one. (d) Amount of drug present in ½ a tsp or 2.5 mL can be calculated using dimensional analysis. % Error ¼

In 98 mL of solution we have In 2.5 mL

)

6g X ¼ 98 mL 2:5 mL

)

X¼

6 g of drug X

2:5  6 mg ¼ 0:153 g ¼ 153 mg 98

If the prescription had been executed without any error in the volume measurement, )

6g X ¼ 100 mL 2:5 mL

)

X¼

2:5  6 mg ¼ 0:150 g 100

or

150 mg

The actual dose contains an excess of 3 mg of drug (2% error).

Example 1.2 You weighed on a class A prescription balance 600 mg of HCTZ to prepare 120 mL of syrup for a patient. When checked on a high precision analytical balance, the true weight was found to be 0.570 g. Calculate the error and % error in the weighing. What is the SR of the balance? Solution We first need to decide which one is the actual or measured value and which one is the desired or theoretical value. You thought that you weighed 600 mg because that is what the prescription balance measured, but the real value is 570 mg. Thus, the actual value ¼ 570 mg and the desired value ¼ 600 mg. We then use Eqs. (1.1) and (1.4) to calculate the error and % error, respectively. ) Error ¼ 570 mg  600 mg ¼ 30 mg Using Eq. (1.4), (continued)

1.5 Practice Examples

11

Example 1.2 (continued) 30 mg  100 ¼ 5 6 Sometimes % error is reported as an absolute number. This is allowed because, in principle, uncertainties are
deviations from the true value. The sensitivity requirement (SR) of the balance cannot be estimated just from the information given above because, as discussed in Sect. 1.2, SR is defined as the smallest weight required to deflect the pointer one division from the rest point of the scale. Since we do not know for certain that 30 mg will actually cause a deflection of exactly one division from the zero point of the readout, we cannot really compute the SR of the instrument. %Error ¼

Example 1.3 Students A, B, C, and D were asked to weigh five samples of 120 mg of a drug powder on a class II prescription balance with SR ¼ 6 mg. All balances were calibrated and verified to operate at maximum capacity by qualified technicians. Their samples were taken and measured by certified technicians on more accurate analytical balances, and the results of the actual weights are shown in the table below. A 140 135 119 112 94

(a) (b) (c) (d)

B 122 123 126 125 119

C 56 90 101 169 199

D 102 98 100 102 99

Classify the measurements as accurate and/or precise. Which student has the best technique? Comment on student D. Comment on the results of students A and C.

Solution We first need to calculate the average and the standard deviation associated with the measurements for each student. You could do that manually using formulas, but I recommend using a spreadsheet program like Microsoft Excel. It is quicker and much more accurate. We can characterize the measurements as “accurate” if the average value is close to the ideal one (120
6 mg) and (continued)

12

1 Measurement

Example 1.3 (continued) precise if the standard deviation is very small compared to the SR of the balance or if the difference of the highest from the lowest value is 2SR.

Average Stdev

A 140 135 119 112 94 120 18.5

B 122 123 126 125 119 123 2.7

C 56 90 101 169 199 123 59.1

D 102 98 100 102 99 100.2 1.8

(a) From our results above, we can see that students A, B, and C were accurate because their average was in between 114 and 126 mg. Students B and D were precise. (b) Only student B was accurate and precise. Essentially all his measurements were executed with deviation 6 mg or the highest from the lowest measurement is 126–119 ¼ 7 mg < 12 mg (which is equal to 2SR). (c) Student D is precise but very inaccurate. Since all balances were previously calibrated by certified technicians, a possible reason for being inaccurate similarly all the time is that the student was weighing 100 mg drug powder instead of 120 mg. Too bad that he did not read the exercise properly. If this exercise was about filling five compounded prescriptions, five patients would have gotten products with drug amount significantly less than the dose. (d) The results of students A and C are also unacceptable. Although the average of their measurement comes exactly and very close to the ideal value, respectively, every single one of the individual dosage forms (powders) is containing either too much or too little of the prescribed drug quantity.

Exercises 1.1 How is accuracy different from precision? 1.2 Write a definition of the sensitivity requirement of a balance equal to 6 mg. 1.3 Mark the statements below as True or False: (a) (b) (c) (d)

A precise measurement is always accurate. An accurate measurement is always precise. Only accurate measurements are precise. Only precise measurements are accurate.

Exercises

13

(e) Measurements that have exactly the same value are precise. (f) Measurements that have exactly the same value are accurate. (g) Precise measurements have values close to each other even if none of them is near the true or correct value. (h) Accurate measurements have values close to each other even if none of them is near the true or correct value. (i) An experimental measurement can be precise or accurate but not both. (j) An average value of a set of measurements is said to be precise if it is near the true (or ideal or correct) value. (k) An average value of a set of measurements is said to be accurate if it is near the true (or ideal or correct) value. 1.4 Fill in the blanks: (a) A measurement which is very close to the true value is designated as _____________. (b) _____________ is defined as the closeness of measurements to each other. 1.5 The sensitivity requirement of a balance is 6 mg. Calculate the smallest weight that would lead to a maximum potential error of 5%. (Answer: 120 mg) 1.6 The minimum allowable weight of a balance is 135 mg. What is the SR of the balance if the maximum allowable error is 5%? (Answer: cannot be determined) 1.7 A student measured five times the same tablet (true weight ¼ 90 mg) using the same analytical balance (SR ¼ 1.4 mg). Given the set of values below, choose the correct statement: 90.1

88.1

90.8

88.8

91.3

This set of measurements can best be described as: (A) Accurate and precise (B) Accurate but not precise (C) Precise but not accurate 1.8 A student measured five times the same tablet (true weight ¼ 90 mg) using the same analytical balance (SR ¼ 1.4 mg). Given the set of values below: 90.1

88.1

90.8

88.8

91.3

(a) How many measurements should be discarded to render her data precise? (b) How many measurements can be considered as accurate? (Answer: (a) only one; either the 88.1 or the 91.3 but not both; (b) all of them but the 88.1)

14

1 Measurement

1.9 A prescription balance weighing 0.120 g of drug is more precise than a balance weighing 0.1208 g. True or False? 1.10 A student measured the weight of a capsule (true value ¼ 69.70 g) on a balance with SR ¼ 0.2 g, three times, 66.79 g, 66.70 g, and 66.69 g. Can these measurements be designated as accurate? 1.11 You are preparing an o/w emulsion for oral administration. Addition of 1 mL (23 drops) of vanillin per 10 L of emulsion is recommended to mask the bad odor of the drug. Due to a stream of air from the ventilation system, the drops are prematurely cutoff from the glass pipet. As a result, 23 drops are equivalent to 0.83 mL. Calculate the error and % error in the measurement. (Answer: 0.17 mL; 17%) 1.12 You need to make 1 L of 9% NaCl. You weigh out 90 g NaCl but instead of dissolving the salt with water and diluting it up to 1 L of solution, you dissolved it with 1 L of water, thus making a total volume of 1035 mL. Calculate (a) the error and % error incurred in the volume measurement, (b) the error and % error in NaCl concentration. (Answer: (a) 35 mL; 3.5%; (b) –3.04 g/L,  3.4%) 1.13 A prescription calls for 8.940 g of KCl dissolved in 120 mL of syrup for a patient who is treated for hypokalemia. The signa of the prescription is “i tbsp p.o. b.i.d. p.c.” If potassium chloride was weighed out with a 0.5% error, what is the true weight of KCl? How much KCl is contained in each dose? (Answer: 8.9847 g; 1.1231 g) 1.14 You are preparing 20 doses of codeine phosphate for cough relief. Each dose is 15 mg of codeine phosphate in one teaspoonful of cherry syrup. Codeine phosphate was weighed with a 1% error. What is the sensitivity requirement of the balance? (Answer: 3 mg) 1.15 A dermatologist prescribed 50 g of 0.5% w/w hydrocortisone ointment. You used a class III prescription balance that has a sensitivity of 6 mg to weigh hydrocortisone powder. Calculate the % error introduced in the measurement. (Hint: 0.5% w/w hydrocortisone ¼ 0.5 g of hydrocortisone in 100 g of ointment) (Answer: 2.4%) 1.16 ℞ Menthol Phenol liquefied ZnO Talc

0.15 0.5 aa

4.0 (continued)

Exercises Propylene glycol 2-Propanol

15 11.0 3 fξ

(Assume: 1 fξ ¼ 30 mL) Specific gravity of: 2-propanol ¼ 0.79; phenol ¼ 0.95; propylene glycol ¼ 1.25. Ingredient quantities are in grams unless otherwise specified. In preparing the antiseptic lotion above, you weighed ZnO with a 1.5% error. What is the actual amount of ZnO in the lotion? (Answer: 4.06 g) 1.17 Calculate the actual volume transferred if the error associated with measuring 24 mL using a 100 mL graduated cylinder is 3%. (Answer: 24.72 mL) 1.18 A prescription for earache relief calls for 8.33 mg of ethyl aminobenzoate per mL of glycerin. You are using a balance that has an SR of 4 mg. How many mL of the ear drop formulation should you prepare if the error in the active ingredient should not exceed 10%? (Answer: 4.80 mL) 1.19 Calculate the desired volume if the resultant error after “accurately” transferring 6.7 mL were 0.30 mL. (Answer: 7.00 mL) 1.20 Thirty doses of a drug were weighed on a balance that has an SR of 5 mg. The error associated with the measurement was 2%. How much drug is in a single dose? (Answer: 8.33 mg)

Additional Exercises 1.21 What is the error and % error in the measurement if the actual volume measured is 0.3 mL smaller than the theoretical value? (Answer: 0.3 mL) 1.22 Using a high precision spectroscopic method, the actual amount of drug W was found to be 0.15 times bigger than the desired one in a compounded prescription. What is the error and % error incurred in the original weighing? (Answer: 15%)

16

1 Measurement

1.23 Calculate the SR of a balance if a 10 mg weight deflects the pointer two divisions from the zero point of the index (neglect human error). What would you do to ensure that the SR you calculated is the correct one? (Answer: 5 mg) 1.24 A medical dropper delivers 0.05 mL/drop of purified water when is held vertically. At 20 angle, the dropper delivers 0.045 mL/drop. Unfortunately, you calibrated the dropper at a vertical position but you measured the water for the prescription holding the dropper at a 20 angle. (a) What is the volume error and % error for 3.4 mL volume? (b) Which number is not affected by the value of the desired total volume, the error or the % error? (c) What is the error and % error for 13.6 mL volume? (d) Calculate the error and % error associated with 34 drops. (Answer: (a) –0.34 mL, 10%; (b) % error; (c) –1.36 mL, 10%; (d) – 0.17 mL, 10%) 1.25 Guaifenesin was weighed on a balance that has a sensitivity requirement of 7 mg and dissolved in 140 mL of cherry syrup. The error in the weighing was 10%. Calculate the concentration in mg/mL and mol/L of the drug in the syrup (MW ¼ 198.21). (Answer: 0.5 mg/mL; 0.00252 mol/L) 1.26 You have prepared 120 mL of 4.2% w/v boric acid eyewash solution. When the solution was tested using an HPLC (high-performance liquid chromatography) instrument, the concentration of boric acid was found to be 4.3% w/v. (a) How much boric acid did you use to prepare the wash solution? (b) Assuming no error in the HPLC analysis, what is the error and % error associated with the weighing of boric acid? (Hint: 4.2% w/v boric acid ¼ 4.2 g of boric acid in 100 mL of solution) (Answer: (a) 5.16 g; (b) 0.12 g, 2.38%) 1.27 Purified water was placed in a graduated cylinder and weighed on a balance. Using as density of purified water 1 g/mL, it was found that a 1.25% error occurred for measuring 23 mL of water. What is the actual volume of water in the graduated cylinder? (Answer: 23.29 mL) 1.28 Isotonic saline was prepared by diluting 100 mL of 9% w/v NaCl stock solution to 1 L with water. (a) What is the actual volume of 9% NaCl used if the % error associated with the transfer of the stock solution was 3.33? (b) What is the actual concentration of the final solution?

Exercises

17

(Hint: 0.9% w/v NaCl ¼ 0.9 g of NaCl in 100 mL of solution) (Answer: (a) 96.67 mL; (b) 0.87%) 1.29 You were asked to dissolve 1 g of lyophilized drug powder with 1 mL sterile saline solution. Because you ran out of 1 mL syringes, you used the much higher capacity 10 mL syringe and you actually transferred 1.1 mL. Calculate (a) the % error in the volume measurement; (b) the final concentration in g/mL; (c) % error in drug concentration. (Answer: (a) 10%; (b) 0.909 g/mL; (c) 9.09%) 1.30-32 Instructions for preparing antibiotic solution from lyophilized powder are usually supplied in a package insert as shown below. AMPICILLIN SODIUM PACKAGE INSERT DIRECTIONS FOR USE Use only freshly prepared solutions. Intramuscular or intravenous injections should be administered within 1 h after preparation, since the potency may decrease significantly after this period. Intramuscular Administration—Reconstitute as directed in the table below with sterile water for injection, USP, or bacteriostatic water for injection. Shake well until dissolved. Vial size 125 mg 250 mg 500 mg 1 ga 2g

Diluent to be added 1.2 mL 1.0 mL 1.8 mL 3.5 mL 6.8 mL

Approximate available volume 1.0 mL 1.0 mL 2.0 mL 4.0 mL 8.0 mL

Average concentration 125 mg/mL 250 mg/mL 250 mg/mL 250 mg/mL 250 mg/mL

a

The 1 g vial should be reconstituted only with sterile water for injection or bacteriostatic water for injection

1.30 The directions for intramuscular injection required reconstitution of a 2 g vial with 6.8 mL of sterile water for injection. You did not carefully read the table and added 8 mL which is the volume written in the adjacent column. What is the error and % error introduced in the measurement of volume? (Answer: 1.2 mL; 17.65%) 1.31 Addition of 1 mL of sterile water in a 250 mg ampicillin sodium vial should make the concentration of antibiotic equal to 250 mg/mL. The concentration in the vial analyzed by high-performance liquid chromatography (HPLC) was found to be 232 mg/mL. (a) Calculate the actual volume of the solution in the vial. (b) What is the error and % error in ampicillin sodium concentration? (Answer: (a) 1.078 mL; (b) 18 mg/mL, 7.2%)

18

1 Measurement

1.32 During reconstitution of 250 mg vial of ampicillin sodium, backpressure caused 0.22 mL of sterile water for injection to spill out. What is the concentration of the antibiotic in the reconstituted vial? (Answer: cannot be calculated)

Chapter 2

Error Assessment of Drug Dose in Pharmaceutical Mixtures

Abstract This chapter describes a unique method to calculate the maximum potential error associated with drug dose in pharmaceutical mixtures and unit dosage forms, composed of an infinite number of ingredients that are measured on multiple balances of variable sensitivities, and thus provides instant guidance on the pharmaceutical ingredient preparation. The method also determines the sensitivity of the instruments measuring each of the components and the least allowable weight (LAW) of the drug within the predetermined assay standard. A great variety of solved examples with capsules, creams, and ointments as the finished products are used to demonstrate a step-by-step application of the method. The method was written in such a way to help you better understand the principles of balance operation, comprehend the relationship of the sensitivity requirement (SR) of a balance and the maximum allowable error in the dose, appreciate that sequential measurements carry a complicated error propagation on the result, and sharpen up your critical thinking by having to consider all possible consequences of any action involved in prescription preparation prior to actually performing relevant calculations. What follows is the description of the aliquot method of weighing based on the LAW as a method that allows weighing a drug quantity smaller than the sensitivity of the balance. Although the concept of the aliquot method of weighing based on the least allowable weight has been around for more than 50 years, this is really the first time that the theory of the method is developed and applied correctly. The method as described elsewhere ignores errors associated with multiple weighing of drugexcipient mixtures, and as a result, the dose error in the final product is underestimated. The topics are advanced but extremely important. Keywords Pharmaceutical measures · Maximum allowable error · Least allowable weight · Sensitivity requirement · Scale-up/down method · Aliquot method · Geometric dilution

Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_2) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_2

19

20

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Learning Objectives After reading this chapter, you should be able to: • Calculate the maximum potential error associated with the measurement of any component in a mixture • Evaluate the required sensitivity of any of the balances for a given MAE • Estimate the LAW of the drug or any other component in the mixture for a known MAE • Apply the aliquot method of weighing to prepare compounded prescriptions that call for a drug amount smaller than the LAW at a given MAE In the previous chapter, we have discussed the calculation of the potential error that might incur in a drug measurement. More specifically, we have learned that using Eq. (1.5), we can calculate one of the three parameters, that is, the maximum potential error associated with a measurement, the SR of the balance and the mass of the drug, if two of those are known. % Error ¼

SR  100 true value

ð1:5Þ

More often, the drug we measure is to be mixed with other ingredients called excipients. Excipients like lactose, glucose, starch, cellulose, and derivatives are important in drug formulation and dosage form manufacturing. Their uses are more extensively discussed in Chap. 9. One of the rules of the Good Manufacturing Practice (GMP) is that drugs and excipients should be measured on different balances, sometimes located in different rooms, in an effort to minimize cross-contamination of products. In this chapter, we discuss how to calculate the error in drug dose in pharmaceutical mixtures where drug and multiple excipients are measured on same or different balances and then mixed together. Additionally, we demonstrate how to calculate the potential error in the dose when an aliquot of a stock mixture containing the drug is weighted, mixed with excipients, and used to fabricate various dosage forms. Undoubtedly, the development of a method capable of calculating the maximum potential error associated with the drug concentration in pharmaceutical mixtures finds direct application in drug formulation and manufacturing of dosage forms. Throughout the world, the Federal Drug Administration (FDA) approves only the dosage units in which the drug content is consistent with the amount written in the approval papers (label claim). The maximum allowable deviation in drug content of a dosage form, from its production until its expiration, depends on many factors such as the therapeutic class and therapeutic index of the drug, its degradation mechanism, the route of administration, that is, intravenous versus transdermal, and the type of dosage form. Industrial solid dosage forms are usually subjected to two tests, the weight variation test and the test of content uniformity. Although these tests are considered successful if not more than one out of 30 dosage units deviates more than 15% from the label claim, the permissible deviation from the specified value of drug content in industrial dosage forms is usually less than 2%. In extemporaneous

2.1 A Method to Determine Uncertainties of Drug Concentration. . .

21

preparations prepared from prefabricated industrial dosage forms (sold in the pharmacies), the permitted deviation from the specified drug amount is reported as 10% (USP Chapter 795 2006). Compounded drug products are not FDA-approved. Rather, the oversight and regulation of compounding pharmacies is conducted by the state boards of pharmacy with the FDA maintaining sufficient authority over their operations. A survey organized by the Department of Pharmacovigilance and Rational Drug of the Federal Drug Administration, on compounded pharmacy preparations in 2001, showed that 31% of the products contained only 59–89% of the drug dose (Statement by the Acting Director of the Center for Drug Evaluation and Research 2005). In sharp contrast, in the same survey, the failure rate of commercial industrial pharmaceutical products was found to be close to 2%. The survey results are alarming and undoubtedly endanger the future role of compounding pharmacy as an integral part of the country’s modern health care system. The possible errors that occur after multiple measurements of drug, excipients and the pharmaceutical mixture in various balances of same or different sensitivity, are discussed next.

2.1

A Method to Determine Uncertainties of Drug Concentration in Pharmaceutical Mixtures (Savva 2006, 2016)

When two or more measured independent quantities combine arithmetically, the individual errors associated with each measurement propagate through the calculations to produce errors in the results. The key idea for determining the uncertainty related to the concentration of drug in a mixture is based on the creation of a function which describes the variation of the drug amount in the total mixture, as shown below: f ¼

x x þ y þ z þ 

ð2:1Þ

x ¼ amount of drug y, z ... ¼ all other components of the mixture except drug x f ¼ fraction or the weight concentration of the drug in the mixture Equation (2.1) is continuous in the domain of our interest since x, y, z. . . are always real positive independent numerical variables. The error (df) associated with the drug concentration in the mixture f can be found from the total differential of the above function, as shown below: f ¼ f ðx; y; z; . . .Þ

22

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

∂ f ) df ¼ ∂x

y, z, ...

∂ f dx þ ∂y

x, z, ...

∂ f dy þ ∂z

x, y, ...

dz þ   

ð2:2Þ

Generally, when a function contains more than one independent variable as in Eq. (2.1), the differential is called the total differential of the function, and it can be determined by partial differentiation of the equation. Partial differentiation entails taking the (partial) derivative of each one of the independent variables, one after another, while keeping all other independent variables constant. For example, varying the x by dx while holding all other variables y, z... constant (written as a subscript) will cause a change in f equivalent to ∂∂xf dx. Similarly, the second y, z, ...

term is the contribution of the change of the component y (all other ingredients including the drug are kept constant) to the total error of the drug concentration in the mixture. We are using absolute values of the partial derivatives because we wish to determine the maximum (positive) error, and changes can be positive and negative. Something similar would be the investigation of how volume (V ) and pressure (P) affect the temperature (T ) of a gas. The dependent variable in this case is T, and the independent variables are V and P. In order to investigate how the independent variables affect the dependent variable, we must vary the value of V while holding P constant and then vary P while holding V constant. The ideal gas equation is P  V ¼
n  R 
T and its total differential (n and R are constants) is ∂T dT ¼ ∂V dV þ ∂T dP. The first term shows the rate of change of the tempera∂P V P ture with volume (keeping the pressure constant) while the second term represents the rate of change of the temperature as a function of pressure (keeping the volume constant). The sum of our two rates is the total temperature change when the volume and pressure are varied independently. Returning to our own function, it is important to note that x, y, z, ... are real independent variables, since they represent different substances in the mixture and are weighed on a balance separately. How the measurement of each ingredient affects the concentration of drug in the mixture f depends on the quantity of each ingredient and the sensitivity of the balance that is going to be used for the measurement. Obviously, the smaller the quantity and the less sensitive the weighing instrument, the bigger is the maximum potential error associated with the measurement. From the practical point of view, the total differential gives a linear approximation of a change in the dependent variable f, which is derived from infinitesimal changes in all of the independent variables. In other words, df, dx, dy, and dz are the uncertainties or maximum possible errors associated with the measurement of the respective quantities f, x, y, and z. More specifically, dx, dy, and dz are the sensitivities of the weighing balances used to measure the respective quantities x, y, and z. Executing the algebra of partial derivatives of Eq. (2.1): df y þ z þ    dx dy þ dz þ    ¼  þ f x x þ y þ z þ  x þ y þ z þ  where dff is the relative error of the drug concentration f.

ð2:3Þ

2.2 Applications of the Method

23

Essentially, Eq. (2.3) indicates the fractional or relative error in drug concentration (by mass) in the mixture. We can therefore write Eq. (2.3) as df uncertainty in f ¼ f ideal value of f

ð2:4Þ

To better investigate the influence of the number of components of the mixture n on the maximum potential error in drug concentration, Eq. (2.5) was derived from Eq. (2.3), under the assumption that all ingredients were measured on balances of same sensitivity, that is, dx ¼ dy ¼ dz ¼ . . .. )

df dx ¼  ½ 1 þ ð n  1Þ  f  f x

ð2:5Þ

n ¼ the number of all ingredients except the drug x The main conclusion that can be summarized from Eq. (2.5) is the following: If the drug is mixed with only one excipient (n ¼ 1) and both (drug and excipient) are weighed on the same balance or on balances of equal sensitivity, then dff ¼ dxx. This result, which is obtained by substituting n ¼ 1 in Eq. (2.5), suggests that in a two-component mixture, we do not need to compute the potential error associated with the measurement of the second component (the excipient) if both the components are measured on balances of same sensitivity. We need only to calculate the error associated with the measurement of the drug. Indeed, in this case, Eq. (2.5) is reduced to Eq. (1.5), with dff ¼ Error, dx being the SR of the balance and x the ideal or true value (Table 2.1).

2.2

Applications of the Method

Equation (2.3) can be used to calculate the following: A. The maximum potential error associated with the concentration of any component in the mixture (not just the drug) B. The required sensitivity of any of the balances for a given maximum allowable error (MAE) Table 2.1 Comparison of Eqs. (1.5) and (2.5) with n ¼ 1 in Eq. (2.5) SR %Error ¼ desired value  100 SR Relative Error ¼ desired value

)

Relative Error ¼

df f

Eq. (2.5) ,

df f

¼ dxx

Eq. (2.5)

SR ¼ dx, desired value ¼ x

Note that the relative error is not the same as the error being equal to SR as defined in Eq. (1.1)

24

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

C. The least allowable weight (LAW) of the drug or any other component for a known MAE D. All of the above (A, B, and C) for the preparation of a stock mixture E. A, B, and C for the preparation of a pharmaceutical product Example 2.1 Calculate the % probable error in drug concentration x in the following clinical batch manufactured in a GMP certified laboratory. Drugs x and y and lactose were measured on pharmaceutical scales of sensitivity 0.01 g, 0.05 g, and 1 g, respectively, that is, dx ¼ 0.01 g, dy ¼ 0.05 g, and dl ¼ 1 g. Drug x Drug y Lactose, l

0.5 g 2.5 g 67 g

Solution The maximum possible error in drug x concentration in the mixture can be calculated by the partial derivative of the Eq. (2.1) C¼

x xþyþl

C ¼ fraction or the weight concentration of the drug in the mixture (equal to f) Using the total differential of the function, that is, Eq. (2.3), dC ðy þ lÞ=x dy þ dl ¼  dx þ C xþyþl xþyþl )

dC ð2:5 þ 67Þ=0:5 0:05 þ 1 2:44 ¼  0:01 þ ¼ ¼ 0:035 C 70 70 70 ) Relative error ¼ 0:035 or

 3:5%

But C ¼ 0:5 70 ¼ 0:007, ) absolute error, dC ¼ 0:035  0:007¼ 0:000245 g Conclusively, the error associated with the drug x concentration in the mixture is: C  dC ¼ 0:007  0:000245 g ¼ 0:007  3:5%: Had we used Eq. (1.5) to calculate the error in drug x concentration, the result would have been seriously underestimated. (continued)

2.2 Applications of the Method

25

Example 2.1 (continued) %Error ¼

SR dx 0:01 g  100 ¼  100 ¼  100 ¼ 2% desired value x 0:5 g

Example 2.2 Calculate the sensitivity of the balance that drug x needs to be weighed on if the MAE of drug x concentration in the following mixture is 1%. The SR of the other balances are dy ¼ 0.05 g and dl ¼ 0.1 g. Drug x Drug y Lactose, l

0.5 g 2.5 g 67 g

Solution Using Eq. (2.3), dC ð2:5 þ 67Þ=0:5 0:05 þ 0:1 ¼ 0:01 ¼  dx þ C 70 70 ) 0:7  0:15 ¼ 139  dx ) dx ¼ 0:004 g Therefore, drug x should be measured on a balance with SR  4 mg, so that the mixture is prepared with a maximum possible error in drug x  1%.

Example 2.3: Batch Process Scale-Up Calculate the least allowable weight (LAW) of drug x which can be weighed on the specified electronic balances, with a maximum permissible error (MAE) of 1%. Drug x, y, and lactose were measured on balances of sensitivity 0.01 g, 0.05 g, and 1 g, respectively, that is, dx ¼ 0.01 g, dy ¼ 0.05 g, and dl ¼ 1 g. Drug x Drug y Lactose, l

0.5 g 2.5 g 67 g

(continued)

26

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.3 (continued) Solution As shown in Example 2.1, the error associated with drug x in the mixture is 3.5 > 1%. Obviously, in order to reduce the error associated with the measurement of drug x, we need to increase the amount of drug x weighed on the specified balance. However, changing the amount of drug x in the mixture would require proportional changes in the quantities of the other two components, so that the ratio of drug x to the other two ingredients remains the same. This is a batch process scale-up example. Thus, apart from Eq. (2.3), we must build an additional equation, which is actually the ratio of drug in the final mixture (see equation in the second condition). First condition: dx ðy þ lÞ=x dy þ dl ¼ 0:01 ¼  dx þ x xþyþl xþyþl

ð2:3Þ

Second condition: x 50 mg 1 ¼ ¼ x þ y þ l 70,000 mg 140

ð2:6Þ

Carrying out the calculus, ) 140x ¼ x þ y þ l

)

y þ l ¼ 139x

It is important to understand that Eq. (2.6) of the second condition ensures preparation of a larger quantity mixture, in which the proportion of drug x to drug y and lactose remains constant to 1=140 . The actual values of x, y, and l in the mixture will certainly be different from the original, but their ratio will stay the same at 1=140 . Substituting the result of the second condition in Eq. (2.3), 0:01 ¼

139 1:05  0:01 þ ) x ¼ 1:74 g, 140x 140x y ¼ 8:70 g and l ¼ 233:56 g ) Total mixture ¼ 244 g

2.2 Applications of the Method

27

Example 2.4 Calculate the % maximum allowable error of drug concentration x in capsules prepared on a GMP-certified laboratory of a compounding pharmacy. Capsules, drugs x and y, and lactose were measured in pharmaceutical scales of sensitivity 0.005 g, 0.01 g, 0.05 g, and 1 g, respectively. Drug x Drug y Lactose, l

0.5 g 2.5 g 67 g

Mix and make 280 capsules. Solution



Each capsule contains 250 mg of the mixture

70,000 mg 280 cap

 ¼ 250 mg=cap .

The composition of each capsule is: Drug x Drug y Lactose, l

1.786 mg 8.929 mg 239.286 mg

Apart from the error associated with the measurement of each component, the error associated with the measurement of the dosage form must also be calculated. This error will be added to the error associated with the measurement of all components in the mixture to calculate the total error associated with the final product. The error associated with the weighing of each capsule (250 mg) is equal to: % Error ¼

5 mg  100 ¼ 2 250 mg

The error associated with drug x in the mixture was calculated by Eq. (2.3) in Example 2.1. dC ¼ 0:035 C

)

relative error ¼ 0:035

or

 3:5%

Therefore, the total maximum potential error associated with the concentration of drug x in the capsule is 2 + 3.5 ¼ 5.5%.

28

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.5 Calculate the % error associated with drug x concentration in capsules prepared in a compounding pharmacy, as shown below. The mother or stock mixture, the capsules, drugs x and y, and lactose were measured in pharmaceutical balances of sensitivity 1 g, 0.005 g, 0.01 g, 0.05 g and 1 g, respectively. Drug x Drug y Lactose, l

1g 5g 134 g

Mix by the method of geometric dilution to make 280 capsules, each one containing 250 mg of powder. Solution As in Example 2.4, each capsule contains 250 mg mixture. Drug x Drug y Lactose, l Total

1.786 mg 8.929 mg 239.286 mg 250 mg

Loading 280 capsules requires only half of the mixture (70 g). The remaining of the stock or mother mixture will be stored for future use. Apart from the error associated with the measurement of each ingredient, there is an error associated with the measurement of the quantity of the stock mixture (called aliquot) that is used to fill the capsules. However, this error does not affect the ratio of the drug in the mixture, and therefore, it does not affect the accuracy of the dose. It can only affect the amount of powder available to fill the capsules. If the error is negative, there will not be enough powder to fill the capsules. If the error is positive, some powder will remain after you filled all the capsules. On the contrary, the error associated with the drug in the dosage form, that is, the capsule directly affects the drug dose since, if you add less powder mixture in a capsule, the patient will be taking less dose, whereas if more powder mixture is added in a capsule, the patient will be taking more drug in the dose. The error associated with the measurement of the capsule (250 mg) is equal to: %Error ¼

5 mg  100 ¼ 2 250 mg

The error of drug x concentration in the mixture using Eq. (2.3) is equal to: (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

29

Example 2.5 (continued) 5þ134 dC ðy þ lÞ=x dy þ dl 0:05 þ 1 2:44 ¼  dx þ ¼ 1  dx þ ¼ ¼ 0:0174 C xþyþl xþyþl 140 140 140 dC ¼ 0:0174 C

)

relative error ¼ 0:0174

or

 1:74%

Therefore, the total maximum potential error associated with the concentration of drug x in the capsule is 2 þ 1:74 ¼ 3:74%

2.3

The Aliquot Method of Weighing (Savva 2015)

The method is useful when you have to weigh a drug amount smaller than the smallest permissible weight (LAW) for a given maximum allowable error (MAE), and the active ingredient is to be administered or may be administered in combination with inert ingredients such as lactose or starch. It is important to understand that this method cannot be used unless the prescription states clearly that the drug can be mixed without any problem with other inert materials. The reason is that ingredients such as lactose, glucose, sucrose, etc., are inactive when administered alone, but when they are mixed with a drug they may significantly alter its pharmacological activity by affecting the stability, solubility, and bioavailability. The aliquot method of weighing is quite a clever method, and it is the only method that you can use to weigh an amount of drug less than the minimum allowable weight within a given maximum allowable error to fill a compounded prescription. For example, it is impossible to weigh 10 mg of a drug with a maximum permissible error of 5% on a balance that has sensitivity of 6 mg. The minimum allowable amount of drug that can be measured under these conditions can be calculated using Eq. (1.5), and it is equal to 120 mg. You would have to weigh drug quantities larger than what is required for the preparation of the prescription, to achieve the limits of the MAE. Therefore, as the word implies, in the aliquot method of weighing, the drug is initially weighed in quantities much higher than those requested in the prescription and then mixed with an appropriate amount of vehicle (excipient) to make the stock or maternal mixture. The amount of excipient added should be such that the ratio of the drug to the stock mixture is the same as the ratio of the drug to the total powder specified in the original prescription. From that stock mixture, an aliquot or a portion that was calculated to provide the requested quantity of the drug is weighed again and used to prepare the prescription. The method is quite flexible as to the amount of the maternal mixture we make, provided that at the end we use the right portion (aliquot) from the stock mixture containing the desired dose of the drug, so that the error in drug concentration does not exceed the

30

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

maximum allowable error. We next present an example of this lengthy but very useful method. Example 2.6 Describe how to prepare a compounded prescription that calls for a single powder dose that requires a total of 24 mg of prednisone, with a 10% maximum potential error on a balance that has SR equal to 6 mg. Drug and lactose are available in powder form. Solution Step 1: We first determine the error associated with the drug dose when the amount of drug specified in the prescription is weighed directly on the balance. If the error is greater than the permissible one, then we must use the aliquot method of weighing. Using Eq. (1.5), 6 mg Maximum allowable error, %MAE ¼ 24 mg  100 ¼ 25 > 10%. Therefore, you should use the aliquot method of weighing method. Any attempt to weigh directly the drug dose on a balance will result in an error >10%. Step 2: It can be shown that the errors resulting from independent measurements are additive in nature. Therefore, we need to identify the number of weighings that involve the drug. Assign symbols to the measurements using capital letters and to the corresponding errors using small letters. Set the cumulative error to be equal to the Maximum Allowable Error (MAE) and perform relevant calculations. The total number of drug weighings is three: (1) the drug that is going to be mixed with lactose to form the stock powder, (2) the aliquot, and (3) the mixture contained in the single powder dosage form. Let Q: Drug quantity that is going to be mixed with pure lactose, L, to form the stock powder, S; q is the error associated with Q. A: the aliquot quantity that will be taken from the stock powder, S; a is the error associated with A. M: the mass of the mixture in the powder dose; m is the error associated with M. MAE ¼ q þ a þ m ¼ 10%

ð2:7Þ

Because the weight of the dosage form is not specified in the prescription, we can set it to be the aliquot weight. In other words, we eliminate a (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

31

Example 2.6 (continued) measurement by setting the aliquot to be our dosage form. Since M ¼ A, Eq. (2.7) is reduced to MAE ¼ q þ a ¼ 10%: To make it even easier for us, I am going to set q ¼ a, and therefore, MAE ¼ q þ a ¼ 2q ¼ 10%

)

q ¼ 10=2 ¼ 5% ¼ a:

Next, we calculate the least allowable weight (LAW) within a maximum permissible error of 5%. The LAW is equal to the amount of drug to be measured on the balance (Q), and it is equal to the aliquot (A) which in turn is equal to M. ) Q ¼ A ¼ M ¼ LAW ¼

6 mg  100 ¼ 120 mg 5

Let us summarize what we did before proceeding to the next step. We have two independent measurements. The first one is from the bottle containing pure drug to make the maternal mixture, and the second measurement relates to the weighing of the aliquot (from the stock mixture) containing the drug mixed with lactose. For convenience, we have chosen each measurement to be associated with same maximum permissible error of 5%, and therefore, the amount of drug and aliquot amount are equal to 120 mg. Step 3: The stock mixture is prepared under two conditions: the maximum potential error associated with the drug and the aliquot measurements is 5% (for each measurement), and the drug quantity in the aliquot (120 mg) contains the dose of the drug (24 mg). These conditions are expressed in the following ratio: Desired weight of drug LWQ of drug ¼ LWQ of aliquot mixture X mg of stock mixture

or

D Q ¼ A S

ð2:8Þ

Substituting the values in Eq. (2.8), 24 mg of drug 120 mg of drug ¼ 120 mg of aliquot mixture S ) S ¼ 600 mg of stock mixture

)

(continued)

32

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.6 (continued) The stock mixture (as well as the aliquot) consists of lactose and prednisone. Therefore, the 600 mg of the stock powder consists of 120 mg prednisone and 480 mg lactose. In conclusion, in order to prepare a 24 mg powder dose of prednisone with a 10% maximum allowable error, we had to make 600 mg of stock mixture after weighing 120 mg of prednisone and 600–120 ¼ 480 mg of lactose. The two ingredients are mixed using the method of geometric dilution. From the stock powder, we only use an aliquot of 120 mg since it contains 24 mg of prednisone (the dose). The remaining 480 mg of stock mixture are put aside for future use. Priority in the aliquot method is to identify the number of weighings that involve the drug, assign symbols to them using uppercase letters and to the corresponding potential errors using lowercase letters. The total cumulative error in drug dose is set to be equal to the MAE. The total number of drug measurements involved in the preparation is three: (1) the drug (Q) that is going to be mixed with the excipient to form the stock powder, (2) the aliquot (A), and (3) the mixture (M) contained in the powder dosage form. In our special case, we only have to measure Q and A. We conveniently set the two quantities equal, and therefore, the error associated with the measurement of the powder dosage (m equal to a) is equal to the error (q) associated with the measurement of Q. Thus, a ¼ q ¼ 5%. The total error should be equal to the maximum allowable error. MAE ¼ a þ q ¼ 10% Notice how efficiently we have constructed the ratio in Eq. (2.8). The 600 mg of stock mixture (denominator of the right-hand side) contains 120 mg drug (numerator of the right-hand side), so that 120 mg from the stock mixture (aliquot; denominator of the left-hand side) contains proportionally 24 mg of prednisone drug, that is, the dose (numerator of the left-hand side). Also, the numerical Eq. (2.8) is easy to remember, since the numerator of the right side is equal to the denominator of the left side with their values set to equal the LAW, while the numerator of left side is equal to the desired amount of the drug (dose) in the formulation. Of course, we are not obliged to always use this particular mathematical proportion. Depending on the cost of the drug and excipients, we can appropriately change the quantities of the ratio 2.8 to reduce the cost of the prescription. Answer verification We can verify our answer in multiple ways. Having in mind that we performed two independent drug weighings, each 120 mg, we can use Eq. (1.5) to calculate the maximum potential error associated with each weighing. The sum of the two errors should yield a total or cumulative error of 10%. (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

33

Example 2.6 (continued) The error in drug concentration in the stock powder can be calculated with either Eq. (1.5) or Eq. (2.3). Remember that for a single excipient, in addition to the drug, Eq. (2.3) is simplified to Eq. (1.5). Using Eq. (2.3), we find the relative error   dC l=x dx l dx ¼  dx þ ¼ þ1  ¼ C xþl xþl x xþl   lþx dx 6 mg ¼ ¼ 0:05 or  5%  x x þ l 120 mg The dC C ¼ 0:05 or  5% is called relative error. It is unitless, and it is related to the absolute error with Eq. (1.4), Relative Error ¼

absolute error ideal or true or desired value of drug

The absolute error is dC and the ideal value of the drug is C. Absolute error ¼ relative error  ideal value of the drug ) dC ¼

dC  C ¼ 0:05  C C

But, C¼

x 120 1 ¼ ¼  ¼ 0:2 x þ l 120 þ 480 5

)

dC ¼ 0:05  0:2 ¼ 0:01:

Alternatively, the absolute error in drug concentration can be calculated by substituting x C ¼ xþl in Eq. (2.3)       lþx dx x lþx dx ¼ ¼ dC ¼ C     x xþl xþl x xþl dx 6 mg ¼ ¼ 0:01 or  0:01 or 10% x þ l 600 mg In order to make sense of our answer, we need to recognize that C and dC have units of concentration, weight of drug per weight of mixture, that is, g/g, mg/mg, μg/μg, etc. Accordingly, the weight concentration of the drug in the stock mixture is (continued)

34

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.6 (continued) C  dC ¼ C ave  dC ¼ ð0:2  0:01Þ w=w ¼ ð0:2  0:01Þ mg of drug=mg of mixture The drug quantity in the final mixture is found by multiplying the drug concentration with the aliquot (A). Therefore, the minimum and maximum values of C would be C min ¼ Cave  0:01 ¼ 0:19 < Cave ¼ 0:2 < C max þ 0:01 ¼ 0:21 mg=mg: Similarly, the minimum and maximum quantities of the drug, D and D+, are D ¼ A  C min ¼ 120 mg 0:19 mg=mg ¼ 22:8 mg Dþ ¼ A  Cmax ¼ 120 mg 0:21 mg=mg ¼ 25:2 mg Dave ¼ A  C ave ¼ 120 mg 0:20 mg=mg ¼ 24 mg The percentage errors associated with these two quantities are %error  ¼

22:8  24  100 ¼ 5:0%, 24 ) %error ave ¼

%errorþ ¼

25:2  24  100 ¼ 5:0% 24

j%error j þ j%errorþ j ¼ 5:0% 2

Clearly the error in drug concentration in the stock powder is the same as that in the aliquot since there is no dilution involved from the stock powder to the aliquot. We simply weigh 120 mg from stock powder and use it as one single drug powder dose. However, because the aliquot in this case is the final mixture that we are using as the dose, what matters is actually the drug quantity present in this final mixture and not the drug concentration. Therefore, to calculate the total error in the dose, we need to account for the error in the measurement of the aliquot (A) which is equal to the weight of the dosage form (powder). Solving Eq. (1.5) with a desired value and a SR equal to 120 mg and 6 mg, respectively, we get a ¼ m ¼ %Error ¼

6 mg  100 ¼ 5 120mg

) Total error in the dose ¼ m þ q ¼ 5 þ 5 ¼ 10%: (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

35

Example 2.6 (continued) We could prove that the total error in the drug dose is 10% by first considering the minimum and maximum aliquot mass that can be weighed on the given balance 120  6 < 120 < 120 þ 6

or

114 < 120 < 126

Correspondingly, the minimum and maximum quantities of the drug, D and D+, are D ¼ Amin  C min ¼ 114 mg

0:19 mg=mg ¼ 21:66 mg

Dþ ¼ Amax  Cmax ¼ 126 mg 0:21 mg=mg ¼ 26:46 mg: The concentrations Cmin ¼ 0.19 mg/mg and Cmax ¼ 0.21 mg/mg were obtained by dividing the minimum and maximum drug quantity Qmin ¼ 114 mg and Qmax ¼ 126 mg (in our special case A ¼ Q ¼ 120 mg) with the stock powder S ¼ 600 mg. The percentage error associated with these two quantities (D and D+) considering that the desired dose is 24 mg can be calculated by Eq. (1.4). %error ¼

21:66  24  100 ¼ 9:75, 24 ) %errorave ¼

%errorþ ¼

26:46  24  100 ¼ 10:25 24

j%error j þ j%errorþ j ¼ 10% 2

Alternatively, we can check the reliability of the method, by looking at the two extreme drug concentration values from opposite ends using Eq. (2.1). C¼

120  6 ð120  6Þ þ ð480  6Þ

The two extreme error values, that is, maximum negative (left) and maximum positive (right) are referring to the final second weighing (aliquot measurement). The maximum negative error corresponds to an aliquot mass of 120 – 6 ¼ 114 mg, while the maximum positive error corresponds to an aliquot equal to 126 mg. C¼

120  6 120 þ 6 and C ¼ ð120  6Þ þ ð480 þ 6Þ ð120 þ 6Þ þ ð480  6Þ )

114 126 and 600 600 (continued)

36

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.6 (continued) The amount of drug in the aliquot that corresponds to the two potential error extremes can be calculated from the following ratios: )

114 D 126 Dþ ¼ ¼ and 600 114 600 126

D stands for the amount of drug in the aliquot in the two extreme cases. )D ¼ 21.66 mg and D+ ¼ 26.46 mg. The % relative errors in both cases can be calculated using Eq. (1.4). 21:66  24  100 ¼ 9:75% 24 26:46  24 %errorþ ¼  100 ¼ 10:25% 24

%error  ¼

and

Taking the absolute values, we find the average value of the maximum potential error of the drug concentration. ) %error ave ¼

j9:75j þ j10:25j ¼ 10% 2

Example 2.7 Fill the prescription below for cluster headaches, with a maximum allowable error of 5%, using a prescription balance of SR ¼ 6 mg. ℞ Prednisone Lactose

q.s. ad

0.002 0.300

M. et. Ft. caps no. i, D.T.D. caps no. xii Sig.: Cap. i. t.i.d. p.c.

Solution Step 1: Read the instructions carefully to properly understand how to prepare and dispense the medication order. Then calculate drug needed and (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

37

Example 2.7 (continued) determine whether the aliquot method needs to be employed to fill the prescription. Statement Lactose q.s. ad 0.300

M. et. Ft. caps no. i D.T.D. caps no. xii

Explanation Add sufficient quantity of lactose to make exactly 0.3 g of mixture. In other words, enough lactose (298 mg) should be added to and mixed with 2 mg of prednisone to make a total of 300 mg of mixture (for conventions regarding “missing” units in prescriptions see Appendix A) Mix the written quantities to make one capsule Give 12 such doses, that is, make 12 capsules, each containing 2 mg of prednisone and 298 mg of lactose

Drug needed to fill the prescription ¼ 2 mg/caps ∙ 12 caps ¼ 24 mg < LAW since the LAW for an MAE of 5% and a balance SR of 6 mg is 120 mg. Thus, the aliquot method needs to be employed to fill the prescription. The total mixture which is needed to fill the prescription, F (see step 2) is 3.6 g. F ¼ ð300 mg=capsuleÞ  ð12 capsulesÞ ¼ 3600 mg ¼ 3:6 g Step 2: Identify the number of weighings that involve the drug. Assign symbols to the measurements using capital letters and to the corresponding errors using small letters. Set the cumulative error to be equal to the maximum allowable error (MAE) and perform relevant calculations as shown below: The total number of drug weighings to fill the prescription is three: (1) the drug that is going to be mixed with lactose to form the stock powder, (2) the aliquot, and (3) the mixture contained in each capsule. Let Q: Drug quantity that is going to be mixed with pure lactose, L, to form the stock powder, S; q is the error associated with Q. A: the aliquot quantity that will be taken from the stock powder, S; a is the error associated with A. F: the final mixture that is used to fill the capsules, that is, 3.6 g. M: the contents of each capsule, that is, 300 mg; m is the error associated with M. (continued)

38

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.7 (continued) MAE ¼ q þ α þ m ¼ 5%

ð2:7Þ

M ¼ 300 mg and the error incurred in the measurement, m using Eq. (1.5) is equal to 2%. )m¼

6 mg  100 ¼ 2% 300 mg

Considering that the cost of the drug is much higher than lactose, it is best to set the mass of the aliquot, A, equal to the final mixture F needed to fill 12 capsules. Thus, A ¼ F ¼ 3.6 g. Since A ¼ F, there is no dilution involved in the preparation of the aliquot, the drug concentration is the same as that in the stock powder from which it is taken from and thus the aliquot measurement does not contribute to the dose error (a ¼ 0). Substituting this result in Eq. (2.7), ) 5% ¼ q þ 0 þ 2

)

q ¼ 3%:

Using Eq. 1.5, once more we can calculate the LAW of the drug Q¼

6 mg  100 ¼ 200 mg 3

Step 3: Use Eq. (2.8) to write the mathematical ratio and determine the stock mixture quantity, S. D is the total dose of the drug (24 mg). D Q ¼ A S 24 mg of drug 200 mg of drug ¼ ) 3600 mg of mixture S )S¼

ð2:8Þ

A  Q 3600 mg  200 mg ¼ ¼ 30,000 mg ¼ 30 g of stock mixture D 24 mg

But S¼LþQ

)

L ¼ S  Q ¼ 29,800 mg (continued)

2.3 The Aliquot Method of Weighing (Savva 2015)

Example 2.7 (continued) Therefore, we make 30 g of total mixture by weighing and mixing 200 mg of drug and 29,800 mg of lactose. From this mixture, we use an aliquot equal to 3600 mg, which contains exactly 24 mg of prednisone. The remaining stock mixture is stored for future use. We aim to fill each capsule with 300 mg of powder mixture that contains 2 mg of drug. The mathematical relationship of this concept is shown below: 2 mg of drug 24 mg of drug ¼ 300 mg of mixture in capsule 3600 mg of mixture in aliquot 200 mg of drug ¼ 30 g of stock mixture The capsules are opened and filled by hand one by one, using a small spatula or by tapping the bottom of the empty shell slightly and quickly in the pile of the powder mixture, after “taring” the weight with an empty capsule shell. When the amount of powder in the capsule appears to the eye as satisfactory, we close the capsule and weigh it. If the weight of the capsule is less than or greater than 300 mg, open the capsule and add or remove powder, respectively. The process is repeated until the weight of the capsule is 300 mg. What if we wanted to fill some more capsules? You do not have to do anything else other than weigh more A or F. Do not change F in the ratio 2.8; just weigh more. I repeat; carry out your calculations with A ¼ F ¼ 3600 mg, but weigh more than 3.6 g if you need to prepare more capsules. By setting the aliquot to be equal to the final mixture, we have ensured to always have 24 mg of drug in 3600 mg of mixture or 2 mg of drug in 300 mg of mixture (1 capsule). Equation (2.8) is a theoretical mathematical relationship with a practical utility. All you have to do to prepare more capsules is to weigh more mixture from the stock powder mixture. For example, if you wish to fill 15 capsules, A ¼ F ¼ (15 capsules)∙(30 mg/capsule) ¼ 4.5 g. The mass of the drug will change proportionally to the powder mixture according to the ratio 2:300 or 1:150. Furthermore, since the Q/S stays constant, the maximum potential error in the drug dose stays constant, as well, whether you wish to fill 12 capsules or just one capsule (regardless of F < 3.6 g). Changing the numbers of Eq. (2.8) will result in a new drug dose-to-powder mixture ratio, and it will throw off your calculations.

39

40

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.8 Bacitracin ointment is indicated for minor cuts, scrapes, and burns to prevent infection. Describe how you could prepare the prescription below with an accuracy not less than 95% if the sensitivity requirement of the prescription balance is 12 mg. Bacitracin USP is supplied at strength 50 units/mg. ℞ Bacitracin Pet. Album Sig: apply a small amount of the product to the affected area 1 to 3 times daily

q.s. ad

200 units/g 1 oz

Note: Petrolatum Album (white petrolatum) is a purified mixture of semisolid oleaginous hydrocarbons obtained from petroleum. Because of its hydrophobic properties and oleaginous nature (jelly appearance), it is commonly used as a base for ointments. Semisolids are always measured on a balance due to their high viscosity

Solution Step 1: Calculate the amount of bacitracin needed for the prescription and determine if the drug is less than the least allowable weight (LAW). ) LAW ¼

SR 12 mg  100 ¼  100 ¼ 240 mg % MAE 5

Amount of bacitracin needed: 1 ounce ¼ 28:35 g ) 200 units=g  28:35 g ¼ 5670 units=50 units=mg ¼ 113:4 mg < LAW Thus, aliquot method has to be used. Before proceeding to the next step, let us think of how to implement the aliquot method. Remember that you are NOT supposed to mix the drug with lactose or starch unless it is clearly stated in the subscription of the medication order. Step 2: You decided to mix bacitracin with petrolatum (vazeline®) to make the stock ointment from which an aliquot (1 oz) will be used to compound the prescription. The dosage form is a small jar containing 1 oz of the jelly mixture. The aliquot is the final mass of the product. The error associated with 28.35 g should then be calculated and subtracted from the maximum allowable error to determine the new adjusted MAE. (continued)

2.4 Calculating the Error Associated with the Drug Dose When Two. . .

Example 2.8 (continued) 12 mg  100 ¼ 0:04% %Error ¼ 28,350 mg

)

41

adjusted MAE ¼ 5  0:04

¼ 4:96% ) LAW ¼

12 mg  100 ¼ 242 mg 4:96

Step 3: Set up the appropriate proportion for the determination of the stock mixture, as shown below: 113:4 mg of Bacitracin 242 mg of Bacitracin ¼ ) S ¼ 60,500 mg 28,350 mg of mixture S of stock mixture Pet:Album:needed ¼ 60,500 mg  242 mg ¼ 60,258 mg

Before we close this chapter, I would like to include one more example to demonstrate the significance of the error propagation theory in calculations related to the uncertainty of the drug concentration and drug dose in a mixture.

2.4

Calculating the Error Associated with the Drug Dose When Two-Component Formulations Are Measured on Balances of Different Sensitivity

Example 2.9: (Savva 2016) Fill the prescription given in Example 2.7 with a maximum allowable error of 5%, this time however using more sensitive balances for the measurement of drug, lactose, and capsule, dx ¼ 1 mg, dl ¼ 100 mg, and dm ¼ 1 mg, respectively. ℞ Prednisone (x) Lactose (l ) M. et. Ft. caps no. i, D.T.D. caps no. xii Sig.: Cap. i. t.i.d. p.c.

q.s ad

0.002 0.300

(continued)

42

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.9 (continued) Solution The quantities needed to fill 12 capsules are 24 mg, 3576 mg, and 3600 mg of drug, lactose, and mixture, respectively. We first determine the error associated with the capsule measurement and subtract it from the MAE to calculate the adjusted MAE related to the drug concentration in the powder mixture. The error incurred in the measurement, m with the balance SR ¼ dm ¼ 1 mg using Eq. (1.5) is m¼

1 mg  100 ¼ 0:33% 300 mg

) adjusted MAE ¼ 5  0:333 ¼ 4:667% We next use the same equation to determine the error associated with the drug dose if the drug is weighed directly on the balance. ) % Error ¼

1 mg  100 ¼ 4:167% < adjusted MAE 24 mg

According to our calculations there is no need to use the aliquot method to fill the capsules within the MAE. However, since lactose is mixed with the drug and it is measured on a balance with a different sensitivity, we need to use Eq. (2.3) to determine the uncertainty of the drug concentration in the powder mixture. 3576 dC l=x dl 100 ¼  dx þ ¼ 24  1 þ ¼ 0:04139 þ 0:02778 ¼ 0:06917 C xþl x þ l 3600 3600

) %Error ¼

dC  100 ¼ 6:917% > adjusted MAE C

According to the error propagation method, we have to use the aliquot method of weighing if we wish to fill the prescription within the MAE. Although, I am certain that Eq. (2.3) yielded the correct error associated with the drug concentration in the mixture, we can always check the reliability of the method as we did in a prior example. 24  1 Let C ¼ ð24  1Þ þ ð3576  100Þ The maximum negative (left) and maximum positive (right) are (continued)

2.4 Calculating the Error Associated with the Drug Dose When Two. . .

43

Example 2.9 (continued) 24  1 24 24 þ 1 < < ð24  1Þ þ ð3576 þ 100Þ ð24 þ 3576Þ ð24 þ 1Þ þ ð3576  100Þ )

23 24 25 < < 3699 3600 3501

or

0:006218 < 0:006667 < 0:007141

The % relative errors associated with each concentration value 0:006218  0:006667  100 ¼ 6:73% 0:006667 0:007141  0:006667  100 ¼ þ7:11% %error C þ ¼ 0:006667

%error C  ¼

) %error C ave ¼

j6:73j þ j7:11j ¼ 6:92% 2

Undoubtedly, the error associated with the drug concentration in the mixture is larger than the maximum allowable one, and therefore, we need to use an aliquot method of weighing. However, if you still doubt the result, we can calculate the drug amount in each capsule and prove our point again. D, D+, and Dave represent the minimum, maximum, and average drug mass in each capsule and are calculated by multiplying the mass of powder mixture (300 mg  1 mg) with the corresponding drug concentration. Dave ¼ M ideal  f ave ¼ 300 mg  0:006667 mg=mg ¼ 2:000 mg D ¼ M min  f  ¼ 299 mg  0:006218 mg=mg ¼ 1:859 mg Dþ ¼ M max  f þ ¼ 301 mg  0:007141 mg=mg ¼ 2:150 mg The % relative errors below are calculated using Eq. (1.4). %error D ¼

1:859 mg  2 mg  100 ¼ 7:04% 2 mg

%error Dþ ¼

2:150 mg  2 mg  100 ¼ þ7:47% 2 mg

) % error Dave ¼

j7:04j þ j7:47j ¼ 7:26% 2 (continued)

44

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

Example 2.9 (continued) Notice that the average error equals the error associated with the capsule measurement (m ¼ 0.33%) plus the error associated with the drug concentration in the stock mixture (q ¼ 6.94%). We can now proceed to fill the prescription using the aliquot method of weighing. Since the balances that measure the ingredients are different, we follow the procedure described in Example (2.5) to calculate the LAW of the drug x within the adjusted MAE of 4.667%. The system of equations that we need to solve are: l  dx þ dl df l dx dl ¼  þ ) x þ l ¼ x df f x xþl xþl f

x 24 mg 1 ¼ ¼ x þ l 3600 mg 150

)

150x ¼ x þ l and

149x ¼ l

Substituting the bottom equation into the top,  1 mg þ 100 mg 0:04667 ¼ 35:57 mg

150x ¼

149x x

)

150x ¼

249 mg 0:04667

)

LAW ¼ x

Carrying out the calculus, 149x ¼ l ) l ¼ 5299.93 mg and the total mixture which is also the stock powder S ¼ x + l ¼ 5335.5 mg. We could have also calculated the value of S using the proportion DA ¼ QS, where D ¼ 24 mg, A ¼ F ¼ 3600 mg, and Q ¼ LAW ¼ 35.57 mg, but it was much easier to use the equation we derived. Answer proof 35:57  1 Let C ¼ ð35:57  1Þ þ ð5299:93  100Þ The maximum negative (left) and maximum positive (right) are 35:571 35:57 35:57þ1 < < ð35:571Þþ ð5299:93þ100Þ ð35:57þ5299:93Þ ð35:57þ1Þþ ð5299:93100Þ )

34:57 35:57 36:57 < < 5434:5 5335:5 5236:5

or

0:006361 < 0:006667 < 0:006984

The % relative errors associated with the drug concentration in the mixture is (continued)

2.5 The Method of Geometric Dilution

45

Example 2.9 (continued) 0:006361  0:006667  100 ¼ 4:58% %error C  ¼ 0:006667 0:006984  0:006667  100 ¼ þ4:75% %error C þ ¼ 0:006667 ) %error C ave ¼

j4:58j þ j4:75j ¼ 4:67% 2

The error associated with the drug in each capsule is Dave ¼ M ideal  Cave ¼ 300 mg 0:006667 mg=mg ¼ 2:000 mg D ¼ M min  C  ¼ 299 mg 0:006361 mg=mg ¼ 1:90 mg Dþ ¼ M max  Cþ ¼ 301 mg 0:006984 mg=mg ¼ 2:10 mg The % relative errors below are calculated using Eq. (1.4). %error D ¼

1:90 mg  2 mg  100 ¼ 4:90% 2 mg

%error Dþ ¼

2:10 mg  2 mg  100 ¼ þ5:10% 2 mg

) %error Dave ¼

2.5

j4:90j þ j5:10j ¼ 5:0% 2

The Method of Geometric Dilution

The method of geometric dilution is useful for mixing two or more solids or semisolid materials (cream, ointment, gel, etc.) that are present in very different quantities. The method is most common in small-scale laboratory preparations where a small amount of powder has to be mixed with a larger quantity of an ointment or a cream base. Incorporation of the smaller mass into a larger one is achieved by adding the larger mass material, gradually in small portions, into the smaller amount one. Each time, we add to the solid or semisolid present in smaller quantity, approximately equal amount of the larger mass substance. We mix thoroughly the two portions, and we repeat this process until the entire larger mass material is added to the mixture. The purpose is to achieve uniform dispersion of the drug present in much smaller amount in the large mass of the diluent. We must emphasize that although the method of geometric dilution is very old, it does not warrant the spontaneous and uniform mixing of the ingredients. Hundreds

46

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

of years ago, where mixtures were formed by hand, it was not easy to conduct analytical tests to assess the drug concentration, and the process of geometric dilution was a rational and convenient method. We know today that mixing of solids is a complex exercise that depends on multiple factors such as the mixing vessel, the granules porosity, size and moisture content, the blending mode, and mixing time. Generally, no method is infallible. Being scientists, we can never be left to chance and we always have to have complete control of the running processes. Uniform dispersion of the drug in the mixture has to be confirmed by analytical methods. Powders are placed into tumbling blenders, and at regular time intervals, samples are taken from different depths of the mixture mass, and the drug concentration is determined by sensitive analytical methods. When the drug concentration in all samples is equal to the desired value, we stop the mixing process and we proceed to the next step of dosage from development. Can this be done in a regular pharmacy store where analytical equipment and trained personnel are missing?

2.6

Concluding Remarks

The increasing complexity of the extemporaneous preparations coupled with high potency and narrow therapeutic window of many drugs imposes a more stringent quality checks by the Ministry of Health. Dosage forms manufactured by pharmaceutical companies are subjected to the strict rules of Good Manufacturing Practice (GMP). One of those rules requires weighing each different ingredient on designated separate scales. The United States Pharmacopeia-National Formulary (USP-NF) recommends that appropriate weights be chosen to limit the error to 0.1% for analytical assays (USP Chapter 41 2006). This is translated to a least weighable quantity (LWQ) equivalent to 6 g using a traditional prescription balance that has a sensitivity requirement (SR) equivalent to 6 mg. When the measurement is conducted on a more sensitive electronic balance with SR equal to 0.1 mg, LWQ becomes 100 mg. It is to be noted that, for extemporaneous compounded preparations, the USP permits errors as high as 10% (USP Chapter 795 2006). From the educational point of view, the aliquot method is an extremely useful tool for you to: 1. Understand the principles of balance operation 2. Understand the relationship between the SR of a balance and the MAE in a dose 3. Realize that successive independent measurements of the drug and excipients lead to complex error in the final drug concentration of a mixture 4. Refine your critical thinking, by having to consider all possible consequences of any action involved in the preparation of compounded medications Whenever the desired mass of a drug is less than the least allowable weight, you have two choices: buy a more sensitive balance or weigh a multiple of the requested drug quantity (USP Chapter 1176 2006). Every lab must have high sensitivity analytical balances, but getting a more sensitive balance will not solve the problem

Exercises

47

for long because sooner or later there will be lesser amounts of drugs to be measured. It is therefore a fact that the aliquot method is an indispensable tool for extemporaneous compounding and formulation research of active ingredients. In terms of formulation research and development, pharmaceutical scientists have to apply the rules that analytical chemistry has adopted, that is, using balances with sensitivity 1000 times the mass to be weighed in order to limit the maximum possible error to 0.1%. Regardless of the implementation of this measure, given the complexity of the compounded prescriptions, often containing more than five ingredients, the application of error propagation theories to calculate the potential uncertainties of the drug concentration in pharmaceutical mixtures is a necessity.

Exercises 2.1. Calculate the % maximum probable error in drug concentration x in the following clinical batch manufactured in a GMP-certified laboratory that should produce 12 unit doses, each weighing 300 mg. Drug x and excipients y, w, and z were measured on pharmaceutical scales of sensitivity 1 mg, 0.05 g, 60 mg, and 4 mg, respectively, that is, dx ¼ 1 mg, dy ¼ 50 mg, dw ¼ 60 mg, and dz ¼ 4 mg. The 12 capsules were measured on a balance of sensitivity dm ¼ 1 mg. Drug x Excipient, y Excipient, w Excipient, z

2 mg 90 mg 200 mg 8 mg

Mix and make one capsule. Dispense a total of 12 capsules. (Answer: 7.64%) 2.2. Calculate the least allowable weight (LAW) of drug x in Exercise 2.1 that can be weighed on the specified electronic balances, with a maximum permissible error (MAE) of 2%. (Answer: x ¼ 105 mg, y ¼ 4724.6 mg, w ¼ 10,499 mg, z ¼ 420 mg) 2.3. Determine the sensitivity of the balance that drug x needs to be weighed on if the MAE of drug x concentration in the mixture of Exercise 2.1 is 4%. The SR of the other balances are as described in that exercise. (Answer: 121.6 μg) 2.4. Determine the sensitivity of the balance that drug x needs to be weighed on if the MAE of drug x concentration in the mixture of Exercise 2.1 is 2%. The SR of the other balances are as described in that exercise. (Answer: The formulation cannot be prepared within a MAE of 2%)

48

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

2.5. A GMP-certified lab prepared the prescription shown in Example 2.7 with a 5% maximum allowable error. Drug was measured on a balance with an SR ¼ 0.1 mg, lactose was measured on a balance of SR equal to 0.1 g, and the dosage form was measured on a balance with an SR equal to 6 mg. (a) Calculate the % maximum possible error associated with Prednisone concentration in the compounded prescription. (b) Calculate the minimum allowable drug quantity that conforms to the maximum allowable error. (c) Determine the sensitivity of the balance that drug x needs to be weighed on to achieve an MAE in drug x concentration of 5%. (d) Determine the sensitivity of the balance that lactose needs to be weighed on to achieve an MAE in drug x concentration of 5%. (Answer: (a) 3.2%; (b) 25.23 mg; (c) 0.0537 mg; (d) 93.1 mg) 2.6. Fill the prescription of Example 2.6 with 8% maximum potential error. What is the LAW, the aliquot amount, the stock mixture, and the lactose mass? (Answer: 150 mg, 150 mg, 937.5 mg, 787.5 mg) 2.7. Explain how you should weigh 32 mg of a drug with an error equal to 5%, using starch as the diluent, on a balance that has a sensitivity equal to 8 mg. The ratio of drug to starch in the prescription is 1/75. (Answer: Aliquot, A ¼ 2432 mg; Q ¼ 171.27 mg, S ¼ 13016.3 mg, starch ¼ 12,845 mg) 2.8. You filled a prescription that called for 30 capsules of drug D using a balance of sensitivity equal to 9 mg. You weighed an aliquot quantity that was equal to the total powder mixture needed to fill the 30 capsules. The error associated with the drug q is 5%. The total weight of the stock powder made from drug and lactose was 41.539 g. After the capsules were filled using a capsule filler machine, it was found that the maximum error in the dose is 8%. (a) (b) (c) (d)

What is the weight of each capsule? What is the aliquot amount? How much drug is needed to fill the prescription? Calculate the drug quantity per capsule. Answer: (a) 300 mg; (b) 9 g; (c) LAW (Q) ¼ 180 mg; (d) 1.3 mg)

2.9. ℞ Drug D Lactose M. et ft. caps no i, d.t.d caps no XL Sig. Caps i b.i.d. p.c. and h.s.

q.s. ad

0.350

Exercises

49

The prescription above was prepared using the aliquot method of weighing with a 6% maximum potential error. The total weight of the stock powder was 44.3 g, and all ingredients were weighed on a class III prescription balance that had an SR equal to 7 mg. (a) (b) (c) (d)

Calculate the total powder mixture needed to fill 40 capsules. How much drug is needed to fill the prescription? Calculate total lactose used to fill the prescription. What is the daily dose? (Answer: (a) 14 g; (b) 175 mg; (c) 44.125 g; (d) 4.15 mg)

2.10. Pindolol could be used to treat hypertensive patients that also suffer from heart failure because it is a beta-blocker with intrinsic sympathomimetic activity (partial agonist), and as such, it lowers the blood pressure without decreasing cardiac output. Given the following prescription: ℞ Pindolol Lactose M. et ft. caps no i, d.t.d caps no XX Sig. i cap b.i.d., 10 days

ad

0.005 0.150

(a) Describe how you would fill the prescription above with a maximum potential error of 10% using a balance that has sensitivity requirement equal to 5 mg. (b) What is the daily dose? (c) What is the maximum potential error in the daily dose? (Answer: (a) aliquot preparation is not needed; (b) pindolol 10 mg; (c) 8.33%) 2.11. Therapeutic doses of clonidine for cases of hypertension are usually between 0.2 and 1.2 mg daily. Explain how you could fill the prescription below, with a 3% error using a balance that has sensitivity requirement equal to 6 mg. ℞ Clonidine Lactose M. et ft. caps no i, d.t.d caps no XVI Sig. i cap b.i.d., 8 days

ad

0.001 0.150

(Answer: Prescription cannot be filled within the allowable error. A more sensitive balance is needed to fill the prescription) 2.12. The suppository form prescribed below is known to provide a longer duration of pain relief as compared to the corresponding oral administration of capsules mainly due to a better absorption of the drug and a less extensive metabolism (absence of a first pass effect). Describe how you would fill the prescription with a 5% error in the dose if the sensitivity requirement of the balance is 3 mg.

50

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures

℞ Hydromorphone HCl Cocoa butter M. et ft. supp. No. i, d.t.d. supp. # XII Sig: Insert 1 suppository rectally every 12 h

q.s. ad

0.15% w/w 2g

(Answer: LAW ¼ 61.86 mg)

Additional Exercises 2.13. Prepare the prescription below using an electronic balance that has an SR of 0.1 mg, with a maximum potential error in the dose of 1%. ℞ Drug D Infuse 1 L in D5W

4 μg/mL

2.14. Combination of thiazide and potassium-sparing diuretics is usually indicated to patients with hypertension or congestive heart failure who develop hypokalemia when thiazides or other kaliuretics are used alone. Explain how you should fill the prescription below, using the aliquot method of weighing on a prescription balance that has a SR ¼ 4 mg with a 4.5% max allowable error in the dose. ℞ Amiloride HCTZ Lactose M. et. Ft caps # i, d.t.d caps no 14 Sig: i caps daily

q.s. ad

5 mg 50 mg 200 mg

(Answer: Amiloride LAW ¼ 164 mg) 2.15. The formula below contains a decongestant, antihistaminic, and an opioid analgesic. It is commonly used to treat cough and to relieve congestion due to common cold or flu. ℞ Phenylephrine Chlorpheniramine Hydrocodone bitartrate

5 mg/tsp 2 mg/tsp 2.5 mg/5 mL (continued)

Exercises ℞ Sucrose Purified water Sig: i tsp. b.i.d and h.s.

51

60 g 120 mL

ad

(a) Describe how you would fill the prescription with a 5% maximum potential error in the dose, if the sensitivity requirement of the balance is 6 mg? (b) How much sucrose did the pharmacist use all together? (c) How long is the medication going to last? (Answer: (a) LAW for chlorpheniramine and hydrocodone bitartrate ¼ 240 mg; (b) 60 g; (c) 8 days) 2.16. You prepare the prescription shown below for a diabetic patient who has a cough and suffer from nasal congestion, symptoms of a common cold. ℞ Phenylephrine Chlorpheniramine Hydrocodone bitartrate Purified water Sig: i tsp. b.i.d and h.s.

5 mg/tsp 2 mg/tsp 2.5 mg/5 mL 120 mL

ad

(a) Describe how you would fill the prescription below with 5% error in the dose, if the sensitivity requirement of the balance was 6 mg. (b) What is the error in the dose? (Answer: (a) Use a liquid aliquot for the chlorpheniramine and hydrocodone; (b) 5%) 2.17. The combination medicine below is used to relieve moderate to severe pain. Describe how you would fill the prescription with a 2% error in the dose if the sensitivity requirement of the balance was 5 mg. Is this method cost-effective? Hydrocodone bitartrate Acetaminophen M. et ft. caps# i, d.t.d caps no 40 Sig: i caps q.6 h. for the pain

2.5 mg 500 mg

(Answer: LAW ¼ 497.5 mg, aliquot ¼ 20,100 mg, total acetaminophen ¼ 99,502.5 mg) 2.18. (a) Fill the prescription below with a 3% maximum potential error in the dose using a balance that has a sensitivity requirement of 6 mg.

52

2 Error Assessment of Drug Dose in Pharmaceutical Mixtures ℞ Oxycodone HCl Acetaminophen M. et div. Caps# 30 Sig: i caps q.8 h. for the pain

0.105 9.75

(b) What is the oxycodone daily dose? (c) How long is the medication going to last? (Answer: (a) LAW ¼ 511.28 mg of oxycodone HCl; (b) 10.5 mg; (c) 10 days) 2.19. You planned an experiment on animals to investigate whether there is a synergistic action of anti-inflammatory drugs against osteoarthritis. You have to prepare the formulation given below, mix each dose with food and then feed the five arthritic rats (average weight ¼ 350 g) that participate in the study. Etoricoxib Indomethacin Starch M. et ft. powders # i, d.t.d powders # 25

q.s. ad.

286 μg/kg 0.25 mg 0.250 g

Assuming that the sensitivity of the analytical balance available in the laboratory is 0.1 mg, answer the following if the maximum allowable error in the dose is 1%. (a) Calculate the LAW in order to prepare the powders within the given maximum allowable error. (b) How much starch is needed to prepare the formula? (c) What is the % error associated with the measurement of indomethacin before and after scaling-up the formula? (Answer: (a) 10.42 mg; (b) 26.015 g; (c) 1.6% and 0.4%) 2.20. Having reviewed your plan, your advisor suggested that you should include “control” experiments to provide a basis for comparison. Since the goal of the original plan was to investigate possible synergistic anti-inflammatory action (the total effect should be greater than the sum of the individual effects), the control experiments have to include each drug alone at twice as much dose compared to that in Exercise 2.7. The weight and number of the animals included in each study, the SR of the balance and percentage error are as stated in Exercise 2.7. Formulation #2: Etoricoxib Starch M. et ft. powders # i, d.t.d powders # 25

q.s. ad.

572 μg/kg 0.250 g

References

53

Formulation #3: Indomethacin Starch M. et ft. powders # i, d.t.d powders # 25

q.s. ad.

0.50 mg 0.250 g

(a) What is LAW for formulation #2? (b) Calculate the starch needed to prepare formulation #2. (c) What is % error associated with indomethacin powders in formulation #3? (Answer: (a) 10.42 mg; (b) 13,002 mg; (c) 0.84%)

References M. Savva, Uncertainty analysis of drug concentration in pharmaceutical mixtures. Anal. Chem. Insights 2, 1–3 (2006) M. Savva, A scientific analysis of the aliquot method of weighing. Int. J. Pharm. 5, 675–679 (2015) M. Savva, Error assessment of drug concentration in multicomponent pharmaceutical mixtures. Pharmaceut. Technol. 40-2, 26–29 (2016) Statement by the Acting Director of the Center for Drug Evaluation and Research, U.S. Food and Drug Administration, Department of Health and Human Services on Hearing: Federal and State Role in Pharmacy Compounding and Reconstitution: Exploring the Right Mix to Protect Patients, before the Senate Committee on Health, Education, Labor, and Pensions. http:// www.fda.gov/ola/2003/pharmacycompound1023.html. Accessed 28 Jan 2005 USP Chapter 1176: Prescription balances and volumetric apparatus. In the The United States Pharmacopeia, 29th rev., Rockville, MD, The United States Pharmacopeial Convention, 2006: p. 3020 USP Chapter 41, Weights and balances, in The United States Pharmacopeia, 29th rev edn., (The United States Pharmacopeial Convention, Rockville, MD, 2006), p. 2499 USP Chapter 795, Pharmaceutical compounding—nonsterile preparations, in The United States Pharmacopeia, 29th rev edn., (The United States Pharmacopeial Convention, Rockville, MD, 2006), p. 2733

Chapter 3

Density and Specific Gravity

Abstract The use of specific gravity and density from a pharmaceutical point of view is described in this chapter. You will learn how to measure, calculate, and apply the specific gravity to conveniently determine relative quantities of ingredients that are part of a liquid or a semisolid pharmaceutical dosage form. The importance of units as part of measurement was discussed in Chap. 1. In this chapter, I demonstrate the importance of units from a different perspective by showing how relative numbers of a pharmaceutical parameter change when the corresponding units are changed. I also describe density as a material property that defines the mass of a material in its own volume. Through this definition, you are better equipped to understand the concept of drug concentration discussed in Chap. 4. Keywords Specific gravity · Density · Gravimetric units · Apothecary units

Learning Objectives After reading this chapter, you should be able to: • Understand the difference between density and specific gravity • Convert density of materials from gravimetric to apothecary units • Use the density of pure ingredients to determine the specific gravity of liquid dosage products

In order to calculate the density of any particular material, we divide its weight by the volume it occupies. Although the System International (SI) unit of density is kilogram per cubic meter (kg/m3), in pharmaceutical sciences, density is most commonly expressed in gram per cubic centimeter or per milliliter.

Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_3) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_3

55

56

3 Density and Specific Gravity

The density of gases is largely dependent on temperature and pressure. Increasing temperatures would cause an expansion of gaseous substance and a lowering of their density, whereas increased pressures would lead to more dense gases per unit volume. On the contrary, the density of other relatively non-compressible materials such as solids and liquids is not significantly affected by small changes in temperature and pressure. Specific gravity, also called relative density, is the ratio of the weight of a unit volume of some substance to the weight of an equal volume of purified water or some other standard substance at a stated temperature. Since the density of purified water is 1 g/cm3, the difference between density and specific gravity is one of units. In other words, the specific gravity is unitless, and it is equal to density when the later is expressed in metric units (g/cm3 or g/mL).

3.1

Mathematical Definitions and Equations

density ðd Þ ¼

mass ðmÞ volume ðV Þ

ð3:1Þ

The mass (m) could be expressed as kilograms (kg), pounds (lb), ounces (oz), grains (gr), grams (g), milligrams (mg), micrograms (μg), nanograms (ng), etc. The volume of the substance (V ) could be expressed in cubic meters (m3), liters (L), cubic centimeters (cm3), milliliters (mL), microliters (μL), nanoliters (nL), etc. The specific gravity of a substance (s) is expressed by Eq. (3.2). ms s ρs ¼ mVH2O ¼

V H2O

ds dH2O

¼

 ds g  ¼ ds in mL 1 g=mL

ð3:2Þ

ρs ¼ specific gravity of substance ms ¼ mass of substance Vs ¼ equal volume of substance to that of purified water (VH2O)

3.2

Specific Gravity of Liquids and Solids

The specific gravity of liquids can automatically be determined with a hydrometer or manually in several steps using a pycnometer. A pycnometer is typically a vessel usually made of glass or stainless steel that is calibrated to hold a given volume of a liquid with high accuracy. The specific gravity of a liquid can be measured in few easy steps as follows: First, the empty dry pycnometer is weighed on an analytical balance (Wempty). Second, the pycnometer is filled with purified water up to the

3.3 Application of Density and Specific Gravity

57

calibration mark and weighed again (WH2O). The difference between the two is the mass of the water. The volume of the water in the pycnometer is calculated after substituting the density of the water at the particular temperature in Eq. (3.1). The procedure is then repeated with equal volume of a liquid whose specific gravity is to be found. Equation (3.2) allows one to determine the specific gravity of the unknown liquid. The two equal volumes cancel out, and therefore, ρs ¼ ms/mH2O.

3.3

Application of Density and Specific Gravity

Density or specific gravity can be used to determine the volume or mass of a material if a liquid is written in units of mass or more rarely a solid is written in units of volume. Knowing the volume that a particular mass occupies could help us determine the appropriate container to store or dispense the medication. Always remember that if we have two equal masses of a substance, the one with the bigger density will occupy less space. Example 3.1 Calculate the volume occupied by 60 g of olive oil if the specific gravity of olive oil is 0.92. Solution ρs ¼

moil V oil mH2O V H2O

¼

d oil doil g ¼ 0:92 ¼ mL dH2O 1 g=mL

)

d oil ¼ 0:92

g mL

But d¼

m d oil ¼ V d H2O

)

V oil ¼

moil 60 g ¼ g ¼ 65:2 mL d oil 0:92 mL

As mentioned previously, the specific gravity is not equal to the density when density is expressed in units other than g/cm3. To make sure we understand the difference, let us express the density of water in grains (gr) per fξ. Example 3.2 Express the density of water in gr per fξ. Solution dH2O ¼ 1 g/cm3 ¼ 1 g/mL. 1 gr ¼ 65 mg ¼ 0.065 g. 1 fξ ¼ 29.57 mL 1 gr

) d H2O

1 g 0:065 g mH2O 29:57 gr gr ¼ 455 ¼ ¼ ¼ 1fξ 0:065 f ξ fξ V 1 mL 29:57 mL

58

3 Density and Specific Gravity

Example 3.3 Calculate the specific gravity of the lotion whose formula is given below if exactly 5.4 mL of water was used to prepare the lotion. ℞ Drug X Olive oil Talc Purified water M. et ft. lotion

25% w/w 10 g 4 15 mL

q.s. ad

Solution Let us first try to understand and summarize the information given above: Total volume of lotion ¼ 15 mL (read interpretation of “q.s. ad” in Appendix A) Mass of water (dH2O ¼ 1 g/mL) added to make the lotion ¼ 5.4 g Drug X 25% w/w ¼ 25 g of drug X in 100 g of lotion (including drug X) The specific gravity of the lotion can be obtained by dividing the mass of lotion by the corresponding volume. The volume of the lotion is given (15 mL), but we have a little bit of trouble to calculate the total mass of the lotion as drug X is written in units of concentration. On the other hand, the mass of drug X in the lotion cannot be calculated because we do not know the total mass of the lotion. However, we know that the mass of oil, talc, and water is 19.4 g. If we just think of what 25% w/w means, we may come up with a solution. The numerator 25 g is equal to the amount of drug in the lotion (X), whereas the denominator 100 g is equal to the amount of all ingredients in the lotion, that is, 19.4 g + X. We can then construct the equation shown below: X 25 ¼ X þ 19:4 g 100 The above equation says that by adding X amount of drug, the total mass of the lotion becomes X + 19.4 g, and the concentration of drug in the lotion, X 25 which is Xþ19:4 g, is in turn equal to 100. Solving the above equation, 100 X ¼ 25 X þ 485

)

X ¼ 6:467 g:

Total mass of the lotion ¼ 6.467 g + 19.4 g ¼ 25.867 g ρlotion ¼

25:867 g g ¼ dlotion ¼ 1:725 15 mL mL

Exercises

59

Exercises 3.1. What size containers would you use to dispense 1.4 oz of essence oil that has a specific gravity equal to 0.71, used for aromatherapy. (Answer: 2 fξ capacity bottles) 3.2. Calculate the density of skimmed milk if 1 gallon of it weighs 3974 g. (Answer: 1.05 g/mL) 3.3. How many ounces of methyl iodide are present in 1 L if the specific gravity of methyl iodide is 2.28? (Answer: 80.42) 3.4. The density of ethanol is 0.726 g/cm3. Express this value in units of ounces per m 3. (Answer: 25,608.5 oz/m3) 3.5. The weight of 2 fluid ounces of water is 910 grains. What is the weight in grams of 1 fluidounce of peanut oil if the specific gravity of peanut oil is 0.92? (Answer: 27.21 g) 3.6. Calculate the weight in pounds of 2 pints of petrolatum if the specific gravity of petrolatum is 0.83. (Answer: 1.73 lb) 3.7. Calculate the volume difference in milliliters of 2 ounces of purified water and 2 fluid ounces of absolute ethanol (ρH2O ¼ 1; ρethanol ¼ 0.726). (Answer: 2.44 mL) 3.8. The following medication order was given for wart removal. ℞ Salicylic acid Mineral oil (ρmo ¼ 0.867) Zinc oxide Talc Purified water M. et ft. lotion Sig. Apply on each wart ½ mL with a syringe

40% w/w 9 mL aa q.s. ad

2 15 mL

(a) Calculate the mass of mineral oil in grams used to fill the prescription. (b) What would be the specific gravity of the lotion if exactly 4.23 mL of purified water were used to prepare it? (Answer: (a) 7.8 g; (b) 1.78)

60

3 Density and Specific Gravity

3.9. Given the following formula: Salicylic acid Mineral oil Zinc oxide Talc Ethanol 70%

20% w/w 18 mL aa q.s. ad

4 30 mL

(a) Calculate the amount of mineral oil in grams used to fill the prescription (ρmo ¼ 0.867). (b) Calculate the specific gravity of the lotion if exactly 8.45 mL of 70% ethanol was used to prepare it (ρethanol 70% ¼ 0.925). (Answer: (a) 15.606 g; (b) 1.31)

Chapter 4

Units of Concentration and the Salt Factor

Abstract Twelve different units of concentration that are commonly used in pharmacy, chemistry, and biology, plus the salt factor, are defined both descriptively and mathematically in this chapter. Conversions of concentration units are demonstrated in great detail with practice examples. A vast number of practice exercises extracted from the area of protein research and other ingenious uses of drugs in hospitals to treat patients unresponsive to conventional treatments were created for you to become proficient in concentration unit conversions. Also, practice problems describing prescriptions that can be used to treat diseases such as psoriasis, arthritis, cholesterolemia, skin hyperpigmentation, and acne will keep you locked on a very interesting text that relates units of concentration with formulation creativity, pharmacological activity and toxicity. Keywords Concentration · Density · Salt factor · Solute · Solvent · Solution · Mass per unit volume · Percent by weight · Percent weight per volume · Percent by volume · Molarity · Molality · Normality · Ratio strength · Parts per million · Parts per billion · Mole fraction · Activity

Learning Objectives After reading this chapter, you should be able to: • • • • •

Understand the difference between density and concentration Define solute, solvent, and solution Define and differentiate between units of concentration Perform various conversions of concentration units Determine relative quantities of ingredients in compounded prescriptions from their corresponding concentrations and vice versa • Define and use the salt factor to normalize drug dosage

Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_4) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_4

61

62

4 Units of Concentration and the Salt Factor

Concentration of a substance is traditionally defined as the amount of the substance in a homogeneous mixture over the amount of mixture or the amount of solvent (Eq. 4.1). Concentration ðCÞ ¼

mass ðmÞ volume ðV Þ

ð4:1Þ

Notice the striking similarity of Eq. (4.1) to Eq. (3.1). Density ðdÞ ¼

mass ðmÞ volume ðV Þ

ð3:1Þ

The difference between the two is that density refers to how much of the pure substance is present in a unit volume of that pure substance. For example, how many grams of water are present in 1 cm3 of pure water? Concentration on the other hand refers to how much of a substance is present in a unit volume of a mixture with other substances. For example, how much water is present in 1 cm3 of 5% w/v aqueous glucose solution? Furthermore, unlike density which is almost always expressed in units of g/mL, concentration is commonly expressed in multiple units. In pharmaceutical sciences, we are mostly dealing with solid dosage forms like tablets and capsules and semisolids like creams and ointments. It is important to define few terms before we proceed discussing the units of concentration.

4.1 4.1.1

Solute, Solvent, and Solution Solute

Solute is the component of the solution or the homogeneous mixture usually present in the lesser amount. In pharmacy, solute is usually the drug, and it could be a solid, liquid, and very rarely gas.

4.1.2

Solvent

Solvent is the component of a homogeneous mixture present in the greatest amount, and it could be solid, liquid, or semisolid. This component is the dispersion medium, or as we say in pharmaceutics, it forms the basis of the dosage form. For example, white petrolatum is the basis of an ointment, that is, the drug is mixed with white petrolatum as the ointment-forming component.

4.2 Concentration Units in Pharmaceutical Sciences

4.1.3

63

Solution

Solution is a homogeneous mixture consisting of the solute and the solvent. Solute and solvent could be solid or liquid. Units of mass are used when the solvent is solid or semisolid, whereas units of volume are used for liquid solvents. Examples of dosage forms where the solute is “uniformly” dispersed in a solid medium are powders, capsules, and tablets. Examples of pharmaceutical dosage forms in which the solute is “homogeneously” dispersed in a semisolid medium are creams, ointments, and gels. Frequently, solid-solutions are simply called physical mixtures or homogeneous mixtures. It is important to note that the concept of concentration does not apply to heterogeneous mixtures since the composition of the solute is not uniform throughout. Note: The terms uniform and homogeneous are synonymous. The only true spontaneous homogenous mixture that has content uniformity 100% is a liquid solution. All other “homogeneous” solid and semisolid mixtures are not true spontaneous homogeneous mixtures. The content uniformity of those mixtures could be close to 100%, but it can never be 100%. For our calculation purposes, we will consider pharmaceutical dosage forms like tablets, capsules, creams, and ointments, as non-spontaneous (some work needs to be done to be mixed) uniform mixtures with the concentration of the drug being constant throughout the mixture.

4.2 4.2.1

Concentration Units in Pharmaceutical Sciences Mass per Unit Volume

This kind of unit, for example, g/mL, g/dL, mg/cm3, μg/L, and gr/tsp, can be used to express the solubility of a drug in a solvent, the concentration of a drug in the blood of a patient or the drug dose. For example, the solubility of a drug in isotonic saline is 1 g/L, or the concentration of glucose in the blood of a diabetic patient is 600 mg/dL, or the concentration of a drug in a 5 mL dose is 2 gr/tsp.

4.2.2

Percent by Weight (% w/w)

The unit denotes the grams of solute in 100 g of mixture (or solution). It is commonly used to express the concentration of drugs or the strength of powder mixtures in ointments, gels, creams, lotions, powders, etc. For example, the strength of betamethasone ointment is 10%.

64

4 Units of Concentration and the Salt Factor

Betamethazone 10%w=w ¼

10 g of betamethazone 100 g of mixture

This unit is also used in chemistry to express the concentration of strong acids and strong bases. For example, HCL 37%w=w ¼

37 g of hydrochloric acid 100 g of solution

Notice that although concentrated hydrochloric acid (>10 M) is supplied as a liquid, its concentration is expressed as % w/w. To convert the grams of solution into milliliters, one needs to know the density or the specific gravity of the solution. How concentrated should an acid or a base be in order to be expressed in units of % w/w? There is really no rule. It all depends on manufacturer’s preferences, but usually concentrations higher than 10 M are expressed as % w/w.

4.2.3

Percent Weight per Volume (% w/v)

The unit denotes the grams of solute in 100 mL of solution. For example, 2% aspirin solution is equivalent to 2 g of aspirin in 100 mL of solution.

4.2.4

Percent by Volume (% v/v)

The unit denotes the milliliters of solute per 100 mL of solution. It is commonly used for liquid solutes mixed with a liquid solvent to form a solution. For example, alcohol USP is a 95% mixture of absolute ethanol in water, that is, Ethanol 95%v=v ¼

95 mL of ethanol 100 mL of solution

Recognize that in this case, the liquid solute ethanol is present in higher quantity than water (it does not always have to be).

4.2.5

Molarity (M)

The unit expresses the moles of solute in a liter of solution. For example,

4.2 Concentration Units in Pharmaceutical Sciences

NaCl 0:150 M ðisotonic salineÞ ¼

65

0:150 mol of NaCl 1 L of solution

The unit of molarity is sensitive to temperature because the density of a solution may change slightly with temperature due to an expansion of the volume of the solution. Notice also that when we use it in formulas as a unit, we write it as mol, whereas in a sentence we write it as moles.

4.2.6

Molality (m)

The unit is used to express moles of solute in a kilogram of solvent. For example, HCl 16 molal ¼

16 mol of HCl 1 kg of water

The unit of molality is frequently used in problems dealing with calculating colligative properties of solutions such as osmotic pressure and freezing point depression. This is very useful, as we shall see later, during the preparation of sterile solutions for injection or infusion. Careful adjustment of the tonicity of the liquid solutions to be infused systematically is required in order to avoid causing hemolysis (hypotonic solution) or crenation (hypertonic solution) of cells. The advantage of using molality instead of molarity is that unlike molarity, molality is independent of temperature, because mass does not change with temperature.

4.2.7

Normality (N)

The unit denotes the equivalents of solute per liter of solution. It is very important to understand that equivalents (Eq; the unit of EW) are always defined in terms of some chemical reaction (titration). For acid–base reactions, the equivalent (Eq) or the equivalent weight (EW) is defined as the weight in grams of an acid or a base required for furnishing or accepting exactly 1 mol of protons (H+), respectively. For oxidation–reduction reactions, the equivalent weight (EW) is defined as the weight in grams of oxidizing or reducing agent required for accepting or furnishing exactly 1 mol of protons (H+), respectively. For example, NaCl 150 mN ¼

150 mEq of NaCl 0:150 Eq of NaCl ¼ 1 L of solution 1 L of solution

The EW of a substance is called the mass of 1 equivalent (Eq) just as the MW of a substance is the mass of 1 mole. EW and MW are related through the following equation:

66

4 Units of Concentration and the Salt Factor

g  MW mol g EW ¼ Eq n


ð4:2Þ

where n is the number of moles of protons furnished by 1 mole of acid or accepted by 1 mole of base participating in the reaction. From Eq. (4.2), it is easy to recognize that n is expressed in units of Eq/mol. If the molecule is a strong electrolyte, that is, a strong base, a strong acid, or a salt, made by a weak base and a strong acid or a weak acid and a strong base, n can be figured out by the maximum valence of the molecule multiplied by the stoichiometric coefficient of the particular atom involved. For example, n for H2SO4, HCl, and MgCl2 is 2, 1, and 2, respectively (see Chap. 10 for more details on electrolytes). The relationship between molarity (M) and normality can be derived from Eq. (4.2) as follows: First solve Eq. (4.2) for 1/EW, )

1 1 ¼n EW MW

or

Eq Eq mol ¼  g mol g

Multiplying both sides by g/L, Eq g Eq mol g Eq Eq mol  ¼   ) ¼  g L mol g L L mol L But, Eq/L ¼ N

and

mol/L ¼ M ) N¼nM

4.2.8

ð4:3Þ

Ratio Strength

The unit is used to express 1 part of the solute per part of the solution. Parts of solids and semisolids are expressed in grams while parts of liquids are expressed in milliliters. For example, 1 g of hydrocortisone 100 g of cream 1 g of aspirin (b) Aspirin solution 0:1 : 100 ¼ 1000 mL of solution (a) Hydrocortisone cream 1 : 100 ¼

Notice how the aspirin solution in (b) was written to conform with the definition of the unit, that is, the numerator has to be unity (1).

4.2 Concentration Units in Pharmaceutical Sciences

4.2.9

67

Parts per Million (ppm) by Mass

The unit denotes the parts of solute in a million parts of solution. It is commonly used to express the concentration of substances in very dilute solutions, that is, levels of pollutants in air or amount of contaminants in water or concentration of toxic substances in body fluids. Parts of solids are expressed in grams, whereas parts of liquids are expressed in milliliters. For example, the mercury concentration of potable water in New Jersey was found to be 3 ppm. 3 parts of Hg 3 g of Hg 3 g of Hg ¼ ¼ 106 parts of water 106 g of water 106 mL of water 3 g of Hg ¼ 6 10 mL of solution

3 ppm Hg ¼

Notice that the density of the solution can be approximated by the density of the solvent, since the concentration of the solute is very small.

4.2.10 Parts per Billion (ppb) by Mass The unit denotes the parts of solute in a billion parts of solution. For example, the lead contamination of soil in an industrial area was found to be equal to 3 ppb. 3 ppb Pb ¼

3 parts of Pb 3 g of Pb ¼ 9 9 10 parts of soil 10 g of soil

Soil is a mixture of about 50 % inorganic matter and 50 % water, air and organic matter.

4.2.11 Mole Fraction (x) The unit denotes moles of solute per mole of all the components that constitute the mixture. xsolute ¼

moles of solute nsolute ¼ moles of all components nsolute þ nall other components

In a more general form, Eq. (4.4) can be written as

ð4:4Þ

68

4 Units of Concentration and the Salt Factor

xi ¼

ni ni ¼ ni þ    þ nk ntotal )

k X

because

k X

ni ¼ ntotal

i¼1

xi ¼ x1 þ x 2 þ    þ xk ¼ 1

ð4:5Þ

i¼1

The sum of all the mole fractions of a mixture is always equal to 1 (Eq. 4.5).

4.2.12 Activity Units This unit is mostly used for proteins, smaller peptides, and enzymes, but also for other smaller synthetic or natural molecules such as antibiotics and anticancer agents. The actual definition of the activity unit is dependent on the physiological function of the molecule. For example, the catalytic activity of an enzyme is defined as the number of substrate molecules transformed per unit time by a single enzyme molecule under standard conditions when the enzyme is the rate limiting factor. Similarly, the unit activity of the antibiotic bacitracin is defined as the amount of bacitracin needed to kill a certain population of a particular bacterial strain. The unit of activity provides information additional to purity. For instance, why would the same enzyme from two different suppliers that is provided at identical purity, have different activity? The enzyme from these two sources maybe of identical purity, but the conditions under which it was purified and the composition of solutions used to reconstitute them may be affecting their active conformation during renaturation. Also, the conditions under which the drug activity is evaluated by the two manufacturers maybe different. Pharmaceutical scientists and other health practitioners have to be very careful when reading unit activity information of drugs. For example, it could happen that a particular antibiotic is 99+% pure presenting a very high unit activity for a particular bacterium but a very moderate activity against a different bacterial strain. It wouldn’t be wise to use this particular antibiotic to treat patients that suffer an infection from a bacterium which is not very sensitive to the antibiotic. It is therefore, important to read not only the unit of activity, but also the conditions under which the drug activity is evaluated. Conversion of some to the common units used in pharmaceutical sciences can be found in Appendix C.

4.3

Practice Examples

Example 4.1 Calculate the mole fraction of KCl in 100 mL of 0.9% solution. Specific gravity of the solution is 1.03 (AWs: K ¼ 39; Cl ¼ 35.5; H ¼ 1; O ¼ 16). (continued)

4.3 Practice Examples

69

Example 4.1 (continued) Solution The formula for the mole fraction of KCl is: xKCl ¼

nKCl nKCl þ nH2O

Step 1: Calculate first the grams of KCl in 100 mL of 0.9 % w/v solution and divide your answer with the MW of KCl to convert the grams into moles of KCl. 0:9 g  100 mL ¼ 0:9 g 100 mL

)

0:9 g ¼ 0:0121 mol 74:5 g=mol

Step 2: Use the specific gravity of the solution to calculate the mass of 100 mL of solution. Then calculate the grams of water in the solution and divide by the MW of the water to find out the # of moles of water in the solution. 100 mL  1:03 g=mL ¼ 103 g of solution KCl þ water ¼ 103 g

)

0:9 g þ water ¼ 103 g

)

water ¼ 103  0:9 ¼ 102:1 g nH2O ¼ xKCl ¼

102:1 g ¼ 5:672 mol 18 g=mol

nKCl 0:0121 mol ¼ 0:00213 ¼ nKCl þ nH2O ð0:0121 þ 5:672Þ mol

It can be shown that (Eq. 4.5) xKCl þ xH2O ¼

nKCl nH2O nKCl þ nH2O þ ¼ ¼1 nKCl þ nH2O nKCl þ nH2O nKCl þ nH2O

Therefore, xH2O ¼ 1  xKCl ¼ 1  0.0213 ¼ 0.99787

Example 4.2 Convert 5 ng/μL into: (a) (b) (c) (d) (e)

ng/mL μg/μL μg/mL mg/μL ng/mL (continued)

70

4 Units of Concentration and the Salt Factor

Example 4.2 (continued) (f) mg/L (g) g/μL (h) oz/mL (i) g/L (j) mg/fξ (k) oz/mL (l) % w/v (m) Ratio strength (n) ppm Solution The best way to solve these problems is to take note of the initial units and the final units and multiply the initial units with a conversion factor, so that you end up with the final units. (a) The initial unit is μL and the final unit is mL. Recognize that 1 μL ¼ 103 mL. The conversion factor is therefore 3 10 mL/μL. Go ahead and multiply the denominator by the conversion factor (103 mL/μL). )5

ng 5 ng ng ng ¼ ¼ 5000 ¼ 5  103 μL μL  103 mL=μL mL mL

(b) Recognize that 1 ng ¼ 103 μg

)

conversion factor ¼ 103 μg=ng

ng 5 ng  10 ¼ )5 μL μL

3 μg ng

¼ 0:005

μg μL

5 ng  103 μg ng μg ng ¼ (c) 5 ¼5 μL μL  103 mL=μL mL (d) Recognize that 1 mg ¼ 106 ng )5

(e) 5

ng 5 ng  10 ¼ μL μL

6 mg ng

ng 5 ng  10 ¼ μL 103 mL

)

¼ 5  103

1 ng ¼ 106 mg

6 mg ng

¼ 5  106

mg μL

mg mL (continued)

4.3 Practice Examples

71

Example 4.2 (continued) 6 mg ng

ng 5 ng  10 ¼ (f) 5 μL 106 L (g) Recognize that

¼5

mg L

1 g ¼ 109 ng )5

)

1 ng ¼ 109 g

ng g ¼ 5  109 μL μL

(h) Recognize that 1 dL ¼ 0:1 L ¼ 0:1  106 μL ¼ 105 μL )5

)

1 μL ¼ 105 dL

ng 5  109 g g ¼ ¼ 5  104 5 μL dL 10 dL

(i) Recognize that 1 L ¼ 106 μL ) 1 μL ¼ 106 L )5

ng 5  109 g g ¼ ¼ 0:005 μL L 106 L

(j) Recognize that 1 fξ ¼ 29.57 mL ) 1 mL ¼ (1/29.57) fξ Conversion factor ¼ (1/29.57) fξ/mL Since fluidounces are expressed in terms of milliliters, convert μL to mL first.

)5

ng 5  106 mg 5  103 mg 5  103  29:57 mg mg ¼ ¼ 0:1478 ¼ ¼ 3 fξ μL f ξ fξ 10 mL mL  1 29:57 mL

(k) Recognize that 1 oz ¼ 28.35 g ) 1 g ¼ (1/28.35) oz ) Conversion factor ¼ (1/28.35) oz/g Convert first ng into g, since ounces are expressed in terms of grams. (continued)

72

4 Units of Concentration and the Salt Factor

Example 4.2 (continued) )5

6 1 oz ng 5  109 g 5  106 mg 5  10 mg  28:35 g ¼ 3 ¼ ¼ ¼ 1:76  107 oz=mL μL 10 mL mL mL

(l) Recognize that % w/v ¼ g of drug in a 100 mL of solution Convert ng of drug into g and μL of solution into mL. )5

ng 5  109 g ¼ ¼ 5  106 g=mL μL 103 mL

Multiplying and dividing at the same time by 100, )

5  106 g  102 5  104 g ¼ 0:0005%w=v ¼ 100 mL mL  102

One could also use the answer in part (e) 0.005 mg/mL ¼ 5106 g/mL and set up a ratio to determine % w/v of drug as shown below:

)

5  106 g X ¼ mL 100 mL

)

X ¼ 0:0005 g

or

0:0005%w=v

(m) Recognize that ratio strength of drug is to be expressed as 1 g/mL of solution. )5

ng 5  109 g ¼ ¼ 5  106 g=mL μL 103 mL

Then set the numerator equal to 1 g and determine the value of the denominator in mL. )

5  106 g 1g ¼ mL X

)

X ¼ 2  105 mL and ratio strength ¼ 1 : 2  105

We can easily verify our answer as shown below: 1g 5  106 g ¼ mL 2  105 mL (continued)

4.3 Practice Examples

73

Example 4.2 (continued) (n) Recognize that ppm of drug is grams of drug per million milliliter of solution. Since this is a very dilute aqueous solution of the drug, we can consider the density of the solution equal to the density of purified water, that is, 1 g/mL. Convert ng of drug into g and μL of solution in mL. )5

ng 5  109 g ¼ ¼ 5  106 g=mL μL 103 mL

Multiplying and dividing by 106, )

5  106  106 g 5g ¼ 6 ¼ 5 ppm 6 10 mL 10 mL

Example 4.3 Calculate the quantity of betamethasone present in 8 oz of 2.45% betamethasone cream. Solution Step 1: Recognize that 2.45% of betamethasone cream (semisolid) is equivalent to 2.45 g of betamethasone in 100 g of cream. Betamethazone 2:45% w=w ¼

2:45 g of betamethazone 100 g of cream

Step 2: Convert the ounces into grams and multiply with the % w/w. 8 oz ¼ 8 oz∙28.35 g/oz ¼ 226.8 g )

2:45 g  226:8 g ¼ 5:56 g 100 g

Thus, 5.56 g of betamethasone are present in 8 oz of 2.45% w/w cream.

74

4 Units of Concentration and the Salt Factor

Example 4.4 Convert HCl 37% into % w/v. Specific gravity of the hydrochloric acid solution is 1.23. Solution Recognize that HCl 37% is a solution of concentrated acid and is therefore given as % w/w. We are going to use the density to convert the mass of the solution into volume. Also, remember that specific gravity is for the solution (NOT for the solute), and it equals the density when the latter is expressed in grams per cubic centimeters or per milliliters, that is, dHCl solution ¼ 1.23 g/mL. ) %w=w ¼

37 g 100 g 1:23 g=mL

¼

37 g  1:23 g=mL 45:51 g of HCl ¼ ¼ 45:51%w=v 100 g 100 mL

Alternatively, set up a ratio to calculate grams of hydrochloric acid in 100 mL of solution (% w/v). 37 g of HCl X ¼ 81:3 mL 100 mL

)

X ¼ 45:51 g of HCl

or 45:51% w=v

Example 4.5 Calculate how many grains (gr) of antipyrine are present in 1 fξ of 2.5% w/v antipyrine solution in 97% glycerin. Solution Step 1: Recognize that antipyrine 2.5% w/v is equal to 2.5 g in a 100 mL of solution. The problem also states that the solution is made of 97% glycerin. This last information is redundant. All we need to know is the volume (100 mL), not the constistency of the antipyrine solution. Convert fξ into mL and multiply the result with the concentration of antipyrine. )

2:5 g  29:57 mL=f ξ ¼ 0:73925 g=f ξ 100 mL

Step 2: Convert g into gr. 1g  0:73925 g ¼ 11:37 gr 0:065 g Thus, 11.37 gr of antipyrine are present in 1 fξ of 2.5% w/v antipyrine solution.

4.3 Practice Examples

75

Example 4.6 Convert antipyrine 2.5% w/v to antipyrine w/w. The specific gravity of the antipyrine solution is 1.25. Solution Antipyrine 2.5% w/v is equivalent to 2.5 g of drug in a 100 mL of solution. To convert % w/v to % w/w, we need to multiply the solution’s volume (denominator) by the specific gravity. This is equivalent to dividing the numerator by the specific gravity. It is very important to understand that the specific gravity refers to the solution (NOT to the solute). Thus, all arithmetic manipulations are applied to the denominator which expresses the volume of solution. 2:5 g

% w=v ¼

2:5 g 2g 1:25 ¼ 2%w=w ¼ g ¼ 100 mL  1:25 mL 100 g 100 g

Thus, 2.5% w/v of antipyrine is equivalent to 2% w/w antipyrine.

Example 4.7 Convert H2SO4 96% into (a) molality, (b) % w/v, (c) molarity, (d) normality, and (e) mole fraction. The specific gravity of the sulfuric acid solution is 1.84 (AWs: H ¼ 1; O ¼ 16; S ¼ 32). Solution (a) Step 1: H2SO4 96% w/w (strong acid). First, find out the grams of sulfuric acid and grams of water that comprise the mixture (100 g). H2 SO4 þ water ¼ 100 g

)

96 g þ water ¼ 100 g

)

water ¼ 100 g  96 g ¼ 4 g Step 2: Divide grams of sulfuric acid by the MW and set up a ratio to calculate the moles of sulfuric acid in 1 kg of solvent. MWH2SO4 ¼ 2 þ 32 þ 64 ¼ 98 g=mol nH2SO4 ¼ )

96 g ¼ 0:98 mol 98 g=mol

0:98 mol X ¼ 4 g of solvent 1000 g of solvent

)

X

¼ 245 mol of H2 SO4 in 1 kg of water or 245 molal or 245 m (continued)

76

4 Units of Concentration and the Salt Factor

Example 4.7 (continued) (b) H2SO4 96% is a solution of concentrated acid and is therefore % w/w. The first step is to use the density to convert the mass of the solution into volume. Also, remember that specific gravity equals density when the latter is expressed in grams per cubic centimeters or per milliliters, that g . is, d H2SO4 ¼ 1:84mL % w=w ¼

96 g 100 g g 1:84mL

¼

96 g  1:84 176:6 g of H2 SO4 ¼ ¼ 176:6%w=v 100 mL 100 mL of solution

) H2 SO4 96%w=w is equal to 176:6%w=v:

(c)

176:6 g=98 g=mol X ¼ 100 mL of solution 1000 mL of solution ) X ¼ 18:02 M

or

18:02 Molar

or

18:02 mol=L

(d) N ¼ n M. But n (Eq/mol) ¼ 2 ) N ¼ 2 Eq/mol∙18 mol/L ¼ 36 Eq/L (e) Calculate separately the moles of sulfuric acid and the moles of water in the mixture. nH2SO4 ¼

96 g ¼ 0:98 mol 98 g=mol

xH2SO4 ¼ xH2O ¼

and

nH2O ¼

4g ¼ 0:222 mol 18 g=mol

nH2SO4 0:98 mol ¼ 0:815 ¼ nH2SO4 þ nH2O ð0:98 þ 0:22Þ mol

nH2O 0:222 mol ¼ 0:185 ¼ nH2SO4 þ nH2O ð0:98 þ 0:22Þ mol ) xH2SO4 þ xH2O ¼ 1

Example 4.8 The concentration of a drug in an ointment is 1:200 by weight. (a) Calculate how many grams of drug are there in 1 oz of ointment. (b) Convert the ratio strength into % w/w. (continued)

4.3 Practice Examples

77

Example 4.8 (continued) Solution (a)

1 g of drug X ¼ 200 g of ointment 28:35 g of ointment

(b)

1 g of drug X ¼ 200 g of ointment 100 g of ointment

)

X ¼ 0:142 g of drug

) X ¼ 0:142 g of drug in 100 g of ointement or 0:5%w=w

Example 4.9 According to WHO (World Health Organization), the concentration of arsenic (As) in drinking water should be no higher than 10 ppb. Calculate nanograms of As in a glass of water (250 mL). Solution Since 10 ppb is a very dilute concentration, we may assume that the density of the solution is equal to that of water, that is, 1 g/mL. 10 g of As 10 g  109 ng=g of As X ng ¼ ¼ 9 9 250 mL of water 10 g of water 10 mL of solution

) X

¼ 2500 ng ¼ 2:5 μg of As How do you feel about the WHO guidelines with regard to the concentration of As in drinking water now?

Example 4.10 A patient who suffers from arthritis was prescribed the formula shown below: ℞ Glucosamine Vitamin C Chondroitin α-lipoic acid Sucrose M. et ft. powder i, D.T.D. powders no. XVI Sig. Place contents in a glass and add 5 oz of water. Drink every other day

q.s. ad.

0.25 0.5 0.2 1 mg 0.05 oz

(continued)

78

4 Units of Concentration and the Salt Factor

Example 4.10 (continued) (Specific gravity of the solution (dose) is 1.01; MWs: vitamin C ¼ 176 g/mol; sucrose ¼ 342 g/mol) (a) Calculate sucrose required to make one powder. (b) Calculate the molality (m) of vitamin C per dose. (c) Calculate the normality of sucrose in a dose. Solution (a) Subtract from the mass of 1 powder (0.05 oz) the mass of all ingredients except sucrose. Mass of 1 powder ¼ 0:05 oz  28:35 g=oz ¼ 1:4175 g ) Sucrose for 1 powder ¼ 1:4175 g  0:25 g  0:5 g  0:2 g  0:001 g ¼ 0:4665 g (b) You are asked to calculate the concentration of vitamin C in a glass of water. Divide the gram quantity of vitamin C by its MW and normalize the answer for 1 kg. This concentration is different from the concentration of vitamin C in the powder. 0:5 g 176 g=mol

5 oz 

g 28:35oz

¼

X 1000 g of water

) X ¼ 0:02 moles per 1000 g of solvent ¼ 0:02 mol=kg of solvent ¼ 0:02 molal (c) Sucrose is a non-ionic molecule ) n ¼ 1 and N ¼ M Normality is defined as the equivalents of solute in a liter of solution. In this particular case, 1 Eq ¼ 1 mol. The first task is to use the density to convert the mass of one dose into the volume of the solution for one dose. ) Solution volume of one dose ¼ ð5 oz  28:35 g=ozÞ=1:01 g=mL ¼ 140:35 mL Sucrose Normality ¼ ð0:4665 g=342 g=EqÞ=0:14035 L ¼ 0:00972 Eq=L or 0:00972 N

4.4 Salt Factor, S

79

Example 4.11

℞ 1,25-Dihydroxy-D3 Coal tar solution Pet. Album Sig. Apply on the affected area q.d. h.s.

3 μg/g 0.8 mL 4 oz

(a) Calculate grams of coal tar solution required to fill the prescription (ρcoal tar ¼ 1.13). (b) Calculate the 1,25-dihydroxy-D3 needed to fill the prescription. Solution (a) 0.8 mL ∙ 1.13 g/mL ¼ 0.9 g. (b) Let the 1,25-dihydroxy-D3 quantity be X. The total quantity of the lotion ¼ 0.9 g + 4 oz∙28.35 g/ oz + X ¼ 114.3 g + X )

X 3 μg ¼ X þ 114:3 g g

Before cross multiplying, you have to make the units uniform; otherwise, the denominator will be multiplied by a unit which is a million times smaller. )

X 3 μg ¼ 6 X þ 114:3 g 10 μg

X 3 The units now cancel out and ) Xþ114:3 g ¼ 106

) 106 X  3 X ¼ 342:9 g

4.4

)

X ¼ 0:0003429 g ¼ 342:9 μg

Salt Factor, S

Drug dosages are always written with respect to the drug’s free form, but the drug itself could be synthesized and supplied as a salt. Since the salt moiety is part of the molar mass of the drug, but is not pharmacologically active, adjustments have to be made to calculate the effective dose of the drug (Eq. 4.8). In other words, more drug has to be administered in its salt form in order to elicit the same pharmacological response as a smaller dose of the corresponding free form of the drug. These calculations are made

80

4 Units of Concentration and the Salt Factor

possible via the use of the salt factor. The salt factor serves no purpose, and its use is wrong and misleading for drugs that the salt moiety contributes to their therapeutic efficacy by enhancing or diminishing their pharmacological activity. The salt factor, S, is defined as the fraction of the free form of an acid/base drug present in its corresponding salt structure. To calculate S, you need to divide the mass of the free form of drug present in the salt by the molar mass of the corresponding salt (Eq. 4.6). This can be done by multiplying the molecular weight (MW) of the free form of the drug with the number of drug molecules present in the salt chemical structure and divide by the MW of the salt form of the drug (Eq. 4.7). Since the MW of the salt form of the drug is always bigger than that of the free form, the salt factor is always 0:9%isotonic saline Therefore, the 5%w/v glycerin exerts osmotic pressure higher than that of 0.9% NaCl given that a semipermeable membrane separates these two solutions. The reality though is that tissue and other cell plasma membranes are permeable to glycerol, and therefore, there is no concept of a stable tonicity with glycerol solutions. My opinion is that there is nothing wrong with the prescription above given as an ophthalmic lubricant. Also, a 50%w/v glycerin solution (very painful) is used for the transient management of corneal edema and for reducing intraocular pressure. However, you may wish to make the above solution “isotonic,” since we do not know how long it will take to be diluted with lacrimal fluid. You may also wish to avoid either permeation of excess glycerol through the corneal membranes or a temporal dehydration of corneal epithelium. Step 2. Use Eq. 5.1 to calculate the volume needed from the 5%w/v glycerol solution to make the final solution isotonic. Set the 30 mL of 0.9% NaCl as the volume and the concentration, respectively, of the final isotonic solution. Let Ci ¼ 1.65% NaCl, Cf ¼ 0.9% NaCl and Vf ¼ 30 mL ) Vi ¼

C f  V f 0:9%  30 mL ¼ 16:36 mL ¼ 1:65% Ci

Thus, the isotonic solution is prepared by diluting 16.4 mL of 5%w/v glycerol with 13.6 mL of purified water. Alternatively, you may convert the whole 30 mL isotonic by setting the initial volume to be equal to 30 mL. Let Ci ¼ 1.65% NaCl, Cf ¼ 0.9% NaCl and Vi ¼ 30 mL C i  V i 1:65%  30 mL ¼ 55 mL ¼ 0:9% Cf ) Vwater ¼ 55–30 ¼ 25 mL

) Vf ¼

172

7 Isotonic Solutions

Exercises 7.1. A molecule has shown unparallel biological activity when tested in vitro in cell cultures. Does the molecule have potential to become a drug or is it already a drug? 7.2. Which of the following substances have the potential to passively diffuse through plasma membrane? Substance A B C D E

MW 160 55 3000 76 1250

Characteristic property Nonionic, hydrophobic Nonionic, amphiphilic Nonionic, hydrophobic Ionic Nonionic, hydrophilic

7.3. Choose the statement that best expresses the concept of tonicity. (a) Tonicity refers to the osmolarity of a solution in relation to the osmolarity of intracellular fluid. (b) Tonicity refers to the hydrostatic pressure of cells. (c) Tonicity refers to the osmolarity of a solution. (d) Tonicity refers to the diffusion of solutes through plasma membranes by osmosis. (e) The relative solute concentration of two solutions. 7.4. Which of the following solutions is hypertonic to intracellular fluid? Pick all that apply. (a) (b) (c) (d) (e)

1% w/v glucose 6% w/v glucose 5% w/v NaCl 0.95% w/v dextrose 0.9% w/v NaCl

7.5. The sketch below shows a conical flask containing a solution (C2) embedded within a beaker also containing a solution (C1).

Exercises

173

Assuming that water, but not solute, can go through the walls of the conical flask, the sketch above is drawn at equilibrium (initially at time zero, the level of the two solutions was identical). Choose the correct statements about the two solutions at time zero when the conical flask was first placed in the beaker unless if otherwise stated: (a) The two solutions were isosmotic. (b) The solution in the beaker was hyperosmotic as compared to the solution in the conical flask. (c) The solution in the beaker was hypoosmotic as compared to the solution in the conical flask. (d) The solution in the beaker was isotonic. (e) The solution in the beaker was hypotonic. (f) The solution in the conical flask was hypotonic. (g) The solute concentration was C2 > C1. (h) At equilibrium, water moves out of the conical flask into the beaker. (i) The water concentration in the beaker was higher than the water concentration in the conical flask. 7.6. A solution in which the solute concentration is higher than the solute concentration inside the cell is called (a) (b) (c) (d) (e) (f) (g)

Isotonic Hypotonic Hypertonic Isosmotic Hypoosmotic Hyperosmotic Cytotonic

7.7. Calculate osmolarity for the following solutions: (a) (b) (c) (d) (e)

0.25 M ZnCl2 0.4%w/v Al2(SO4)3 0.1 M tris hydroxyethyl amine (free base) 0.1 M tris hydroxyethyl amine hydrochloride 0.02%w/v Mg(OOCCH2)2N-CH2CH2-N(CH2COO)2Mg (Answer: (a) 0.75; (b) 0.0585; (c) 0.1; (d) 0.2; (e) 0.00179)

7.8. Calculate the sodium chloride equivalent (E-values) for: (a) Epinephrine (C9H13O3N) OH

.

HO

HO

CH3 NH

174

7 Isotonic Solutions

(b) Epinephrine hydrochloride (C9H14O3NCl) CH3

OH HO

HCl

NH

HO

(c) Pralidoxime chloride (C7H9ON2Cl)

Cl

+

N

-

CH3

N

OH

(d) Dicyclomine HCl (C19H36O2NBr) CH3 HBr N O

CH3 O

(e) Acetazolamide (C4H6O3N4S2) O H3C

N NH

N S

SO 2 NH2

(f) Sodium bicarbonate (NaHCO3) (g) Polyethyleneglycol (MW 5000), HO–[CH2CH2O]113–H (h) Sodium carbonate (Na2CO3) (i) Disodium cromoglycate (C23H14O11Na2) O

O +

Na

O

-

O

O

O

-

+

Na

OH O

O

O

O

(Answer: (a) 0.17; (b) 0.28; (c) 0.36; (d) 0.16; (e) 0.14; (f) 0.73; (g) 0.006; (h) 0.87; (i) 0.18)

Exercises

175

7.9. Use the sodium chloride equivalent method to prepare the IV infusion solution shown below: Epinephrine sulfate Na2HPO4 Water for injection

0.01%w/v q.s. q.s. ad.

1.0 L

How many grams of sodium phosphate dibasic are required to make 1 L of the solution above isotonic? MW (epinephrine sulfate) ¼ 281 g/mol, E-value (epinephrine sulfate) ¼ 0.22, MW (Na2HPO4) ¼ 142 g/mol, MW (NaCl) ¼ 58.5 g/mol, E-value (sodium phosphate dibasic) ¼ 0.65. (Answer: 13.81 g) 7.10. Classify the following commercial products according to their tonicity. (a) (b) (c) (d) (e) (f)

Dextrose 5%w/v (E-value ¼ 0.17) Sodium bicarbonate 5%w/v (E-value ¼ 0.73) Neomycin sulfate 8.2%w/v (C23H46N6O13 ∙ 3H2SO4; MW ¼ 908, i ¼ 4) Diazepam 5 μg/mL (C16H13ClN2O, i ¼ 1, E-value (diazepam) ¼ 0.11) Valproic acid 125 mg/tsp ((CH3CH2CH2)2CHCOOH; i ¼ 1) Human serum albumin 25%w/v (MW 66 kDa; hint: use ΔTf ¼  i  kf  m, where kf is the cryoscopic constant (1.86) and m is the molality of the protein solution) (Answer: (a) isotonic; (b) hypertonic; (c) hypertonic; (d) hypotonic; (e) hypotonic; (f) hypotonic)

7.11. You wish to prepare 250 mL of HBS (HEPES buffer saline) for a protein stability experiment. The HBS is an isotonic solution of HEPES and NaCl. Based on the concentration of the therapeutic protein, you decided to use 25 mM HEPES (2-hydroxyethylpiperazino-N0 -2-ethanosulfonic acid, monosodium salt; MW ¼ 261; i¼ 2). How much sodium chloride you should use in order to make the solution isotonic in the absence of protein? (Answer: 1.865 g) 7.12. Use the sodium chloride equivalent method to compound the following ophthalmic isotonic solution. Rx Drug Z (hydrochloride salt) NaCl Sterile deionized water M.et. ft. sterile, isotonic solution Sig. 2 gtt o.u. q.4 h for 5 days

0.55%w/v q.s. q.s. ad.

25 mL

176

7 Isotonic Solutions

Empirical formula of drug Z: C17H18FN3O3∙HCl, E-value ¼ 0.17 (a) How much sodium chloride is needed to fill the prescription? (b) Calculate the dose (in μg) if the dropper delivered 20 drops/mL? (Answer: (a) 0.2 g; (b) 550 μg) 7.13. Local anesthetics like proparacaine produce local anesthesia by blocking the sodium channels of the excitable membrane of nerve axons, thus stabilizing the neuronal membrane and preventing the transmission of nerve pulses. Proparacaine hydrochloride isotonic solution 0.5%w/v (Sig. 2 gtt q.15 min during eye examination) is indicated for topical anesthesia during eye examination. Local anesthesia begins within 30 s after instillation and lasts about 15 min. The empirical formula of the drug is C16H27N2O3Cl and MW ¼ 331 g/mol. Proparacaine HCl CH3

O O H3C

N

HCl CH3

O NH2

(a) How much water should you use to prepare 45 mL of isotonic proparacaine HCl solution if 1.8%w/v NaCl, pH 6 is available in your lab. (b) If the eye examination lasted 50 min, how much proparacaine hydrochloride had the patient actually received during the examination (assume 19 drops/mL)? (Answer: (a) 24.826 mL; (b) 2.11 mg) 7.14. Among other uses, the benzodiazepine midazolam is useful for patient premedication before an operation. You received a medication order for 10 mL of midazolam 6 mg/mL for IM administration. You have midazolam and midazolam hydrochloride available in powder form. Describe how you should prepare the solution; what form of midazolam you should use? Midazolam empirical formula: C18H13ClFN3∙HCl; MW ¼ 362; Evalue ¼ 0.17. (Answer: use midazolam hydrochloride and 78.66 mg of NaCl)

Exercises

177

7.15. Given the following prescription: Rx Fluconazole Lactose Sterile water M.et.ft. infusion solution, D.T.D # 4 Sig. Infuse 1400 mg on first day, then 600 mg daily for 3 days.

10 mg/5 mL q.s. q.s. ad.

1L

Fluconazole: C13H12F2N6O, MW ¼ 306; i ¼ 1; E-value ¼ 0.10 Lactose: C12H22O11 (a) Calculate the amount of drug needed to fill the prescription. (b) How many grams of lactose are there in 1 L of isotonic solution? (c) How many milliliters must be infused at the end of the second day? (Answer: (a) 8 g; (b) 97.78 g; (c) 1000 mL) 7.16. A 33-pound pediatric patient was prescribed the medication order below for respiratory tract infection: Rx Cefuroxime ZnCl2 Sterile water M.et.ft. isotonic solution Sig. 60 mg/kg/day IM in three divided doses

125 mg/mL q.s. q.s. ad.

7 mL

Cefuroxime: C20H22N4O10S, MW ¼ 510; i ¼ 1; E-value ¼ 0.06. (a) Calculate the osmolarity of cefuroxime in the solution. (b) What is the contribution of zinc chloride to the total osmotic strength of the solution? (c) How many milliliters should the patient receive per dose? (Answer: (a) 0.245 osmol/L; (b) 48.7 mOsm/L; (c) 2.4 mL) 7.17. Given the following formula: Scopolamine HCl NaCl Boric acid (5%w/v) Sterile water M.et. ft. sterile, isotonic solution

0.25%w/v 0.45%w/v q.s. q.s. ad.

15 mL

Calculate the volume of 5%w/v boric acid needed to make the above solution isotonic, using the freezing point depression data given below:

178

7 Isotonic Solutions

ΔTf 1%w/v boric acid ¼ 0.29  C ΔTf 0.9%w/v sodium chloride ¼ 0.52  C ΔTf 0.25%w/v scopolamine HCl ¼ 0.08  C (Answer: 1.86 mL) 7.18. Given the following: Dextrose (60% w/v) Sterile water for injection

q.s. ad.

1L

E-value (dextrose) ¼ 0.17 (a) How many milliliters should be used from the 60%w/v dextrose stock solution to make the final solution isotonic? (b) What is the concentration of dextrose in the final isotonic solution? (Answer: (a) 88.2 mL; (b) 5.3%w/v)

Additional Exercises 7.19. You were asked to prepare 1.5 L of 0.12%w/v epinephrine sulfate isotonic solution for injection. The epinephrine sulfate chemical formula is C9H13O3NH+, HSO4 (MW ¼ 281 g/mol). How much of NaCl should you use to adjust the tonicity of the solution? (Answer: 13.1 g) 7.20. Ciprofloxacin is a broad spectrum fluoroquinolone-based antimicrobial agent. Explain how you should compound the following prescription for eye infections. Rx Ciprofloxacin HCl monohydrate NaCl Sterile water M. et. Ft. sterile isotonic solution Sig. 2 gtt o.u. q.1 h for 8 h. Then 2 gtt o.u. q.8 h for 6 days.

0.45%w/v q.s. q.s. ad.

15 mL

Empirical formula of the drug is C17H18FN3O3∙HCl H2O, MW ¼ 386 g/mol, i ¼ 2. The structural formula of ciprofloxacin HCl is shown below:

Additional Exercises

H2O . HCl

179 H N N

N O

F O

OH

(a) How much sodium chloride should you use to make the solution isotonic? (b) How much ciprofloxacin has the patient received after 8 h? (assume 20 drops/mL) (c) How much ciprofloxacin has the patient received after 6 days if he started taking the medication on day 1 at 8 am? (assume 20 drops/mL) (Answer: (a) 0.124 g; (b) 7.2 mg; (c) 22.5 mg)

7.21. The following formula can be used to prepare isotonic ciprofloxacin solution (pH 6) for intravenous infusion. Ciprofloxacin lactate HCl Dextrose Sterile water

0.25%w/v 0.006 M q.s. q.s. ad.

1L

In stock: Hydrochloric acid is 37% (ρ ¼ 1.2) Chemical formula of lactic acid: CH3CH(OH)COOH, MW ¼ 90, pKa (lactic acid) ¼ 3.85 Ciprofloxacin lactate (MW ¼ 422, i ¼ 2, E-value ¼ 0.146) (a) Calculate the osmolarity of the solution with respect to ciprofloxacin HCl. (b) How many microliters of HCl 37% are there in the formula? (c) How many grams of dextrose are needed to make the solution isotonic? (Answer: (a) 0.012 Osm/L; (b) 493 μL; (c) 48.52 g) 7.22. Pneumococcal pneumonia can be effectively treated with intravenous administration of the first-generation cephalosporin, cefazolin. Rx Cefazolin sodium NaCl Dextrose Bacteriostatic water M. et. Ft. isotonic solution Sig. Inject 10 mL IV slowly, within no less than 3 min, q.12 h for 2 days.

2.25 g 0.2%w/v q.s. q.s. ad.

45 mL

180

7 Isotonic Solutions

The chemical formula of cefazolin sodium salt is shown below (MW ¼ 476 g/mol): N

N

N

NH

N

S

O

N

N

S

O Na

+

O

-

N S

CH3

O

(a) Use the E-value for cefazolin sodium to calculate the NaCl osmotically equivalent to 2.25 g of drug. (b) How many milligrams of dextrose are needed to fill the prescription? (c) How much of cefazolin sodium will the patient receive in a day? (Answer: (a) 0.291 g; (b) 141 mg; (c) 1 g)

7.23. Given the following formula: Epinephrine Sodium metabisulfite Sodium chloride Sterile water M. et. Ft. sterile isotonic solution so that when 12.5 mL of the solution is diluted to 0.5 L yields epinephrine concentration of 1:10,000 g/mL.

0.25 g q.s. q.s. ad.

0.5 L

Epinephrine: MW ¼ 183 g/mol, i ¼ 1, E-value ¼ 0.168 Sodium metabisulfite: Na2S2O5, i ¼ 3. (a) Calculate the osmolarity of epinephrine. (b) How many grams of sodium chloride are needed to compound the formula? (Answer: (a) 0.022 Osm/L; (b) 4.0425 g)

Chapter 8

Diffusion

Abstract I consider this chapter to be one of the most important chapters in the book. We know that a drug has to form a solution in an aqueous medium if it will be administered IV. We also know that drugs can cross membrane barriers only as molecular dispersions. Therefore, they can get absorbed and distributed in the body only if they mix and dissolve in biological fluids. Drug dissolution and membrane crossing occur by diffusion. Through nice descriptions and activities related to the phenomenon of diffusion and osmosis, you will understand the process of dissolution and you will relate it to drug absorption and distribution into tissues. You will correlate the physicochemical properties of drugs to their ability to cross cell plasma membranes and will understand the effect of food and gastrointestinal motility on drug absorption. Keywords Solution · Molecular dispersion · Absorption · Dissolution · Convection · Advection · Diffusion · Diffusion coefficient · Noyes–Whitney equation · Concentration gradient · Permeability · Osmosis · Flux · Drug transport · Transporters

Learning Objectives After reading this chapter, you should be able to: • Understand that only molecular dispersions (solutions) of drugs can be absorbed • Understand that dissolution involves a chemical interaction between solute and solvent • Define convection, advection, and diffusion • Remember and describe the characteristics of diffusion • Describe the process of drug dissolution • Calculate the diffusion coefficient using the Noyes–Whitney equation Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_8) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_8

181

182

8

Diffusion

• Employ the concept of concentration gradient to explain the rate or speed of particle diffusion • Describe passive transmembrane diffusion as the process of drug absorption • Define diffusivity and permeability • Apply Fick’s first law to calculate drug mass flux and permeability coefficient from permeability studies • Describe the main two categories of protein transporters and their role in drug transport • Calculate water concentrations • Describe the difference between diffusion and osmosis

8.1

Drug Dissolution

Drugs can be absorbed only as molecular dispersions or solutions. Molecules can go through plasma membranes only as individual separate entities. If they are lumped together as aggregates, they will not go through because their size is too big. A drug can form a solution in a solvent when the two different molecules are miscible. In other words, there must be a similarity in their physicochemical properties if the two are to be mixed. The greater the similarity in their physicochemical properties, the stronger will be the solute–solvent attractive interactions (adhesive forces) and the higher the solubility of the solute. If the adhesive forces are weaker than solute– solute and solvent–solvent attractive forces (cohesive forces), not only the solute solubility will be very small, but the two molecules may separate into two distinct phases. Depending on their density, solute aggregates may submerge, precipitate, or float on the solvent surface. At the molecular level, the atoms of molecules are at constant vibrational, rotational, and translational motion due to their potential energy. When a solid is damped into water, the water molecules will try to dissolve the solute molecules by associating with them. As the top panel of Fig. 8.1 demonstrates, during the process of dissolution, the ordered structure of liquid water (empty circles) due to hydrogen bonding is perturbed around the dry solute particles (bigger solid circles). The water system around the solid is in “crisis” seeking ways to incorporate the new solute molecules in its ordered liquid structure. Upon collision, the kinetic energy of a solvent molecule is converted to potential energy, and it is used to break solute–solute and solvent–solvent bonds. Formation of new solvent–solute bonds results in an energy release and the stabilization of the solution. Solute molecules will be hydrated by water molecules by forming the maximum number of hydrogen and other van der Walls bonds for the ultimate goal of minimizing the energy of the system (remember that bond breaking requires energy, whereas bond formation releases energy). Since the solvent–solute bonds are stronger than solute–solute and solvent– solvent bonds, solute molecules will be effectively surrounded and insulated by a maximum number of water molecules, and they will be spaced, by diffusion, as far apart as possible from each other as the most stable configuration of the solution (Fig. 8.1, bottom panel).

8.1 Drug Dissolution

183

Fig. 8.1 The dissolution process of a crystalline drug (solid circles). Water molecules are represented as smaller empty circles in an ordered liquid arrangement due to mainly hydrogen bonds. In this sketch, every solute is shown to be surrounded by eight water molecules; six in-plane (solid line, hexagon) and two out-of-plane (not shown)

The dissolution process at the macroscopic level is shown in Fig. 8.2. At the top panel, the drug pellet is placed in a die cavity, so that the dry surface area (A) of the drug exposed to the solvent stays constant during the process of dissolution. Studies in which the drug surface area exposed to the solvent is maintained constant are called intrinsic dissolution studies. Water molecules collide with the pellet surface and drag drug molecules away from the pellet. The solvent layer closest to the pellet is called stagnant layer (h). This layer is very viscous due to the high number of partially hydrated solute molecules. With time, solute molecules get fully hydrated and diffuse deeper into the bulk phase. The graph in the lower panel shows that the concentration of the drug in the stagnant layer is highest or supersaturated (Cs) and drops to its stable solubility value (Cb) as it diffuses away from the pellet surface deeper into the bulk water phase. The rate of dissolution of the drug can be quantitatively expressed with the Noyes–Whitney equation, which is derived from the Fick’s first law of diffusion. dm D ¼ ∙ A ∙ ðC s  C b Þ dt h

ð8:1Þ

184

8

Diffusion

Fig. 8.2 Top: A representation of drug dissolution from a constant pellet surface. Dissolution experiments that maintain the dry drug surface constant are called intrinsic dissolution studies. Bottom: The drug concentration in solution as a function of distance from the pellet (or time)

where dm dt :

is the mass of solute dissolved per unit time D: is the diffusivity or diffusion coefficient of the solute in the solvent in units of area per time, for example, cm2/s h: is the thickness of the solvent layer in which the drug exists at the highest drug concentration Cs A: is the constant surface area of the dry solute exposed to the solvent Cs: is the supersaturated concentration of the drug Cb: is the concentration of the drug in solution in the bulk When using the above equation, you may need to convert units of volume from L or mL to m3 or cm3. For example, drug concentration may be given as mol/L, but applying Eq. 8.1 to determine the rate of drug transport will require conversion of L into cm3 if the surface area of the drug is measured in cm2. Basically, 1 L ¼ 1000 mL ¼ 1000 cm3.

8.2 Transmembrane Diffusion

185

Fig. 8.3 A membrane of thickness Δx has many of each of two kind of channels that allow only the passage of solvent and only the passage of solute particles (solid circles), respectively. At time zero, the solution on the right side of the membrane is more concentrated than the one on the left side. The direction of diffusion indicates the diffusion of solute particles. C1 and C2 are solute concentrations in water

8.2

Transmembrane Diffusion

As discussed in the previous section, drugs can cross membrane barriers only when they exist as molecular dispersions or solutions. We have also seen that the process of dissolution involves hydration and diffusion of the solute molecules into the solvent. What is diffusion? Diffusion is defined as the movement of solute particles, without bulk flow, from an area of high concentration to an area of low concentration, resulting in the uniform distribution of solute particles within a solvent (hence a solution). Figure 8.3 shows a tube that is separated by a thin permeable membrane of constant area A and width Δx. The membrane has selective channels that allow the passage of only solute and only solvent particles, respectively. At t ¼ 0, the solute concentration is higher on the right side of the membrane which is permeable to both solute and solvent particles. Which direction should the solute and solvent particles move to establish equilibrium? Following the rules of diffusion and assuming that we have aqueous solutions, the solute will diffuse through the membrane from right to left (higher to lower solute concentration), whereas water molecules will diffuse through the membrane channels also from higher to lower water concentration (left to right). The process of diffusion of both particles will stop when the solute concentrations on both sides of the membrane are equal. Equal solute concentration implies equal water concentration. The rate of transfer of solute between two compartments that are separated by a thin membrane, as described above, is given by the Fick’s law of diffusion.


 dm dC ΔC C1  C2 ¼ D∙A∙ ¼ D∙A∙ ¼ D∙A∙ dt dx Δx Δx where dm dt :

is the mass of solute transferred across the membrane per unit time

ð8:2Þ

186

8

Diffusion

D: is the diffusivity or diffusion coefficient of the solute in the solvent in units of cm2/s A: is the membrane surface area Δx: is the thickness of the membrane that separates the two solutions dC dx : is the concentration gradient for the solute C1 and C2: are the solute concentrations in the two compartments Some of the characteristic properties of molecular diffusion are: 1. It is random in nature, and it does not have to follow the movement of the fluid or solvent. 2. It always proceeds down a concentration gradient from high to low concentration (a concentration gradient is the change of concentration with distance). 3. At best, it dominates only on a millimeter length scale. 4. It is affected by the phase, that is, it is faster in gases, slowest in solids. 5. It is adversely affected by the viscosity of the liquid phase. 6. It increases with temperature. 7. Smaller molecules diffuse faster. 8. Each individual substance has its own concentration gradient. We have seen the simultaneous diffusion of solute and water molecules through the membrane in Fig. 8.3. It turns out that every substance has its own, so-called chemical potential, and therefore, every ion or molecule can develop its own concentration gradient independently and can diffuse through cell membranes simultaneously and independently of other substances. Unlike diffusion, the passive transport of solute molecules with the flow of a fluid is called advection. For example, when drug molecules are transferred from the side of absorption to distant target organs via the systemic blood circulation, they are transferred by advection. Convection is defined as the passive transport of solute particles by advection and diffusion. Equation 8.2 can be modified to incorporate two more important parameters: the permeability coefficient Pm and the mass flux, J. The permeability coefficient or permeability is defined as the speed or relative ease with which a particle diffuses through a membrane. Essentially, it is the diffusivity divided by the thickness of the membrane, and it is expressed in velocity units, that is, cm/s. Its magnitude depends on the characteristics of the membrane and the physicochemical properties of the solute. For example, solute permeability decreases with the lipophilicity of the membrane, solute polarity, and particle size. Mass flux, J, is defined as the mass of particles that flows through a unit area per unit time. The representative units of J are mol ∙ cm2 ∙ s1. dm ¼ Pm ∙ A ∙ ΔC dt

ð8:3Þ

8.2 Transmembrane Diffusion

187

J¼

1 dm ∙ ¼ Pm ∙ ΔC A dt

ð8:4Þ

where ΔC ¼ C1  C2 Example 8.1 A new drug was subjected to permeability studies against Caco-2 cells that were grown in monolayers at 37  C for about 3 weeks, in order to evaluate its potential for oral administration. The concentration of the drug in the donor compartment was adjusted to 0.1 mM, and the rate of mass transfer of the drug from the apical to the basolateral side of the Caco-2 monolayer was studied as a function of time through a cross-sectional area or monolayer transwell area of 1 cm2 (Fig. 8.4). Given the table below: Time (min) 0 10 20 30 40 50 60 70 80 90

Drug recovery in basolateral side (nmol) 0 0.05 0.26 0.54 0.67 0.78 0.71 0.67 0.66 0.67

Drug cumulative recovery in basolateral side (nmol) 0 0.05 0.31 0.85 1.52 2.3 3.01 3.68 4.34 5.01

Fig. 8.4 Transwell assay for measuring epithelial permeability. Cell monolayers are grown on a porous membrane support in growth media (pink solution). Drug solution is added in the donor compartment. Samples for analysis are taken from the acceptor compartment

(continued)

188

8

Diffusion

Example 8.1 (continued) (a) What is the cell apical-to-basolateral flux of the drug? (b) What is permeability coefficient of the drug? (c) What is the diffusivity of the drug if the thickness of the Caco-2 membrane monolayer is 3.4 nm? (d) Does the drug have a potential to be administered orally? (high permeability drugs exhibit permeability coefficients between 5106 and 104 cm/s). Solution (a) From Eq. 8.4, J ¼ A1 ∙

dm dt

or

dm A

¼ J ∙ dt , the flux can be calculated

from the slope of the plot of the drug cumulative mass versus time.

The best fit of the linear part of the data is shown to be ¼ 0:0696 ∙ t  1:22. Therefore, the apical-to-basolateral flux of the drug, J ¼ 0:0696 ∙ cmnmol 2 ∙ min. (b) The permeability coefficient of the drug can be calculated using the same equation. The concentration difference ΔC can be approximated by the concentration in the donor compartment since the concentration in the receiver compartment (basolateral side) is always comparatively very small. Thus, ΔC ¼ C1 ¼ 0.1 mM. m A

) Pm ¼

0:0696 cmnmol J 2 ∙ min ¼ ¼ 0:000696 cm=min C1 100 nmol cm3 (continued)

8.2 Transmembrane Diffusion

189

Example 8.1 (continued) (c) The diffusion coefficient can be calculated using D ¼ Pm ∙ Δx ¼ 0:000696

cm ∙ 3:4 ∙ 107 cm ¼ 2:37 ∙ 1010 cm2 =min min

(d) We have to convert the units of permeability to cm/s. ) Pm ¼

0:000696 cm=min ¼ 1:16 ∙ 105 cm=s 60 s=min

The drug is a high-permeability drug as determined by the in vitro monolayer studies. The drug could exhibit a high absorption rate upon in vivo administration, and therefore, it may have a potential to be administered orally.

Example 8.2 The new drug described in the previous example is a highly soluble, highly permeable drug. It is indicated for acute migraine headaches, and a fast onset of action is desired. Its mode of action involves a 24-h irreversible inhibition of a subclass of serotonin receptors. (a) Would you formulate the product as an immediate or as a sustained release product? (b) Would you recommend to take it with food or an empty stomach? Solution (a) For fast onset of action, it would be best to formulate the drug as an immediate release product. As shown in Fig. 8.5, the drug released from the stomach to the upper intestine is of much higher concentration with an immediate release product (C1) than with a sustained release product (C2). Using Eqs. 8.2–8.4,

 dm C1  Cblood dm C 2  C blood ¼ D∙A∙ ¼ D∙A∙ as opposed to dt dt Δx Δx the rate of drug absorption per unit membrane area will be higher from the immediate released product because of the concentration gradient which drives the rate of diffusion of the drug (C1  Cblood) > (C2  Cblood). (continued)

190

8

Diffusion

Fig. 8.5 The inset represents an upper portion of the small intestine. Broken arrows show the direction of diffusion (absorption) of drug molecules through the intestinal epithelium of thickness Δx. Drug solution reaching the upper intestine is of much higher concentration from an immediate release dosage form (C1) than from a sustained release product (C2)

Example 8.2 (continued) It is to be noted that the drug concentration on the basolateral side of the intestinal membrane is very, very small because upon its absorption, it is constantly diluted by the systemic circulation in the whole volume of the blood (~5 L). When the concentration gradient can be approximated by the concentration in the donor compartment, that is, C1  Cblood  C1 and C2  Cblood  C2, we say that we have sink conditions. (b) The presence of food may delay the dissolution of the drug in the stomach. Its transfer to the intestine may be more gradual and as a result, its concentration gradient as the driving force for absorption, will be (continued)

8.2 Transmembrane Diffusion

191

Example 8.2 (continued) much smaller. The recommendation should be to take the drug on an empty stomach for maximum and quicker drug absorption.

Figure 8.6 shows a representative epithelial membrane. Cell membranes are typically composed of 30–35% lipids of which ~65% are phospholipids. Drug molecules can cross plasma membranes into the blood either transcellularly or paracellularly. Hydrophilic molecules can only cross the intestinal membrane paracellularly through the water-filled tight junctions between cells, and they do that by convection (advection and diffusion). A hydrophilic drug present in an aqueous solution goes through the aqueous paracellular pathways as a solution. The drug does not have to strip off the solvent to slide through in-between the cells. Their transport depends on their particle size. Typically, only polar substances of MW < 200 with a diameter smaller than 5 Å (0.5 nm) can go through the aqueous paracellular pathway. In general, most of the orally administered drugs cannot go through paracellularly, and

Fig. 8.6 (a) A representative membrane barrier composed of epithelial cells joined together by water-filled tight junctions, adherens junctions, gap junctions, and desmosomes. (b) The inset shows a piece of the membrane consisting of phospholipids as its main component. (c) A representative phospholipid structure

192

8

Diffusion

as a pathway, paracellular transport does not contribute significantly to drug absorption. The dimensions of water molecules are about 3 Å and exhibit a significant rate of paracellular transport. Glycerol with an MW of 92 has approximate dimensions of 6 Å and fails to enter this pathway. Instead it easily crosses the epithelial barrier transcellularly and actively through a protein transporter. It is important to know that the paracellular pathway varies in different tissues. For example, the capillary endothelium is very leaky, and all small molecules are transported by convection paracellularly from blood into interstitial tissues. On the contrary, hydrophobic molecules cross membrane barriers by diffusion through the cell matrix transcellularly. In general, transcellular transport is the major component of drug absorption and is the same in all tissues. The composition of plasma membranes is similar (but not identical) with respect to % proteins, lipids, and carbohydrates regardless of tissue and cell type. At the molecular level, hydrophobic drugs have more affinity for the lipids of plasma membrane than for water. They will gladly leave the unfavorable environment of water that is solubilized to a small extent to associate with plasma membrane lipids. An initial collision with the phospholipid headgroups will provide the energy needed to break some of the watersolute non-covalent forces for a split second. This brief moment of time is enough to allow a positive interaction with the hydrophobic moieties of phospholipids. Because of their hydrophobicity they will immediately form many van der Waals bonds with the phospholipid hydrophobic chains and slide deeper by diffusion within the plasma membrane bilayer. Before you even realize it, one solute after another, driven by the concentration gradient on the two sides of the membrane, has reached the other side of the membrane. After all, the thickness of plasma membranes is only about two phospholipid molecules ( 1, contrary to nonionic crystalloids like glucose or colloids, like the human serum albumin, that have i ¼ 1. Therefore, the osmolarity and hence the osmotic strength of ionic compounds will always be higher that nonionic solutes and colloids, assuming that the comparison of these solutions is done with the concentrations expressed in molar units. For example, the osmolarity of 5 mM solutions of glucose and human serum albumin is the same, but the osmolarity of 5% w/v solutions of glucose (MW ¼ 180) and albumin (MW ¼ 67,000) is not. Both molecules have i ¼ 1, but when you normalized the % w/v concentration with the MW of each compound, their molar concentrations are very different. 50 g

5%w=v glucose ¼

50 g glucose 180 g=mol ¼ ¼ 0:278 M 1L 1L 50 g

50 g albumin 67;000 g=mol ¼ 5%w=v albumin ¼ ¼ 0:000746 M 1L 1L

10.3

Conditions and Diseases That Cause Abnormal Fluid Shifts

Compartment fluid volumes can increase, decrease, or stay the same depending on a number of factors as described below. Mathematically one can calculate at least 27 different scenarios resulting from fluid shifts (three hypothetical compartments combined with three different outcomes, yields 33 different scenarios). Our goal is not to examine every different scenario as not all 27 could exist in real life. Instead, our goal is to get familiar with the common causes of fluid imbalances, establish some guidelines for clinical intervention, and give some meaning to the calculation problems included in the chapter.

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Loss of fluid can occur from the vascular compartment (internal or external hemorrhage), the gastrointestinal tract (vomiting, diarrhea), the kidney (diuresis), and the skin (burns). Hypovolemia is a state of decreased circulating volume of blood plasma in the body due to significant blood or fluid losses as described above. On the other hand, fluid gain (hypervolemia) more frequently occurs with organ failure, that is, heart, kidney, and liver failure. For example, when impaired liver function causes reduced albumin synthesis, plasma can escape from the central circulation and accumulate in places such as abdomen causing ascites. What is important for us to understand is that serious electrolyte disturbances will lead to fluid shifts. Putting it in different words, big losses of osmotically active solutes from a particular compartment will lead to fluid shifts out of that compartment, whereas a concentration increase of osmotically active species will lead to a fluid shift into that compartment. An additional point that needs to be emphasized is that there is no analytical test that could directly measure the volume of the ECF and compare it to the volume of the ICF. In order to more accurately determine the fluid imbalance and effectively reestablish homeostasis, clinical practitioners rely on a number of diagnostic tests and physical examinations. The information from those tests is then used in empirical equations to determine appropriate clinical interventions. Excessive fluid loss can be mainly determined from the weight loss of the patient (60% of total body weight is water). Additional symptoms of hypovolemia could be thirst, tachycardia, weakness, dizziness, low blood pressure, and negligible urine output. Inorganic ion and other osmotically active solute deficits are identified by direct determination of their levels in the blood. Clinical practitioners have to identify to the best of their abilities the nature of the imbalance and administer the right electrolyte quantities in appropriate fluids to restore homeostasis.

10.4

Rules of Physics That Could Prove Useful in Clinical Intervention

Although the choice of the fluids, the rate of infusion, and the actual dose are very complicated issues, I have attempted to formulate few simple rules to follow for fluid replacement. These rules listed below are not based on the actual condition of the patient, rather they are rules of physics that govern the movement of fluid between compartments, in general. The hypothesis is that since osmosis is the driving force of water movement from one compartment to another, addition of substances that selectively partition into a particular compartment should drive more water from other compartments into that one. Thus, parenteral administration of an isotonic colloid solution should result in plasma volume expansion (Fig. 10.2), whereas parenteral administration of an inorganic ion that preferentially partitions into the ECF, for example, Na+, should increase both IVF and ISF (Fig. 10.3). Table 10.2 lists some

10.4

Rules of Physics That Could Prove Useful in Clinical Intervention

225

Infuse 1 L of a colloid

3L

5L

10 L 9.5 L 26 L

25.5 L

Initial state

Final state

Fig. 10.2 Administration of 1 L of an isotonic colloidal solution resulted in a 2 L increase of the IVF volume (the actual volume increase is hypothetical). The presence of additional colloids in the blood caused fluid movement of 0.5 L out of each of the other two compartments

Infuse 1 L of 3 % saline

3L

4L

10 L 11 L 26 L

Initial state

25 L

Final state

Fig. 10.3 Parenteral administration of 1 L of a hypertonic 3% w/v NaCl solution should result in an ECF increase. Sodium chloride can freely diffuse out of the blood to the ISF, but cannot passively permeate cell membranes. Increased solute concentration in the interstitial spaces draws some water out of the cells. Again, the values for volume shifts are hypothetical

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commercially available hypotonic colloidal solutions made in pure water. However, due to their high MW and high viscosity, these colloidal solutions have to be made isotonic with the addition of NaCl prior to their IV infusion. Also, the pharmacology of the molecules used in the preparation of resuscitation fluids cannot be neglected. In fact, the pharmacology of colloids and crystalloids coupled with the pathophysiology of the disease state or condition should be a priority in the process of selecting a therapeutic fluid. In Chap. 8, we discussed the passive diffusion of drugs and other nonionic molecules, and we learned that ions cannot passively cross the cell plasma membrane. However, protein channels and transporters can mediate the intracellular translocation of many ions and other polar solutes, creating alternative ways for the cell to balance the intracellular water activity to the extracellular one. Consider for example a case where the ICF is of lower osmolarity than the ECF because the serum sodium concentration is above physiological levels. Cells have two choices: either they dispose of water and shrink or they may activate the Na+ channels and increase the intracellular Na+ concentration to match the osmolarity of ECF and eliminate cell volume changes. Rapid uptake or outflow of electrolytes and other organic solutes by cells allow them to regulate their volume without significant water movement. Diseases, conditions, and pharmacotherapy that affect the functionality of these transporters will affect fluid shifts among the different body compartments. For example, serious hypokalemia (lower K+ serum concentration) can be treated with IV infusion of KCl solutions. The tonicity of these solutions could be adjusted with the addition of NaCl, but never with dextrose. The reason is a pharmacological one. Basically, the presence of dextrose in the blood stimulates secretion of insulin. Binding of insulin to its receptor increases the activity of Na,K-ATPase that mediates increased transmembrane transport of extracellular K+ ions into skeletal muscle cells. As a result, the K+ serum levels will remain low. This is exactly the reason why acute hyperkalemia is treated with insulin injection followed by an IV infusion of 50% w/v dextrose solution to avoid a hypoglycemic event. There is a big difference in infusing a hypertonic solution like that shown in Fig. 10.3 from infusing a 0.9% NaCl solution. Infusion of an isotonic saline solution should result in ECF volume increase, but not to the expense of ICF. In other words, no fluid redistribution between the two compartments, that is, ECF and ISF, should occur because, first, the osmolarity of this solution is physiological and, second, the Na+ concentration (0.9% NaCl ¼ 154 mEq/L) is very close to physiological levels of the ECF (see Table 10.1). Also, no convection of ions should take place from the central circulation to other body compartments. Of course, if there is no plasma deficit to start with, the kidney will remove the excess volume infused with time from the body. Lastly, administration of a solute that is preferentially taken up by cells should increase the ICF volume. Diffusion-driven partition of a solute in the cells is dependent on its concentration in the ECF. Therefore, administration of that particular solute should increase both ECF and ICF volumes (Fig. 10.4). Glucose is taken up by cells through a process called facilitated diffusion. It is stored inside the cells

10.4

Rules of Physics That Could Prove Useful in Clinical Intervention

227

Infuse 1 L of D5W

3L 10 L

26 L Initial state

3.25 L 10.25 L 26.5 L

Final state

Fig. 10.4 ICF increase and general rehydration after parenteral administration of a 5% w/v glucose solution

for a very short period of time until it gets metabolized. This process is dynamic in nature. The difference between an in vitro experiment and the real in vivo situation is that in vitro the volumes of the three compartments reach equilibrium after some time. In vivo equilibrium is more dynamic because of the metabolism of glucose that causes its redistribution in the body within the same time period. Facilitated diffusion does not require energy, and it always proceeds down the concentration gradient. In other words, ECF glucose concentration will always be higher than its ICF concentration, but as soon as some of it is used up, more of it gets inside the cells. Also, remember that unlike isotonic saline, isotonic glucose concentration is not equivalent to the physiological glucose concentration in the plasma or ECF. Administration of normal saline could also work for some type of cells under the hypothesis that some Na+ will passively diffuse through the membrane channels into the cells, that the Na-K pumps will slow down their operation, and that water will flow inside the cells to restore the osmolarity of intracellular fluids to physiological levels. In the absence of electrolyte imbalances, the following rules could be established for fluid replacement: 1. Increase plasma volume by administering an isotonic or hypertonic colloid solution. 2. Increase ECF with isotonic or hypertonic saline solutions. 3. Increase ICF and rehydrate cells by the administration of isotonic glucose and hypotonic sodium chloride solutions. In addition, a wise practitioner must be able to understand the physiology of the condition and to use common sense whenever possible. For example, if the patient

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has lost a lot of blood, the best way to replace blood is to administer blood. Most of the synthetic plasma expanders have no capacity to carry oxygen. Restoration of the cardiac output using these solutions is not equivalent to tissue oxygen restoration, and organ shut down could eventually occur. In the presence of electrolyte imbalances, the above rules can be modified, refined, or totally changed. Therapeutic fluids must be selected based on the pathophysiology of the disease state and the biochemistry of the drug and the crystalloid. Regardless of the therapeutic fluid, IV infusion should be done at a slow rate, especially if the condition responsible for electrolyte imbalance has been there for a while. Brain cells, for example, are particularly sensitive to abrupt osmotic changes. In chronic electrolyte disturbances, brain cells adapt in the hyposmolar or hyperosmolar environment by losing or storing organic solutes, respectively. Attempts to correct these chronic electrolyte imbalances by rapid infusion of crystalloid solutions (hypotonic or hypertonic) can lead to cerebral edema, herniation, seizures, coma, and death. If you must administer a therapeutic solution by an IV bolus, it has to be a very small volume. An exception to this last rule is the administration of osmotic diuretics (see Sect. 10.6). Example 10.2 Based on the numbers given in Fig. 10.1, calculate how much volume out of 650 mL of isotonic crystalloid electrolyte solution administered intravenously will theoretically remain in the vascular compartment? What is the volume percentage increase of the interstitial fluid? Solution The most common isotonic electrolyte solution is aqueous 0.9 % w/v NaCl. Na+ and Cl
are small charged atoms that flow out from the vascular compartment to the interstitial fluid and vice versa, but their intracellular concentration is tightly and actively regulated by ATP-dependent pumps. In addition, isotonic crystalloid solutions are not osmotically different from body fluids, and thus, no fluid shifts should take place between the extracellular and intracellular compartments. The infusate will partition between the intravascular (plasma) and extravascular (interstitial fluid) compartments and dilute them both proportionally as if they were a single compartment. This process takes place spontaneously in order to keep the chemical potential or the water activity of both compartments equal. The key idea to solve the problem is that the ratio of the ISF to the IVF stays constant after adding isotonic saline. Subsequently, the ratio of the volume is each compartment to the total volume of both compartments will also stay constant after the addition of saline solution. The easier way to solve this problem would be to multiply the ratio of each compartment to the total volume by the added volume. From Fig. 10.1, the volume of the two compartments is 3 L + 10 L ¼ 13 L (continued)

10.4

Rules of Physics That Could Prove Useful in Clinical Intervention

229

Example 10.2 (continued) 3 10 ∙ 650 ¼ 150 mL; V ISF ¼ ∙ 650 ¼ 500 mL V IVF ¼ 13 13 Alternatively, V ISF 10 ¼ : This is the preserved property before and after the addition 3 V IVF of normal saline V ISF þ V IVF ¼ 13:650 L, final state Substituting the preserved property into the final state, )

10 13:650 ∙ 3 ∙ V IVF þ V IVF ¼ 13:650 ) V IVF ¼ ¼ 3:150 L 3 13 ) V saline distributed in IVF ¼ 3:150
3 ¼ 0:150 L V saline distributed in ISF ¼ 0:650
0:150 ¼ 0:500 L

Example 10.3 The ECF bathing the cell in the sketch has suddenly become hypertonic due to an excess Na+ concentration. Describe at least two processes using the plasma membrane accessories—phospholipids, Na+ channels, aquaporins, and Na,KATPase—shown below by which the cell maintains homeostasis.

(continued)

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Example 10.3 (continued) Assume that the Na+ channels are unidirectional allowing selectively the passage of Na+ ions inside the cell, aquaporins are bidirectional selectively conducting water in and out of the cell, phospholipid bilayers can allow passive diffusion of organic hydrophobic solutes and some water at a very slow rate, and the Na,K-ATP transports 2 K+ into the cell while 3 Na+ are transported outside the cell against their concentration gradient. Solution The cell has two choices that can be implemented very fast to maintain homeostasis. It either must transport water out via the aquaporins or allow the rapid diffusion of Na+ inside the cells through the Na+ channels. Most probably the cell will initiate both the processes simultaneously to counteract more efficiently the osmotic pressure of the ECF with smaller water loses. There is no reason to employ the Na,K-ATPase pump as operation of the pump will actually decrease the osmolarity of the ICF and increase the ECF Na+ concentration. This is an excitable nerve or muscle cell. The Na+ channels play a central role in action potential. If their flow inside the cells is not tightly regulated, the (resting) membrane potential will be higher (more positive) and the timing and duration of the action potential will be irregular with perhaps life-threatening consequences. There is another slower mechanism that cells may use to counteract increased osmotic pressure of the ECF. They may allow the passive diffusion of organic solutes like amino acids through the phospholipid bilayer of the plasma membrane. Increased intracellular concentration of these organic solutes effectively increases cytosol osmolarity. This mechanism is commonly used by brain cells in chronic hypernatremia where the uptake and storage of organic solutes like taurine, betaine, and myo-inositol facilitates water pulling inside the cells, long-term cell volume restoration, and functionality in the presence of increased serum Na+ concentration. Chronic hypernatremia must be corrected slowly as removal of the stored organic solutes from brain cells under physiological ECF osmolarity takes days. Attempts to treat hypernatremia by a quick infusion of a hypotonic NaCl solution could result in cell swelling, cerebral edema, and coma.

10.5

Maintenance and Replacement Fluids

Some of the common maintenance and replacement solutions used in clinical practice are listed below in Tables 10.2 and 10.3.

10.5

Maintenance and Replacement Fluids

231

Table 10.2 Composition of some of the prescribing maintenance fluids for parenteral administration Solution Crystalloids Isotonic saline (NS) 0.2% Saline 3% Saline D5W D10W D5NS D5½NS 2.5% dextrose, 0.45% NaCl 4% dextrose, 0.18% NaCl 5% dextrose, 0.3% KCl 0.3% KCl, 0.9% NaCl Hartmann’s (H) Ringer’s (R) Ringer’s lactate (LR) D5LR 2.5% dextrose, ½LR

Plasmalyte A Plasmalyte 56 Plasmalyte 56 in 5% dextrose WHO Colloids 5% Human serum albumin (HSA) 20% HSA Dextran 40 Dextran 40 in 5% dextrose Dextran 70 Dextran 70 in 0.9% NaCl 6% Dextran 70–7.5% NaCl

Composition 0.9% w/v NaCl 0.2% w/v NaCl 3% w/v NaCl 5% w/v dextrose 10% w/v dextrose 5% w/v dextrose, 0.9% w/v NaCl 5% w/v dextrose, 0.45% w/v NaCl 2.5% w/v dextrose, 0.45% w/v NaCl 4% w/v dextrose, 0.18% w/v NaCl 5% w/v dextrose, 0.3% w/v KCl 0.3% w/v KCl, 0.9% w/v NaCl Na+ ¼ 131 mmol/L, Cl
¼ 111 mmol/L, K+ ¼ 5 mmol/L, Ca2+ ¼ 2.4 mmol/L, lactate ¼ 28 mmol/L Na+ ¼ 148 mmol/L, Cl
¼ 156 mmol/L, K+ ¼ 4 mmol/L, Ca2+ ¼ 2.25 mmol/L Na+ ¼ 130 mmol/L, Cl
¼ 109 mmol/L, K+ ¼ 4 mmol/L, Ca2+ ¼ 2 mmol/L, HCO3
¼ 29 mmol/L, lactate ¼ 28 mmol/L 5% w/v dextrose, Na+ ¼ 130 mmol/L, Cl
¼ 109 mmol/L, K+ ¼ 4 mmol/L, Ca2+ ¼ 1.5 mmol/L, lactate ¼ 28 mmol/L 2.5% w/v dextrose, Na+ ¼ 65 mmol/L, Cl
¼ 54 mmol/L, K+ ¼ 2 mmol/L, Ca2+ ¼ 1 mmol/L, HCO3
¼ 15 mmol/L, lactate ¼ 14 mmol/L Na+ ¼ 140 mmol/L, Cl
¼ 98 mmol/L, K+ ¼ 5 mmol/L, Mg2+ ¼ 1.5 mmol/L, gluconate ¼ 23 mmol/L, acetate ¼ 27 mmol/L Na+ ¼ 40 mmol/L, Cl
¼ 40 mmol/L, K+ ¼ 13 mmol/L, Mg2+ ¼ 1.5 mmol/L, acetate ¼ 16 mmol/L Na+ ¼ 40 mmol/L, Cl
¼ 40 mmol/L, K+ ¼ 13 mmol/L, Mg2+ ¼ 1.5 mmol/L, acetate ¼ 16 mmol/L, 5% w/v dextrose 0.4% w/v NaCl, 0.1% w/v KCl, 0.65% w/v CH3COONa, 0.8% w/v glucose 5 g of HSA/100 mL; MW of HSA ~67,000 Da 20 g of HSA/100 mL 100 g of dextran MW 40 kDa per liter of water 100 g of dextran MW 40 kDa in 1 L of D5W 60 g of dextran MW 70 kDa per liter of water 60 g of dextran MW 70 kDa per liter of isotonic saline

(continued)

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Table 10.2 (continued) Solution Haemaccel® Voluven® eloHAES® Hemohes®

Composition 3.5% w/v of modified gelatin MW 30 kDa, Na+ ¼ 145 mmol/L, Cl
¼ 145 mmol/L, K+ ¼ 5.1 mmol/L, Ca2+ ¼ 6.25 mmol/L 6% w/v Hydroxyethyl starch MW ~130 kDa in 0.9% w/v NaCl 6% w/v Hexastarch MW ~200 kDa in 0.9% w/v NaCl 6% w/v Pentastarch MW ~200 kDa in 0.9% w/v NaCl

Table 10.3 Composition of some hydration fluids for oral administration Rehydrating solutions Crystalloids WHO (World Health Organization) Rehydralite Maintenance solutions Pedialyte liquid nonflavored Resol

Ricelyte

10.6

Composition 0.4% w/v NaCl, 0.1% w/v KCl, 0.65% w/v CH3COONa, 0.8% w/v glucose Na+ ¼ 75 mmol/L, Cl
¼ 65 mmol/L, K+ ¼ 20 mmol/L, citrate ¼ 30 mEq/L in 2.5% w/v dextrose Na+ ¼ 45 mmol/L, Cl
¼ 35 mmol/L, K+ ¼ 20 mmol/L, citrate ¼ 30 mEq/L in 2.5% w/v dextrose Na+ ¼ 50 mmol/L, Cl
¼ 50 mmol/L, K+ ¼ 20 mmol/L, Ca2+ ¼ 2 mmol/L, Mg2+ ¼ 2 mmol/L, H2PO4
¼ 5 mEq/L, citrate ¼ 34 mEq/L in 2% w/v dextrose Na+ ¼ 50 mmol/L, Cl
¼ 25 mmol/L, K+ ¼ 45 mmol/L, citrate ¼ 34 mEq/L in 3% w/v rice syrup solids

Osmotic Diuretics

Osmotic diuretics are commonly used to treat disease states that are associated with (a) decreased urine volume or (b) accumulated fluid in the interstitial spaces of organs that their capillary endothelium cells are tightly knit, such as the blood–brain barrier. They are commonly used to reduce intracranial pressure, to reduce intraocular pressure before surgery for glaucoma, to increase urine flow in acute renal failure, or to accelerate renal excretion of water soluble toxins. Unlike the other drug diuretics that usually work in small doses by blocking renal transporters, osmotic diuretics are pharmacologically inert, and they exert their effects by virtue of their physicochemical properties. Essentially, osmotic diuretics work (1) by increasing the osmolarity of plasma and (2) by increasing the osmolarity of renal filtrate or urine. It is important to understand that osmotic diuretics are very different from plasma expanders because in addition to increasing the plasma osmolarity, they have the capacity to be freely filtered through at the glomeruli and increase the osmolarity of the urine. Therefore, the initial effect of plasma expansion of osmotic diuretics is dissipating rapidly by their ability to induce diuresis and eliminate plasma volume by increasing the rate of urination.

10.6

Osmotic Diuretics

233

The desired physicochemical and biophysical properties of osmotic diuretics are: • • • • • • • • • • • • •

They have very high polarity. They have very high water solubility. Their MW is small enough (so that they can be filtered through at the glomerulus). They remain in the extracellular space (ECF). They are not able to passively cross cell plasma membranes. They are not substrates of protein transporters. They do not cross the blood–brain barrier. They are not reabsorbed or they are poorly reabsorbed in the renal tubules. They are not actively secreted in the urine. They are not metabolized to a large extent. They are freely filtered through at the glomerulus. They are primarily excreted in the urine as the mode of elimination from the body. They are pharmacologically inert (I know that we have said that before).

The properties listed above are not only important; they are all necessary for different reasons. First, they have to be pharmacologically inert because induction of diuresis is due to a physical phenomenon and not to a pharmacological event, and because they are administered at large quantities. Imagine the magnitude of the anticipated pharmacological response if these agents with doses given in gram quantities were substrates of biological receptors. Second, their capacity to achieve high osmolar in plasma is due to their enormous solubility in water. Third, their ability to increase the volume of IVF is due to (a) their high aqueous solubility, (b) their high polarity which restricts their passive diffusion through plasma membranes and contributes to their confinement in the vascular compartment, (c) the fact that they are not substrates of protein transporters and therefore they cannot be carried intracellularly, and (d) that their osmotic activity is not compromised by enzymatic destruction of the molecule. Furthermore, their ability to induce diuresis is due to, in addition to the aforementioned factors, their small particle size that allows free passage into the renal filtrate at the glomerulus, none or very little reabsorption in the renal tubules, and none or very little active secretion into the renal tubules. Well-known osmotic diuretics are mannitol, isosorbide, and glycerol. The most commonly used is, by far, mannitol (Fig. 10.5). Fig. 10.5 Mannitol molecular and structural formula and aqueous solubility

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Example 10.4 Could cholesterol be an osmotic diuretic? Compare its structure with mannitol and provide explanations to your answer. Solution The structures of mannitol and cholesterol are shown below:

OH

OH HO

OH OH

OH HO

Mannitol

Cholesterol

Mannitol carries six hydroxyl groups (OH) in its molecule that can hydrogen bond with water and six carbon atoms. As a result, mannitol is a very hydrophilic molecule. Cholesterol on the other hand (C27H46O, MW ¼ 386.6) carries 27 apolar carbon atoms and only a single hydroxyl group. It is an extremely hydrophobic molecule that is almost never free in water and is always associated with the hydrophobic tails of phospholipids in plasma membranes. The aqueous solubility of mannitol and cholesterol are 1.2 M and 6.5 nM, respectively, making mannitol’s solubility in water 185 million times larger than that of cholesterol. Putting aside all other important differences, a molecule that does not sorb water cannot be an osmotic diuretic. At the molecular level, water molecules can hydrogen bond with every –OH group and surround the whole molecule of mannitol, whereas water can only approach the molecule of cholesterol only in the proximity of that single –OH group.

Example 10.5 A 154 lb, male adult presented to the emergency department with acute renal failure. He is to be infused 200 mg/kg over 10 min with mannitol. What is the rate of infusion in mL/min if the mannitol solution available is 10% w/v? Solution 154 lb ¼ 70 kg 2:2 lb=kg

(continued)

10.7

The Gram-Equivalent Weight (EW)

235

Example 10.5 (continued) ) 200 ) )

10.7

mg ∙ 70 kg ¼ 14 g kg 14 g 10 g 100 mL

¼ 140 mL

140 mL ¼ 14 mL=min 10 min

The Gram-Equivalent Weight (EW)

Electrolyte deficit or excess can be accurately verified only after measuring their levels in the blood. In addition, as we have discussed previously, one of the main tests that is used in the diagnosis of fluid imbalances is the analysis of levels of electrolytes in the blood. The equivalent unit (or milliequivalent) per liter is almost used exclusively by physicians, pharmacists, and nurses to express the concentration of electrolytes in solutions and to further calculate relative amounts needed for therapy. This unit is the established unit of measure of electrolyte concentration because it takes note of the valence of ions, thus relating the total number of ionic charges in solution with the chemical, osmotic, and physiological activities. The equivalent weight of an ionic compound can be calculated using Eq. 4.2, EW ¼ MW n , assuming that the MW of the compound and the reaction describing the complete dissolution of a salt in an aqueous medium are known. Calculating the EW of relatively dilute salts in solution, assuming full dissociation of the salt into the corresponding ions, is quite an easy task. Example 10.6 Calculate the EW of CaCl2 (MW ¼ 111 g/mol) Solution Using Eq. 4.2, EW

  g Eq

¼

MW

g ðmol Þ

n

where n (expressed in units of Eq/mol) is the total # of electrons or protons that participate in the reaction, that is, the dissolution of calcium chloride. It can be calculated by multiplying the number of moles of an ion (stoichiometric coefficient in the chemical reaction) with the ionic charge or valence of the corresponding ion. CaCl2 ! Ca2þ þ 2 Cl


(continued)

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Example 10.6 (continued) n ¼ 1∙2 ¼ 2, if the calculation is based on calcium ions Ca+2, or the # of positive charges n ¼ 2∙1 ¼ 2, if the calculation is based on chloride ions Cl
, or the # of negative charges Substituting the values in Eq. 4.2, ) EW ¼

g mol Eq mol

111 2

¼ 55:5

g Eq

Therefore, 1 Eq of CaCl2 equals 55.5 g.

10.7.1 Converting Mass Expressed in Grams (g) into Equivalents (Eq) and Vice Versa

Example 10.7 How many mEq of calcium and chloride are there in 1 L of 0.1% w/v CaCl2? (MW ¼ 111 g/mol) Solution 0.1% w/v ¼ 0.1 g/100 mL ¼ 1 g/L EW (CaCl2) ¼ 55.5 g/Eq (from Example 10.6) Set up a ratio to calculate the Eq in 1 g of solution. )

55:5 g 1 g ¼ ) X ¼ 0:018 Eq ¼ 18 mEq of CaCl2 1 Eq X

Since calcium chloride dissociates completely in water, dissolution of 18 mEq of CaCl2 will yield 18 mEq of Ca+2 and 18 mEq of Cl
ions. We could have reached the same answer by first converting the % w/v into molarity and then converting molarity into normality using Eq. 4.3.

Example 10.8 How many grams of sodium chloride (NaCl) are needed to prepare 1 L of solution that contains 154 mEq of sodium and 0.154 Eq of chloride? (MW ¼ 58.5 g/mol) (continued)

10.7

The Gram-Equivalent Weight (EW)

237

Example 10.8 (continued) Solution Determine the EW of sodium chloride and then set up a ratio to calculate the grams that are equivalent to the Eq of NaCl. NaCl ! Naþ þ Cl
154 mEq of Na+ and 154 mEq of Cl
came from dissolution of 154 mEq of NaCl salt. n ¼ 1 ∙ 1 ¼ 1 ) EW ¼

g mol Eq mol

58:5 1

¼ 58:5

g ) 1 Eq of NaCl equals 58:5 g: Eq

Since we now have 0.154 Eq of NaCl, )

58:5 g X ¼ ) X ¼ 9:009 g of NaCl 1 Eq 0:154 Eq

Example 10.9 Compound the following prescription for a patient who is on a loop diuretics therapy. ℞ Potassium chloride Purified water M. et ft. solution Sig: 1 tsp in 6–8 fξ of fluid b.i.d.

q.s. ad.

10 mEq/tsp 90 mL

Calculate how many grams of KCl are needed to compound the prescription. Solution Step 1: Calculate total mEq in 90 mL 90 mL  10 mEq=5 mL ¼ 180 mEq of KCl Step 2: Use the EW of KCl to convert Eq into grams (continued)

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Example 10.9 (continued) KCl ! Kþ þ Cl
n ¼ 1 (based on the total # of positive or negative charges of K+ or Cl
) EW ¼

g mol Eq mol

74:5 1

¼ 74:5

g ) 1 Eq of KCl equals 74:5 g Eq

Since we now have 0.18 Eq of KCl, ) 74:5 g=Eq  0:18 Eq ¼ 13:41 g of KCl is required to fill the prescription

Example 10.10 A 22-year-old male shows up in the ER with a cardiac arrhythmia and ECG changes. Blood analysis verified that his potassium levels were below 3 mEq/ L. The physician orders an infusion of 2 L of 30 mEq/L KCl within 2 h. (a) How many grams of KCl are needed to prepare the infusion? (b) What is your choice of solvent? Solution (a) Step 1: Calculate total mEq of KCl in 2 L ) 2 L  30 mEq=L ¼ 60 mEq of KCl Step 2: Use the EW of KCl to convert Eq to grams EW ¼

g 74:5 mol

1

Eq mol

¼ 74:5

g ) 1 Eq of KCl equals 74:5 g Eq

Since we now have 60 mEq of KCl, ) 74:5 g=Eq  0:06 Eq ¼ 4:47 g of KCl is required for the infusion (b) Calculate the osmolarity of the solution (see Chap. 7) (continued)

10.7

The Gram-Equivalent Weight (EW)

239

Example 10.10 (continued) KCl ! Kþ þ Cl
) i ¼ 2 n ¼ 1 ) Normality ¼ Molarity ¼ 30 mmols=L ) Osmolarity ¼ i  C ¼ i  M ¼ 2  ð0:030Þ ¼ 0:060 Osmolar The solution is hypotonic and needs to be made isotonic with the addition of NaCl (definitely not D5W; see Sect. 10.4, hypokalemia and insulin). In cases like this, where the instructions are incomplete, it is always recommended for the pharmacist to consult with the physician.

Example 10.11 A 22-year-old female was diagnosed with hypernatremia (elevated sodium levels in the serum) accompanied with a relatively normal intravascular volume, but reduced ICF. What possible replacement fluids would you choose from Table 10.2 to reestablish homeostasis of the patient? Solution The fact that the patient has elevated sodium serum levels, cell dehydration, but normal blood volume suggests that the osmotic force exerted by sodium excess in the blood is causing intracellular fluid shift toward the EC compartments. The patient is dehydrated and should be given slow infusions of hypotonic replacement fluids to reestablish physiological serum sodium levels and replenish body water of the intracellular compartment by osmosis. An appropriate replacement fluid is 0.2–0.45% w/v saline solution. Avoid administering 5% D5W. Secretion of insulin could upregulate the Na,KATPase that will in turn increase the outflow of Na+ ions from cells causing a further increase of the already elevated serum Na+ concentration. Also, the patient can be given water orally. If the patient is unable to tolerate oral water, avoid IV infusion of plain water as it could cause swelling and rupture of RBC. Lastly, the rate of infusion of the hypotonic solution should be very slow. The exact rate of infusion can be obtained from published medical guidelines.

Example 10.12 A 23-year-old cancer patient (154 lb) is transported to a hospital emergency room with seizures. His sodium plasma levels were within physiological range yesterday, but current blood analysis showed sodium plasma levels of 109 mEq/L, indicating severe acute hyponatremia. (continued)

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Fluids and Electrolytes

Example 10.12 (continued) (a) How much is his total sodium deficit? (Assume ECF density of 1 g/cm3) (b) How much volume from a 0.9% w/v sodium chloride solution do you have to infuse the patient with? Solution (a) Information that is needed to answer this question: Blood density ¼ 1 g/mL ¼ 1 kg/L Average normal Na+ levels in the ECF ¼ 140 mEq/L (from Table 10.1) Fluid content of average human ¼ 60% of body weight ECF ¼ 40% of total body fluid Based on the information above, the approximate volume of extracellular fluid of the patient, VECF ¼ (154 lb/2.2 (lb/kg)) ∙ 0.6 ∙ 0.4 ¼ 16.8 kg But density of ECF ¼ 1 kg/L ) V ECF ¼ 16:8 L Total body sodium deficit ¼ (135–109) mEq/L ∙ 16.8 L ¼ 436.8 mEq The most commonly used empirical equation that calculates total body sodium deficit uses total body water and not the ECF. I have not used the intracellular volume simply because the intracellular sodium concentration is very small, and it is not even certain that there is an intracellular sodium deficiency. (b) 0.4368 Eq Na+ ¼ 0.4368 Eq Cl
¼ 0.4368 Eq NaCl NaCl is a monovalent salt, and thus, n ¼ 1 and EW ¼ MW ¼ 58.5 g/Eq Mass of NaCl needed ¼ 0.4368 Eq ∙ 58.5 g/Eq ¼ 25.5645 g Volume of 0.9% NaCl needed to be infused ¼ 25.5645 g/(0.9 g/ 100 mL) ¼ 2840.5 mL

10.7.2 Calculating Equivalent of Ions in Solutions That Are Made by Multiple Common Ion Salts In general, the equivalent of a common ion is an additive property similar to the gram quantity of substances. Thus, all we need to do is to figure out the gram quantity of the common ion contributed from each salt, sum them up, and convert the total mass into equivalents of that particular ion. Example 10.13 The World Health Organization recommends the following oral solution for rehydration: (continued)

10.7

The Gram-Equivalent Weight (EW)

241

Example 10.13 (continued) NaCl KCl CH3CO2Na Glucose Purified water q.s. ad

4.0 g 1.0 g 6.5 g 8.0 g 1L

Calculate the mEq of sodium ions present in a 250 mL dose. Solution Only sodium chloride and sodium acetate contribute to the sodium ion concentration in solution. Let us first calculate grams in 250 mL of Na+ from each of these salts and add them together. This can easily be accomplished by multiplying the mass of the salt present in 250 mL with the ratio of the AW (Na+)/MW(salt). For NaCl, 4

g 23 mol Naþ g NaCl ∙ 0:25 L ∙ ¼ 0:393 g Naþ g L 58:5 mol NaCl

For CH3CO2Na (MW ¼ 82), 6:5

g 23 mol Naþ g CH3 CO2 Na ∙ 0:25 L ∙ ¼ 0:456 g Naþ g L 82 mol CH3 CO2 Na

) Total Naþ ¼ 0:849 g The EW of Na+ is equal to its AW because the valence of sodium (# of positive charges) is 1. Using Eq. 4.2, we can convert grams of Na+ to mEq. )

23 g Naþ 0:849 g Naþ ¼ ) X ¼ 0:037 Eq or 37 mEq of Naþ 1 Eq X

Alternatively, you could have calculated the moles of Na ions directly. For NaCl, 4Lg ∙ 0:25 L NaCl ¼ 0:0171 mol NaCl ¼ 0:0171 mol Naþ g 58:5 mol NaCl For CH3CO2Na (MW ¼ 82), (continued)

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Fluids and Electrolytes

Example 10.13 (continued) 6:5Lg ∙ 0:25 L CH3 CO2 Na ¼ 0:0198 mol CH3 CO2 Na ¼ 0:0198 mol Naþ g CH3 CO2 Na 82 mol ) Total Naþ ¼ 0:037 mol or 0:037 Eq of Naþ

Note: It is always a good habit to verify your answers if possible. For example, students usually have doubts of the mass of individual ions composing the two salts. In principle, the mass of individual ions should always add to the mass of the corresponding salt. In our case, the mass of Na+ coming from NaCl plus the mass of Cl
should sum up to the mass of NaCl. Similarly the mass of Na+ plus the mass of acetate ion should add up to the mass of sodium acetate. Let us verify that for the NaCl. We have already found that the mass of Na+ is 0.393 g. Using the same equation, the mass of chloride coming from NaCl is g 35:5 mol Cl
g ¼ 0:607 g Cl
) 4 ∙ 0:25 L NaCl ∙ g L 58:5 mol NaCl

The two numbers, 0.393 g + 0.607 g, add up to exactly 1 g of NaCl. You may verify the mass of sodium acetate.

Exercises 10.1. Which organ plays the most significant role in the restoration of water and electrolyte imbalances? (a) (b) (c) (d) (e)

Liver Stomach Skin Kidneys Lungs

10.2. Choose the correct statement? (a) (b) (c) (d) (e)

IVF is part of ICF ECF is part of IVF IVF is part of ECF Blood is part of plasma Lymph is part of IVF

Exercises

243

10.3. Which of the following substances is a crystalloid? (a) (b) (c) (d) (e)

Glucose Albumin Alpha 1 antitrypsin Gamma globulin High-density lipoprotein

10.4. Without calculating the osmotic pressure, rank the following solutions according to their osmolarity: 0.9% w/v NaCl, 0.9% w/v dextrose, 5% w/v albumin. The MW of NaCl, glucose, and albumin are 58.5, 180, and 67,000, respectively. (a) (b) (c) (d) (e)

0.9% NaCl > 5% albumin > 0.9% glucose 0.9% NaCl ¼ 0.9% glucose > 5% albumin 5% Albumin > 0.9% NaCl > 0.9% glucose 0.9% NaCl >0.9% glucose >5% albumin 0.9% NaCl > 0.9% glucose ¼ 5% albumin

10.5. Which fluid composition has the greatest lasting plasma volume supportive effect? (a) (b) (c) (d) (e)

Isotonic saline Hypertonic dextrose (6% w/v) Isotonic saline + 6% w/v colloid hypertonic saline (1% w/v) isotonic dextrose + 6% w/v colloid

10.6. Which of the following is not within the normal physiological serum concentration range? (a) (b) (c) (d) (e)

Mg2+ 1 mmol/L Ca2+ 5 mmol/L K+ 5 mmol/L Na+ 140 mmol/L Cl
105 mmol/L

10.7. Which of the following solutions are indicated for the treatment of severe hypokalemia ( 0) or exponential decay (k < 0). In order to make sense of the significance of rate constants, we need to go back and study their respective equations. We can see from Eq. 12.2 that the zero-order rate constant is the rate of change of drug concentration with time. It carries units of concentration over time, and its magnitude can be used to evaluate the rate of change of drug concentration with time in the whole time domain. For example, if the rate constant of a linear decay process is k ¼ 0:1 L mg ∙ min, 0.1 mg/L of drug is decaying per minute. Careful examination of Eq. 12.3 reveals additional useful information. We know that the product of the zero-order rate constant with time is the drug concentration eliminated because, first, the product has units of concentration and, second, when it is subtracted from C0, it yields the drug concentration remaining (Fig. 12.9). On the contrary, the first-order rate constant of an exponential decay is derived from Eq. 12.5, dC dt ¼ k ∙ C, and it clearly indicates that the rate of change of drug concentration is dependent on the first-order rate constant and the drug concentration at each instant. The larger the k the bigger the rate of change, but since k is not the slope of the equation, it cannot be used just by itself to evaluate the rate of exponential growth or decay in a system. In other words, you can only obtain instantaneous rates accurately since the slope of the curve changes at every instant. Instantaneous slopes can be determined graphically by drawing tangents on the

Fig. 12.9 Description of the parameters of the zero-order equation

12.6

The Importance of the Leading Coefficients of Special Exponential. . .

281

curve at specific time points, or they can easily be determined by substituting Eq. 12.6 in Eq. 12.5 as shown below: dC dC ¼ k ∙ C ¼ k ∙ C 0 ∙ ek ∙ t ) ¼ k ∙ C 0 ∙ ek ∙ t dt dt

ð12:11Þ

Most important is that the first-order rate constant can instead be used to determine how long the exponential decay process will last. Whenever we need to compare the exponential decay rates between two or more drugs, we compare their half-lives using Eq. 12.8. Since the half-life of a first-order process is a constant concentration-independent parameter, it can be used as a means of comparison of the duration of the exponential decay process. The bigger the first-order rate constant k, the shorter the half-life of the decay process, and the shorter the duration of the decay process of the drug, in general. A drug that has highest first-order elimination rate constant (or smallest t1/2) will be eliminated from the body earlier. However, keep in mind that a drug that has a shorter t1/2 or a shorter decay process will not necessarily have a faster rate of concentration change at all times. Example 12.5 Determine the rate of change of drug concentration at time zero and at half-life for a first-order decay process. Solution Using Eq. 12.11, the initial rate of change of drug concentration at t ¼ 0 is 

dC dt



¼ k ∙ C 0 ∙ ek ∙ 0 ¼ k ∙ C0 initial

When t ¼ t1/2 and C ¼ C20 , Eq. 12.11 becomes 

dC dt

 ¼ k ∙ t¼t 1=2

C0 2

both in units of concentration over time. It is easy to describe the significance of a zero-order rate constant, for example, k ¼ 0:1 L mg ∙ min, but what is the meaning, for example, of a first-order rate constant of 0.1 per hour? Rates usually measure something over time. Without the concentration parameter, what do we actually measure; inverse time? Eq. 12.6 can help us fully explain the significance of the first-order rate constant. We know that the exponent ek∙t carries no units. We also know that the product C 0 ∙ ek ∙ t ¼ eCk 0∙ t is equal to the drug concentration remaining, and we also know that ek∙t is always a fraction with a value less than 1 for all times bigger than zero.

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Rates and Orders of Reactions

Fig. 12.10 Parameter description of the first-order equation

Therefore, ek∙t or ek1∙ t must be the fraction of the drug remaining, always relative to the initial concentration C0 (Fig. 12.10). This definition suggests that I need to specify the time in order to properly interpret k. For example, what is the meaning of a rate constant k ¼ 0.1 h1 at t ¼ 1 h? When k ¼ 0.1 h1 and t ¼ 1 h, ek∙t ¼ e0.1 ¼ 0.9 and C ¼ 0.9C0. In other words, 0.9C0 remained and 0.1C0 has decayed or eliminated after 1 h. Please remember that this relationship is not linear with time. In other words, it does not mean that after 10 h the decay is going to be 1 or 100 % of C0. After 10 h, the initial concentration will be decreased to e0:1 ∙ 10 ∙ C0 ¼ e1 ∙ C 0 ¼ Ce0 or 0.37 C0 or 37 % C0. This is the concentration of drug remaining after 10 h during a decay process that is characterized by a rate constant of k ¼ 0.1 h1 (63% C0 eliminated). Accordingly, in order to relate the first-order rate constant to the rate of change of drug concentration, you need to specify the time. The first-order rate constant–time product will yield the relative reduction of the initial concentration in an exponential decay process within a specified time period. For example, the reduction in the initial 1 drug concentration with k ¼ 0.1 h1 after 20 h is e0:1 ∙ 20 ¼ e2 ¼ 7:39 times its C0 original value or 7:39 ¼ 0:135 C 0 . Again, this is the drug concentration remaining in the system. To calculate the concentration of drug decayed or eliminated, you need to subtract the concentration remaining from the initial concentration (Eq. 12.12). C ¼ C 0  C 0 ∙ ek ∙ t ¼ C 0 ∙ 1  ek ∙ t




ð12:12Þ

where C is the drug concentration decayed or eliminated. Example 12.6 The equations below represent the elimination profile of two new drugs that were studied in adult female patients: Drug A: C ¼ 3 ∙ e0.5∙t Drug B: C ¼ e0.6∙t Assuming that concentrations are in mg/L and time is in hours, calculate the following: (continued)

12.6

The Importance of the Leading Coefficients of Special Exponential. . .

283

Example 12.6 (continued) (a) Which drug is eliminated earlier from the body, drug A or drug B? (b) Which drug presents faster change of concentration with time at time zero and at half life? (c) How long will it take for 95% of the drug to be eliminated from the body? Solution (a) t 1=2, A ¼

ln ð2Þ 0:693 ¼ 1:39 h ¼ kA 0:5

t 1=2, B ¼

ln ð2Þ 0:693 ¼ 1:16 h ¼ kB 0:6

Therefore, drug B is eliminated earlier from the body. (b)

  dC mg ¼ k ∙ C 0 ∙ ek ∙ 0 ¼ k ∙ C0 ¼ 0:5 ∙ 3 ¼ 1:5 dt initial, A L∙h   dC mg ¼ k ∙ C0 ¼ k ¼ 0:6 dt initial, B L∙h   dC C0 mg ¼ 0:75 ¼ k ∙ dt t¼t1=2, A L∙h 2   dC C0 mg ¼ 0:3 ¼ k ∙ dt t¼t1=2, B L∙h 2

Evidently drug B is eliminated earlier (shorter t1/2), but its initial rate of concentration change with time is slower than that of drug A. Drug A has a much higher initial concentration in the blood. Its concentration in the blood declines faster than drug B initially but as its concentration approaches that of drug B, its rate of elimination becomes smaller than the rate of elimination of drug B since kA < kB and drug B is cleared out of the blood earlier. (c) When 95% of the drug is eliminated, only 5% or 0.05 is remaining. )

C ¼ 0:05 C0

t 0:95, A ¼

ln ð0:05Þ 3 ¼6h ¼ kA 0:5

(continued)

284

12

Rates and Orders of Reactions

Example 12.6 (continued)

Fig. 12.11 Drug A (top) and drug B (bottom; broken line) elimination profiles. Notice that although the initial rate of change of the concentration with time is faster (steeper) with drug A, it takes longer for drug A to be cleared out of the body because kA < kB

t 0:95, B ¼

ln




1 0:05

kB

¼

ln ð20Þ 3 ¼5h ¼ kB 0:6

It takes 5 and 6 h for 95% for drug B and drug A to be eliminated from the blood, respectively. In agreement with their half-lives, it takes longer for drug A to be eliminated from the body, although its initial rate of its concentration change with time is a little bit faster than B (Fig. 12.11).

12.7

Horizontal and Vertical Shifts of Exponential Functions

Horizontal shifts of exponential functions can take place by adding or subtracting a constant number to the input variable of the exponential function x or t, whereas vertical shifts can be brought about by adding or subtracting a constant number to the function. The parent exponential function y ¼ 4 ∙ e0.2∙x along with its vertical upward y ¼ 4 ∙ e0.2∙x + 1 and vertical downward y ¼ 4 ∙ e0.2∙x  1 shifts, by 1 unit, are presented in Fig. 12.12 (top panel). The bottom panel of the figure displays the parent function along with the left- and right-shifted functions by 1 unit, y ¼ 4 ∙ e0.2∙ (x+1) and y ¼ 4 ∙ e0.2∙(x1), respectively.

12.7

Horizontal and Vertical Shifts of Exponential Functions

285

Fig. 12.12 Vertical (top) and horizontal (bottom) shifts of the parent exponential function y ¼ 4 ∙ e0.2∙x

Vertically and horizontally translated functions can be used to describe delay of drug release in dosage forms. Example 12.7 The same drug was administered intravenously by a student to two male monozygotic twin mice to study the drug elimination profile. Blood sample was drawn at variable time intervals, and the results were plotted in Fig. 12.13. One of the monoexponential functions graphed below is C ¼ 3.5 ∙ e0.1∙t (a) What is the equation of the other curve? (b) Determine the elimination half-life of each curve. (c) Provide a possible explanation for the second profile. (continued)

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12

Rates and Orders of Reactions

Example 12.7 (continued)

Fig. 12.13 The exponential function C ¼ 3.5 ∙ e0.1 ∙ t

Solution (a) The bottom curve must be C ¼ 3.5 ∙ e0.1∙t because it appears to intercept the y-axis at t ¼ 0 at 3.5 (back extrapolation of the graph to the y-axis). ) C0 ¼ 3:5

and k ¼ 0:1 h1

(b) Careful examination of the two curves reveals that the concentration achieved by the bottom curve was achieved by the top curve 1 h later, consistently. The equation of the top curve is therefore C ¼ 3.5 ∙ e0.1∙(t1). The elimination half-life is the same in both curves. They both appear to be reaching zero concentration with 1 h difference but at the same rate. t 1=2 ¼

ln ð2Þ 0:693 ¼ ¼ 6:93 h k 0:1

(c) The student may have made a mistake in keeping the time correctly. He may have collected the first blood sample immediately after the IV administration for the second mouse, but he wrote down that he has collected the first sample after 1 h. Consequently, the concentrations were shifted rightward by 1 h. This is not a real story, but mistakes happen. Good students will analyze their data carefully, and they will realize that the two curves that phenomenologically appear different, they actually have identical leading coefficient (C0) and the same rate constant. Always keep a good logbook.

12.7

Horizontal and Vertical Shifts of Exponential Functions

287

Example 12.8 The data shown in the table below and in Fig. 12.14 describe the release of a drug from a controlled release product with time, in simulated gastric media. If the drug release follows a monoexponential profile, can you determine the release rate constant of the drug? t (h) C (mg/L)

0 0

1 18.1

3 45.1

6 69.9

10 86.5

15 95.0

20 98.2

Solution At first glance, you may be thinking that having a first-order kinetic profile, the way to go about it is to convert C into ln C and determine k from the slope of the now linear form of the ln C-t graph. Alas, the ln C-t plot of the drug release data will not be a line. The main problem is that the data in the table are drug concentrations released and not drug concentration remaining in the dosage form. Therefore, we are going to first discuss how to model our data with an exponential equation that describes the drug release profile and then verify our answer by converting our drug release concentration into drug concentration remaining, plotting ln C-t and determining the slope of the line. Our first task is to use our creativity to model the profile with an exponential equation. We have the information that the profile is exponential, but we need to figure out an equation that will best fit our data. The general form of simple exponential equations such as C ¼ a ∙ ek∙t and C ¼ a ∙ ek∙t with a ¼ k ¼ 1, where a ¼ C0, is shown in Fig. 12.15. The two interesting features shown in Fig. 12.15 are: first, both curves cross the y axis at 1, that is, when t ¼ 0, C ¼ 1, and second, the curve C ¼ a ∙ ek∙t is the exact reflection of the curve C ¼ a ∙ ek∙t in the y-axis. Reflecting the curve C ¼ a ∙ ek∙t in the y-axis is equivalent to changing the sign of the exponent of the function C ¼ a ∙ ek∙t. Eliminating unrealistic values of negative time from (continued)

Fig. 12.14 Drug release profile from a controlled release product

288

12

Rates and Orders of Reactions

Example 12.8 (continued) Fig. 12.15 yields the curve in Fig. 12.16, which is similar to that shown in Fig. 12.7. Changing the sign of the leading coefficient would result in the equation C ¼  a ∙ ek∙t that is represented by the reflection of the curve C ¼ a ∙ ek∙t in the x-axis (Fig. 12.17). In order to shift the curve C ¼  a ∙ ek∙t upward by 1 unit, I only need to add 1 to the function. As Fig. 12.18 shows, the curve C ¼ (1  a ∙ ek∙t) ¼ (1  1 ∙ e1∙t) goes through the origin and really resembles in shape to that of the drug release profile shown in Fig. 12.14. The difference is only in the absolute numbers. It appears that drug release reaches a maximum drug concentration of 100 mg/L. All I have to do to shift my curve upward 100 units by changing the leading coefficient from 1 to 100 and adding 100 to the function. The final equation that describes the drug release profile is C ¼ (100  100 ∙ ek∙t) or C ¼ 100 ∙ (1  ek∙t). We have derived this equation before (Eq. 12.12). Now that we have the exponential equation that describes the release of the drug, we are ready to calculate the rate constant of drug release k. Reorganizing my equation, C ln 1  100 C 1 ¼ ek ∙ t ) k ¼ 100 t




Substituting any C-t pair should give you a similar value of k. For t ¼ 1 h,
ln ð1 C Þ 1 C ¼ 18.1 mg/L, k ¼ t 100 ¼  ln 1  18:1 100 ¼ ð0:2Þ ¼ 0:2 h For the second method, first notice that the equation that describes the drug released C ¼ 100  100 ∙ ek∙t is composed of two factors, and only the second one is exponential. The first one is just a number, 100. It is for this reason that the natural logarithm of C is not linear with time. To convert the concentration of drug released to concentration remaining in the dosage form, we need to subtract every concentration from 100. The drug remaining in the dosage form and its natural logarithm as a function of time are shown in the table below, and the corresponding plots are shown in Fig. 12.19. t (h) 0 1 3 6 10 15 20

C (mg/L) 100  0 ¼ 100 100  18.1 ¼ 81.9 100  45.1 ¼ 54.9 100  69.9 ¼ 30.1 100  86.5 ¼ 13.5 100  95.0 ¼ 5 100  98.2 ¼ 1.8

ln C 4.6 4.4 4.0 3.4 2.6 1.6 0.6

(continued)

Example 12.8 (continued)

Fig. 12.15 Graph of C ¼ a ∙ ek∙t (solid line) in comparison with the C ¼ a ∙ ek∙t (dashedline) with a ¼ k ¼ 1, where a ¼ C0

Fig. 12.16 Graph of C ¼ a ∙ ek∙t, where a ¼ k ¼ 1, on the domain [0,+1) and range (0,1]

Fig. 12.17 Graph of C ¼ a ∙ ek∙t and C ¼  a ∙ ek∙t, where a ¼ k ¼ 1, on the domain [0,+1) and range (0,1] and [-1,0), respectively

(continued)

290

12

Rates and Orders of Reactions

Example 12.8 (continued)

Fig. 12.18 Graph of C ¼  a ∙ ek∙t and C ¼ (1  a ∙ ek∙t) (dashed-curve), where a ¼ k ¼ 1, on the domain [0,+1) and range [-1,0) and [0,1), respectively

Fig. 12.19 Drug concentration remaining in the dosage form was plotted as a function of time

The rate constant of drug release can be calculated from the slope of the ln C-t plot (Fig. 12.19, right). Make sure that you use the lnC in the numerator and not C.

k ¼ slope ¼

ln ðC 2 Þ  ln ðC 1 Þ 4:4  4:6 ¼ 0:2 ) k ¼ 0:2 h1 ¼ t2  t1 10

The slope could have been determined more accurately from the equation of the ln C-t trend line, y ¼  0.2 x + 4.6, using the spreadsheet software. The initial concentration C0 ¼ e4.6 ¼ 100 and k ¼ 0.2.

Exercises

291

Exercises 12.1. When does the average rate Select all that apply. (a) (b) (c) (d) (e) (f)

ΔC Δt

equal the instantaneous rate of reaction

dC dt ?

When the reaction is a first-order process When the reaction is a second-order process When the time interval of study is infinitesimally small When the concentration varies exponentially with time When the reaction is a zero-order process When the average rate of reaction varies linearly with time

12.2. What is the order of reaction of the esterification of acetic acid with ethanol to ethylacetate, as shown below? CH3 COOH ðlÞ þ C2 H5 OH ðlÞ⟶CH3 COOC2 H5 ðlÞ þ H2 O ðlÞ (a) (b) (c) (d) (e)

0 1 2 4 Cannot be determined

12.3. Given the following graph, the initial concentration has units of:

(a) (b) (c) (d) (e)

mg/day mol/L g/day g/L Cannot be determined

292

12

Rates and Orders of Reactions

12.4. Given the following graph, the initial concentration has a value of:

(a) (b) (c) (d) (e)

257 0.257 2.57 0 100

12.5. Given the following first-order graph, the initial concentration has units of:

(a) (b) (c) (d) (e)

mg/day mol/L g/day g/L Cannot be determined

12.6. Given the following graph, the initial concentration has a value of:

Exercises

(a) (b) (c) (d) (e)

293

0.276 0.0276 4 54.6 Cannot be calculated

12.7. Given the following graph, the rate constant of the process has units of:

(a) (b) (c) (d) (e)

mg/day/L mol/L/day g/day g/L Cannot be determined

12.8. Given the following graph, the rate constant of the process has a value of:

(a) (b) (c) (d) (e)

257 0.257 2.57 0 100

294

12

Rates and Orders of Reactions

12.9. Given the following graph, the rate constant of the process has units of:

(a) (b) (c) (d) (e)

h mol/L/min min 1/h Cannot be determined

12.10. Given the following graph, the rate constant of the process has a value of:

(a) (b) (c) (d) (e)

0.1 1.56 0.44 0.22 Cannot be calculated

Exercises

295

12.11. What is your best estimate for the half-life in days from the data in the firstorder plot below?

(a) (b) (c) (d) (e)

50 100 250 350 450

12.12. Choose the correct statement given a drug elimination process from the blood described by the equation C ¼ C0 ∙ ek∙t. (a) (b) (c) (d) (e)

C is the initial drug concentration in the blood. C0 is the drug concentration remaining in the blood. k is expressed in units of concentration over time. ek ∙ t is the fraction of the drug concentration remaining in the blood. k ∙ t is the drug concentration eliminated from the blood.

12.13. A formulation containing 3.2 g/mL of a drug was subjected to stability studies. Its degradation constant k was found to be 0.05 mg/mL/h at 25  C. What is the shelf-life of the drug formulation? (Answer: t90 ¼ 7.3 years) 12.14. A drug follows first-order degradation kinetics and has a shelf-life of 180 days at 25  C. What is the approximate half-life (in years) at the same temperature? (Answer: t90 ¼ 3.24 years) 12.15. A 100 mg strength controlled release dosage form releases the encapsulated drug in the GI with a rate of 6.24 mg/h. How long would it take to release 70% of the drug? (Answer: 11.2 h) 12.16. The metabolism of a drug is characterized by a rate constant of 0.06 h1: Choose all correct statements.

296

12

(a) (b) (c) (d) (e)

Rates and Orders of Reactions

6% of the drug has been metabolized during the first hour. 0.06 of the drug is metabolized every hour. 0.06 of the drug is metabolized within an hour. 60% of the drug is metabolized after the first 10 h. 94% of the drug has been metabolized within the first 1 h.

12.17. The release of a drug from its dosage form is characterized by a rate constant of 0.42 mg/h: Choose all correct statements. (a) (b) (c) (d) (e)

0.42 mg of the drug was released during the first hour. 0.42 mg of the drug is released every hour. 0.42 mg of the drug is released within an hour. 4.2 mg of the drug is released after the first 10 h. 4.2% of the initial drug amount was released within the first 1 h.

12.18. Choose the correct statement given a drug degradation process described by the equation C ¼ C0  k ∙ t. (a) (b) (c) (d) (e)

C is the degraded drug concentration. C0 is the final drug concentration. k is expressed in units of inverse time. C is the initial drug concentration. k ∙ t is the degraded drug concentration.

12.19. The metabolism of a drug in the liver follows a zero-order process with a rate constant of 0:34 mg h . If the initial amount of the drug in the organ is 120 mg, how much drug is metabolized after 8 h? (a) (b) (c) (d) (e)

117.28 mg 7.9 mg 112.1 mg 2.72 mg 120 mg

12.20. A drug is eliminated from the body by a first-order process with a rate constant of 0.09 h1. How much drug is eliminated from the body 1 h after IV administration? (a) (b) (c) (d) (e)

0.09 its initial concentration 91% its initial concentration Cannot be calculated 99% its initial concentration 0.09 mg/L

12.21. A drug is eliminated from the body by a first-order process with a rate constant of 0.62 h1. How much drug is eliminated from the body 1 h after its IV administration? (a) 0.62 of its initial concentration (b) 38% of its initial concentration

Exercises

297

(c) Cannot be calculated (d) 0.54 of its initial concentration (e) 46% of its initial concentration 12.22. The decay of radioisotopes is a first-order process. A patient is taking radioactive iodine (I-131) as part of her therapy after thyroidectomy. The recommended dose is 100 mCi. Assuming that the compounding pharmacist in the hospital prepared the liquid formulation 3 days in advance, how much radioactivity remains in the dosage form after 3 days. The half-life of I-131 is 8.02 days. What is the % decrease in the dose? (Answer: 77.2 mCi, 22.8% dose decrease) 12.23. The initial amount of a drug in a dosage form is 1000 mg. During drug release studies, it was found that the dosage form released 250 mg after 2.5 h. Assuming that the drug release is a first-order process, how long will it take for the amount of drug in the dosage form to drop to 600 mg? (Answer: 4.44 h) 12.24. The equations below represent the elimination profile of two new drugs that were studied in adult female patients: Drug A: C ¼ 120 ∙ e0.25∙t Drug B: C ¼ 212 ∙ e0.63∙t Assuming that concentrations are in mg/L and time is in hours, calculate the following: (a) Which drug is eliminated earlier from the body, drug A or drug B? (b) Which drug presents a faster change of concentration with time at time zero and at half-life? (c) How long will it take for 99% of the drug to be eliminated from the body? (Answer: Drug A: (a) 2.77 h; (b) 30 Lmg∙ h, 15 Lmg∙ h ; (c) 18.4 h; drug B: (a) 1.1 h; (b) 133:56 Lmg∙ h ,  66:8 Lmg∙ h; (c) 7.3 h)

C (M) 1 2 3 4 5

Rate of reaction 0.5 4 13.5 32 62.5

dC dt




298

Fig. 12.20 Plots of variable order processes

12

Rates and Orders of Reactions

Exercises

299

12.25. Determine the order of reaction from the information in the table below. C represents the reactant or drug concentration when the rate of reaction was estimated. (Answer: Third order) 12.26. Which of the plots in Fig. 12.20 are zero order and which are first-order decay processes? Select all that apply. 12.27. The concentration of the active ingredient in an extended release gastroresistant liquid capsule is 25% w/v. The capsule volume capacity is 1 mL. (a) How much drug is present in the capsule? Write the concentration of the active ingredient in the capsule in units of g/mL. (b) What would be the drug concentration (% w/v) if the liquid capsule was instantaneously ruptured in patient’s intestine (250 mL) upon ingestion? Neglect the 1 mL of the capsule. (c) The drug is released in a controlled fashion, as intestinal fluid leaches into the capsule through capsule pores and transfers the drug into the intestinal fluid. Capsule and intestinal volumes exchange, but the two remain constant. Also, assume that no drug is released into the intestine within the first 30 min. What would be the drug concentration in the capsule after 1 h if the capsule releases the drug at a rate 0.32 h1? (d) Derive the equation that describes the drug remaining in the liquid capsule and plot the graph C versus time. (e) Derive the equation that describes the drug released in the intestinal fluid and plot the graph C versus time, assuming that the drug in not absorbed or eliminated for the duration of your study and the volume of the intestinal fluid stays constant to 250 mL. (Answer: (a) 250 mg, 0.25 g/mL; (b) 0.1% w/v; (c) 213 mg/mL: (d) C ¼ C0  ek(t0.5)) (e) C¼C0(1ek(t0.5))

Chapter 13

Fundamental Concepts of Dosage Calculations

Abstract This chapter demonstrates how one calculates dosage regimens using certain pharmacokinetic parameters and the simplest of the pharmacokinetic models, the one-compartment model. Prior to describing the one-compartment model, the chapter discusses the difference between drug selectivity and receptor biological event specificity, relates drug selectivity with therapeutic efficacy and toxicity, and explains how to relate the physicochemical properties of drugs to their extent of tissue distribution. It also discusses the difference between excretion, clearance, and elimination and defines the concept of bioavailability which is critical in understanding the difference between dose and effective dose. Ultimately, you will learn how to calculate effective drug doses using the concept of bioavailability and salt factor. The second part of the chapter describes the operational principles of the one-compartment pharmacokinetic model, defines the therapeutic index of drugs, the concepts of drug volume of distribution, and clearance, and demonstrates how to calculate their values from plasma drug concentrations. Finally, it demonstrates how to apply these pharmacokinetic parameters and the concepts of drug accumulation and fluctuation to calculate optimum drug loading and maintenance dosages that achieve plasma drug levels within therapeutic range. After reading Chaps. 11–13, you are definitely ready to study the main course of pharmacokinetics. Keywords Selectivity · Specificity · Absorption · Distribution · Elimination · Clearance · Metabolism · Excretion · Dose · Dosage interval · Volume of distribution · Therapeutic range · Accumulation · Fluctuation · Loading dose · Maintenance dose

Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_13) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_13

301

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Learning Objectives After reading this chapter, you should be able to: • Explain the difference between drug selectivity and receptor biological event specificity • Relate drug selectivity with therapeutic efficacy and toxicity • Relate the physicochemical properties of drugs to the extent of tissue distribution • Discuss the difference between excretion, clearance, and elimination • Define the concept of bioavailability • Explain the difference between dose and effective dose • Use the concept of bioavailability and salt factor to calculate effective drug dose • Describe the operational principles of the one-compartment pharmacokinetic model • Define the therapeutic index of drugs • Define the concepts of drug volume of distribution and clearance and calculate their values from plasma drug concentrations • Determine the relative distribution of a drug in the plasma and tissues from its volume of distribution • Describe the concepts of drug accumulation and fluctuation • Apply pharmacokinetic parameters to calculate optimum drug loading and maintenance dosages that achieve plasma drug levels within therapeutic range Dose is defined as the specified amount of a therapeutic agent administered to a patient for a therapeutic effect to take place and last for as long as possible; the longer the better. The two main determinants of a dosage regimen are the dose size and the dosing frequency. For a therapeutic effect to be produced as quickly as possible without adverse effects, an appropriate dose size must be chosen, aka loading dose, and for the clinical effectiveness to last, drug levels in the body have to be replenished and sustained relatively constant by proper dosing intervals, aka maintenance dose. A loading dose is always different than the maintenance dose but not all drugs have a loading dose, that is, a maintenance dose may be given from the beginning. A real understanding of the math involved in dosage calculations can only take place after you get familiar with the way drugs function in the body to produce a pharmacological effect—a study called pharmacodynamics—and the way the body handles drugs—a study known as pharmacokinetics.

13.1

Understanding the Drug-Mediated Therapeutic Effect

The functionality of cells within a tissue or an organ depends on their ability to effectively communicate with neighboring cells and function collectively to maintain homeostasis. For cells to stay active, they have to be able to identify and use correctly the various chemicals present in their external environment through a series of biochemical processes that produce energy. This energy is in turn utilized to synthesize complex molecules and to manufacture and maintain complicated organelles.

13.1

Understanding the Drug-Mediated Therapeutic Effect

Signal transduction

Signal transduction

Biological event 1

Biological event 2

303

Fig. 13.1 Drug binding to cell membrane receptors will enhance, reduce, or completely inhibit the transduction of an already existing physiological signal that translates to a particular biological event. The drug can bind to more than one similar receptors, but each receptor induces a uniquely specific biological event

How do cells recognize chemicals in the extracellular fluid (ECF)? How do they sort them in different compartments for immediate or future use? It turns out that cells carry specific “sensor” devises. These “sensors” known as receptors are usually protein macromolecules that lie either on the plasma membrane or inside the cell. Their role is to detect signals communicated to cells in a chemical manner and convert these signals into meaningful biological responses. The natural endogenous ligands of receptors are known as messengers, and they include hormone molecules and neurotransmitters. A particular messenger will selectively bind to a particular receptor generating a signal that is conducted into the cell’s interior through a sequence of biochemical events in a process called signal transduction. Every signal will eventually be translated into a single specific biological action (Fig. 13.1). It is widely accepted that receptors are only selective in signal recognition, but they are very specific in the resultant action. That is, various chemicals of similar structure may selectively bind to a particular receptor to produce the specific biological event that occurs naturally upon binding of messenger molecules to those particular receptors. It is also true that one messenger may bind to structurally similar (homologous) receptors present in the same or different tissues to produce a distinctively different outcome. Medicinal chemists take advantage of receptors’ selectivity and design drugs that mimic the structure of messenger molecules. As a result of this resemblance, drugs bind to specific receptors in place of the messenger molecule, and they either enhance the magnitude of the signal conducted by them or inhibit it. Drugs do not usually create new signals, and they certainly cannot change the nature of an existing physiological response. Most drugs are selective but not specific in their actions. In other words, the chemical structure of a drug allows it to bind to a particular receptor with highest affinity, but it does not preclude its binding to a number of structurally related

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receptors present in the target or other tissues (Fig. 13.1). For example, vasopressin interacts with the V1-receptors of smooth muscle cells in arteries causing the vessels to contract while its binding to V2-receptors that reside in collecting duct cells of the kidney triggers water reabsorption with a completely different mechanism. This lack of specificity in drug action is the primary reason for toxicity although cases where toxic effects are mediated by the same receptor-effector mechanism are not rare. For example, administration of a high insulin dose to a diabetic patient will increase the rate of glucose transport to the muscle and the cells of adipose tissue, leaving very little glucose in the blood. Glucose is the major fuel source of the brain and nervous tissues, and as a result of low glucose levels in the blood, the patient may fall into hypoglycemic coma. Obviously, one would like to restrict the presence of the drug to the vicinity of the target tissue in cases where the drug binds to more than one kind of receptors present in different tissues or if the same biological effect mediated by the receptors in other organs is not beneficial. For example, the neurotransmitter epinephrine is known to bind to several adrenergic receptors, that is, α1, α2, and β2 which exist in practically all organs. Binding of the messenger to the α1 receptors in the vascular compartment would cause vasoconstriction of arteries leading to elevated blood pressure. Interestingly, epinephrine binding to α1 receptors in the liver would activate glycogenolysis, whereas α1 activation in the eye would cause pupil dilation. On the other hand, activation of the β2 receptors in the lung will cause muscle relaxation, an effect that could be of benefit to asthma patients. It is not surprising that therapeutic use of epinephrine is limited, and it does not involve systemic administration. Dentists routinely use epinephrine-containing solutions of local anesthetics for analgesia during painful dental procedures. Epinephrine causes vasoconstriction at the site of the injection, thus minimizing the loss of the anesthetic from the gums promoting an increased duration of analgesic action during dental procedure. Accidental injection of the solutions in patient’s veins could lead to serious unwanted effects from the cardiovascular system. In conclusion, the magnitude of the pharmacological response elicited by a particular drug is dependent on the concentration of the drug at the site of action and the availability of the receptors at the target tissue. In an ideal world, calculations involving the amount of drug needed to produce a therapeutic effect would only be dependent on the number of the available target receptors that bind the drug and elicit that particular pharmacological response. Unfortunately, this is never the case due to primarily our inability to directly deliver drugs selectively to the site of action, that is, the receptor site, and the lack of convenient noninvasive methods of measuring drug concentrations at the site of action in patients. Drugs are administered by various routes, that is, oral, intravenous, intramuscular, subcutaneous, transdermal, nasal, and buccal, at sites quite distant from the target tissues. Successful delivery of drugs is dependent on the ability of our body to transfer them to the sites of action, but also on the physicochemical properties of the drug that affect its transfer and biological activity, that is, its solubility in body fluids and membrane permeability, and its pharmacological specificity, respectively. Do we really have control over the factors that regulate the pharmacological action of drugs? The answer is no. How are we then going to accurately calculate effective

13.2

The Process of Drug Absorption, Distribution, Metabolism. . .

305

drug dosages? We are in no position to calculate absolute doses for every individual patient, but fortunately we are able to calculate clinically safe average dosages for patients of average characteristics with respect to their organ functionality, body weight, and height. We next discuss the dependency of the drug concentration at the site of action on the dose and the volume of distribution of the drug in the body.

13.2

The Process of Drug Absorption, Distribution, Metabolism and Excretion (ADME)

13.2.1 Drug Absorption Drug absorption is the passage of drug molecules from the site of administration through multiple membrane barriers, mainly by a passive diffusion mechanism, into the systemic circulation. Membranes are composed of epithelial cells that are tightly connected with each other to form usually more than one layer. The crossing of drug molecules through membrane barriers happens mostly transcellularly. These cellmade barriers are so effective that only single molecules of small dimensions can penetrate through them. Therefore, a prerequisite of drug absorption for all extravascular routes of administration such as oral, rectal, intramuscular, and subcutaneous is the formation of a drug molecular dispersion (solution). Two points that need to be mentioned about the extravascular route of administration are: first, it cannot be used if an immediate action is required because the process of absorption is slow, and second, it always leads to large fluctuations of drug levels in the blood, thereby usually rendering this route of administration unsuitable for drugs of low therapeutic index. Highly polar drugs do not get absorbed well because although they form a solution, they do not partition within the apolar lipidic environment of the membranes. Drugs that are way too hydrophobic also fail to pass across cell membranes because although they partition easily into cell membranes, they do not dissolve into biological fluids to form molecular dispersions. Quick and full absorption is therefore exhibited by drugs that have mainly a hydrophobic character but possess some hydrophilic groups in their structure that facilitate their dissolution and are of low molecular weight (MW), so that they can go through the cell plasma membrane.

13.2.2 Drug Distribution With the exception of the intravenous route of administration, drugs have to permeate through multiple membrane barriers before they eventually reach into the blood circulation. Drugs will be transferred to various tissues via the systemic circulation and will escape through the capillary pores to the interstitial spaces of the target cells. This last process that takes place after the drug absorption is called drug distribution. The higher the tissue vascularization, the faster is usually the process of drug

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distribution. Only free drug is able to permeate paracellularly the capillary endothelium. Drug bound to serum or plasma proteins will not be able to escape out of the vascular compartment due to the big size of the proteins. Also, low MW polar drugs cannot diffuse through plasma membranes intracellularly, and they will be excreted unmodified by the kidney in the urine. Lipophilic drugs that partition into tissues outside the blood circulation or drugs that display tissue-specific binding are anticipated to stay longer in the body. In general, drug concentration in the plasma or blood after a single dose is dependent on the dose size and the volume of distribution of the drug. The more polar the drug, the smaller is its volume of distribution because it is confined in the ECF compartment. The more lipophilic the drug, the higher is its volume of distribution and the higher the required dose to achieve a specific drug concentration in the plasma because a significant portion of the dose partitions into the tissues. Calculated dosages, especially for drugs with narrow therapeutic index, are often normalized per body weight mainly due to the volume of distribution of these drugs.

13.2.3 Drug Metabolism Liver is the major organ where drug biotransformation or metabolism takes place although other organs rich in enzymes such as spleen, gastrointestinal tract, lung, blood, and skin can also metabolize drugs. The enzymes mostly responsible for the biotransformation of drugs and other foreign molecules or xenobiotics belong to the superfamily of cytochrome P450 that are present predominantly in the smooth endoplasmic reticulum of hepatocytes and other tissues. Other enzymes can metabolize drug molecules as well to a smaller extent. Biotransformation of drugs routinely involves the attachment of polar hydrophilic groups such as –OH, –SH, – COOH, and –NH2. This kind of structure modification renders drug molecules more polar and severely limits their tissue intracellular distribution, thus their renal excretion is greatly facilitated. This is why biotransformation is sometimes misinterpreted as a conscious drug inactivation or detoxification process of our body. Some books may be referring to the process of biotransformation as the defense mechanism of our body against drugs. A question that I always had as a student was how does the body recognize these molecules as drugs to defend against them? It really does not. As far as we know, our body does not sort small molecules at first contact into harmful and useful chemicals to treat them differently. Even immune cells may not recognize them as foreign molecules. Interaction between drugs and our body is mutual and partially blind. While the drug exerts its effect on the body, the body is trying to use the drug to its advantage; in other words, the body will try to modify the drug molecule or even completely break it down in order to be able to use it as a building block or as an energy source. This process is done on a trial-and-error basis. A first line of evidence that argues in favor of this conclusion is that all enzymes involved in drug biotransformation have endogenous ligands as substrates, suggesting that their primary function is to help cells establish a constant supply of energy for growth, differentiation, and homeostasis through the process of

13.3

The Concepts of Bioavailability, Effective Dose, and Dose Conversions. . .

307

metabolism. Drug biotransformation is only a minor function of these enzymes. Cells will take up small molecules from their surroundings and try to modify them and use them randomly. This is the reason why sometimes biotransformation products could turn out to be more active or toxic than the original drug. In summary, biotransformation and metabolism are synonymous terms and refer to the process of chemical modification of drugs in the body. This process could yield a single product or multiple products called metabolites. The metabolites are usually less active than the parent drugs or completely inactive, although cases exist where biotransformation resulted in more active compounds. In general, biotransformation of the drugs involves changes in their molecular structure that essentially alter their physicochemical properties resulting in changes in their body distribution and biological activity. To this end, calculations of doses have to take into account all factors that affect drug metabolism in the body.

13.2.4 Drug Excretion Drug elimination is defined as the process of drug metabolism and/or drug excretion. In other words, drugs can be eliminated from the body by excretion unmodified, by metabolism alone, or by both processes. After metabolism, the drug ceases to exist in the body, and excretion of the metabolite, pharmacologically active or not, is irrelevant to the elimination of the parent drug. The main two organs of elimination are kidney and liver. Drugs that are in the vascular compartment pass through the kidney where depending on their physicochemical characteristics, they will be reabsorbed into the blood or excreted into urine. The process of excretion is facilitated with polar drugs, but it is less efficient with hydrophobic drugs. Hydrophobic drugs are usually metabolized in the liver prior to their renal excretion, or they pass via biliary excretion into the intestines where they are eliminated from the body by fecal excretion.

13.3

The Concepts of Bioavailability, Effective Dose, and Dose Conversions Between Dosage Forms

In general, factors that affect the drug’s ADME will affect plasma drug concentration, which in turn affects drug concentration at tissue target receptors and the pharmacological response. For example, polar drugs that are not substrates of uptake transporters are absorbed inefficiently, thereby achieving relatively low blood and tissue drug concentrations. Drugs that are substrates of the cytochrome P450 superfamily will also have an incomplete absorption and will also achieve relatively low plasma and tissue drug concentration due to their intestinal and liver metabolism. Polar drugs administered IV will achieve momentarily high concentration in the

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plasma but their failure to be distributed into the tissues will inevitably lead to their rapid excretion by the kidney. How much of a drug will be distributed throughout the body and bind to the target receptors to elicit the desired pharmacological activity is depended on its ADME profile. These factors have to be taken into account in calculations of drug dosages because if they are neglected, situations ranging from inefficient therapy (too little dose) to lethal effects (dose too high) could arise. In pharmacokinetics, drug absorption refers specifically to the uptake of drug in the systemic circulation (after passing through the liver) and not just its entry into the body. Drug absorption is intimately related to the concept of bioavailability or bioavailability factor (F) which is defined as the fraction of the administered dose that reaches the systemic circulation, and it can take values from 0 to 1. For example, a bioavailability factor, F, of 0.56 means that only 56% of the extravascularly administered dose will reach the systemic circulation. The remaining 44% of the drug molecules failed to get absorbed into the systemic circulation due to formulation issues of the dosage form, unfavorable physicochemical properties of the drug, or biological factors such as enzymatic destruction or substrate affinity for efflux transporters. The effective dose is defined as the product of the bioavailability factor of the drug, the dose, and the salt factor (Eq. 13.1). Effective dose ¼ S ∙ F ∙ Dose

ð13:1Þ

Example 13.1 A 50-year-old male with chronic bronchitis is to receive 200 mg of theophylline (F ¼ 1) tid in the form of regular oxtriphylline tablets (F ¼ 0.78). The chemical structures of theophylline and oxtriphylline (choline theophyllinate anhydrous) are shown in Fig. 13.2. What is the oxtriphylline dose of the patient? Solution The effective theophylline dose is 200 mg. You need to figure out the oxtriphylline dose equivalent to 200 mg oral theophylline dose. The salt factor of oxtriphylline, S ¼ 180 283 ¼ 0:64 O O H3C O

N N CH3

H3C H N

-O

N

CH3

N H

Theophylline

N

CH3

H N

N H

+

HO

N H3C

CH3

Oxtriphylline

Fig. 13.2 Structures of theophylline (MW 180) and oxtriphylline (MW 283)

(continued)

13.4

One-Compartment Pharmacokinetic Model

309

Example 13.1 (continued) Effective dose ¼ S ∙ F ∙ Dose ) Effective dose 200 mg 200 mg ¼ ¼ ¼ 400 mg Oxtriphylline dose ¼ S∙F 0:64 ∙ 0:78 0:5 Essentially, administering an oral dose of 400 mg oxtriphylline tablet is equivalent to an oral administration of 200 mg theophylline. A formula similar to Eq. 13.1 can be used to carry out dose conversions between dosage forms. Example 13.2 A 50-year-old male with atrial fibrillation receives 0.2 mg of IV digoxin daily. The patient is stable with a plasma drug concentration equal to 1.3 ng/mL. Compute an oral tablet dosage (F ¼ 0.8) that will maintain the same therapeutic plasma digoxin concentration. Solution DoseIV ¼ F ∙ DosePO

)

DosePO ¼

DoseIV 200 μg ¼ 250 μg ¼ 0:8 F

Thus, the daily oral dose for the patient is 0.25 mg.

13.4

One-Compartment Pharmacokinetic Model

There are many pharmacokinetic models that can be used to provide quantitative estimates of the ADME parameters which are used to calculate specific dosages aiming to achieve therapeutic plasma drug concentrations. The most commonly used in the clinic is the one-compartment model. A pharmacokinetic compartment is an imaginary unit that consists of a group of tissues that display similar rates of drug distribution. The hypothesis of the one-compartment model is that the drug is distributed instantaneously in all tissues of the body (Fig. 13.3) within seconds after its IV administration. Thus, all the tissues are lumped together in one compartment, and they are indistinguishable from each other. Due to practical difficulties associated with measuring drug levels in tissue organs and cells, drug concentrations are only measured in the vascular compartment by sampling the patient blood. Since the vascular fluid volume is always at equilibrium with all other tissues, it is logical to assume that the concentration of drug in the plasma is some multiple of the drug concentration in the tissues. Also, in the one-compartment pharmacokinetic model, equilibrium between the vascular compartment and the surrounding tissues is considered to be instantaneous

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Fundamental Concepts of Dosage Calculations

Drug Dose Absorption (% of drug entering the blood per unit time)

Vascular Compartment Cp

Distribution

Tissues Ct 1. Pharmacological effect 2. Biotransformation/inactivation

Drug Elimination Fig. 13.3 Schematic representation of the fundamental hypothesis that there exists a mathematical relationship between the drug concentration in the blood (Cp) and the concentration of the drug in the tissues (Ct), which governs the magnitude of the pharmacological response. Blood and all other extravascular tissues take up the drug rapidly and are treated as a single compartment

and fully reversible. The reversibility refers to drug redistribution, that is, the release of the tissue-partitioned drug back into central circulation. Pharmacokinetic parameters are estimated by plotting the drug plasma concentration with time and fitting the kinetic profile of drug elimination to a first-order equation that calculates the first-order elimination rate constant k (Eq. 12.6). The first-order elimination rate constant is related to the specific pharmacokinetic parameters of the apparent volume of distribution of the drug, Vd, and the drug clearance, CL. Essentially, drug absorption and drug elimination (hence drug accumulation), in a one-compartment model, are first-order kinetic processes that have extensively been discussed in the chapter of Rates and Orders of Reactions. In summary, the essential features of the one-compartment model are (1) drug distribution is so fast that there is no discernible distribution phase and (2) drug elimination is a first-order process characterized by a first-order rate constant k. C ¼ C 0 ∙ ek ∙ t

ð12:6Þ

13.5

13.5

Drug Apparent Volume of Distribution (Vd)

311

Drug Apparent Volume of Distribution (Vd)

We have previously encountered the general equation that relates the drug concentration with the mass of drug in solution and the volume of the aqueous solution
C ¼ mV . We use the same equation to relate drug dosage with the concentration of the drug in the serum. Vd ¼

Ab IV dose ¼ C0 Cp

ð13:2Þ

where Vd, Ab, Cp, and Cp, 0 or C0 are the drug apparent volume of distribution, the amount of drug in the body, the plasma drug concentration at the same time when Ab was determined, and the plasma drug concentration at zero time. It is important to remember that the amount of drug in the body and the plasma drug concentration must be determined at exactly the same time. For example, to calculate Vd from plasma drug concentration that was determined 1 h after an IV administration, I need to know the amount of drug in the body after exactly 1 h. At time zero, the numerator represents an IV dose because it involves direct administration of the drug into the bloodstream and removes any ambiguities about the actual amount of drug that entered the blood due to incomplete absorption. Notice that Eq. 13.2 is a mass preservation equation encountered in the chapter of Dilution. It denotes that the mass of the drug added to the system in which the drug is homogeneously distributed equals the drug concentration in the system times the volume of the system. Mathematically, the higher the plasma drug concentration (Cp), the lower is the volume of distribution of the drug (Vd). That is, drugs with a low Vd exhibit a relatively small distribution of the drug to surrounding tissues effectively lowering the tissue drug concentration (Ct). On the other hand, drugs that are extensively distributed to surrounding tissues are characterized with high Ct, low Cp, and large Vd values. This pharmacokinetic parameter has a drug dosageindependent value that depends on the physicochemical properties of the drug and the disease state of the patient. Thus, dosage increase results in a proportional increase of the Cp and the Ct (since Vd remains constant), resulting in greater pharmacological response. Drastic dosage decreases could lower the plasma drug concentration to the extent that the drug concentration in the tissues is so low that the magnitude of the pharmacological response is not big enough to be of any therapeutic relevance. The fraction of the drug in the plasma can be found by dividing the patient’s plasma volume by the drug’s Vd. Fraction of drug in the plasma ¼

Vp Vd

ð13:3Þ

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Example 13.3 The IV bolus administration of 0.06 g of a new drug to a 70 kg adult male resulted in a serum concentration at zero time of 1.5 mg/dL. Calculate the volume of distribution of the new drug. Where the drug is mostly located? Solution IV dose ¼ 60 mg C p,0 ¼ 15 mg=L Vd ¼

IV dose 60 mg ¼4L ¼ C p ,0 15 mg=L

In order to assess where the drug is mostly located, we need to relate the drug’s Vd to the volume of plasma, Vp. From the chapter of Fluids and Electrolytes, we know that for a 70-kg adult male, Vp ¼ 3 L. Fraction of drug in the plasma ¼

Vp 3 L ¼ 0:75 ¼ Vd 4 L

Essentially, 75% of the drug is in the vascular compartment. In reality, determining the volume of distribution of a drug is not as easy as it is implied in Example 13.3. C0 is never given to you in the clinic. You are going to have to administer the drug, draw blood from the patient at multiple time intervals, and estimate C0 from a Cpt plot. The cardiac output is estimated at the rate of 72 beats/min with a 70 mL/beat, to be approximately equal to 5.0 L/min. Thus, all the blood is pumped through the heart once every minute. Determining the drug concentration at time zero is practically impossible. This term can be calculated by measuring Cp at multiple time periods and then back-extrapolating the regression line of the Cp  t plot to the y-axis to obtain the value of C0 (Example 13.4). Example 13.4 A single 450 mg IV bolus of an investigational anticancer drug was administered to a 40-year-old, 70 kg female patient. Blood was drawn at predetermined time intervals and drug concentration of each sample was determined in plasma as shown below. C (mg/L) Time (h)

2.9 1

1.8 2

1.2 3

0.74 4

0.3 6

0.1 8

0.05 10

(continued)

13.5

Drug Apparent Volume of Distribution (Vd)

313

Example 13.4 (continued)

Fig. 13.4 Cp  t (left) and ln Cp  t (right) plots

Curve fit the data to a one-compartment model, interpret the best-fit parameters and determine the apparent volume of distribution of the drug? Solution We have the IV dose ¼ 450 mg, but in order to determine the Vd, we need to know the C0. We can calculate the C0 from either the best fit equation of Cp  t plot or the ln Cp  t plot (Fig. 13.4). Conversion of the concentration into ln C values is shown below. ln C t (h)

1.06 1

0.59 2

0.18 3

0.30 4

1.20 6

2.3 8

3 10

Comparison of the trendline equation y ¼ 4.612 ∙ e0.461∙x with the prototype first-order equation C ¼ C0 ∙ ek∙t yields a C0 ¼ 4.612 mg/L and k ¼ 0.461 h1. We get the same numbers if we use the trendline of the right plot C0 ¼ e1.5284 ¼ 4.612 mg/L and k ¼ 0.461 h1. In the absence of a trendline, you could approximate C0 from the intercept graphically and k from the slope of the line. Vd ¼

IV dose 450 mg ¼ 97:6 L ¼ C p,0 4:612 mg=L

In common practice, Ct is never known, rather the Cp is conveniently related to the therapeutic response. Hence, plasma drug concentrations that elicit therapeutic effects are called therapeutic drug levels, and they lie between the mean effective concentration (MEC) and the mean toxic concentration (MTC) (Fig. 13.5). The ratio between the toxic and the therapeutic dose is called the therapeutic index (TI) of the drug.

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Fig. 13.5 Schematic representation of drug therapeutic index. Traces from top to bottom, represent overdose, optimum dose and inefficient dose

TI ¼

TD50 ED50

ð13:4Þ

TD50: The median toxic dose is the dose that 50% of the patients experience toxicity. ED50: The median effective dose is the dose at which 50% of the patients experience therapeutic effects. The bigger the value of therapeutic index, the safer is the drug for the obvious reason of the mean therapeutic dose being much lower than the mean toxic dose. Drugs that have a very narrow therapeutic window (low therapeutic index) require a minimum fluctuation of their concentration in the blood by carefully selecting the route of administration, the dose, and dosing interval.

13.6

Drug Clearance (CL)

Clearance (CL) is defined as the volume of plasma cleared of drug per unit time and has units of volume/time. It is a parameter that measures the efficiency with which the kidney and liver can extract or eliminate the drug from the plasma. It relates the rate of drug elimination from an organ with the concentration of the drug in the plasma that enters the organ. The two primary organs for drug elimination are the kidney and the liver. Clearance is additive in nature, that is, the total body clearance is the result of drug being eliminated from the body by various organs. By definition, the total clearance (CL) can mathematically be expressed by Eq. 13.5.

13.6

Drug Clearance (CL)

315

CL ¼

rate of drug elimination plasma drug concentration

ð13:5Þ

The rate of drug elimination is a first-order process equal to k ∙ Ab, and therefore, for the one-compartment pharmacokinetic model, Eq. 13.5 gets the final form of Eq. 13.7. CL ¼

k ∙ Ab Cp

ð13:6Þ

but Ab ¼ Cp ∙ Vd from the mass balance equation ) CL ¼ k ∙ V d

or

k¼

CL Vd

ð13:7Þ

Example 13.5 Formulate an expression for organ clearance. Solution The drug clearance of an organ can be expressed in a similar way. For example, liver clearance is the plasma volume extracted or cleared of drug per unit time by the liver. Eq. 13.6 can be modified as shown below to include the hepatic clearance (CLh) and the hepatic elimination rate constant (kh). CLh ¼

kh ∙ Ab Cp

Example 13.6 Provide a definition of the renal clearance of a drug being equal to 5 L/h. Solution 5 L of plasma are cleared of drug within 1 h or per hour by the kidney (in the urine). Similar to the additive nature of drug organ clearance, individual organ elimination rate constants are also additive. In other words, summation of the individual organ clearances will yield the total body clearance (CL) in a similar way that addition of the individual organ elimination rate constants will yield the total elimination rate constant.

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13

Fundamental Concepts of Dosage Calculations

CL ¼ CLr þ CLh þ    and k ¼ kr þ kh þ   

Example 13.7 Calculate the clearance of the anticancer drug from the information that was provided in Example 13.4. Solution Using Eq. 13.7, CL ¼ k ∙ V d ¼

0:461 ∙ 97:6 L ¼ 45 L=h h

Therefore, 45 L of plasma is cleared of drug per hour. Drug elimination may be excretion and/or metabolism. This is a relatively high drug clearance. In fact, it can be calculated from the first-order elimination rate constant using Eq. 12.8 that the drug’s half-life in the body is only about 1.5 h.

13.7

Accumulation and Fluctuation of Drug in the Plasma

We have briefly referred to plasma drug concentration fluctuation twice so far without justifying or explaining the details of the event. Fluctuation of the plasma drug concentration occurs when the drug is administered at regular or variable dosing intervals, regardless of the route of administration. The drug accumulates only when the dose is administered before all the drug from the previous dose is eliminated. The drug concentration increases in the plasma for as long as the dose is administered with the drug entering the central circulation at a rate higher than the rate of drug elimination. When dose administration is over or when the rate of the drug entering the systemic circulation becomes smaller than the rate of drug elimination, within each dosing interval, the plasma concentration drops until the next dose. Plasma drug concentration fluctuates between a maximum or peak concentration that usually occurs immediately or a little after each dose and a minimum or trough plasma concentration that is observed just prior to the administration of the next dose. The magnitude of fluctuation is related to the elimination rate constant of the drug, and therefore, it is controlled by the dosing interval. For a constant total dose, the larger the dosing interval, the larger the fluctuation of the drug plasma concentration. Figure 13.6 demonstrates the plasma drug concentration profile of a drug after oral administration of the same daily dose at two separate dosing intervals (τ), 12 and 24 h. The therapeutic range of the drug concentration is marked by the two horizontal lines at 5–12 mg/L. There are at least three distinct features of the profile. First, fluctuation is larger when the whole dose is given all at once than when it is

13.7

Accumulation and Fluctuation of Drug in the Plasma

317

Fig. 13.6 Drug plasma concentration profile after oral administration at dosing intervals τ ¼ 12 h (continuous thick trace) and τ ¼ 24 h (thick dashed curve). The systemic half-life of the drug (t1/2) is about 24 h, and the rate of drug absorption is much higher than the rate of drug elimination. The two demarcation horizontal continuous thin lines denote the therapeutic window of the drug which is 5–12 mg/L. The continuous concave line that reaches a plateau with time represents constant rate IV infusion

split in two portions given twice daily. In other words, for a given total dose, the bigger the dosing interval, the higher the peak, the lower the trough plasma drug concentration, and the larger the fluctuation of the plasma drug concentration. Second, drug accumulation occurs for about 5 t1/2 (96 h), after which time the drug concentration reaches equilibrium in the body or the so-called steady state condition. When steady state conditions are reached, the plasma drug concentration reaches the same peak immediately after each administration and the same trough just prior to administering each dose. Drug accumulation is seen prior to 5 t1/2 when the peaks and the troughs increase with each dose. After about 5 t1/2, there is no more accumulation and the drug plasma concentration fluctuates within each dosing interval between peak and trough levels. As mentioned before, for a given total dose, the smaller the dosing interval, the smaller the fluctuation of the plasma drug concentration. Plasma drug fluctuation is obliterated when the drug is administered via a continuous constant rate IV infusion which achieves constant plasma drug levels without any peak or troughs (Fig. 13.6, continuous concave-shaped line that reaches a plateau). The idea behind the concept of drug accumulation and fluctuation is to use the drug’s pharmacokinetic parameters to optimize the dose and dosing interval, so that the plasma drug concentration is always above the lower boundary of the therapeutic range but it never exceeds the upper boundary (mean toxic or maximum effective concentration) of the therapeutic range.

318

13

Fundamental Concepts of Dosage Calculations

Example 13.8 It was stated that drug accumulation occurs when the dose is administered before the drug from the previous dose is eliminated. Why drug accumulation stops after about 5 t1/2? Shouldn’t drug accumulation continue indefinitely since the drug continues to be administered before the previous dose is eliminated? Solution Recall that in a one-compartment model, the drug is eliminated by a firstorder process, that is, the drug eliminated increases as the accumulated plasma drug concentration increases (Eqs. 12.5 and 13.6). dAb ¼ k ∙ Ab ¼ CL ∙ C p dt As the plasma drug concentration increases due to drug accumulation, b clearance is constant, but the rate of drug elimination dA dt being the product of clearance and the increasing Cp increases as well. At some point in time, the rate of plasma drug increase with each dose equals the rate of drug elimination. When, “rate in” equals “rate out” (rate in ¼ rate out), drug accumulation ceases, and the drug concentration in the plasma will be reaching the same constant peak and trough concentrations with each subsequent dose. This is the basis for calculating drug maintenance doses within therapeutic range.

13.8

Loading (DL) and Maintenance Dose (DM)

Loading doses are administered when maintenance doses need to be administered less frequently due to a relatively long systemic half-life of the drug. When a drug is administered for the first time, two major processes take place: drug distribution and drug elimination. Drug distribution takes place earlier, and depending on the physicochemical characteristics of the drug and its Vd, it can drain a lot of drug out of the blood circulation into tissues. It takes about 5 t1/2 of the drug distribution to reach equilibrium with the tissues. After about 5 t1/2, the drug distribution has reached equilibrium, and the only process draining the drug out of the central circulation is the drug clearance. The frequency and the magnitude of a maintenance dose are adjusted by taking into account the desired therapeutic level of the drug in the plasma and the drug clearance (Eq. 13.8). On the contrary, the loading dose is calculated based on the desired therapeutic level of the drug in the plasma and its volume of distribution (Eq. 13.9), with Css being the desired equilibrium or steady state therapeutic plasma drug concentration.

13.8

Loading (DL) and Maintenance Dose (DM)

319

DM ¼

Css ∙ CL S∙F

ð13:8Þ

DL ¼

Css ∙ V d S∙F

ð13:9Þ

Example 13.9 A 66-year-old, 85 kg, 50 900 tall male is to be initiated on digoxin for moderate congestive heart failure (CHF). His serum creatinine concentration was measured and found to be equal to 0.8 mg/dL, suggesting no kidney impairment. The population average Vd and CL for adults without kidney or liver insufficiency are 6.8 L/kg and 262.5 L/day, respectively, and the digoxin recommended therapeutic levels for CHF are 0.9 ng/mL. Please answer the following given the information in Table 13.1: (a) Why would you administer a digoxin loading dose? (b) Calculate an oral elixir digoxin loading dose to be administered the first day of treatment. (c) Describe how the loading dose should be administered? (d) Compute an oral maintenance dose for a desired plasma concentration of 0.9 ng/mL using tablets. Round the dose to match tablet strength. Solution (a) Loading doses are recommended for drugs that have a relatively long systemic half-life. The goal is to reach steady state therapeutic concentrations at times earlier than 5 t1/2. We can use the total body weight to calculate the Vd because the patient is not obese. V d ¼ 6:8

L ∙ 85 kg ¼ 578 L kg

Next, we convert the CL into units of L/h. (continued) Table 13.1 Digoxin’s dosage forms

Dosage form (bioavailability, F) Tablets (F ¼ 0.7) Oral solution: Elixir 60 mL (F ¼ 0.8) IV/IM solution 1 mL 2 mL

a

Strength 0.0625 mg 0.125 mg 0.250 mg 0.05 mg/mL

b

a

0.1 mg/mL 0.25 mg/mL

It contains 10% v/v of ethanol. It is supplied with graduate droppers with marked divisions of 0.1 mL b It contains 10% v/v ethanol and 40% propylene glycol

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13

Fundamental Concepts of Dosage Calculations

Example 13.9 (continued) CL ¼ k¼

L 262:5 day h 24 day

¼ 10:94 L=h

CL 10:94 L=h ¼ 0:01892 h1 ¼ Vd 578 L

t 1=2 ¼

ln ð2Þ ln ð2Þ ¼ ¼ 36:6 h k 0:01892 h1

It takes about 5 t1/2 or 7–8 days for the drug to reach steady state concentration. Administration of a loading dose is needed to accelerate achieving steady state therapeutic drug concentration in the plasma. (b) DL ¼

mg C ss ∙ V d 0:0009 L ∙ 578 L ¼ 0:650 mg ¼ 1 ∙ 0:8 S∙F

The digoxin strength of the elixir is 0.05 mg/mL. )

0:650 mg mg ¼ 13 mL of elixir 0:05 mL

Administration of a loading dose will help achieve therapeutic levels on the second day of treatment immediately after the administration of the first maintenance dose. (c) Digoxin is a potent drug with a narrow therapeutic index. Loading doses are always much higher than the therapeutic doses, and patients may experience serious toxicity if they are given all at once. A loading dose can be divided into at least three portions, 50%, 25%, and 25% of the total dose. After administration of the first portion of the dose, if no signs of therapy or toxicity are observed within 6–8 h, the second portion of the loading dose is given. The last portion of the loading dose may be given 6–8 h after the administration of the second portion of the dose if no signs of therapy or toxicity are observed. The maintenance dose is administered the second day 6–8 h after the administration of the last portion of the loading dose. (d)

mg

∙ 10:94

L

h L h DM ¼ CssS ∙∙ FCL ¼ ¼ 0:01406 mg h ∙ 24 day ¼ 0:337 mg=day 1 ∙ 0:7 rounded up to 0.375 mg. Dividing by tablet strength of 0.125 mg, 0:0009

(continued)

Exercises

321

Example 13.9 (continued) )

0:375 0:125

mg day mg tablet

¼ 3 tablets per day

In general, avoid calculating dosages with multiple strength tablet or dosages that require breaking a tablet into two, unless there is no alternative. The patient should take three 0.125 mg digoxin tablets the same time every day with or without food (be consistent).

In conclusion, the purpose of including this chapter in the book of Pharmaceutical Calculations is to introduce some important pharmacokinetic concepts early on in your studies that are used to calculate patient’s loading and maintenance doses. These descriptions are by no means clinically complete, but they definitely cover the level of algebra that is needed to successfully provide dosage calculations and adjustments at a clinical level.

Exercises 13.1. The author has defined drug dose as the amount of drug needed to induce a therapeutic effect. Why in most diseases the drug is administered more than once? Select the best answer from below. (a) (b) (c) (d) (e)

Because most drugs alleviate the symptoms instead of curing the disease Because the target receptors desensitize with time Because the drug is eliminated by the kidney with time Because the drug undergoes metabolism in the liver and other tissues Because the drug concentration at the site of action decreases with time below therapeutic levels

13.2. Select the wrong statements from below: (a) (b) (c) (d)

Drugs can create their own biological signals. Drugs can bind to different receptor types with different affinity. Drugs compete with biological messengers. Drugs can never induce a pharmacological response bigger than that elicited by the natural endogenous ligand. (e) The binding site of a drug on a receptor could be different from that of an endogenous ligand.

13.3. Select the statement that provides the best explanation. A drug that binds to more than one kind of receptors has limited therapeutic use because ____________.

322

13

Fundamental Concepts of Dosage Calculations

(a) Binding of a drug to different receptors requires a higher dose. (b) The biological response of one kind of receptor cancels the activity of a different kind of receptor. (c) It is impossible to restrict the distribution of the drug to one kind of tissue receptor. (d) Drug binding to other nontherapeutic receptors may be irreversible. (e) Drug binding to other kind of receptors reduces drug clearance. 13.4. Which of the following statements are wrong about specificity and selectivity of drugs? (a) Specificity is the ultimate of selectivity. (b) Drug selectivity is manifested by their binding affinity to structurally similar receptors. (c) A drug displays binding specificity when it binds only to a single receptor type. (d) The binding specificity of a drug is reflected by irreversible binding to multiple receptor types. (e) Receptor specificity is usually displayed in their unique biological response. 13.5. Which of the following statements are wrong about drug distribution? (a) The rate of drug distribution is faster in highly vascularized tissues. (b) The volume of distribution of polar drugs is higher than that of lipophilic drugs. (c) Drug molecules that are bound to serum albumin cannot be distributed to extravascular tissues. (d) Extracellular distribution of most drugs to the tissues from the systemic circulation is diffusion limited. (e) The extent of tissue distribution of very high MW hydrophilic molecules is very small. 13.6. Select the correct statements about drug elimination. (a) (b) (c) (d)

Drug elimination is defined as drug metabolism and/or drug excretion. Drug excretion is synonymous to drug elimination. Drug elimination is synonymous to drug clearance. Drug that is metabolized is considered being eliminated even if the metabolite is not excreted out of the body. (e) Drug clearance is synonymous to drug excretion.

13.7. If drug solubility in the GI is high enough to dissolve the dose, which drugs have highest oral bioavailability? (a) (b) (c) (d) (e)

Low MW polar drugs that are substrates of efflux transporters Low MW lipophilic drugs Low MW hydrophilic drugs Low MW lipophilic drugs that are substrates of CYP450 Low MW lipophilic drugs that are substrates of efflux transporters

Exercises

323

13.8. Which of the following does not affect the oral bioavailability of nonionic drugs? (a) (b) (c) (d) (e)

Physicochemical properties of the drug pH of the GI Formulation of the dosage form Drug’s affinity for CYP450 enzymes Drug’s affinity for uptake transporters

13.9. What is the equivalent sodium phenytoin IV dose for an IV dose of 250 mg phenytoin? MW (phenytoin) ¼ 252, MW (sodium phenytoin) ¼ 274. (Answer: 291 mg) 13.10. Which of the following is not a property of the one-compartment pharmacokinetic model? (a) (b) (c) (d) (e)

Drug elimination follows first-order kinetics. The one-compartment model is composed only of a central compartment. The plasma concentration is equal to tissue concentration. The elimination rate constant has units of inverse time. Drug distribution is very fast in all tissues.

13.11. Which of the following statements are wrong about Vd? (a) Drugs that exhibit high plasma protein binding affinity have a low Vd. (b) Low MW drugs that have a higher affinity for tissue than plasma proteins have a high Vd. (c) Drugs that are extensively distributed to surrounding tissues have a high Vd. (d) Drug with a low Ct and high Cp have a small Vd. (e) The higher the Vd, the slower the rate of drug distribution. 13.12. What is the relative tissue distribution of a drug in a 70 kg male patient who has a volume of distribution of 4.5 L? (a) (b) (c) (d) (e)

66.67% 0% 100% 87.5% 33.33%

13.13. A new drug was administered at a 15 mg IV bolus injection to a clinical subject. From the Cp–time plot shown below, assuming units of concentration of mcg/L, what is the volume of distribution of the new drug?

324

13

(a) (b) (c) (d) (e)

Fundamental Concepts of Dosage Calculations

2.4 L 0.01 L 100 L 75 L 10 L

13.14. If the elimination rate constant is 0.5 h1, what is the drug’s half-life? (a) (b) (c) (d) (e)

42 min 21 min 0.34 min 1.38 min 83 min

13.15. The rate of drug elimination has units of (a) (b) (c) (d) (e)

g/h g L/h h1 h

13.16. Drugs A and B have similar values of clearance but the elimination half-life of drug A is 10 times bigger than the elimination half-life of drug B, because (a) (b) (c) (d) (e)

t1/2A ¼ CLB kA ¼ 10 kB VdA ¼ 0.1 VdB VdA ¼ 10 VdB VdBkB ¼ 10

13.17. The hepatic clearance of a drug is 2.5 L/h. Choose the statement below that apply. (a) (b) (c) (d) (e)

2.5 L/h of drug is cleared in the body by the liver. 2.5 L of drug is cleared in the body per hour. 2.5 L of drug is cleared from the plasma in 1 h. 2.5 L of plasma is cleared of drug per hour in the urine. 2.5 L of plasma is cleared of drug in 1 h by the liver.

Exercises

325

13.18. The figure below shows the logarithm of drug plasma concentration plotted against time after an IV dose of 135 mg of a drug. Cp is measured in μg/L. Calculate the drug clearance, CL.

(a) (b) (c) (d) (e)

9.4 L/h 0.018 L/h 562.5 L/h 90 L/h 14.4 L/h

13.19. The endogenous substance creatinine is commonly used to quantify renal function because it does not bind to serum proteins and is exclusively excreted in the urine without getting reabsorbed (active secretion of it is small and is usually neglected). Use Eq. 13.6 to derive an expression to calculate the renal clearance (mL/min) of creatinine, given the following information: Name: D.H. Sex: Male Age: 24 Weight: 163 lb Plasma creatinine concentration: 0.8 mg/dL Creatinine production in urine: 1800 mg/24 h Rate of urine production: 1 mL/min (Answer: 156.25 mL/min) 13.20. Given the following information, calculate the renal clearance of the drug: Name: D.H. Sex: Male Age: 24 Drug serum concentration: 2.1 mg/dL Drug urine concentration: 760 μg/mL Rate of urine production: 1 mL/min

Weight: 163 lb

(Answer: 36.19 mL/min)

326

13

Fundamental Concepts of Dosage Calculations

13.21. Which of the following statements are wrong for a drug that is administered at regular dosing intervals at a constant daily dose? (a) Drugs administered at longer dosing intervals reach steady state conditions later. (b) Drug fluctuation is related to the fluctuation of plasma drug concentration between peaks and troughs at steady state. (c) Drug accumulation occurs when the dose is administered before the drug from the previous dose is fully eliminated. (d) Fluctuation of plasma drug concentration is smaller with smaller dosing intervals. (e) Drug accumulation takes place before steady state conditions are reached. 13.22. Given the following patient drug profile, what would you change to make certain that the patient will experience at all times efficient therapy without toxicity? The therapeutic range of the drug is 20–30 mg/L.

(a) (b) (c) (d) (e)

Change the Vd Change the CL Change the t1/2 Change the dosing interval Change the elimination rate constant

13.23. The average therapeutic levels of a new antibiotic (S ¼ 1, F ¼ 1) are in the range of 126 ng/(mL∙kg), and the volume of distribution and the elimination rate constant of the antibiotic are 0.31 L/kg and 0.00025 min1, respectively. (a) Why would it be necessary to administer a loading dose? Calculate a loading dose for an 80 kg adult patient. The patient is not obese. (b) Calculate a maintenance dose for an 80 kg adult patient who is clinically obese. (Answer: (a) 250 mg (b) 90 mg/day)

Exercises

327

13.24. The accumulation profile of a drug in a one-compartment model as shown in Fig. 13.7 can be described by the equation C ¼ Css ∙ (1  ek∙t). Prove that it takes about 5 t1/2 for a drug to achieve stable levels in the blood or plasma by assessing how closely Cp approaches Css after 5 t1/2. Fig. 13.7 Drug plasma concentration profile after constant rate IV infusion

Chapter 14

Dosage Calculations Based on Body Surface Area (BSA)

Abstract In the previous chapter, drug concentration in the blood was correlated to the clearance and drug distribution in the body. In this chapter, I discuss the relation of the body surface area (BSA) to the body weight. Subsequently, a number of complicated BSA equations are derived by applying simple concepts of geometry. Through these derivations you have the opportunity to comprehend and further apply the equations to calculate drug doses. The chapter is focused on the rational use of BSA equations for dosage calculations. Close to the end of the chapter, higher exponential equations are solved, either as they are or after conversion to the more convenient linear form to calculate doses for pediatric populations. In general, not all the drugs can be dosed using the concept of body surface area. Clinicians have to be able to see the connection between drug disposition, organ size, and body size in order to confidently use these equations to calculate doses. This chapter offers a conceptual approach to the use of BSA equations for dosage calculations. It also discusses the advantages and disadvantages of the method. Keywords Body surface area · Dubois and Dubois’s formula · Boyd’s formula · Fujimoto’s formula · Haycock’s formula · Mosteller’s formula

Learning Objectives After reading this chapter, you should be able to: • Describe the difference between body weight-based dosing from BSA-based dosing strategies • Calculate BSA from height and weight using equations • Determine BSA from height and weight using nomograms • Calculate dosage regimens using BSA • Know when to apply BSA-based dosage regimens Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_14) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_14

329

330

14 Dosage Calculations Based on Body Surface Area (BSA)

In the previous chapter, we have briefly explained weight-based dosing on the basis of weight-based apparent volume of distribution of drugs. For most adult patients, normalizing the dose per body weight could be adequate, but for special population groups, such as neonates, infants, obese, and elderly patients, normalizing the drug dose per unit weight may not always be a safe approach. The primary reason for this discrepancy is that these groups of patients have other than just weight differences. For example, neonates and infants cannot be considered simply as little people who weigh less than adults. Reduced body weight and reduced body dimensions imply reduced internal organ size, usually indicative of a reduced capacity for drug metabolism and excretion. In addition, several of their primary tissues and internal organs such as intestinal mucosa, liver, and kidney are not fully mature, and as a result, the processes of drug absorption, metabolism, and excretion are different from adults. Also, due to anatomical immaturity, drug absorption could be enhanced as the intestinal membrane barrier structure is not fully mature. Drug metabolism is usually decreased due to lower number of enzyme populations or enzyme activity while excretion is usually of reduced capacity. The physiology of elderly people is also different from others in the general population. It is well documented that the loss of viable tissue sustained over the time causes physiological changes. When these physiological changes become too extreme, they could cause organ function impairment leading to pathological conditions. The clinical picture with obese individuals is also a unique one. One of their main differences from the general population is the significant increase of adipose tissue present in their body that could serve as a sink storage tissue for lipophilic drugs, thus effectively lowering their concentration in the blood and their concomitant diffusion to the target diseased tissues. Since adipose tissue is so different from any other tissue in the body, it would be wrong to normalize doses based on total body weight for lipophilic drugs. On the other hand, normalizing the dose of polar, less hydrophobic drugs per total body weight could lead to an overdose because although the hydrophilic drugs cannot partition into fat tissue, adipose tissue has been incorrectly included in the total weight of tissue in which the drug could be potentially distributed. In general, weight-based dosing, body surface area-based dosing, age-based dosing, and other special population dosing strategies are dependent on complicated physiological factors, types of medications, and clinical indications. Despite the interest of the topic, the scope of this section is not to introduce the reader to the immense number of formulas used for all kinds of complicated clinical conditions. Rather the intention here is to get the reader familiar with a rational approach to dosage calculations for all kinds of people, both special and ordinary ones. In an attempt to better correlate drug dosage with organ functionality, clinical scientists have looked for other “indices.” One of the most famous and still in use is the body surface area (BSA). It is widely reported that dosage approaches based on body surface area are more accurate than dosage calculations based on body weight, mainly because BSA estimations are based on measurement of both weight and height of the individual patients. Some of the accounts on this matter even mention that BSA approaches account for organ size and function. We do not share this opinion. We can simply say that there will be cases where dosage calculations based on BSA are more accurate than those based on BW and vice versa.

14.1

Fundamental Equation for Surface Area Determination of Cylindrical Objects

331

Fig. 14.1 Height and base radius of a cylinder

H

R

14.1

Fundamental Equation for Surface Area Determination of Cylindrical Objects

Let us start our discussion on body surface area by simulating our body as a cylinder and then use mathematics to determine the surface area of a cylindrical body. Folding our limps and other extremities into our body, we do resemble a cylindrical object. The surface area (SA) of a cylinder can be estimated from the surface area of the sides (2 ∙ π ∙ R ∙ H) plus the surface area of each end or base (π ∙ R2) (Fig. 14.1). SA ¼ 2 ∙ π ∙ R ∙ H þ 2 ∙ π ∙ R2 ¼ 2 ∙ π ∙ R ∙ ðH þ RÞ

ð14:1Þ

The example below is included to demonstrate how to calculate the surface area of a cylindrical body of known dimensions and help us understand how the SA determinations are affected not only by the dimensions but also by the individual’s body weight. Example 14.1 Calculate the SA of cylinder A with RA ¼ 0.178 m and HA ¼ 1.72 m and of cylinder B with RB ¼ 0.159 m and HB ¼ 1.72 m. Notice that HA ¼ HB. Solution Using Eq. 14.1, SAA ¼ 2 ∙ π ∙ RA ∙ ðH þ RA Þ ¼ 2 ∙ 3:14 ∙ 0:178 ∙ ð1:72 þ 0:178Þ ¼ 2:12 m2 SAB ¼ 2 ∙ π ∙ RB ∙ ðH þ RB Þ ¼ 2 ∙ 3:14 ∙ 0:159 ∙ ð1:72 þ 0:159Þ ¼ 1:88 m2

332

14 Dosage Calculations Based on Body Surface Area (BSA)

Analysis and points to remember: 1. Two cylindrical bodies of same height have different SA because of different perimeters. Are these two cylinders of equal weight? If the inside of the cylinders is made of identical material, that is, same density, then the bigger surface area should yield a heavier cylinder. How do we determine the weight or mass of the cylinder? Just calculate the volume of the cylinder, and for our purposes, assume a density equal to 1 g/cm3. The volume of a cylindrical object is given by: V ¼ π ∙ R2 ∙ H

ð14:2Þ

Since, mass or weight, W ¼ d ∙ V ) W ¼ d ∙ π ∙ R2 ∙ H

ð14:3Þ

For d ¼ 1 g/mL, Eqs. 14.2 and 14.3 are identical. For cylindrical body A: ) V A ¼ π ∙ R2A ∙ H A ¼ 3:14 ∙ 0:1782 ∙ 1:72 ¼ 0:171 m3 Given that: 1 m3 ¼ 1000 L, VA ¼ 171 L Also, d ¼ 1 g/cm3 ¼ 1 g/mL ¼ 1 kg/L ¼ 1 kg/103 m3 ) WA ¼ d ∙ VA ¼ 1

kg 10

3

m3

∙ 0:171 m3 ¼ 171 kg

For cylindrical body B: V B ¼ π ∙ R2B ∙ H B ¼ 3:14 ∙ 0:1592 ∙ 1:72 ¼ 0:136 m3 ¼ 136 L ) WB ¼ 136 kg 2. In general, among two bodies made of same material (same density), the larger body is always the heavier one. Notice also that when the density of the body is equal to 1 g/mL, the volume of the body in L is numerically equivalent to the weight of the body in kg. 3. Let us take a closer look of Eq. 14.1 and the SA calculations performed in this example. For cylindrical body A:

14.2

Formulas for Body Surface Area

333

SAA ¼ 2 ∙ π ∙ RA ∙ H þ 2 ∙ π ∙ R2A ¼ 2 ∙ 3:14 ∙ 0:178 ∙ 1:72 þ 2 ∙ 3:14 ∙ 0:1782 ¼ 1:92 þ 0:20 ¼ 2:12 m2 The % surface area of the two bases with respect to the total surface area of the cylindrical body is: %SAA, base ¼

SAA, base 0:2 ∙ 100 ¼ 9:4% ∙ 100 ¼ 2:12 SAA

Similarly for body B: SAB ¼ 2 ∙ π ∙ RB ∙ H þ 2 ∙ π ∙ R2B ¼ 2 ∙ 3:14 ∙ 0:159 ∙ 1:72 þ 2 ∙ 3:14 ∙ 0:1592 ¼ 1:72 þ 0:16 ¼ 1:88 m2 The % surface area of the two bases of the cylindrical body is: %SAB, base ¼

SAB, base 0:16 ∙ 100 ¼ 8:5% ∙ 100 ¼ 1:88 SAB

From the results above, it becomes obvious that the second part of Eq. 14.1 is responsible for only ~9% of the total surface area, and it can be neglected. Eq. 14.1 can then be simplified to SA ¼ 2 ∙ π ∙ R ∙ H

14.2

ð14:4Þ

Formulas for Body Surface Area

Some of the most popular formulas used for dosage calculations are shown in the table below. In the above formulas, BSA is calculated from height and weight, raised to some exponent and multiplied by a particular coefficient. This is quite convenient because height and weight of a patient can easily and accurately be measured in any physician’s office. We have shown in Example 14.1 that height and weight are related through Eqs. 14.2 and 14.3. Our task in the next section is to derive the general form of the BSA equations and further elaborate on the various coefficients and exponents of the BSA formulas.

334

14 Dosage Calculations Based on Body Surface Area (BSA)

Table 14.1 Some of the equations more commonly used to determine BSA in m2 (Kouno et al. 2006) 1. The DuBois and DuBois’s formula (1916): Generated out of nine patients BSA ¼ 0.20247 ∙ H0.725 ∙ W0.425 H ¼ height in meters and W ¼ weight in kilograms 2. The Boyd’s formula (1935): Generated out of 411 patients BSA ¼ 0.017827 ∙ H0.5 ∙ W0.4838 H ¼ height in centimeters and W ¼ weight in kilograms 3. The Fujimoto’s formula (1968): Generated from 201 patients BSA ¼ 0.008883 ∙ H0.663 ∙ W0.444 H ¼ height in centimeters and W ¼ weight in kilograms 4. The Haycock’s formula (1978): Generated from 81 patients BSA ¼ 0.024265 ∙ H0.3964 ∙ W0.5378 H ¼ height in centimeters and W ¼ weight in kilograms 5. The Mosteller’s formula (1987): Generated from 401 patients qffiffiffiffiffiffiffiffi ∙W BSA ¼ H3600 H ¼ height in centimeters and W ¼ weight in kilograms

14.3

Derivation of BSA Formulas

While I was writing this section, I was imagining Dubois and coworkers sitting on their desks trying to figure out a way to develop a convenient formula that calculates BSA from height and weight measurements. There must have been so many questions in their mind. Can BSA be calculated from weight measurements? How is it related to weight? The answer is yes, BSA can be calculated from weight measurements. To find out how SA is related to weight, solve Eq. 14.3 (assuming d ¼ 1) for R and substitute in Eq. 14.4. rffiffiffiffiffiffiffiffiffiffi W R¼ π∙H Substituting the above into Eq. 14.4, rffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi W SA ¼ 2 ∙ π ∙ ∙ H ¼ 2 π ∙ H ∙ W ¼ 3:545 ∙ H 0:5 ∙ W 0:5 π∙H

ð14:5Þ

We can now see the resemblance of the equation above with the BSA formulas listed in Table 14.1. Why do equations for BSA in Table 14.1 have different coefficient and exponents than Eq. 14.5, and why are there so many different equations for BSA? As discussed before, those formulas are all based on Eq. 14.5. However, because our body is not exactly a nicely shaped cylinder or any other geometrical object for that matter, the coefficient and exponents of the biexponential function were properly adapted to match the dimensions and weight of human body with the corresponding body surface areas. The plethora of BSA formulas is due to the fact that the relationship between the BSA, in neonates, infants, children, and adults, with the body height and

14.4

BSA or BW?

335

weight is very different. The larger the number of subjects (see Table 14.1) used to derive a formula, the more accurate is the formula, assuming that the subjects were selected and grouped correctly. Summarizing all steps involved in developing BSA formulas, the first task that the clinical scientist has to overcome is to develop methods to accurately calculate body surface areas. In the old days, physicians and scientists did that by using various coating methods or by determining the area of the paper that was used to wrap up the patient’s body. Nowadays, the task is simplified through the use of 3D anthropometrical scanners (Yang et al. 2003). The second task is to correlate the measured BSAs with the corresponding heights and weights of subjects. Simultaneous fitting of two parameters is not an easy task and is not going to be discussed here. In the old days, this was done by manually plotting the weight and height against BSA and manually fitting the curve into an exponential equation by varying the coefficients. Nowadays, there are mathematical packages that do it for us very accurately.

14.4

BSA or BW?

We have clearly seen that BSA approaches, unlike pharmacokinetic methods, do not take into account metabolic or other organ functionality differences in individual subjects. Differences in body surface area are reflected into weight differences given the same density or composition of the bodies. On the other hand, different body surface areas may not lead to significant differences with respect to weight if the dimensional differences are counterbalanced by differences in body’s composition or density. For instance, a tall, slim individual may have a bigger BSA but approximately equal weight than a shorter, bigger frame, thick-boned individual. By the same token, similar body surface areas may not always indicate similar body weight if the composition or density of the two bodies varies. All these facts suggest that approaches based on BSA have no distinct advantage over methods based on BW. None of these methods can accurately predict, upscaling or downscaling of dosages determined for “normal” individuals, because neither approach takes into account functional parameters involved in the pharmacodynamics and pharmacokinetics. We could therefore conclude that dosage calculation approaches based on BSA or BW should be similarly effective in individuals with body composition close to that of the average normal individual, that is, the relative tissue ratio such as muscle to fat tissue or body fluids remains relatively constant. It is impossible to predict which approach could better approximate drug dosage calculations in all the other cases in which there is an extreme deviation from the normal average. The method of choice for dosage calculation of highly potent drugs with a narrow therapeutic index is decided after extensive animal studies and human clinical trials and the information is usually in the package insert of the corresponding dosage form. In conclusion, you can calculate dosages using the BSA method: 1. When it is clearly stated that you can do so in the package insert of the specific drug.

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Fig. 14.2 A nomogram for the determination of body surface area (BSA) of adults from height and weight based on the DuBois and DuBois’s formula (from Ciba-Geigy Scientific Tables, 2nd Edition, p. 537). BSA values can be read at the point of intersection of the middle axis, with the line connecting two known values of weight and height on the two outermost axes

2. When your patient has the same characteristics (age, gender, ethnicity, height, weight, BSA, etc) as the group of patients that were used in clinical trials to generate the specific BSA formula. 3. When your patient is free of disease states or conditions (not tested during the drug’s clinical trials) that can affect the pharmacokinetics or pharmacodynamics of the drug.

14.6

14.5

Advantages and Disadvantages of BSA Approaches in Dosage Calculations

337

Using BSA Formulas for Dosage Adjustment

The first step is to calculate the BSA of a subject from his/her weight and height using one of the formulas listed in Table 14.1. In the old days, use of biexponential formulas was cumbersome. Graphic representation of those formulas, known as nomograms, was created for easy reading of BSA values (Fig. 14.2). Directions on how to read BSA from the weight and height using a nomogram is described in the legend of Fig. 14.2. Every BSA formula is represented with a different nomogram. BSA values read from a nomogram should be the same as those calculated from the corresponding formula using scientific calculators. Example 14.2 The daily maintenance dose of a narcotic analgesic for a normal individual is 5 mg. Assuming that the BSA for a normal individual is 1.72 m2, calculate the drug dose for a 180 lb, 6 ft, and 3 in. subject using (a) the nomogram in Fig. 14.2 and (b) the DuBois’s formula. Solution Step 1: Calculate the BSA of the subject from the weight and height. (a) Using the nomogram W ¼ 180 lb/2.2 lb/kg ¼ 81.8 kg H ¼ 6 ft, 3 in ¼ 75 in2.54 cm/in ¼ 190.5 cm BSA ¼ 2.10 m2 (b) Using the DuBois’s formula BSA ¼ 0:20247 ∙ ð1:905Þ0:725 ∙ ð81:8Þ0:425 ¼ 2:0999 ¼ 2:10 m2 Step 2: Set up a proportion to calculate the daily dose for 2.10 m2 BSA. 5 mg X ¼ ) X ¼ 6:10 mg 1:72 m2 2:10 m2

14.6

Advantages and Disadvantages of BSA Approaches in Dosage Calculations

Dosage calculations based on BSA have certain advantages and disadvantages, as every other method. We will first start with the disadvantages and finish with the advantages of the method. BSA is not as perceptible a parameter as the weight. The fact that we do not use BSA in the everyday life makes it difficult to judge whether an estimated BSA value would make sense for a particular individual or not; we are definitely more familiar

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with weight measurements. In other words, we would question the weight of an infant reported in its medical chart as 20 kg, but we would not probably question a BSA of 0.65 m2. In addition, extreme caution is recommended to choose the correct formula, presented in the appropriate format, among the plethora of formulas; some of them are dedicated specifically for special populations. Furthermore, many books and manuscripts have the formulas copied down incorrectly. It is always a good idea to consult with a specialist before using BSA formulas to adjust patient’s dosage. Use of nomograms is easier than using the formulas, but they carry the risk of having graphical errors. My advice is to never use any nomogram by itself, even if you have constructed it. It is always safer and more accurate to calculate BSA values using the formulas that were used to construct the particular nomograms and use them to verify the dose determined from nomograms. On the other hand, dosage calculations based on BSA appear to be more flexible than BW. The presence of two adjustable parameters in the equation offers the chance to better optimize dose calculations using various regression techniques. This is perhaps one of the reasons that dosage adjustments based on BSA became a frequent practice in medical areas, such as anesthesiology, cancer chemotherapy, and pediatric medicine where dosage calculations based on the inflexible methods of body weight more frequently fail to give satisfactory results. Example 14.3 The following two formulas were recommended for neonate (1 month old) and infant (1–24 months old) dosage adjustments: Formula for neonates, BSA ¼ 0.0224265 H(cm)0.3964 W(kg)0.5378 Formula for infants, BSA ¼ 0.007184 H(cm)0.725 W(kg)0.425 (a) Is there anything wrong with these formulas? (b) How would you justify a bigger BSA for neonates? Solution (a) The formula for infants is the DuBois’s formula with the height measured in centimeter. The formula for neonates is the Haycock’s formula, but it is copied down incorrectly in another book. The correct formula is: BSA ¼ 0:024265 ∙ H 0:3964 ∙ W 0:5378 In addition, the coefficient of the BSA formula for infants is approximately three times smaller than that of neonates. This could lead to smaller doses for the older and more physiologically mature infants than for the neonates. Figures 14.3 and 14.4 verify that BSA values for neonates are bigger than those for infants. The question is why and for what drugs could these formulas provide therapeutically relevant BSA-based doses? We know from physiology (continued)

14.6

Advantages and Disadvantages of BSA Approaches in Dosage Calculations

339

Example 14.3 (continued)

Fig. 14.3 Body surface area–height plot

Fig. 14.4 Body surface area–weight plot

that the total body water in neonates is 75–80% of the total body weight as opposed to about 70% in infants. The difference is mostly related to the ECF. Although the difference is small, administration of a drug that has a small (continued)

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Example 14.3 (continued) volume of distribution, for example, some of the antibiotics where distribution is restricted mainly in ECF, will result in a lower plasma drug concentration in neonates as compared to infants. If the drug clearance is not affected by the difference in the activity of liver and kidney, neonates will require a higher dose to achieve the same therapeutic levels than infants. Justification for higher dosages in this case is provided by the increased volume of distribution of neonates as compared to the older infants (and other adult patients). For lipophilic drugs with large Vd that are primarily eliminated by metabolism, a higher neonatal dose could be very dangerous because their liver may not be able to handle large doses. In general, as a clinical pharmacist you have a responsibility to understand equations that you use to adjust drug doses to avoid making serious mistakes. In addition, we have seen that understanding pharmacokinetics helps making educated decisions about the validity of BSA-based dosing when dosing instructions are not provided by the manufacturer.

Exercises 14.1. The pediatric recommended dose of an antibiotic drug is 2.5 mg/kg or 32.5 mg/m2. How many milligrams are represented in each dose for a 4.2 lb, 17 in. infant? Compare the dose normalized per body weight with the dose adjusted using the DuBois’s formula. (Answer: 4.78 mg as adjusted per kg versus 4.71 mg as the dose adjusted with BSA) 14.2. Calculate the daily dose for a 30 lb, 36.6 in. child using the Boyd’s formula if the adult normal dose of a drug is 2.5 mg, q.i.d. (Answer: 3.54 mg/day) 14.3. A 136 lb, 1.67 m, 67-year-old male patient is to be infused with 144 (μg/m2)/ h of an anticancer drug over 75 min once a day, 5 days per week for 3 weeks. (a) Convert the dose of the drug in μg/kg/day using the Mosteller’s formula. (b) How many milligrams of drug has the patient received by the end of his treatment? (Answer: (a) 4.92 (μg/kg)/day; (b) 4.56 mg) 14.4. Given the following information for a female patient receiving chemotherapy: Age: 54 Weight: 163 lb Height: 5 ft 8 in. Methotrexate recommended dose: 40 mg/m2 Drug concentration in IV bag: 450 mg/L Dosage regimen: IV infusion once a day, q.i.w. for 6 weeks

Exercises

341

(a) Calculate patient’s BSA using the Fujimoto’s formula. (b) What is the daily volume of infusion? (Answer: (a) 1.83 m2; (b) 162.5 mL) 14.5. Given the information below: Patient: 3-month old infant Weight: 4 lb, 15 ounces Recommended dose: 75 mg/1.72 m2

Height: 20 in

(a) What is the appropriate dose for the infant (use the DuBois’s formula)? (b) If the baby is to receive the recommended dose in two separate injections of 150 μL each, how much drug should the pharmacist weigh to prepare 0.5 mL of solution? (Answer: (a) 7.61 mg; (b) 12.68 mg) 14.6. The recommended “normal” oral dose for an antibiotic in suspension form is 350 mg t.i.d. (a) Use the Boyd’s formula to calculate the dose should a 6-year-old child who weighs 32 lb and is 0.95 m tall receive. (b) Assuming that the antibiotic is supplied as a dry powder, and the dose is to be given in one tsp, how many grams of powder should you weigh out and to what volume should you reconstitute the powder, if the medication is to last 10 days? (c) What is the % w/v of the drug in the suspension? (Answer: (a) 129.1 mg; (b) 3.873 g, 150 mL; (c) 2.6%) 14.7. A premature newborn (W ¼ 4.5 lb, H ¼ 18 in.) who has presented a Pseudomonas aeruginosa infection is to be treated with IM injections of polymyxin B sulfate antibiotic. The recommended antibiotic dose is 2.1106 unit/1.72 m2 per day. Polymyxin B sulfate for IM injections is supplied in ampules containing 500,000 units of antibiotic dissolved in 2 mL of sterile isotonic saline. Each milligram of polymyxin B sulfate is equivalent to 10,000 units of antibiotic salt. (a) Use the DuBois’s formula to calculate the milligrams of antibiotic in each injection if the antibiotic is to be given every 6 h. (b) What is the injection volume? (Answer: (a) 4.75 mg; (b) 190 μL) 14.8. A student who does not like working with exponential equations derived a linear version of Haycock’s equation and plotted the data as shown in the graph below (Fig. 14.5). (a) Write the linear form of the Haycock’s formula. (b) Do the negative values presented in the y-axis make sense to you?

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Fig. 14.5 The weight and height of males were applied on Haycock’s formula to calculate the ln (BSA). The plotted years of age from left-to-right are 0 (newborn), 0.25 (3 months), 0.5, 0.75, 1, 2, 3, 4, 5, 6, 7, 8, and 9

(c) Use the linear form to calculate BSA for a subject whose height and weight is 0.92 m and 14.2 kg, respectively. (Answer: (c) 0.607 m2) 14.9. The BSA of a particular patient was found by Fujimoto’s formula and by Mosteller’s formula to be 1.915441 and 1.993523, respectively. Use this information to determine the weight in kilograms and height in meters of the patient. (Answer: W ¼ 83.98 kg, H ¼ 1.703 m) 14.10. Pick the correct statement. BSA-based dosage regimens are used _____________, (a) When body weight-based dosing did not provide drug therapeutic levels (b) For clinically obese patients (c) When there are instructions from the manufacturer that the drug must be dosed using BSA methods (d) On patients who have a Vd larger than the population average value (e) To dose premature newborns

References T. Kouno, N. Katsumata, H. Mukai, M. Ando, T. Watanabe, Standardization of the Body Surface Area (BSA) formula to calculate the dose of anticancer agents in Japan. Jpn. J. Clin. Oncol. 33, 309–313 (2006) Y.-X. Yang, C.-Y. Yu, W.-K. Chiou, The 3D scanner for measuring body surface area: a simplified calculation in the Chinese adult. Appl. Ergon. 34, 273–278 (2003)

Chapter 15

Intravenous Infusion and Flow Rate

Abstract The advantages of the intravenous infusion compared to the other routes of drug administration were briefly presented in Chap. 13. In this Chapter, we discuss all parameters that clinical scientists should be aware of prior to conducting dose, rate of infusion and frequency of infusion calculations. Finally, the IV infusion accessories are described prior to engaging you in calculation problems that are commonly encountered in the preparation of parenteral infusion solutions. Keywords Infusion · Flow rate

Learning Objectives After reading this chapter, you should be able to: • Calculate infusion flow rates in units of volume and drops per unit time Intravenous infusion allows direct administration of drugs into the central blood stream. Because it bypasses absorption, intravenous infusion of drugs at a constant rate permits precise control of their plasma concentration, thus making intravenous infusion the method of choice for drugs with a narrow therapeutic index. An additional advantage of the intravenous infusion, especially for patients who are in a state of a shock or have lost consciousness, is that drugs can be given with IV fluids such as electrolytes and nutrients. Although the primary scope of this chapter is to familiarize the student with calculations relevant to the flow rate of IV infusion, it should be emphasized that other factors such as non-physiological osmolarity, pH, and temperature of the infusion solutions could cause discomfort and pain to the patient due to serious vessel irritation, tissue infiltration, and extravasation. You should always be aware of the dose, the rate of infusion, the frequency of Electronic supplementary material: The online version of this chapter (https://doi.org/10.1007/ 978-3-030-20335-1_15) contains supplementary material, which is available to authorized users. © Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1_15

343

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Fig. 15.1 Rudimentary sketch demonstrating the assembly of the basic parts of an intravenous infusion set. (1) Primary infusion bag, (2) macro- or microdrip tubing, (3) roller-type flow control clamp, (4) secondary infusion bag (piggyback), (5) primary infusion line, (6) secondary infusion line, (7) Y-type connector, (8) to patient veins. Macro-drip tubing delivers 10–20 drops/mL, whereas micro-drip tubing dispenses 60 drops/mL. The flow rate of primary and secondary infusate is controlled manually using roller-type clamps

administration, and other details of the drug that may affect its plasma levels, its tissue distribution, and patient’s well-being.

15.1

IV Infusion Accessories

Solutions for IV infusion are usually stored in plastic flexible bags although glass containers may also be used especially in situations where some of the ingredients of the solution are incompatible with the plastic bags. For instance, fat emulsions are frequently supplied in glass containers because lipid surfactants may facilitate the release of phthalates from the plastic bags into the infusion solution. The IV bags come in various sizes from 50 mL to 1 L volume capacity to accommodate patient needs. The main accessories of a basic intravenous infusion setting are shown in Fig. 15.1. The primary infusion bag contains the main fluid for patient therapy. Drugs could be added in the primary IV bag at any time during the infusion process or they could be administered from a separate IV bag piggybacked into the primary infusion line as shown in Fig. 15.1. The piggyback method has the advantage of administering the drug at a much faster or slower rate than the primary infusate. Usually, the secondary bag is of smaller volume than the primary infusion bag. The flow rate of primary and secondary infusate is manually adjusted with the aid of a screw- or a roller-type clamp.

15.1

IV Infusion Accessories

345

In Chap. 10, I have discussed IV administration of fluids and electrolytes to treat dehydration and other electrolyte disturbances. I hoped to have made it obvious that careless IV administration of concentrated electrolytes could cause serious health problems and even death. It is your responsibility to master calculations involving electrolyte doses and rates of administration. The practice examples included in this chapter are dealing with calculations that are often encountered in the preparation of IV infusion solutions. Example 15.1 Given the following medication order: Infuse 1750 cc solution of D5W and 10 mEq KCl/L, at a rate of 125 mL/h (a) How much glucose monohydrate and potassium chloride are needed to aseptically prepare the above solution? (b) What would be the drops per minute that the infusion set should be adjusted to, if the infusion tubing delivers 15 drops/mL? (c) How long would it take to infuse 1500 cc of the above solution at the given rate? Glucose (C6H12O6); MW ¼ 180 g/mol Glucose monohydrate (C6H12O6H2O); MW ¼ 198 g/mol. Solution (a) Step 1: Preparation of 5% glucose (C6H12O6) (MW ¼ 180 g/mol) from glucose monohydrate (MW ¼ 198 g/mol). Due to the bigger MW of glucose monohydrate, 5% glucose is equivalent to: 5%glucose ∙

MW ðglucose monohydrateÞ MW ðglucoseÞ

198 ¼ 5:5%glucose monohydrate 180 X 5:5 g=100 mL ¼ ) X ¼ 96:25 g 1750 mL ¼ 5% ∙

Step 2: The grams of KCl required can be calculated as shown below: The EW of KCl is equal to the MW, that is, 74.5 g/Eq To calculate the grams of KCl in 10 mEq, set up the ratio 7:45 mg X ¼ ) X ¼ 0:745 g KCl=L 1 mEq 10 mEq=L ) 0:745 g=L  1:750 L ¼ 1:3 g of KCl (b) To calculate the drops per minute, multiply the rate of infusion with drops/mL. (continued)

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Example 15.1 (continued) 125 mL 15 drops ∙ ¼ 31:25  31 drops=min 60 min mL (c) Divide the given volume with flow rate. 1500 mL ¼ 12 h 125 mL=h

Important Notes: 1. It is important to round off the number of drops to the nearest whole number. Some practitioners round up the number of drops regardless of the math rules, I guess, to make certain that the patient will have all the dose by the end of the dosing period. In this book, all answers are based on math round off rules. I consider that to be more important than infusing few drops less by the end of the specified infusion time. 2. Avoid memorizing any equations. It is strongly recommended to always write down the corresponding units of the numerical answer. Learn to use the units as a guide for the appropriate mathematical operation and the units of the final answer to verify the correctness of your answer. 3. In Chap. 10 (and Example 10.9), we have clearly stated that infusion of IV KCl solutions in dextrose should be avoided in the treatment of severe hypokalemia, as glucose could promote a shift of extracellular K+ into cells. The case described in Example 15.1 is a very mild hypokalemia as it only involves administration of only 17.5 mEq of K+. There are commercially available IV solutions of KCl in D5W and you shouldn’t be surprised if they are still used in hospital protocols. Example 15.2 Given the following medication order:

MODERN VILLAGE HOSPITAL ______________________________________________ Name: S. S. Age: 56 Wt: 64 kg __________________________________________ Drug Z, 6 mg/kg in 50 mL normal saline. Infuse 0.2 mg/kg/min.

Hospital policy requires a 24-h expiration date on compounded IV mixtures.

Drug Z is available as 250 mg/10 mL vial. (a) Calculate the volume of drug solution that must be added to the IV bag. (continued)

15.1

IV Infusion Accessories

347

Example 15.2 (continued) (b) What is the final concentration of drug Z in the IV bag? (c) What is the flow rate in mL/h? (d) How long would it take to infuse the dose? Solution (a) Step 1: Calculate first the dose of the patient. ) 6 mg/kg  64 kg ¼ 384 mg of drug Z Step 2: Divide the dose by the drug concentration of the vial. ) 384 mg/25 mg/mL ¼ 15.36 mL Thus, 15.36 mL (two vials are needed) has to be added to the bag. (b) The answer really depends on the way you prepare the product. (b.1) If you removed 15.36 mL of normal saline solution from the IV bag before adding the drug solution, then the concentration in the final product is: 384 mg/50 mL ¼ 7.68 mg/mL (b.2) Alternatively, you feel very confident with your math skills and you inject the drug volume into the IV bag without removing any of the existing saline solution. The concentration of the final product is 384 mg ¼ 5:875 mg=mL ð50 þ 15:36Þ mL Which of the two ways is more correct? They are both correct as long as the correct drug concentration is used to calculate the drip rate. However, a word of caution here is appropriate. Notice that the volume transferred to the IV bag is significant (almost 31%), and it might create a problem to the patient if the drug solution is not isotonic. Regardless of the method used to add the drug into an IV bag, it is always a good practice to dilute drugs into large volume IV bags, so that the addition of non-isotonic drug solution does not change the osmolarity of the IV infusate. (c) Infusion rate, R ¼ [0.2 (mg/kg)/min]  64 kg ¼ 12.8 mg/min (c.1) Dividing the infusion rate with the drug concentration in the IV bag, 12:8 mg=min ¼ 1:67 mL=min 7:68 mg=mL

(continued)

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Example 15.2 (continued) 12:8 mg=min (c.2) ¼ 2:18 mL=min 5:875 mg=mL (d) Divide the dose with infusion rate 384 mg ¼ 30 min 12:8 mg=min Notice that the duration of the infusion is independent of the method of product preparation.

Example 15.3 You are asked to prepare an IV solution of a drug at a final concentration of 3.84 mg/mL in a normal saline IV bag and infuse it at a rate of 12.8 mg/min. Items in stock: Drug Z is available as 250 mg/10 mL vial. 100 mL normal saline IV bag. (a) Calculate the volume of drug solution that must be added from the vial to the IV bag containing normal saline without removing any solution from the IV bag. (b) How long would it take to infuse the whole solution? Solution (a) We would need to construct our own equation to solve this problem. The first task is to identify known and unknown parameters. Notice that the concentration in the IV bag (Cf) and the drug concentration in the stock vials (Cv) are known. What we seek to find is the volume of the stock solution in the vials that must be transferred to yield the final concentration in the IV bag. Let us call Vv the volume that needs to be transferred from the vials and Vf the volume in the IV bag after adding the drug. We could then use the mass balance Eq. 5.1 to calculate Vv. Cv ∙ Vv ¼ Cf ∙ Vf and Vf ¼ 100 mL + Vv ) 25 ∙ V v ¼ 3:84 ∙ ð100 mL þ V v Þ 21:16 ∙ V v ¼ 384 ) V v ¼ 18:15 mL ) V f ¼ 118:15 mL Verification of the answer: Mass of drug transferred is 25 mg/mL  118.15 mL ¼ 453.75 mg/mL (continued)

Exercises

349

Example 15.3 (continued) )

453:75 mg ¼ 3:84 mg=mL 118:15 mL

(b) Dividing the dose over the concentration of the IV bag mg 3:84 mL ∙ 118:15 mL ¼ 35:4 min 12:8 mg=min

Exercises 15.1. A vial contains 0.750 g of a drug antibiotic as a lyophilized powder. The instructions written in the package insert recommend dissolving the powder with the addition of 3.4 mL of sterile water yielding a 4 mL of final solution in the vial (the powder occupies some volume). (a) What volume of the above solution should be added to a 100 mL NS IV bag, if the patient is to receive 250 mg of the antibiotic over a period of 30 min? (b) What is the final concentration of the antibiotic in the bag? (c) Calculate the flow rate in mL/h. (d) What is the drip rate in drops per minute, if the infusion set dispenses 15 drops/mL? (Answer: (a) 1.33 mL;

(b) 2.49 mg/mL; (c) 202.66 mL/h;

(d) 51 drops/min)

15.2. A medication order calls for an intravenous infusion of 400 mL of 7.5% fat emulsion. (a) Determine the infusion time if the flow rate is adjusted at 12 drops per minute and the infusion set is calibrated at 15 drops/mL. (b) How many calories did the patient receive after 4 h of infusion (1 g of fat ¼ 9.1 kcal)? (Answer: (a) 500 min;

(b) 131 kcal)

15.3. A pediatric patient whose weight is 12.3 kg has suffered severe burns on 30% of her body surface area and has developed hypoproteinemia 24 h after the incident. She was scheduled to receive intravenous infusion of human serum albumin (HSA) solutions as shown below: “Infuse 0.75 g/kg, from a 0.702 mM HSA solution, at a rate of 0.6 mL/min” (a) Assuming that the MW of HSA is 57 kDa, calculate the gram amount needed to prepare 250 mL of 0.702 mM HSA solution.

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(b) How long will it take to infuse the recommended dose? (c) Assuming that 0.5 mL of solution is equivalent to nine drops, what is the drip rate in drops per minute? (Answer: (a) 10.075 g; (b) 6.36 h;

(c) 11 drops/min)

15.4. A 48-year-old male who had undergone an autologous bone marrow transplantation was advised to undergo a Leukine (sargramostim) therapy because he showed signs of engraftment delay. He is 5 ft 3 in. tall and weighs 112 lb. The recommended dose of Leukine is 250 μg/m2/day based on the Mosteller’s equation, for a course of 14 days. (a) If Leukine is supplied in 1 mL vials at 2.8  106 units/mL, calculate the volume of Leukine needed to be transferred to a 500 mL IV bag to make a concentration of 10 μg/mL (1 μg ¼ 5600 units). (b) How much volume should the patient receive in a day? (c) Calculate the rate of flow in mL/h, assuming that the patient will be receiving the daily dose within a 2 h period. (d) What should be the drip rate in drops/min if the IV tubing delivers 20 drops/mL? (Answer: (a) 10.2 mL;

(b) 37.5 mL;

(c) 18.75 mL/h;

(d) 6 drops/min)

15.5. A 6-year-old child who is 0.95 m tall and weighs 38 lb is to receive 6 g/m2 of the diuretic mannitol (MW 182) by intravenous infusion. The infusion is given as a 15% w/v solution over a period of 5 min. (a) Use the DuBois’ equation to calculate the dose of the child. (b) How much volume should the patient receive? (c) Calculate the flow rate in mL/h and drops/mL if the infusion set delivers 10 drops/mL. (d) How many mOsmoles are there in the dose? (Answer: (a) 3.9282 g; (d) 21.6 mOsm)

(b) 26.19 mL;

15.6. Given the following medication order:

(c) 314.28 mL/h, 52 drops/min;

Exercises

351

MODERN VILLAGE HOSPITAL ______________________________________________ Name: S. S. Age: 46 Wt: 176 lb __________________________________________ Aminophylline 10 mg/kg Infuse 0.5 mg/kg/min.

Hospital policy requires a 24-h expiration date on compounded IV mixtures.

Items in stock Aminophylline dihydrate (MW 216) is available as 500 mg/10 mL ampoules. 50 mL normal saline IV bag. (a) (b) (c) (d)

Calculate the volume of drug solution that must be added to the IV bag. What is the final concentration of aminophylline in the IV bag? What is the flow rate in mL/min? How long would it take to infuse the dose? (Answer: (a) 19.2 mL; (d) 20 min)

(b) 11.56 mg/mL;

(c) 3.46 mL/min;

15.7. Drug Z in 540 mL is to be infused within a period of 4.5 h with the flow rate adjusted at 22 drops per minute while the infusion set is calibrated at 11 drops/mL. (a) Calculate the true flow rate if after 2 h only 150 mL has been infused. (b) Calculate a new flow rate (mL/min) and drip rate (drops/min), if you are to infuse the remaining fluid on time (4.5 h total). (c) What is the error introduced after rounding off the drops? (d) How much earlier will the infusion process finish? (Answer: (a) 1.25 mL/min; (c) 1.4%; (d) 2.1 min)

(b) 2.6 mL/min, 29 drops/min;

15.8. Given the following medication order:

MODERN VILLAGE HOSPITAL ______________________________________________ Name: S. H. Age: 44 Wt: 155 lb __________________________________________ K+ 40 mEq in 1 L D5W Infuse over 8 h

Hospital policy requires a 24-h expiration date on compounded IV mixtures.

Items in stock

352

15

Intravenous Infusion and Flow Rate

Potassium monophosphate (MW 136) and potassium diphosphate (MW 174), 1:2 weight ratio available as 30%w/v in 15 mL ampoules. 1 L IV bag D5W. (a) Calculate the volume of potassium phosphate solution that must be added to the IV bag. (b) What is the final potassium concentration in mEq/L of the IV bag? (c) What is the flow rate in mL/h? (d) Calculate the drip rate in drops/min if the infusion set delivers 12 drops/ mL. (Answer: (a) 13.38 mL or 13.18 mL; (b) 40 mEq/L or 39.48 mEq/L; (c) 126.65 mL/h; (d) 25 drops/min) 15.9. A 28-year-old diver who was stung on the neck by a jellyfish was brought to the emergency department to be treated for excruciating pain, agitation, dyspnea, and extreme hypertension. The physician orders an intravenous infusion of 6 mmol/2 h magnesium sulfate. You diluted 3 mL magnesium sulfate from a stock vial to 6 mL with sterile water for injection. You then diluted aseptically 5 mL of that solution to 600 mL with 5% dextrose yielding a final concentration of 1.2 g MgSO4/L. (a) What was the %w/v MgSO4 concentration of the stock vial? (b) How much volume of the final solution should the patient receive in 1 h? (c) What was the adjusted flow rate in mL/min? (Answer: (a) 28.8 %;

(b) 300 mL;

(c) 5 mL/min)

15.10. A patient is going to receive a 650 mL dopamine infusion solution at a rate of 65 μg/min over a time period of 24 h. (a) What is the concentration of the drug in the IV infusion solution? (b) Determine the volume that needs to be added to the IV bag (the final volume is diluted to make exactly 650 mL) if dopamine is supplied as 0.480 g/12 mL vials. (c) What is the drip rate in drops/min if the infusion set is calibrated at 60 drops/mL? (Answer: (a) 144 μg/mL;

(b) 2.34 mL;

(c) 27 drops/min)

15.11. You receive an order to administer a drug IV to a patient at a rate of 65 mL/h from a 500 mL bag that contains 130 mg. How much drug per minute would the patient receive? (Answer: 0.282 mg/min)

Appendices

Appendix A. Interpretation of Prescriptions and Medication Orders The important components of a medical prescription are (see also figure below): 1. 2. 3. 4. 5. 6.

Name, registration number, address, and phone number of the prescriber. Name and address of the patient. Date the prescription is issued. The Superscription which is actually the prescription symbol Rx. The Inscription that describes the medication(s) and corresponding dose(s). The Subscription that includes appropriate compounding instructions to the pharmacist and/or the number of dosage forms to be dispensed. 7. The Signa (or Sig.) that includes instructions for the patient, that is, how to take or apply the medication and dose frequency.

© Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1

353

354

Appendices 718-488-1471

Reg. No. 123456 Chris Ioannou Prescriber information

75 Dekalb Ave. Brooklyn, NY 11201 Name:________________________ Address:____________________

Age:_____

Patient information

Date:______

superscription

Rx

inscription

Norvasc 5 mg

subscription

#21

Signa

Sig. I tab p.o. daily

Whenever ingredient quantities in pharmaceutical formulas and medication orders are written without units, the following rules are followed in this book: 1. Solid ingredients are specified in terms of grams or proportional parts. For example, ℞ Prednisone Lactose

0.002 0.300

q.s. ad

The solid ingredients of the prescription above should be considered as gram quantities. Thus, prednisone ¼ 0.002 g and lactose ¼ 0.298 g. 2. Liquids are usually specified in terms of milliliters. For example, Aconite, extract Aqua

0.75 120

The ingredients in the formula are liquids and their corresponding quantities are 0.75 mL of aconite extract and 120 mL of water. These rules are adapted by some of the old medical dispensatories, but it is definitely not an official rule that everybody should follow.

Appendices

355

Appendix B. Miscellaneous Abbreviations Commonly Used in Pharmacy Abbreviation A aa. a.c. Ad a.d. aq. aq. dist. a.s. a.u. b.i.d. cap. d. disp. div. d.t.d. Et fl. ft. gtt. h.s. IM inj. IV M. NS O. or pt o.d. oint. o.l. o.s. OTC o.u. p.c. p.o. p.r.n. Pulv. q. q.am q.d. q.h.

Meaning Before Of each Before meals Up to Right ear Water Distilled water Left ear Each ear Twice daily Capsule Day Dispense Divide Give of such doses And Fluid Make Drop At bedtime Intramuscular Injection Intravenous Mix Normal saline Pint Right eye Ointment Left eye Left eye Over-the-counter Each eye After meals By mouth As needed Powder Every Every morning Every day Every hour (continued)

356

Appendices Abbreviation q8h q.i.d. q.s. q.s. ad. Rx ss. stat. supp. tab. t.i.d. t.i.w. u.d. ung. x

Meaning Every 8 hours Four times a day A sufficient quantity A sufficient quantity to make You take One-half Immediately Suppository Tablet Three times a day Three times a week As directed Ointment Times

Appendix C. Units of Measurement and Equivalent Conversions 1 kilogram (kg) ¼ 1000 grams (g) ¼ 2.2 pounds (lb) 1 lb ¼ 454 g ¼ 0.454 kg 1 ounce (oz) ¼ 28.35 g 1 dram ¼ 3.54 g 1 grain (gr) ¼ 0.065 g 1 scruple ¼ 20 grains ¼ 1.29 g 1 milligram (mg) ¼ 103 g 1 microgram (μg) ¼ 106 g 1 nanogram (ng) ¼ 109 g 1 cm3 ¼ 1 cc ¼ 1 mL ¼ 103 L 1 minim ¼ 0.06 mL 1 fluid dram ¼ 3.70 mL 1 fluid ounce (fξ) ¼ 29.57 mL 1 deciliter (dL) ¼ 0.1 L 1 pint (pt) ¼ 473 mL 1 quart (qt) ¼ 946 mL 1 gallon (gal) (US) ¼ 3785 mL 1 teaspoonful (tsp) ¼ 5 cc 1 tablespoonful (tbs) ¼ 15 cc

Appendices

357

Appendix D. Selected Roman Numerals Roman numerals i ii iii iv v vi vii viii ix x xiv xvi xix xx xxiv xxx xl l lx lxx xc c

Arabic numerals 1 2 3 4 5 6 7 8 9 10 14 16 19 20 24 30 40 50 60 70 90 100

Appendix E. Specific Gravity of Miscellaneous Liquids Liquid Alcohol USP Coal tar Ethyl alcohol Eucalyptus oil Glycerin HCl 37% w/w Isopropyl alcohol Laureth-4 Mineral oil HNO3 68% w/w Olive oil Propylene glycol

Specific gravity (25  C) 0.86 1.13 0.79 0.93 1.25 1.23 0.785 0.95 0.867 1.4 0.92 1.04 (continued)

358

Appendices Liquid H2SO4 96% w/w Witch hazel

Specific gravity (25  C) 1.84 0.98

Appendix F. Table of Atomic Masses Element Actinium (Ac) Aluminum (Al) Americium (Am) Antimony (Sb) Argon (Ar) Arsenic (As) Astatine (At) Barium (Ba) Berkelium (Bk) Beryllium (Be) Bismuth (Bi) Bohrium (Bh) Boron (B) Bromine (Br) Cadmium (Cd) Calcium (Ca) Californium (Cf) Carbon (C) Cerium (Ce) Cesium (Cs) Chlorine (Cl) Chromium (Cr) Cobalt (Co) Copper (Cu) Curium (Cm) Dubnium (Db) Dysprosium (Dy) Einsteinium (Es) Erbium (Er) Europium (Eu) Fermium (Fm) Fluorine (F) Francium (Fr) Gadolinium (Gd)

Atomic number 89 13 95 51 18 33 85 56 97 4 83 107 5 35 48 20 98 6 58 55 17 24 27 29 96 105 66 99 68 63 100 9 87 64

Atomic mass 227.0 27.0 243.0 121.8 39.95 74.9 210.0 137.3 247.0 9.0 209.0 262.0 10.8 79.9 112.4 40.1 249.0 12.0 140.1 132.9 35.45 52.0 58.9 63.55 247.0 260.0 162.5 254.0 167.3 152.0 253.0 19.0 223.0 157.3 (continued)

Appendices

359 Element Gallium (Ga) Germanium (Ge) Gold (Au) Hafnium (Hf) Hassium (Hs) Helium (He) Holmium (Ho) Hydrogen (H) Indium (In) Iodine (I) Iridium (Ir) Iron (Fe) Krypton (Kr) Lanthanum (La) Lawrencium (Lr) Lead (Pb) Lithium (Li) Lutetium (Lu) Magnesium (Mg) Manganese (Mn) Meitnerium (Mt) Mendelevium (Md) Mercury (Hg) Molybdenum (Mo) Neodymium (Nd) Neon (Ne) Neptunium (Np) Nickel (Ni) Niobium (Nb) Nitrogen (N) Nobelium (No) Osmium (Os) Oxygen (O) Palladium (Pd) Phosphorus (P) Platinum (Pt) Plutonium (Pu) Polonium (Po) Potassium (K) Praseodymium (Pr) Promethium (Pm) Protactinium (Pa) Radium (Ra)

Atomic number 31 32 79 72 108 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 109 101 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88

Atomic mass 69.7 72.6 197.0 178.5 265.0 4.0 164.9 1.008 114.8 126.9 192.2 55.85 83.8 138.9 257.0 207.2 6.9 175.0 24.3 54.9 266.0 256.0 200.6 95.9 144.2 20.2 237.0 58.7 92.9 14.0 253.0 190.2 16.0 106.4 31.0 195.1 242.0 210.0 39.1 140.9 147.0 231.0 226.0 (continued)

360

Appendices Element Radon (Rn) Rhenium (Re) Rhodium (Rh) Rubidium (Rb) Ruthenium (Ru) Rutherfordium (Rf) Samarium (Sm) Scandium (Sc) Seaborgium (Sg) Selenium (Se) Silicon (Si) Silver (Ag) Sodium (Na) Strontium (Sr) Sulfur (S) Tantalum (Ta) Technetium (Tc) Tellurium (Te) Terbium (Tb) Thallium (Tl) Thorium (Th) Thulium (Tm) Tin (Sn) Titanium (Ti) Tungsten (W) Uranium (U) Vanadium (V) Xenon (Xe) Ytterbium (Yb) Yttrium (Y) Zinc (Zn) Zirconium (Zr)

Atomic number 86 75 45 37 44 104 62 21 106 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40

Atomic mass 222.0 186.2 102.9 85.5 101.1 257.0 150.4 45.0 263 79.0 28.1 107.9 23.0 87.6 32.1 180.9 99.0 127.6 158.9 204.4 232.0 168.9 118.7 47.9 183.9 238.0 50.9 131.3 173.0 88.9 65.4 91.2

Answers to Multiple-Choice Questions Chapter 1 1.3. (Answer: (a) F; (b) F; (c) F; (d) F; (e) T; (f) F; (g) T; (h) F; (i) F; (j) F; (k) T) 1.4. (Answer: (a) Accurate; (b) precision) 1.7. (Answer: (b) Accurate) 1.9. (Answer: False)

Appendices

Chapter 7 7.1. 7.2. 7.3. 7.4. 7.5. 7.6.

(Answer: Not a drug yet) (Answer: A and B) (Answer: a) (Answer: b and c) (Answer: b) (Answer: c and f)

Chapter 8 8.1. 8.2. 8.3. 8.4. 8.5. 8.7. 8.8. 8.9. 8.10. 8.11. 8.12. 8.13. 8.14. 8.15. 8.16. 8.17. 8.18. 8.19. 8.20. 8.21. 8.22. 8.24. 8.25. 8.29. 8.30. 8.31. 8.32. 8.33. 8.35. 8.36.

(Answer: a and c) (Answer: c, d, and e) (Answer: c) (Answer: b and c) (Answer: a and b) (Answer: a and b) (Answer: a, h, and i) (Answer: c) (Answer: c) (Answer: e) (Answer: b) (Answer: b, d, and e) (Answer: d) (Answer: c) (Answer: c) (Answer: a) (Answer: e) (Answer: c) (Answer: b) (Answer: a) (Answer: d) (Answer: a) (Answer: d) (Answer: e) (Answer: a) (Answer: a, c, and e) (Answer: e) (Answer: b) (Answer: c) (Answer: d and e)

361

362

Appendices

Chapter 10 10.1. 10.2. 10.3. 10.4. 10.5. 10.6. 10.7. 10.8. 10.9. 10.10. 10.11.

(Answer: d) (Answer: c) (Answer: a) (Answer: d) (Answer: c) (Answer: b) (Answer: a) (Answer: c and e) (Answer: b) (Answer: e) (Answer: True: c, e, g, and i; False: a, b, d, f, h, and k)

Chapter 12 12.1. 12.2. 12.3. 12.4. 12.5. 12.6. 12.7. 12.8. 12.9. 12.10. 12.11. 12.16. 12.17. 12.18. 12.19. 12.20. 12.21. 12.26.

(Answer: c, e, and f) (Answer: c) (Answer: b) (Answer: a) (Answer: e) (Answer: d) (Answer: b) (Answer: b) (Answer: d) (Answer: c) (Answer: d) (Answer: a) (Answer: a, b, c, and d) (Answer: e) (Answer: d) (Answer: a) (Answer: e) (Answer: zero order: B, C, E, and F; first order: A, D, G, H, and I)

Chapter 13 13.1. 13.2. 13.3. 13.4.

(Answer: e) (Answer: a and d) (Answer: c) (Answer: d)

Appendices

13.5. 13.6. 13.7. 13.8. 13.10. 13.11. 13.12. 13.13. 13.14. 13.15. 13.16. 13.17. 13.18. 13.21. 13.22.

(Answer: b and d) (Answer: a and d) (Answer: b) (Answer: b) (Answer: c) (Answer: a and e) (Answer: e) (Answer: e) (Answer: e) (Answer: a) (Answer: d) (Answer: e) (Answer: e) (Answer: a) (Answer: d)

Chapter 14 14.10. (Answer: c)

363

Index

A Absorption, Distribution, Metabolism and Elimination (ADME), 305–307, 309 Acacia, 91 Accumulation, 154, 159, 310, 316, 317, 326, 327 Accuracy, vii, 2–4, 12, 28, 40, 56, 166 Acetaminophen, 51, 214 Acetazolamide, 174 Acne, vii, 86, 87, 217 Activity application, 47, 154 definition, 68 Actual values, 2, 8, 9, 26 Advection, 186, 191, 193, 199, 202 Albumin, 111, 175, 223, 224, 243, 322, 349 Aliquot method definition, 29 description, vii for liquids, 32 rules of the method, 147 usefulness, 99 Alligation alternate, vii, 114, 145–150, 153, 154 Alligation medial, 114, 144 Aminophylline, 85, 351 Arsenic, 77, 358 Ascorbic acid, 89 Atomoxetine, 211–213, 215

B Bacitracin, 40, 41, 68 Balance analytical balances, 5, 9, 13, 46, 52, 56 electronic balances, 5, 6, 25, 46, 47, 50

© Springer Nature Switzerland AG 2019 M. Savva, Pharmaceutical Calculations, https://doi.org/10.1007/978-3-030-20335-1

prescription balances, vi, 2–6, 9, 14, 36, 40, 46, 49, 50, 218 torsion mechanical balances, 6 Benzalkonium chloride, 81, 87, 88, 109, 111 Benzoyl peroxide, 85, 87, 217 Betamethasone, 63, 73, 83, 84, 86 Bioavailability, 29, 307, 308, 322, 323 Body surface area (BSA) advantages, 337, 338, 340 calculations, vii, 330–342 derivation, 334, 335 disadvantages, 337, 338, 340 equations, vii, 334, 338 nomograms, 336–338 “normal” BSA, 330, 335, 337 Boric acid, 16, 90, 159, 167–169, 177 Boyd’s formula, 248, 334, 340, 341

C Calcipotriene, 90, 91 Calibration curves, 100, 102 Capsules, vii, 14, 27, 28, 37, 39, 41–43, 45, 47–49, 62, 63, 210–215, 217, 218, 299, 355 Captopril, 110, 111 Catabolism, 220 Cefazolin, 179 Cefuroxime, 177 Cell culture, 172 Chloroquine, 152 Chlorpheniramine, 50, 51 Cholesterol, 82, 115, 117, 150, 158, 234, 245 Chondroitin, 77, 89 Ciprofloxacin, 178, 179

365

366 Clearance (CL), 252, 310, 314, 315, 318, 322, 324, 325, 340 Coal tar, 79, 81, 87, 357 Colligative properties, 65, 160, 169, 222 Colloids, 219, 222–225, 227, 231, 243, 248 Concentration gradient, 186, 189, 190, 192, 198–201, 205, 227, 230, 245 Concentration of mixtures by increasing the mass of drug, 104 by reducing the volume of solution, 101 Convection, 186, 191–193, 199, 202, 205, 226 Creams, vii, 45, 62, 63, 66, 73, 84, 99, 111, 151, 217 Creatinine, 319, 325 Cromoglycate, 174 Crystalloids, 219, 221, 223, 226, 228, 231, 232, 243, 247 Curve fit, 313 Cytochrome P450, 306, 307

D Dehydration, 171, 239, 345 Density application, vii, 57, 58 definition, 56 equation, 56 unit, 55 Desired value, 2, 8, 9, 34, 46 Dextrose 5% (D5W), 231, 239 Diazepam, 175 Diffusion diffusion coefficient, 184, 186, 189, 200, 202, 203 diffusion rate, 189, 192, 199, 202, 204 diffusivity, 184, 186, 188, 200, 202 passive diffusion, 159, 193, 199–201, 205, 206, 226, 230, 233, 245, 305 Dihydrotachysterol (DHT), 152 Dilution, vii, 28, 32, 34, 38, 45, 94–112, 121, 146, 311 Dilution factor (DF), 99–105, 108, 110 Disintegrant, 210 Dissolution, vi, 182–185, 190, 210, 236, 305 Distribution, vi, vii, 185, 199, 248, 252, 262, 305–307, 309–313, 318, 322, 323, 326, 330, 340, 344 Diuretics, 50, 220, 228, 232–235, 237, 244, 245, 248, 350 DNA (plasmid), 111 Dose effective dose, 79, 307, 308, 314 normalizations, 330

Index Drip rate, 347, 349–352 Drug absorption, vi, vii, 158, 189, 191–193, 200, 206, 262, 275, 305–308, 310, 317, 330 analysis, viii biotransformation, 306 concentration, vi, 9, 17, 20–24, 27, 29, 33–35, 38, 41–44, 46, 47, 95, 96, 98, 159, 166, 184, 190, 194, 204, 206, 251, 254, 256, 257, 262, 263, 271–282, 287, 288, 290, 295, 296, 299, 304, 306, 307, 309–313, 316–318, 320, 321, 326, 340, 347, 348 definition, 21 unit, 68 distribution, 305, 306, 309, 310, 318, 322, 323, 330 efficacy, 80, 302 elimination (excretion), 307, 316, 322 metabolism, 306, 307, 322, 330 toxicity, 304, 314 DuBois’s equation, 350 Dyclonine, 108

E ED50, 314 Edema, 171, 222, 228, 230 EDTA, 163, 164 Effervescent, 89, 246 Electrolytes, 66, 160, 161, 220, 312, 343, 345 Elimination, 220, 233, 262, 272–279, 282, 284–287, 295, 297, 305–307, 310, 314–318, 322–324, 326 Epinephrine, 173, 175, 178, 180, 304 Equivalent weight (EW), 65, 66, 235–242, 345 Error absolute error, 24, 33 application, 20, 47 definition, 3, 33 equations, 28 maximum allowable error, vii, 8, 9, 13, 23, 27, 29, 30, 32, 36, 37, 40, 41, 43, 48, 52 relative error, 22–24, 27, 29, 33, 36, 43–45 Eucalyptus oil, 82, 83, 357 Eucerin cream, 217 E-values, 166–168, 171, 173, 175–180 Excretion, 207, 232, 248, 273, 306–308, 316, 322, 330 Expansion factor, 222, 248

Index Exponential equations horizontal shifts, 284, 285, 287, 288 vertical shifts, 284, 285, 287, 288 Exponents, 253, 255, 262, 271, 280, 281, 287, 333, 334

F Fick’s law, 183, 185 First order reaction, 275–277 Flow rate, 343–352 Fluconazole, 177 Fluctuations, 305, 314, 316, 317, 326 Fluids compartments, 221–224, 228, 247 composition, 196, 208, 221–223, 231, 232, 243 gain, 224 loss, 224 regulation, 221–223 replacement, 224, 227, 230–232, 239 shifts, 223, 224, 226, 228, 239 Flux, 186, 188, 192, 202–204 Fractions, vi, 21, 23, 24, 67, 68, 75, 80, 86, 89, 90, 100, 117–119, 133, 150, 252, 256, 281, 295, 308, 311, 312 Freezing point depression, 65, 160, 169, 170, 177 Fujimoto’s formula, 334, 341, 342

G Geometric dilution, vii, 28, 32, 45 Glucosamine, 77, 89 Glycerin, 15, 74, 91, 171, 217, 357 Good Manufacturing Practice (GMP), 20, 24, 27, 46–48 Graphing equations, 257

H Haycock’s formula, 334, 341, 342 HCTZ, 9, 50, 111 HEPES, 112, 175 Homeostasis, 220, 222, 224, 229, 230, 239, 302, 306 Homogenous mixtures, 63 Hydrocodone, 50, 51, 214 Hydrocortisone, 14, 66, 96, 97, 105, 106 Hydrostatic pressure, 172, 194, 206, 222 Hypernatremia, 230, 239 Hyperosmotic, 160, 173, 206 Hypertonic solutions, 65, 159, 160, 165, 170, 171, 226, 227, 244

367 Hypocalcemia, 246 Hypokalemia, 14, 50, 226, 239, 243, 247 Hyponatremia, 240, 247 Hypoosmotic, 160, 173, 206 Hypotonic solutions, 65, 159, 165–169, 173, 239

I Infusion accessories, 344–352 Isopropyl alcohol (IPA), 81, 91, 108, 123, 145, 357 Isotonic glucose, 162, 227 Isotonic saline, 16, 63, 65, 87, 108, 169, 227, 228, 231, 243, 244, 249, 341 Isotonic solutions, vii, 158–180, 222, 225–228, 247, 347

L Lead (Pb), 46, 67, 86, 224, 304, 308, 330, 335, 338, 359 Least weighable quantity (LWQ), 46 Levigation, 217 Lidocaine, 112, 151 Lipoic acid, 77, 89 Lithium (Li), 111, 246, 248, 359 Loading dose (DL), 302, 318 Logarithms, 251, 260–262, 264, 265, 276, 278, 288, 325 Lotions, 15, 58, 59, 63, 79, 88, 91, 92 Lugol’s solution, 152 Lysine, 112

M Magnesium stearate, 210 Maintenance dose (DM), 302, 318, 337 Mannitol, 233, 234, 245, 248, 350 Mass balance equation, 94–96, 105, 106, 121–123, 125, 148, 315, 348 Mass preservation, 95, 106, 122, 124, 311 Measured values, vi, 2, 7, 9 Measurements, vii, 2–18, 20–23, 26–28, 30, 32, 34, 35, 37, 41, 42, 44, 46, 52, 254, 330, 334, 338, 356 Mefenamic acid, 100, 102, 103, 110, 163–165 mEq, see Equivalent weight (EW) Mercury (Hg), 67, 82, 195, 359 Metabolism, 49, 227, 295, 296, 305–307, 316, 321, 322, 330, 340 Methotrexate, 340 Metronidazole, 152

368 Midazolam, 176 Milk of magnesia, 86 Mineral oil, 59, 60, 88, 91, 357 Minocycline, 92 Mixture problems, 114–155 Molality (M) definition, 64, 65 Mole fraction (X) definition, 67 Molecular dispersions, 96, 182, 185, 305 Mosteller’s formula, 334, 340, 342

N Nafcillin, 83 Nasal, 51, 87, 88, 304 Neomycin, 175 Nifedipine, 83, 85 Nomograms, 336–338 Normality (N) definition, 65 Noyes-Whitney equation, 183

O Obesity, 319, 330, 342 Ointments, vii, 14, 40, 45, 62, 63, 76, 82, 83, 85, 86, 96, 97, 99, 105–109, 112, 151, 355, 356 One compartment model, 309, 310, 313, 315, 318, 323, 327 Ophthalmic solutions, 159, 167, 171, 175 Optical density (OD), 102–104, 110 Order of operations, 252, 255 Order of reaction, 270, 271, 291, 299 Osmolarity definition, 160–161 physiological value, 166 unit, 160 Osmosis, 158, 159, 172, 198, 199, 202, 206, 224, 239 Osmotic pressure, vii, 65, 158–162, 165–171, 194–198, 207, 208, 222, 230, 243

P Parabens (methyl, propyl), 87 Parallax error, 9 Parts per billion (ppb) definition, 67 Parts per million (ppm) definition, 67 Percent by volume definition, 64

Index Percent by weight definition, 63 Percent weight per volume definition, 64 Permeability, 158, 159, 186–189, 193, 200–203, 205, 222, 304 Petrolatum album, 40, 96 Pharmacodynamics, 302, 335 Pharmacokinetics, v–viii, 193, 248, 251–266, 275, 276, 302, 308–311, 315, 317, 323, 335, 340 Pharmacological response, 79, 233, 304, 307, 310, 311, 321 Phosphorous, 249 Physostigmine, 109 Pilocarpine, 109 Plasma expanders, 222, 228, 232, 244, 245 Polymyxin B, 341 Powders, 4, 9, 12, 14, 17, 28–30, 32–35, 37–39, 42, 43, 45, 46, 49, 50, 52, 53, 63, 78, 89, 99, 110, 121–123, 152, 153, 176, 211–214, 216, 217, 249, 341, 349, 355 Pralidoxime, 174 Precision, vii, 2–4, 9, 12, 15 Prednisone, 30, 32, 37, 39, 48, 354 Proparacaine, 176 Propylene glycol, 15, 91, 92, 319, 357 Psoriasis, vii, 84, 90 Pycnometer, 56, 57

Q Quetiapine, 211, 212, 215

R Raoult’s law, 118, 150 Rate constants, 270–278, 280–282, 284, 286, 288, 290, 293, 294, 296, 310, 315, 316, 323, 324, 326 Rate of reactions, 270, 271, 291, 299 Ratio strength definition, 66 Receptors, vi, 158, 189, 193, 226, 233, 303, 304, 307, 308, 321, 322 Risperidone, 211, 212, 215 Rounding, 252, 319, 346, 351

S Salicylic acid, 59, 60, 83, 86, 88 Salt factor, vii, 62–92, 308 Scopolamine, 167–169, 177

Index Selectivity, 303, 322 Sensitivity requirement (SR) definition, 6, 9, 12 equations, 8 Serial dilutions, 99–104, 111 Signal transduction, 303 Slope, 103, 104, 188, 257–259, 262–265, 272–278, 280, 287, 290 Sodium chloride equivalent, 165–171, 173, 175 Solutes, 62–65, 67, 74, 78, 94, 101, 114, 122, 146, 159, 161, 169, 170, 172, 173, 182–186, 192–200, 202, 206, 208, 223–226, 228, 230, 245 Solution, vi, 9, 24, 58, 63, 94, 114, 158, 182, 212, 221, 252, 270, 304, 331, 343 Solvent, 62–65, 67, 75, 94, 105, 114, 122, 146, 159, 170, 182, 184–186, 191, 194, 198–201, 206, 238 Specific gravity application, 57, 58 definition, 56 equation, 56 Specificity, 304, 322 Standard curves, 100, 102, 103, 110 Steady-state concentration, 317–320, 326 Sulfuric acid, 75, 76, 90, 151, 163 Suppository, 49, 218, 356 Suramin, 153 Suspensions, 210, 215, 216, 341 System of equations, 44, 115, 116, 118–121, 124, 132, 133, 150, 151, 153, 256

T Tablets, 2, 62, 63, 119, 205, 210, 246, 308, 356 Tablespoonful (tbs), 82, 356 Tablet strength, 211, 212, 214–218, 249, 320 Tacalcitol, 90, 91 Talc, 14, 58–60, 88, 91, 92 TD50, 314

369 Tetracaine, 151 Therapeutic index (TI), 20, 305, 306, 313, 314, 320, 335, 343 Therapeutic range, 316–318, 326 Tonicity, vii, 65, 159, 160, 166, 169–172, 175, 178, 199, 226, 243, 248, 249 Total body water (TBW), 221, 240, 339 Transdermal, 20, 85, 153, 304 Transport, 158, 184, 186, 191–193, 198, 200, 201, 205, 220, 226, 230, 240, 245, 304 Transporters, 158, 192, 193, 200, 206, 207, 226, 232, 233, 307, 308, 322 Tris, 108, 173 Tumor Necrosis Factor (TNF), 110

V Valproic acid, 175 van’t Hoff dissociation factor, 160–165 equation, 161, 162, 166, 194 Vecuronium, 152 Volume of distribution, 248, 252, 305, 306, 310–314, 318, 330, 340

W Witch hazel, 83, 358

X Xenobiotics, 306 Xylometazoline, 87, 88

Z Zero-order reactions, 272, 273, 277 Zinc oxide (ZnO), 15, 59, 60, 88, 91, 92