THE APhA COMPLETE REVIEW FOR PHARMACY, 7th Ed

1. Pharmacy Math - Hassan Almoazen, PhD

1-1. Units of Measure

Introduction

Calculations in pharmacy may involve four different systems of measure: the metric system, the apothecaries' system, the avoirdupois system, and the household system.

Metric System

The fundamental units of the metric system are the gram, the liter, and the meter. Prefixes are used extensively to express quantities much greater and much less than the fundamental units. Some of the most commonly used prefixes are provided in

Table 1-1.

Apothecaries' System

Although the metric system is the official system of measure for pharmacy today, the apothecaries' system is the traditional system, and some elements might sometimes be found in prescriptions. Units of the apothecaries' system are presented in

Table 1-2.

Avoirdupois System

The avoirdupois system of measure for weight is used in ordinary commerce. Here, the ounce corresponds to 437.5 grains. The avoirdupois grain unit is equal to the apothecaries' grain unit. Sixteen ounces (7,000 grains) correspond to 1 pound. Note that the avoirdupois ounce (473.5 grains) and pound (7,000 grains) measures are not equal to the apothecaries' ounce (480 grains) and pound (5,760 grains) measures.

Household Measures

A tablespoon is equivalent to 15 mL, and a teaspoon is equivalent to 5 mL.

Conversion Factors

A short list of convenient conversion factors follows:

1 inch = 2.54 cm

1 fl oz = 29.57 mL

1 g = 15.4 grains

1 kg = 2.20 lb (avoirdupois)

1 lb (avoirdupois) = 454 g

1 gal (U.S.) = 3,785 mL

1-2. Significant Figures

Introduction

All measured quantities are approximations. The accuracy of a given measurement is conveyed by the number of figures that are recorded. The number of significant figures in a measurement includes the first approximate figure. The last recorded digit to the right of a measured quantity is taken to be an approximation. For example, the weight 13.24 g has four significant figures, and the final digit, 4, is approximate. Calculations should be conducted so as to carry the correct numbers of significant figures.

Frequently, the different quantities in a given calculation have different numbers of significant figures. When that occurs, the following rules apply.

Addition and Subtraction

When adding or subtracting decimal numbers, round all measurements so that they have the same number of decimal places as the least in the set. For example, 13.78 mL and 53.5 mL would be added as 13.8 mL + 53.5 mL = 67.3 mL, using and retaining only one decimal place.

[Table 1-1. Metric System Prefixes]

Multiplication and Division

When multiplying or dividing decimal numbers, round the measurements to include the number of significant figures contained in the least accurate number. For example, 25.678 mL × 1.24 g/mL would be multiplied as 25.7 mL × 1.24 g/mL = 31.9 g, using and retaining three significant figures in each number.

Handling Zero

The digit zero may or may not be counted as a significant figure, depending on where it appears in the measured number. If zero occurs at an interior position in the number (e.g., 3,052 or 2.031), it is significant. If zero occurs as the last digit to the right of the decimal (e.g., 44.50), it is significant. If zero occurs as the first digit to the right of the decimal in a number that is less than 1, it is not significant. For example, in 0.078 there are only two significant figures. If zero occurs as the last digit, or digits, in a whole number (i.e., no decimal is expressed), its significance is unknown without further information. For example, 3,500 might have two, three, or four significant figures.

[Table 1-2. Apothecaries' System of Measure]

1-3. Ratios and Proportions

The most frequently encountered calculations in pharmacy use ratios and proportions. In mathematics, a ratio is the quotient of one quantity divided by another quantity of the same kind, and a proportion is an equality between ratios. Thus, a proportion involves a relationship among four quantities. You can always solve for one of those quantities when the other three are known. If the ratio x/y is equal to the ratio a/b, then the proportion x/y = a/b exists, and x can be obtained by algebraic manipulation (x = ay/b, etc.). Such problems frequently are encountered when adjusting dosages.

One common source of error in proportion problems involves writing one of the ratios upside down (e.g., writing x/y = b/a, when it should be written x/y = a/b). A disciplined approach to setting up such problems can help. For example, you might establish a rule in which you express each ratio as the quotient of like quantities. Then, when you equate the two ratios, if the numerator of one ratio is smaller (or larger) than its denominator, the same should be true of the other ratio as well.

Another source of error involves using mixed units (e.g., using one number expressed in grams and the other in milligrams). To guard against that kind of error, always write the units into the equation along with the numbers. All unit expressions should cancel except those required for the quantity being solved for (dimensional analysis).

Example: If 300 mL of a preparation contains 250 mg of drug, what weight of drug (x) is contained in 1,800 mL of the preparation?

Equate the ratios x:250 mg and 1,800 mL:300 mL to solve for x:

000047

Note that because the new volume is six times greater, the new weight should be six times greater as well.

1-4. Specific Gravity and Density

At times, you will be required to convert a volume measure to a weight measure, or vice versa. To do so, you will need to use either the specific gravity or the density of the material. The specific gravity (SpGr) is a ratio of the weight of the material to the weight of the same volume of a standard material. For liquids, the standard material is water, which has a density of 1 g/mL. Specific gravity is unitless. Density is the quotient of any measure of the weight of a sample of the material divided by any measure of the volume of the sample. The units must be explicitly expressed (e.g., g/mL, lb/gal, etc.). When density is expressed in grams per milliliter, it is numerically equal to specific gravity. Algebraically, density (weight/volume) is easier to work with than the corresponding expression for specific gravity.

Example: What is the weight of 750 mL of concentrated hydrochloric acid (SpGr = 1.20)? From the specific gravity definition, and using the volume of 750 mL, 000085

Rearranging and noting that the density of H2O is 1 g/mL,

weight of 750 mL HCl =

(SpGr)

× (weight of 750 mL H
2O)

= 1.20 × 750 g = 900 g

 

Alternatively, specific gravity is numerically equal to density expressed in grams per milliliter. Thus,

000099

1-5. Percentage Error

Because all measurements are approximations, one must characterize the extent of error involved, or percentage of error, which is defined as

000121

The term error in the numerator indicates the maximum potential error in the measurement (error = larger quantity - smaller quantity), while the term quantity desired in the denominator represents the total amount measured. Percentage of error may be calculated for either a weight or a volume measurement.

Example: A quantity of material weighs 5.810 g on a prescription balance. Using a much more accurate analytical balance, the quantity weighs 5.893 g. What is the percentage of error for the original weighing?

error = 5.893g - 5.810g = 0.083g

The quantity desired is 5.810 g. Thus,

000142

Example: Suppose you wish to weigh out 75.0 mg of an ingredient but mistakenly weigh 65.0 mg instead. Based on the quantity desired, what is the percentage of error?

000157

1-6. Minimum Measurable

By regulation, weighings by a pharmacist cannot exceed a percentage of error greater than 5%, which requires that the sensitivity of the balance be known and limits the smallest quantity that can be weighed. Balance sensitivity is defined in terms of the sensitivity requirement (SR), which is the weight of material that will move the indicator one marked unit on the index plate of the balance. For a class A prescription balance, SR = 6 mg. The minimum weighable quantity for a given balance can be calculated using the percentage of error formula: replace the "error" term with the SR (e.g., 6 mg), replace the percentage of error term with 5%, and replace the "quantity desired" term with "minimum weighable quantity" as follows:

000172

Example: For a balance that has an SR of 4 mg, what is the minimum weighable quantity to ensure a percentage of error no greater than 2%?

000182

Example: What is the SR for a balance with a percentage of error of 5% when weighing 120 mg?

000193

1-7. Patient-Specific Dosage Calculations

Introduction

Drugs with a narrow therapeutic range often are dosed on the basis of patient weight or body surface area. For patients with renal impairment, some drugs are dosed on the basis of creatinine clearance.

Dosing Based on Body Weight

Weight-based dosing might involve using the patient's actual body weight (ABW), ideal body weight (IBW), or perhaps an adjusted ideal body weight that is a function of IBW and ABW. Those weights are invariably expressed in kilograms.

Example: A patient weighing 180 lb is to receive 0.25 mg/kg per day amphotericin B (reconstituted and diluted to 0.100 mg/mL) by intravenous (IV) infusion. What volume of solution is required to deliver the daily dose?

000202

A commonly used equation for calculating IBW is

IBW (kg) =

(sex factor)

+ (2.3 × height in inches over 5 feet),

 

where the sex factor for males is 50.0 and the sex factor for females is 45.5.

Example: The recommended adult daily dosage for patients with normal renal function for tobramycin is 3 mg/kg IBW given in three evenly divided doses. What would each injection be for a male patient who weighs 185 lb and is 5 feet, 9 inches tall?

IBW (kg)

= 50 + (2.3 × 9) = 50 + 21

= 71 kg (ignore ABW of 84 kg)

 

daily dose (tid) = 71 kg × 3 mg/kg = 213 mg (per day)

Thus, each injection = 213 mg/3 = 71 mg

In the absence of other information, the usual drug doses are considered generally suitable for 70-kg individuals. Thus, in the absence of more specific information, an adjusted dosage for a notably larger or smaller individual may be obtained by multiplying the usual dose by the ratio of patient weight to 70 kg (Clark's rule).

Example: If the adult dose of a drug is 100 mg and no child-specific dosing information is available, what would the weight-adjusted dose be for a child who weighs 40 kg?

000088

Dosing Based on Body Surface Area

Dosing based on body surface area requires an estimation of the patient's body surface area (BSA) expressed in square meters (m2). That parameter might be estimated from a nomogram using height and weight or, for adults, by using one of several equations such as:

000066

Example: What is the computed BSA for an adult who weighs 88 kg and is 5 feet, 10 inches tall?

000236

The average adult BSA is taken to be 1.73 m2. That value can be used to obtain an approximate child's dose, given the usual dose for an adult and the child's estimated BSA.

Example: If the adult dose of a drug is 50 mg, what would the BSA-adjusted dose be for a child having an estimated BSA of 0.55 m2?

000249

Dosing Based on Creatinine Clearance

For many drugs, the rate of elimination depends on kidney function. Creatinine clearance (CrCl) is a measure of the volume of blood plasma that is cleared of creatinine by kidney filtration per minute, and it is expressed in milliliters per minute. Creatinine clearance can be calculated using the Cockcroft-Gault equation as a function of patient sex, age, body weight, and serum creatinine.

For males:

000258

For females:

CrCl = 0.85 × CrCl for males

Example: Using the Cockcroft-Gault equation, calculate the creatinine clearance rate for a 76-year-old female weighing 65 kg and having a serum creatinine of 0.52 mg/dL.

000031

Alternatively, creatinine clearance may be estimated using the Jelliffe equations.

For males:

000149

For females:

CrCl = 0.9 × CrCl for males

The normal value for creatinine clearance is taken to be 100 mL/min. Sometimes one must adjust CrCl to the patient's BSA, which is calculated as:

000078

The maintenance dose for some drugs is based on IBW and CrCl. See Chapter 6 for more detailed information.

1-8. Using Batch Preparation Formulas

The relative amounts of ingredients in a pharmaceutical product are specified in a formula. A pharmacist may be required to reduce or enlarge the formula to prepare a lesser or greater amount of product. A given formula might specify either the actual amount (weight or volume) of each ingredient for a specified total amount of product or just the relative amount (part) of each ingredient. In the latter case, the ingredients must all be of the same measure (e.g., weight in grams).

Example: From the following lotion formula, calculate the quantity of triethanolamine required to make 200 mL of lotion.

Triethanolamine

10 mL

 

Oleic acid

25 mL

 

Benzyl benzoate

250 mL

 

Water to make

1,000 mL

 

000097

Example: From the following formula, calculate the quantity of chlorpheniramine maleate required to make 500 g of product.

Chlorpheniramine maleate

6 parts

 

Phenindamine

20 parts

 

Phenylpropanolamine HCl

55 parts

 

Note that the formula will give a total of 81 parts, which will correspond to the desired quantity of 500 g. Then,

000101

1-9. Conventions in Expression of Concentration

Introduction

In pharmacy, one will encounter a diversity of conventions for expressing drug concentrations. One must be prepared to calculate drug concentration units directly from their definitions and to interconvert among them.

Percentage Strength

Percentage, strictly speaking, specifies the number of parts per 100 parts. In pharmacy, percentage comes in three varieties:

• Percent weight-in-weight = %(w/w) = grams of ingredient in 100 grams of product (assumed for mixtures of solids and semisolids)

• Percent volume-in-volume = %(v/v) = milliliters of ingredient in 100 milliliters of product (assumed for solutions or mixtures of liquids)

• Percent weight-in-volume = %(w/v) = grams of ingredient in 100 milliliters of product (assumed for solutions of solids in liquids)

Example: What is the concentration in %(w/v) for a preparation containing 250 mg of drug in 50 mL of solution? Note that %(w/v) is defined as g/100 mL. Thus,

000210

Parts (Ratio Strength)

Concentrations may be expressed in "parts" or ratio strength when the active ingredient is highly diluted. Assumptions concerning (w/w), (v/v), and (w/v) ratios are identical to those for percentages.

Example: What is the concentration in %(v/v) of a solution that has a ratio strength of 1:2,500 (v/v)?

000212

Millimoles

By definition, a 1 molar solution contains 1 gram molecular weight (1 GMW = 1 mole = weight in grams of Avogadro's number of particles) per liter of solution. The molarity expresses the number of moles per liter. The millimolarity (millimoles/liter) is 1,000 times the molarity of a solution.

Example: What is the millimolar concentration of a solution consisting of 0.90 g of sodium chloride (GMW = 58.5) in 100 mL of water? The quantity of 0.9 g in 100 mL corresponds to 9.0 g in 1,000 mL.

000214

Milliequivalents

By definition, the equivalent weight of an ion is the atomic or formula weight of the ion divided by the absolute value of its valence. Thus, the equivalent weight of ferric ion, Fe3+ (atomic weight 55.9, valence 3), is 18.6. A milliequivalent is 1,000th of an equivalent weight (i.e., 1,000 milliequivalent weights equal 1 equivalent weight). For a molecule, the equivalent weight is obtained as the GMW (formula weight) divided by the total cation or the total anion charge. For example, the equivalent weight of MgCl2 (atomic weight of Mg2+ = 24.3, with a valence of +2; atomic weight of Cl1- = 35.5, with a valence of -1) is (24.3 + 2 × 35.5)/2 = 47.7 g. Its milliequivalent weight is 0.0477 g, or 47.7 mg. In the case of a nondissociating (nonionizing) molecule (e.g., dextrose or tobramycin), the equivalent weight is equal to the formula weight.

Example: What is the concentration, in milliequivalents per liter, of a solution containing 14.9 g of KCl (GMW = 74.5 g) in 1 liter? Note that the valence of potassium is 1+, so the equivalent weight equals the molecular weight. Accordingly,

000217

Example: What weight of MgSO4 (GMW = 120) is required to prepare 1 liter of a solution that is 25.0 mEq/L in Mg2+? One must have 25.0 mEq of MgSO4 to obtain 25.0 mEq of Mg. The valence of Mg (and total positive charge) is 2+; therefore, the equivalent weight of MgSO4 is 120/2 = 60 g. Accordingly, 60 mg corresponds to 1 mEq of MgSO4. Then,

000219

Example: How many milliequivalents of Ca2+ are contained in 100 mL of a solution that is 5.0%(w/v) in CaCl2 (GMW = 111, atomic weight of Ca2+ = 40, atomic weight of Cl- = 35.5)? Note that the valence of calcium is 2+. The solution contains 5.0 g CaCl2 per 100 mL, which corresponds to (5.0 g/111) × 2 = 0.090 equivalents of CaCl2, as well as Ca2+. Accordingly, 100 mL of the solution contains 90 mEq of Ca2+.

Milliosmoles

Osmotic concentration is a measure of the total number of particles in solution and is expressed in milliosmoles (mOsm). Thus, the number of milliosmoles is based on the total number of cations and the total number of anions. The milliosmolarity of a solution is the number of milliosmoles per liter of solution (mOsm/L), where:

 

× 1,000 moles

 

number of species =

number of ionic species upon complete dissociation (e.g., dextrose = 1 specie, NaCl = 2 species, MgCl2)

 

The total osmolarity of a solution is the sum of the osmolarities of the solute components of the solution. When calculating osmolarities, in the absence of other information, assume that salts (e.g., NaCl, etc.) dissociate completely (referred to as the ideal osmolarity). You should be aware of the distinction between the terms milliosmolarity (milliosmoles per liter of solution) and milliosmolality (milliosmoles per kilogram of solution).

Example: What is the concentration, in milliosmoles per liter, of a solution that contains 224 mg of KCl (GMW = 74.6 g) and 234 mg of NaCl (GMW = 58.5) in 500 mL? What is the number of milliosmoles per liter of K+ alone?

000221

Milligrams per 100 Milliliters and Milligrams per Deciliter

Traditionally, some lab test values are reported as the number of milligrams per 100 milliliters (mg%) or, equivalently, milligrams per deciliter (mg/dL).

Example: What is the %(w/v) concentration of glucose in a patient with a blood glucose reading of 230 mg/dL? Note that 1 dL = 100 mL and that 230 mg = 0.230 g. Then,

000223

"Units" and Micrograms per Milligram

The concentrations for some drugs whose production involves incomplete isolation from natural sources might be expressed in terms of "units" of activity or micrograms per milligram (mcg/mg) as determined by a standardized bioassay.

Example: A preparation of penicillin G sodium contains 2.2 mEq of sodium (atomic weight = 23, valence = 1+) per 1 million units of penicillin. How many milligrams of sodium are contained in an IV infusion of 5 million units? The 5 million unit dose will contain 5 × 2.2 mEq = 11.0 mEq.

mEq weight of sodium

= 23 mg

 

weight of sodium

= (23 mg/mEq) × 11.0 mEq

= 253 mg

 

Parts per Million and Parts per Billion

Very low concentrations often are expressed in terms of parts per million (ppm) (the number of parts of ingredient per million parts of mixture or solution) or parts per billion (ppb) (the number of parts of ingredient per billion parts of mixture or solution). Thus, ppm and ppb are special cases of ratio strength concentrations.

Example: Re-express 1:25,000 in terms of parts per million.

000224

(i.e., the concentration is 40 ppm)

1-10. Dilutions and Concentrations

Simple Dilutions

In simple dilutions, a desired drug concentration is obtained by adding more solvent (or diluent) to an existing solution or mixture. Mathematically, the key feature of this process is that the initial and final amount of drug present remain unchanged. The amount of drug in any solution is proportional to the concentration times the quantity of the solution. Thus, taking the initial concentration as C1, the initial quantity of solution as Q1, the final concentration as C2, and the final quantity of solution as Q2, one has the relationship: C1 × Q1 = C2 × Q2. When provided with values for any three of those variables, the fourth variable can be calculated. (Because the equation can be rearranged to C1/C2 = Q2/Q1, it sometimes is referred to as an inverse proportionality.)

Example: How much water should be added to 250 mL of a solution of 0.20%(w/v) benzalkonium chloride to make a 0.050 %(w/v) solution?

000225

Alcohol Solutions

The preceding treatment of dilutions assumes that solution and solvent volumes are reasonably additive. For dilutions of concentrated ethyl alcohol in water, that is not the case, because a contraction in volume occurs on mixing. Consequently, you cannot extend the calculation to determine the exact volume of water to add to the initial alcohol solution. That is, the volume of water to be added cannot be obtained simply as Q2 - Q1. Rather, you can specify only that sufficient water be added to the initial concentrated alcohol solution (Q1) to reach the specified or calculated final volume (Q2) of the diluted alcohol solution.

Example: How much water should be added to 100 mL of 95%(v/v) ethanol to make 50%(v/v) ethanol?

000227

Thus, to the 100 mL of 95%(v/v) ethanol add sufficient water to make 190 mL. The sufficient quantity of water will be more than 90 mL because of the contraction that occurs when concentrated alcohol is mixed with water.

Concentrated Acids

Concentrated mineral acids (hydrochloric, sulfuric, nitric, and phosphoric) are manufactured by bubbling the pure acid gas into water to produce a saturated solution. The manufacturer specifies the concentration as a %(w/w). However, when preparing diluted acids for compounding, the pharmacist must express the concentration as a %(w/v), which requires use of the specific gravity of the concentrated acid.

Example: What volume of 35%(w/w) concentrated HCl (SpGr 1.20) is required to make 500 mL of 5%(w/v) solution?

First, determine the weight of HCl required for the dilute solution. Because the dilute solution is 5%(w/v), it will contain 5 g in each 100 mL, or 25 g in 500 mL.

Next, one must determine what weight, x, of the 35%(w/w) solution contains 25 g of HCl. By proportion,

000228

Finally, use the specific gravity of the concentrated solution to convert the weight to volume. Here, recall that specific gravity is numerically equal to density when the latter is expressed in grams per milliliter. Thus,

density = 1.20 g/mL = weight/volume

Rearranging,

000229

Triturations

Triturations are simply 10%(w/w) finely powdered (triturated) mixtures of a drug in an inert substance.

Example: What weight of colchicine trituration is required to prepare 30 doses of 0.25 mg each of colchicine?

For the trituration, 10 mg of the mixture contains 1 mg of drug. Thus,

000232

Alligation Alternate Method

Sometimes a drug concentration is required that is in between the concentrations of two (or more) stock solutions (or available drug products). In that case, the alligation alternate method may be used to quickly obtain the relative parts of each of the stock solutions needed to yield the desired concentration. If stock solutions of concentrations A% and B% (A% > B%) are to be used to make a solution of concentration C%, one sets up the following diagram to obtain the relative parts of solutions A and B.

 

(C% - B%) parts of A

 
 

C%

 
 

B%

 

(A% - C%) parts of B

 

Example: In what proportion should 20%(w/v) dextrose be mixed with 5%(w/v) dextrose to obtain 15%(w/v) dextrose? How much of each is required to make 75 mL of 15%(w/v) solution?

 

10 parts of 20%

 
 

15%

 
 

5%

 

5 parts of 5%

 

Thus, combine the stock solutions in the ratio of 10 parts of 20%(w/v) dextrose:5 parts of 5%(w/v) dextrose (i.e., 2:1). Accordingly, to make 75 mL of 15% solution, mix 50 mL of 20% solution with 25 mL of 5% solution.

Alligation Medial Method

Sometimes one may need to know the final concentration of a solution obtained by mixing specified volumes of two or more stock solutions. In that case, the alligation medial method may be used.

Example: What is the concentration of a solution prepared by combining 100 mL of a 10% solution, 200 mL of a 20% solution, and 300 mL of a 30% solution? Proceed as follows:

000233

1-11. Isotonic Solutions

Introduction

In pharmacy, the preparation of many solutions requires attention to osmotic pressure, a colligative property that is especially relevant for membrane transport. Other colligative properties include freezing point depression and boiling point elevation. Those properties are a function of the total number of particles dissolved in the solution, regardless of the identity of the particles. Here, the term particles corresponds to cations, anions, and neutral undissociated molecules. A solution that has the same osmotic pressure as bodily fluids (blood or tears) is said to be isotonic (and isosmotic). As points of reference, 5.0%(w/v) dextrose, a nondissociating molecule, and 0.9%(w/v) sodium chloride, a dissociating molecule, are isotonic.

Dissociating Solutes

Preparing solutions of specific tonicities requires knowledge of the dissociation properties of the solutes involved. One must know if the solute in question dissociates and, if so, to what extent and into how many particles. For example, in weak solutions, sodium chloride dissociates about 80% into two particles, yielding a solution containing Na+ ions, Cl- ions, and undissociated NaCl molecules. A measure of the extent of dissociation is provided by the dissociation factor, i, which is defined as the ratio of the total number of particles following dissociation to the number of molecules prior to dissociation. For example, 100 molecules of sodium chloride (prior to dissociation) will dissociate 80% to produce 80 particles of Na+, 80 particles of Cl-, and 20 particles of NaCl, or 180 particles in all. The dissociation factor, i, for NaCl, then, is 180/100 = 1.8. A nondissociating molecule (such as dextrose or tobramycin) is assigned a dissociation constant of 1.0. If measured dissociation information is not available, one can assume approximately 80% dissociation for weak solutions of salts. In that case, salts (including drugs) that dissociate into two ions (such as sodium chloride and ephedrine hydrochloride) will have a dissociation factor of 1.8; salts that dissociate into three ions (such as ephedrine sulfate) will have a dissociation factor of 2.6; salts that dissociate into four ions (such as sodium citrate) will have a dissociation factor of 3.4.

Example: What is the dissociation factor (i) for a compound that dissociates 60% into three ions?

For each 100 undissolved molecules, one will obtain the following on dissolution:

60 × 3 =

180 particles of ions + 40 particles of undissociated molecules = a total of 220 particles

 

Thus, the dissociation factor = i = 220/100 = 2.2.

Example: What is the dissociation factor for dextrose, a nondissociating compound?

For each 100 undissolved molecules, one will obtain the following on dissolution:

0 particles of ions + 100 particles of undissociated molecules = a total of 100 particles

Thus, the dissociation factor = i = 100/100 = 1.0.

Sodium Chloride Equivalents or E value

When preparing isotonic drug solutions, one must take the tonicity contribution of the drug into consideration. That can be accomplished by using the sodium chloride equivalent (E value) for the drug, which is defined as the number of grams of sodium chloride that would produce the same tonicity effect as 1 gram of the drug. If the value of the sodium chloride equivalent is not provided, it can be calculated using the molecular weights (MWs) and dissociation factors of sodium chloride and the drug in question:

000234

Example: What is the sodium chloride equivalent of demecarium bromide (GMW = 717, i = 2.6)?

000073

Thus, each gram of demecarium bromide is equivalent to 0.12 g sodium chloride. So how does one proceed to prepare a drug solution that must be made isotonic? Using the total volume of isotonic solution to be prepared, first calculate the hypothetical weight, x, of sodium chloride (alone) that would be required to make that volume of water isotonic (0.9%). Next, using the weight of drug to be incorporated in the solution and its sodium chloride equivalent, calculate the weight of sodium chloride, y, that would correspond to the weight of the drug. Then, calculate the true weight of sodium chloride, z, to be added to the preparation as z = x - y.

Example: What weight of sodium chloride would be required to prepare 50 mL of an isotonic solution containing 500 mg of pilocarpine nitrate (sodium chloride equivalent = 0.23)?

Because isotonic saline requires 0.9 g/100 mL, 50 mL of isotonic saline will require 0.45 g (i.e., x = 0.45 g). The 500 mg of pilocarpine nitrate will correspond to 500 mg × 0.23 = 115 mg sodium chloride (i.e., y = 0.12 g). Thus, the weight of sodium chloride needed to make an isotonic solution = z = x - y = 0.45 g - 0.12 g = 0.33 g.

Example: A (fictitious) new drug, Utopical (MW = 175, dissociation factor i = 3.4), is to be provided as 325 mg in 60 mL of solution made isotonic with sodium chloride. What is the required weight of sodium chloride?

Here, the sodium chloride equivalent of the drug is not given and must be calculated from the information provided.

000213

Because isotonic saline requires 0.9 g/100 mL, 60 mL of isotonic saline will require 0.54 g (i.e., x = 0.54 g). The 325 mg of Utopical will correspond to 0.325 g × 0.63 = 0.20 g of sodium chloride (i.e., y = 0.20 g). Thus, the weight of sodium chloride needed to make an isotonic solution = z = (x - y) = (0.54 g - 0.20 g) = 0.34 g.

1-12. Intravenous Infusion Flow Rates

A physician may specify the rate of flow of IV fluids in drops per minute, amount of drug per hour, or the duration of time of administration of the total volume of the infusion. Therefore, to program an infusion pump to give the medication at the correct rate, one may need to calculate the infusion rate in per minute or per hour increments.

Example: If 250 mg of a drug is added to a 500-mL D5W bag, what should the flow rate be, in milliliters per hour, to deliver 50 mg of drug per hour?

000226

Example: If an infusion flow rate is 100 mL/h and the infusion set delivers 15 drops/mL, what is the rate of flow in drops per minute?

15 drops/mL = × 100 mL/h

= 1,500 drops/h

 
 

= 25 drops/min

 

Example: If 500 mL of an infusion is to be delivered using an IV administration set that delivers 10 drops/mL and the flow rate is set at 1.25 mL/min, how long will it take to deliver the 500 mL?

000235

(The 10 drops/mL is superfluous information.)

1-13. Buffers

Buffer solutions are used to reduce pH fluctuations associated with introduction of small amounts of strong acids or bases. Typical buffer solutions are composed of a weak acid or weak base plus a salt of the acid or base. Solution pH in the presence of a buffer can be calculated using the Henderson-Hasselbalch equations.

For weak acids, pH =

pKa + log(salt/acid)

 

For weak bases, pH =

pKw - pKb + log(base/salt)

 

where pKw =

14

 

Example: What is the pH of a buffer solution prepared to be 0.50 moles (M) in sodium acetate and 0.05 M in acetic acid (pKa of acetic acid = 4.76)?

000237

Example: What is the pH of a buffer solution prepared to be 0.5 M in ammonia (pKb = 4.74) and 0.05 M in ammonium chloride?

Ammonia forms a base in aqueous solution.

pH = pKw - pKb + log(base/salt) = 14.00 -4.74 + log(0.50/0.05) = 14.00-4.74 + 1.00 = 10.26

1-14. Temperature

Frequently, one must convert temperature from Fahrenheit (F) to Centigrade (C), and vice versa. The following formula can be used: 9°C = 5°F - 160.

Example: A patient has an oral temperature of 100°F. What is that temperature in °C?

000239

1-15. Questions

1.

If 100 capsules contain 340 mg of active ingredient, what is the weight of active ingredient in 75 capsules?

A. 453 mg

B. 340 mg

C. 255 mg

D. 128 mg

E. 75 mg

 

2.

What is the weight of 500 mL of a liquid whose specific gravity is 1.13?

A. 442 mg

B. 565 g

C. 442 g

D. 885 mg

E. 221 g

 

3.

A pharmacist weighs out 325 mg of a substance on her class A prescription balance. When she subsequently checks the weight on a more sensitive analytical balance, she finds it to be only 312 mg. What is the percentage of error in the original weighing?

A. 4%

B. 5%

C. 6%

D. 10%

E. 12%

 

4.

What is the minimum weighable quantity for a maximum of 5% error using a balance with a sensitivity requirement of 6 mg?

A. 80 mcg

B. 100 mg

C. 120 mg

D. 150 mg

E. 240 mg

 

5.

A patient weighing 175 lb is to receive an initial daily intramuscular (IM) dosage of procainamide HCl (500 mg/mL vial) of 50 mg/kg (ABW) to be given in divided doses every 3 hours. How many milliliters should each injection contain?

A. 3.98 mL

B. 0.49 mL

C. 8.23 mL

D. 1.87 mL

E. 0.99 mL

 

6.

What is the ideal body weight of a female patient whose height is 5 feet 8 inches?

A. 68 kg

B. 64 kg

C. 150 lb

D. 121 lb

E. 53 kg

 

7.

What is the approximate BSA of an adult patient who weighs 154 lb and is 6 feet tall?

A. 1.73 m2

B. 3.15 m2

C. 1.89 m2

D. 0.70 m2

E. 2.67 m2

 

8.

If the adult dose of a drug is 125 mg, what is the dose for a child whose BSA is estimated to be 0.68 m2?

A. 485 mcg

B. 318 mg

C. 85 mg

D. 49 mg

E. 33 mg

 

9.

What is the creatinine clearance for a 65-year-old female who weighs 50 kg and has a serum creatinine level of 1.3 mg/dL?

A. 34 mL/min

B. 40 mL/min

C. 26 mL/min

D. 82 mL/min

E. 100 mL/min

 

10.

Using the formula that follows, determine how much zinc oxide is required to make 750 g of mixture:

 

Zinc oxide

150 g

 

Starch

250 g

 

Petrolatum

550 g

 

Coal tar

50 g

 
 

A. 200 g

B. 188 g

C. 413 g

D. 113 g

E. 38 g

 

11.

Using the formula that follows, determine the weight of kaolin that would be required to produce 500 g of mixture:

 

Kaolin

12 parts

 

Magnesium oxide

3 parts

 

Bismuth subcarbonate

5 parts

 
 

A. 83 g

B. 300 g

C. 208 g

D. 333 g

E. 250 g

 

12.

How much dextrose is required to prepare 500 mL of an aqueous 10% solution?

A. 250 mg

B. 500 mg

C. 10 g

D. 25 g

E. 50 g

 

13.

What weight of hexachlorophene should be used in compounding 20 g of an ointment containing hexachlorophene at a concentration of 1:400?

A. 25 mcg

B. 50 mcg

C. 50 mg

D. 80 mg

E. 5 g

 

14.

What weight of magnesium chloride (MgCl2, formula weight = 95.3) is required to prepare 200 mL of a solution that is 5.0 millimolar?

A. 191 mg

B. 95.3 mg

C. 19.1 mg

D. 477 mcg

E. 95 g

 

15.

What weight of magnesium chloride (MgCl2, formula weight = 95.3; Mg2+, atomic weight = 24.3; Cl1-, atomic weight = 35.5) is required to prepare 1,000 mL of a solution that contains 5.0 mEq of magnesium?

A. 238 mg

B. 4.76 g

C. 1.19 g

D. 60.7 mg

E. 476 mcg

 

16.

What is the milliosmolarity (ideal) of normal saline (NaCl, formula weight = 58.5)?

A. 100 mOsm/L

B. 154 mOsm/L

C. 254 mOsm/L

D. 287 mOsm/L

E. 308 mOsm/L

 

17.

How much water for injection should be added to 250 mL of 20% dextrose to obtain 15% dextrose?

A. 333 mL

B. 83 mL

C. 250 mL

D. 166 mL

E. 58 mL

 

18.

What volume of a 5% dextrose solution should be mixed with 200 mL of a 20% dextrose solution to prepare 300 mL of a 15% dextrose solution?

A. 150 mL

B. 200 mL

C. 100 mL

D. 50 mL

E. 250 mL

 

19.

What is the final concentration obtained by mixing 200 mL of 20% dextrose with 100 mL of 5% dextrose?

A. 10%

B. 15%

C. 7.5%

D. 12.5%

E. 17.5%

 

20.

Magnesium chloride (MgCl2) is a 3-ion electrolyte that dissociates 80% at the relevant concentration. Calculate its dissociation factor (i).

A. 1.8

B. 2.2

C. 2.4

D. 2.6

E. 3.2

 

21.

Tobramycin (formula weight = 468) has a dissociation factor of 1.0. What is its sodium chloride equivalent?

A. 0.069

B. 0.0092

C. 0.117

D. 0.286

E. 0.782

 

22.

What weight of sodium chloride should be used in compounding the following prescription for ephedrine sulfate (formula weight = 429, dissociation factor = 2.6, sodium chloride equivalent = 0.23)?

 

Ephedrine sulfate

Sodium chloride

Purified water ad

Make isoton. sol.

0.25 g

qs

30 mL

 
 

A. 1.22 g

B. 784 mcg

C. 212 mg

D. 527 mcg

E. 429 mg

 

23.

A patient is to receive an infusion of 2 g of lidocaine in 500 mL D5W at a rate of 2 mg/min. What is the flow rate in milliliters per hour?

A. 2.0 mL/h

B. 6.5 mL/h

C. 15 mL/h

D. 30 mL/h

E. 150 mL/h

 

24.

What is the pH of a buffer solution prepared with 0.05 M disodium phosphate and 0.05 M sodium acid phosphate (pKa = 7.21)?

A. 4.55

B. 5.23

C. 6.18

D. 7.05

E. 7.21

 

25.

Convert 104°F to Centigrade.

A. 22°C

B. 34°C

C. 40°C

D. 46°C

E. 54°C

 

26.

Calculate the amount of water (in grams) in 100 mL of 65% (w/w) syrup that has a density of 1.313.

A. 30 g

B. 75 g

C. 45.95 g

D. 65.25 g

E. 35 g

 

27.

How much boric acid will be needed to prepare an isotonic solution of the following prescription?

 

Phenacaine HCl

1%

E value for phenacaine HCl is 0.20

 

Chlorobutanol

0.5%

E value for chlorobutanol is 0.24

 

Boric acid

qs

E value for boric acid is 0.52

 

Purified water ad

60mL

 
 
 

A. 0.55 g

B. 0.67 g

C. 0.75 g

D. 1.2 g

E. 2.2 g

 

28.

How many milligrams of sodium chloride are required to make 30 mL of a solution of 1% dibucaine HCl isotonic with tears?

Note: The freezing point depression of 1% dibucaine HCl solution is -0.08°C, and the freezing point depression of an isotonic solution is -0.52°C.

A. 1.2 g

B. 0.950 g

C. 0.450 g

D. 0.228 g

E. 0.850 g

 

29.

How many milliequivalents of Na+ are contained in a 30 mL dose of the following solution?

 

Disodium hydrogen phosphate

18 g

Na2HPO4 7H2O (MW 268)

 

Sodium biphosphate

48 g

NaH2PO4 4H2O (MW 138)

 

Purified water ad

100 mL

 
 
 

A. 144.63 mEq

B. 104.34 mEq

C. 40.29 mEq

D. 100 mEq

E. 52 mEq

 

30.

How many grams per liter are needed to prepare 2N (normality) of H2SO4?

A. 25 g

B. 35 g

C. 78 g

D. 49 g

E. 98 g

 

31.

How many milliequivalents of magnesium sulfate are represented in 1 gram of anhydrous magnesium sulfate (MgSO4) (MW 120)?

A. 122 mEq

B. 16.67 mEq

C. 12 mEq

D. 10 mEq

E. 19 mEq

 

32.

What is the concentration (g/mL) of a solution containing 4 milliequivalents of CaCl2 2H2O per milliliter (MW 147)?

A. 0.345 g/mL

B. 0.986 g/mL

C. 0.389 g/mL

D. 0.294 g/mL

E. 0.545 g/mL

 

Questions 33-37 are related to the following problem:

An IV infusion for a patient weighing 132 lb calls for 7.5 mg of drug/kg of body weight to be added to 250 mL of 5% dextrose injection solution.

33.

What is the patient's weight in kilograms?

A. 75 kg

B. 25 kg

C. 120 kg

D. 55 kg

E. 60 kg

 

34.

How much drug is needed?

A. 250 mg

B. 350 mg

C. 150 mg

D. 450 mg

E. 900 mg

 

35.

What is the total number of milliliters the patient receives per day if the IV solution runs at 52 mL/h?

A. 1,350 mL

B. 1,248 mL

C. 256 mL

D. 1,000 mL

E. 1,500 mL

 

36.

What is the infusion rate in drops per minute (1 mL = 20 drops)?

A. 25 drops/min

B. 22 drops/min

C. 17 drops/min

D. 30 drops/min

E. 15 drops/min

 

37.

How many IV bags does the patient receive per day?

A. 10 bags

B. 12 bags

C. 2 bags

D. 5 bags

E. 8 bags

 

38.

If 50 mg of drug X is mixed with enough ointment base to obtain 20 grams of mixture, what is the concentration of drug X in ointment (expressed as a ratio)?

A. 1:300

B. 1:400

C. 1:200

D. 1:600

E. 1:100

 

39.

When 23 mL of water for injection are added to drug-lyophilized powder, the resulting concentration is 200,000 units/mL. What is the volume of the dry powder if the amount of drug in the vial was 5,000,000 units?

A. 2 mL

B. 4 mL

C. 1 mL

D. 5 mL

E. 9 mL

 

40.

If 20 grams of salicylic acid are mixed with enough hydrophilic petrolatum to obtain a concentration of 5%, how much ointment was used to prepare the prescription?

A. 400 g

B. 380 g

C. 250 g

D. 480 g

E. 280 g

 

1-16. Answers

1.

C.

x mg: 340 mg = 75 cap: 100 cap

x = 340 mg × 75 cap/100 cap = 255 mg

 

2.

B. A specific gravity of 1.13 corresponds to a density of 1.13 g/mL.

 
 

density =

weight/volume; thus, weight

 
 

=

density × volume = 1.13 g/mL

 
   

× 500 mL = 565 g

 

3.

A.

000238

 

4.

C.

000242

 

5.

E.

000244

 

6.

B.

IBW = 45.5 + (2.3 × 8) = 64 kg

 

7.

C.

000245

 

8.

D.

000105

 

9.

A.

000243

 

10.

D. Note that the formula is designed to produce a total of 1,000 g of the mixture. Then, by proportions:

000111

 

11.

B. Note that the formula will produce a total of 20 parts of the mixture. Then, by proportions:

000116

 

12.

E. Note that this will be a solution of a solid in a liquid; thus, the concentration will be %(w/v).

000118

 

13.

C. By proportions, x g hexachlorophene: 20 g ung. = 1 part hexachlorophene: 400 parts ung.

000123

 

14.

B. A 1.0 molar solution will contain 95.3 g in 1,000 mL. A 5.0 molar solution will contain 95.3 g × 5 = 477 g in 1,000 mL. A 5.0 millimolar solution will contain 477 mg in 1,000 mL. Thus, 200 mL of a 5 millimolar solution will contain 477 mg/5 = 95.3 mg in 200 mL.

 

15.

A. Because magnesium has a valence of 2, a formula weight of MgCl2 will contain two equivalent weights of magnesium (and chloride, for that matter). Thus, 5 equivalents of magnesium are contained in 5 × 95.3/2 g = 238 g MgCl2. Accordingly, 5 mEq of magnesium are contained in 238 mg MgCl2.

 

16.

E. Normal saline is 0.90%(w/v), or 0.90 g/100 mL = 9.0 g/1,000 mL.

000128

 

17.

B.

000132

 

18.

C.

 
 

20% (concentration of stock A)

15 - 5 = 10 = parts of A

 
 

15% (desired concentration)

 
 

5% (concentration of stock B)

20 - 15 = 5 = parts of B

 
 

Relative volumes are 10:5, or 2:1. Thus, 200 mL of a 20% dextrose solution (A) will require 100 mL of a 5% dextrose solution (B) to produce 300 mL of a 15% dextrose solution.

 

19.

B.

000134

 

20.

D. Each 100 molecules will provide

80 Mg ions

160 Cl ions

20 undissociated molecules

260 particles total

Thus, the dissociation factor = 260/100 = 2.6

 

21.

A.

000108

 

22.

C. Because 900 mg of sodium chloride in 100 mL is isotonic,

000129

which is the amount of sodium chloride alone needed to make 30 mL isotonic. But 1 g of ephedrine sulfate is equivalent to 0.23 g of sodium, thus:

 
 

y

= 0.25 g × 0.23 = 0.058 g

= 58 mg of sodium chloride

 
   

Accordingly, the amount of sodium chloride to add (z = x - y) is (270 mg - 58 mg) = 212 mg.

 

23.

D. The bag contains 2,000 mg in 500 mL, or 4 mg/mL. Therefore, a rate of 2 mg/min corresponds to 0.5 mL/min, which corresponds to 30 mL/h.

 

24.

E.

000138

 

25.

C.

000141

 

26.

C. Note that 65 %w/w means 65 g in 100 g syrup. The total weight of the 100 mL syrup is 100 mL × 1.313 (density) = 131.3 g. Every 100 g of syrup contains 35 g of water, so

000143

 

27.

B. The 1% phenacaine HCl equals 0.6 g, and the 0.5% chlorobutanol equals 0.3 g. Because we know the E value of phenacaine HCl, we can write the ratio:

000146

Because we know the E value of chlorobutanol, we can write the ratio:

000148

The amount of sodium chloride needed to make 60 mL of solution isotonic is calculated as follows:

000151

amount of sodium chloride needed

= 0.54 g - (0.12g + 0.072g)

= 0.348 g NaCl


Because we know the E value for boric acid, we can write the ratio:

000152

 

28.

D. The weight of 1% of drug equals 0.3 g. The needed change in freezing point depression to make the solution isotonic is 0.52°C - 0.08°C = 0.44°C, so the ratio is as follows:

000156

The amount of sodium chloride needed to make 30 mL of water isotonic is calculated as follows:

000158

 

29.

A.

1 mEq of disodium hydrogen 
000162 1 mEq of disodium biphosphate 000154

We can write the ratio as follows:

000147

We can also write the ratio as follows:

000165

To adjust for volumes, we can write the ratios as follows:

000167

Thus, x
1 = 40.29 mEq disodium hydrogen phosphate, and x2 = 104.34 mEq sodium biphosphate. The total Na+ mEq = 104.34 + 40.29 = 144.63 mEq.

 

30.

E. Normality is the number of equivalents in one liter of solvent. The expression "2N normal" means two equivalents dissolved in one liter. Because one equivalent of sulfuric acid equals 49 g, 2N is equivalent to dissolving 98 g of sulfuric acid in 1 L of solvent.

 

31.

B.

000103

 

32.

D.

000109

 

33.

E.

000113

 

34.

D.

7.5 mg/kg × 60 kg = 450 mg

 

35.

B.

52 ml/h × 1 h × 24 = 1,248 mL

 

36.

C.

000117

 

37.

D.

000119

 

38.

B.

000125, so X = 400

 

39.

A.

000169

so x = 25 mL, and the volume of powder is 25 - 23 = 2 mL

 

40.

B.

000211, so x = 400g, and the amount of ointment is 400 - 20 = 380 g

 

1-17. References

Ansel HC, Stoklosa MJ. Pharmaceutical Calculations. 12th ed. Philadelphia: Lippincott Williams & Wilkins; 2006.

Khan MA, Reddy IK. Pharmaceutical and Clinical Calculations. 2nd ed. Lancaster, Pa.: Technomic Publishing Co; 2000.

O'Sullivan TA. Understanding Pharmacy Calculations. Washington, D.C.: American Pharmaceutical Association; 2002.


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