Calculations Quantities, Dilutions and Concentrations Pharmacy Technician Training Systems Passassured, LLC
Calculations, Quantities, Dilutions and Concentrations PassAssured's Pharmacy Technician Training Program Calculations Quantities, Diutions and Concentrations Click Here for Glossary Index! Click Here to Print Topic Help File,.pdf (Internet Access is Required for this Feature) Educational Objectives Expressions of Quantity Milliequivalents (meq) Millimoles Expressions of Concentration Examples of Concentration Expressions Ratio Strength Example 1 Example 2 Pass Assured, LLC, Pharmacy Technician Training Systems Copyright Pass Assured, LLC, Web Site - http:// -o- p2
PassAssured's Pharmacy Technician Training Program Calculations Quantities, Diutions and Concentrations Click Here for Glossary Index! Click Here to Print Topic Help File,.pdf (Internet Access is Required for this Feature) Educational Objectives Review the units of measurement for drugs and expressions of quantity and concentration for drugs in drug products. Illustrate different methods for determining quantities of ingredients and concentration of drugs when preparing or dispensing drug products. Calculate the amounts of two solutions of different strengths, which must be combined to get a third solution of a specified strength. Calculate the final strength of a diluted or mixed solution when given the original strength(s) and volume(s) and the final volume. Expressions of Quantity The Method of Expression of Quantity is Determined by The most accurate representation of the amount of drug convenience Route of Administration: Systemic versus topical administration Tradition - (How the doctor writes his prescription.) Useful units of quantity used in pharmacy: Units of Quantity Table ( 02070010 ) p3
Milliequivalents (meq) A milliequivalent is equal to the number of univalent counter ions (H+ or OH) which will be needed to react with one molecule of the substance. When an atom has a valence of one ( e.g. Na+, K+, CI, etc.) a meq of that ion is equal to the atomic weight of the atom in milligrams. For example, one meq of sodium (Na+) with an atomic weight of 23 (AW) weighs 23mg. When an atom has a valence of two or more (e.g. Mg+2, Ca+2, Al+3) a meq of that ion is equal to the atomic weight of the atom in milligrams divided by the atom's valence. For example, one meq of calcium (Ca+2) with an atomic weight of 40 weighs 20 mg. Because in chemical compounds or salts ions always come in pairs, (positive and negative ions are always together in salts like NaC1 and CaC12) we can never weigh or measure just one ion. When both ions have a valence of one (e.g. NaC1), one millimole (the molecular weight (MW) of the compound in milligrams) of the compound contains one meq of each ion. If one or both of the ions in the pair has a valence of two or more (e.g. CaC12), one millimole of the compound contains two or more meq's of each ion. Two examples are given below to illustrate. It should be noted that calculations with milliequivalents can be complex. What is presented here is a simplified version. For a more complete explanation, consult a book of chemistry or pharmacy. Example 1 A prescription order calls for Potassium Chloride (KC1) 20 meq. Calculate the number of milligrams of Potassium Chloride, which will give this amount. The atomic weight (AW) of K+ = 39 and the AW of C1 = 35.5. The molecular weight of (MW) of KC1 = 74.5 mg Since both ions have a valence of one, 74.5 mg KC1 (the weight of one millimole) contains 1 meq of K and 1 meq of C1. Therefore, 20 meq of KC1 is 20 x 74.5 mg = 1490 mg. Example 2 A medication order calls for Calcium Chloride (CaCl2) 4 meq. Calculate the number of milligrams of Calcium Chloride, which will give this amount. AW of Ca +2 = 40 and the AW of CI = 35.5 MW of CaCl2 = 40 + 35.5 +35.5 = 11mg Since Ca+2 has a valence of two, 111mg of CaCl2 (the weight of one millimole) contain 2mEq of Ca+2 and 2 meq of Cl. Therefore, 4 meq of CaCl2 is 2 x 111 mg = 222 mg Millimoles p4
Used in pharmacy and medicine to express the dose for certain ions which can have different valences depending on the conditions. Phosphate is one example of such an electrolyte. A millimole of a compound is the molecular weight of the compound in milligrams. Expressions of Concentration Concentration gives the quantity of drug per amount (weight or volume) of product. Always used for expressing doses for topical preparations because concentration is the driving force for transfer of the drug across a membrane or barrier, such as the skin. It may be used for some systemic products. In reality, since a pure drug is almost never dispensed, quantities, when used in reference to a product, are technically concentrations. Example: Acetaminophen 500mg weighs more than 500mg. The active ingredient (Acetaminophen) weighs 500mg. Examples of Concentration Expressions Weight in Weight - Tobramycin Ophthalmic Ointment 3 mg/g is 3 mg of Tobramycin in each 1 g of ointment. Weight per Volume - Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension Percent - grams or milliliters of drug per 100 g or per 100 ml of product This definition specified by the USP is very important Solids in solids are percent weight in weight (% w/w): Hydrocortisone 1% Ointment is 1 gram of Hydrocortisone in 100 grams of ointment Solids in liquids are percent weight in volume (% w/v): Dextrose 5% in Water (D5W) is 5 grams of Dextrose in 100 ml of solution Liquids in liquids are percent volume in volume (5% v/v): Isopropyl Alcohol 70% is 70 ml of Isopropyl Alcohol in 100 ml of solution Ratio Strength 1 gram (for solids) or 1 ml (for liquids) per given weight or volume of product As with percents, ratio strength can be weight in weight (solids in solids), weight in volume (solids in liquids), or volume in volume (liquids in liquids). Example: Weight in Volume Potassium Permanganate Solution 1:5000 is 1 g of Potassium p5
Permanganate in 5000 ml of solution Other Amount per Volume or per Weight Example: Units = Nystatin 100,000 units/ g of ointment meq = Potassium Chloride 20 meq/5 ml oral solution The following are examples of calculating amounts of ingredients and concentrations of finished products when preparing solutions and other pharmaceutical preparations. In some cases four methods are illustrated for each problem. Proportion Alligation Percent Dimensional analysis When performing calculations like this, you should use the method which makes the most sense to you for the type of problem presented. Often a combination of methods is the most convenient. In the examples we use liquids and solutions. The same general principles apply to calculations for solids. Calculate the amounts of two solutions of different strengths which must be combined to get a third solution of a specified strength. The technician should follow these general steps Calculate the quantity of drug or chemical needed in the third solution. Calculate the volume(s) of each of the two given solutions that will give you this amount. Example 2 The technician is asked to make 4 oz of a 50% solution of Isopropyl Alcohol (IPA) Ingredients available: Isopropyl Alcohol 70% and water How many milliliters of each will you need? Solve By Proportion: (Remember the definition of percent, a 50% solution of a liquid contains 50 ml of that liquid in 100 ml of solution.) First, calculate the number of milliliters of pure IPA needed for the third solution: Proportion Equation/1 - Ex. 2 ( 02070040 ) p6
Then calculate the number of milliliters of 70% IPA required to give this amount of pure IPA: Proportion Equation/2 - Ex. 2 ( 02070050 ) Solve By Percent Rate (Percent) x Whole = Part 50% x 120 ml = Part Convert 50% to a decimal and multiply: 0.5 x 120 ml = 60 ml of IPA How many milliliters of 70% IPA will give this amount? 70% x (X ml of 70% IPA Soln.) = 60 ml of pure IPA 70% (x) = 60 Convert 70% to a decimal and solve for X: Move the decimal point two places to the left and drop the percent sign, 70% = 0.7 Percent Formula - Ex. 2 ( 02070070 ) Measure 86 ml of 70% OPA and add sufficient Purified Water to give 120 ml of solution. p7
Solve By Dimensional Analysis Dimensional Analysis Formula - Ex 2 ( 02070080 ) Measure 86 ml of 70% Isopropyl Alcohol and add sufficient Purified Water to give 120 ml of solution Solve By Alligation Alligation Worksheet, Step 1 - Ex. 2 ( 02070090 ) Alligation Worksheet, Step 2 - Ex. 2 ( 02070095 ) Example 1 You have 5 ml of Gentamicin Ophthalmic Solution 0.3%. You dilute it with 2 ml of sterile normal saline. What is the percent of Gentamicin in the final solution? Calculate the amount of drug in the original solution(s) from the concentration and volume. p8
By the definition of percent 0.3% Gentamicin is 0.3 g per 100 ml Percent Equation - Ex. 1 ( 02070200 ) Calculate the final volume of the new solution 5 ml + 2 ml = 7 ml Divide the amount, 0.015 g, by the final volume, 7 ml Formula - Ex. 1 ( 02070300 ) Remember percent is gm per 100 0.002 g/ml x 100 ml = 0.2% Example 2 You have an IM injection of Rocephin 2 Gm per 10 ml vial. You put 1 ml in an empty vial and dilute with 3.3 ml of Lidocaine HCI 2% for injection. What is the final concentration of Rocephin in mg/ml? 2 Gm = 2000 mg Equation - Ex. 2 ( 02070400 ) p9
What is the final concentration of Lidocaine HCI in percent? 2% = 2 gm/100 ml Equation - Ex. 2 ( 02070500 ) Pass Assured, LLC, Pharmacy Technician Training Systems Copyright Pass Assured, LLC, Web Site - http:// -o- p10