Altering product strength use of stock solutions and

Altering product strength, use of stock solutions, and problem solving by allegation by Ali Khidher Alobaidy

� The strength of a pharmaceutical preparation may be increased or decreased by changing the proportion of active ingredient to the whole. A preparation may be strengthened or made more concentrated by the addition of active ingredient, by admixture with a like preparation of greater strength, or through the evaporation of its vehicle, if liquid. The strength of a preparation may be decreased or diluted by the addition of diluent or by admixture with a like preparation of lesser strength.

Special considerations of altering product strength in pharmaceutical compounding � The dilution of a liquid dosage form, as a solution or suspension, may be desired to provide a product strength more suitable for use by a particular patient (e. g. , pediatric, elderly, those in disease states). Thediluentisselectedbasedonitscompatibilitywithth evehicleoftheoriginal product; that is, aqueous, alcoholic, hydroalcoholic, or other. The dilution of a solid dosage form (as a powder or the contents of a capsule) or a semisolid dosage form (as an ointment or cream) also may be performed to alter the dose or strength of a product.

� The concentration of a liquid preparation, as through the evaporation of a portion of its solvent or vehicle, rarely is performed nowadays. However, the fortification of a liquid , solid , or semisolid dosage form, by the addition of a calculated quantity of additional therapeutic agent, remains a viable practice in pharmacy compounding.

Relationship between strength and total quantity � The percentage or ratio strength (concentration) of a component in a pharmaceutical preparation is based on its quantity relative to the total quantity of the preparation. If the quantity of the component remains constant, any change in the total quantity of the preparation, through dilution or concentration, changes the concentration of the component in the preparation inversely.

� If a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one half. By contrast, if a mixture is concentrated by evaporation to one half its original quantity, the active ingredient (assuming that none was lost by evaporation) will be contained in one half as many parts of the whole, and the strength will be doubled.

� If, then, the amount of active ingredient remains constant, any change in the quantity of a solution or mixture of solids is inversely proportional to the percentage or ratio strength; that is, the percentage or ratio strength decreases as the quantity increases, and conversely. This relationship is generally true for all mixtures except solutions containing components that contract when mixed together. Problems in this section generally may be solved by any of the following methods: � 1. Inverse proportion. � 2. The equation: (1 st quantity) x (1 st concentration) = (2 nd quantity) x (2 nd concentration), or Q 1 x C 1 = Q 2 x C 2. � 3. By determining the quantity of active ingredient (solute) present or required and relating that quantity to the known or desired quantity of the preparation.

Dilution and concentration of liquids If 500 m. L of a 15% v/v solution are diluted to 1500 m. L, what will be the percentage strength (v/v)? � 1. (inverse proportion) 1500 ml 15% 500 ml x% � X = 5% 2. Equation Q 1 (quantity) x C 1 (concentration) = Q 2 (quantity) x C 2 (concentration) 500 (m. L) x 15 (%) = 1500 (m. L) x x% X = 5% 3. By determining the quantity of active ingredient (solute) present or required. 500 x o. 15 = 75 1500 ml 100% 75 ml x% X = 5%

If 50 m. L of a 1: 20 w/v solution are diluted to 1000 m. L, what is the ratio strength (w/v)? � Note: A student may find it simpler in solving certain problems to convert a given ratio strength to its equivalent percentage strength. 1 20 � X 100 X = 1: 20 = 5% Q 1 (quantity) x C 1 (concentration) = Q 2 (quantity) x C 2 (concentration) 50 (m. L) x 5 (%) = 1000 (m. L) xx (%) = 0. 25% = 1: 400 → 0. 25 100 1 x X = 100 ÷ 0. 25 = 400

� If a syrup containing 65% w/v of sucrose is evaporated to 85% of its volume, what percentage (w/v) of sucrose will it contain? Any convenient amount of the syrup, for example, 100 m. L, may be used in the calculation. If we evaporate 100 m. L of the syrup to 85% of its volume, we will have 85 m. L. � C 1 x V 1 = C 2 x V 2 � 100% x 65 ml = C 2 x 85 ml � C 2 = 76. 47%

� How many grams of 10% w/w ammonia solution can be made from 1800 g of 28% w/w strong ammonia solution? � Q 1 x. C 1 = Q 2 x C 2 � 1800 (g) X 28 (%) = x (g) X 10% = 5040 g

� How many milliliters of a 1: 5000 w/v solution of the preservative lauralkonium chloride can be made from 125 m. L of a 0. 2% solution? 1: 5000 = 0. 02% 125 (m. L) x 0. 2 (%) = x (m. L) x 0. 02 (%) x = 1250 m. L

� If 1 gallon of a 30% w/v solution is to be evaporated so that the solution will have a strength of 50% w/v, what will be its volume in milliliters? 1 gallon = 3785 m. L 30 % x 3785 = 50% x v 2 V 2 = 2271 ml

Strengthening of pharmaceutical product � As noted previously, there is occasion in which a pharmacist may be called upon to strengthen an existing pharmaceutical product. This may be accomplished by the addition of active ingredient or by the admixture with a calculated quantity of a like product of greater concentration.

� If a cough syrup contains in each teaspoonful, 1 mg of chlorpheniramine maleate and if a pharmacist desired to double the strength, how many milligrams of that ingredient would need to be added to a 60 -m. L container of the syrup. Assume no increase in volume? 1 mg 5 ml X 60 ml � X = 12 mg chlorpheniramine maleate in original syrup. � To double the strength, 12 mg of additional chlorpheniramine maleate would be required.

Stock solutions � Stock solutions are concentrated solutions of active (e. g. , drug) or inactive (e. g. , colorant) substances and are used by pharmacists as a convenience to prepare solutions of lesser concentration.

� How many milliliters of a 1: 400 w/v stock solution should be used to make 4 liters of a 1: 2000 w/v solution? � 1: 400 = 0. 25% � 1: 2000 = 0. 05% � Q 1 x C 1 = Q 2 x C 2 � 0. 05% x 4000 = Q 2 x 0. 25% � Q 2 = 800 ml

� How many milliliters of a 1: 400 w/v stock solution should be used in preparing 1 gallon of a 1: 2000 w/v solution? � 1 gallon = 3785 m. L � 1: 400 = 0. 25% � 1: 2000 = 0. 05% � Q 1 x C 1 = Q 2 x C 2 � 3785 x 0. 05 = Q 2 x 0. 25 � Q 2 = 757 ml

� How many milliliters of a 1% stock solution of a certified red dye should be used in preparing 4000 m. L of a mouthwash that is to contain 1: 20, 000 w/v of the certified red dye as a coloring agent? � 1: 20, 000 = 0. 005% � Q 1 x C 1 = Q 2 x C 2 � 4000 x 0. 005 = Q 2 x 1 � Q 2 = 20 ml

� How many milliliters of a 1: 16 solution of sodium hypochlorite should be used in preparing 5000 m. L of a 0. 5% solution of sodium hypochlorite for irrigation? � 1: 16 = 6. 25% � Q 1 x C 1 = Q 2 x C 2 � 5000 x 0. 5 = 6. 25 x Q 2 � Q 2 = 400 ml

� How many milliliters of a 1: 50 stock solution of phenylephrine hydrochloride should be used in compounding the following prescription? Rx. Phenylephrine HCl 0. 25% Rose Water ad 30 m. L Sig. For the nose. � 1: 50 = 2% � Q 1 x C 1 = Q 2 x C 2 � 30 x 0. 25% = Q 2 x 2% � Q 2 = 3. 75 ml

� Some interesting calculations are used in pharmacy practice in which the strength of a diluted portion of a solution is defined, but the strength of the concentrated stock solution used to prepare it must be determined. The relevance to pharmacy practice may be explained, for example, by the need of a pharmacist to prepare and dispense a concentrated solution of a drug and direct the patient to use a specific household measure of a solution (e. g. , 1 teaspoonful) in a specified volume of water (e. g. , a pint) to make of solution of the desired concentration (e. g. , for irrigation or soaking). This permits the dispensing of a relatively small volume of liquid, enabling a patient to prepare relatively large volumes as needed, rather than carrying home gallons of a diluted solution from a pharmacy.

� How much drug should be used in preparing 50 m. L of a solution such that 5 m. L diluted to 500 m. L will yield a 1: 1000 solution? 1 1000 X 500 � X = 0. 5 g 0. 5 5 ml X 50 ml � X = 5 g

� How many grams of sodium chloride should be used in preparing 500 m. L of a stock solution such that 50 m. L diluted to 1000 m. L will yield a ‘‘ 1⁄3 normal saline’’ (0. 3% w/v) for irrigation? 0. 3 g 100 ml X 1000 ml � X = 3 g 50 ml 3 g 500 ml x � X = 30 g

How many milliliters ofa 17% w/v concentrate of benzalkonium chloride should be used in preparing 300 m. L of a stock solution such that 15 m. L diluted to 1 liter will yield a 1 : 5000 solution? � � 1 liter = 1000 m. L � 1 : 5000 means 1 g of benzalkonium chloride in 5000 m. L of solution � 1 g 5000 ml � X 1000 ml X = 0. 2 g of benzalkonium chloride in 1000 m. L of diluted solution (15000), which is also the amount in 15 m. L of the stronger (stock) solution to be prepared, and: � � 0. 2 g 15 ml � X 300 ml � X = 4 g of benzalkonium chloride needed. � because a 17% w/v concentrate contains 17 g per 100 m. L, then: � 17 g 100 ml � 4 g x � X = 23. 5 ml

� A solution of known volume and strength may be diluted with water to prepare a solution of lesser strength. In such calculations, first calculate the quantity of diluted solution that may be prepared from the concentrated solution. Then, subtract the volume of the concentrated solution from the total quantity that may be prepared to determine the volume of water needed.

How many milliliters of water should be added to 300 m. L of a 1: 750 w/v solution of benzalkonium chloride to make a 1: 2500 w/v solution? � � 1 g 750 ml � X 300 ml � X = 0. 4 g � 300 m. L of a 1: 750 (w/v) solution contains 0. 4 g of benzalkonium chloride. � 1 g 2500 ml � 0. 4 g x � X = 1000 ml The difference between the volume of diluted (weaker) solution prepared and the volume of stronger solution used represents the volume of water (diluent) to be used. � � 1000 m. L 300 m. L = 700 m. L

� How many milliliters of water should be added to a pint of a 5% w/v solution to make a 2% w/v solution? � 1 pint = 473 m. L 5 g 100 ml X 473 ml � X = 23. 65 g 2 g 100 ml 23. 65 g x � X = 1182. 5 ml – 473 = 709. 5 ml

� If the quantity of a component is given rather than the strength of a solution, the solution may be diluted to a desired strength as shown by the following example. � How many milliliters of water should be added to 375 m. L of a solution containing 0. 5 g of benzalkonium chloride to make a 1: 5000 solution? 1 g 5000 ml 0. 5 g X � X = 2500 ml of 1: 5000 (w/v) solution containing 0. 5 g of benzalkonium chloride. � 2500 m. L - 375 m. L = 2125 m. L

� If 15 m. L of a 0. 06% ATROVENT (ipratropium bromide) nasal spray were diluted with 6 m. L of normal saline solution, what would be the final drug concentration? � 15 m. L x 0. 06% = 0. 009 g of ipratropium bromide � 15 m. L + 6 m. L= 21 m. L, new total volume 0. 009 g 21 ml X 100 ml � X = 0. 043%

Allegation Alligation is an arithmetical method of solving problems that involves the mixing of solutions or mixtures of solids possessing different percentage strengths. � Alligation Medial. Alligation medial is a method by which the ‘‘weighted average’’ percentage strength of a mixture of two or more substances of known quantity and concentration may be easily calculated. By this method, the percentage strength of each component, expressed as a decimal fraction, is multiplied by its corresponding quantity; then the sum of the products is divided by the total quantity of the mixture; and the resultant decimal fraction is multiplied by 100 to give the percentage strength of the mixture. Of course, the quantities must be expressed in a common denomination, whether of weight or volume. �

Example Calculations Using Alligation Medial � What is the percentage strength (v/v) of alcohol in a mixture of 3000 m. L of 40% v/v alcohol, 1000 m. L of 60% v/v alcohol, and 1000 m. L of 70% v/v alcohol? Assume no contraction of volume after mixing. � 0. 40 x 3000 m. L = 1200 m. L � 0. 60 x 1000 m. L = 600 m. L � 0. 70 x 1000 m. L = 700 m. L � � Totals: 5000 m. L 2500 m. L � 2500 (m. L) ÷ 5000 (m. L) = 0. 50 x 100 = 50%

� What is the percentage of zinc oxide in an ointment prepared by mixing 200 g of 10% ointment, 50 g of 20% ointment, and 100 g of 5% ointment? � 0. 10 x 200 g = 20 g � 0. 20 x 50 g = 10 g � 0. 05 x 100 g = 5 g � � Totals: 350 g 35 g � 35 (g) ÷ 350 (g) = 0. 10 x 100 = 10%

In some problems, the addition of a solvent or vehicle must be considered. It is generally best to consider the diluent as of zero percentage strength, as in the following problem. � What is the percentage strength of alcohol in a mixture of 500 m. L of a solution containing 40% v/v alcohol, 400 m. L of a second solution containing 21% v/v alcohol, and a sufficient quantity of a nonalcoholic third solution to make a total of 1000 m. L? � � 0. 40 x 500 m. L = 200 m. L � 0. 21 x 400 m. L = 84 m. L � 0 x 100 m. L = 0 m. L � � Totals: 1000 m. L 284 m. L � 284 (m. L) ÷ 1000 (m. L) = 0. 284 x 100 = 28. 4%