WEIGHTS AND MEASURES Lecture – 13 03/04-02-2012 DR. SHAHNAZ USMAN Associate Professor Dept. of Pharmaceutics RAKMHSU

Dilution and concentration of the liquids Stock solutions can be diluted to make a product that has a lower concentration; also mixtures of powders or semisolids (e.g, ointments) can be diluted to give a product of lower concentration of the drug(s). The diluent is an inert solid or semisolid or base that does not contain any active ingredients. Mixtures also may be concentrated by adding pure drug or mixing with a product containing a higher concentration of the drug.

STOCK SOLUTIONS To facilitate the dispensing of certain soluble substances, the pharmacist frequently prepares or purchases solutions of high concentration. Portions of these concentrated solutions are diluted to give required solutions of lesser strength. These concentrated solutions are known as stock solutions. This procedure is satisfactory if the substances are stable in solution or if the solutions are to be used before they decompose.

1 teaspoonful = 5ml; 1 qt = 2 pint; 1 pint = 473ml Q. How much of a drug is needed to compound 120 ml of a prescription order such that when 1 teaspoonful of the solution is diluted to 1 qt, a 1:750 solution results. 1 teaspoonful = 5ml; 1 qt = 2 pint; 1 pint = 473ml 1st step: 1 : 750 :: x : 946 x = 946 x 1 = 1.26g 750 2nd step: 5ml : 1.26g :: 120ml : x g x = 1.26 x 120 = 30.24g 5 Answer Q. An ampule of solution of an anti-inflammatory drug contains 4 mg of drug/ml. What volume of the solution is needed to prepare a liter of solution that contains 2 µg of the drug/ml? 1 mg = 1000µg ; 1 liter = 1000ml

Problems related to Dilution and concentration 1st step: 2µg : 1 ml :: x : 1000ml x = 2 x 1000 = 2000µg = 2mg 1 2nd step: 4mg : 1 ml :: 2 mg : x ml x = 1 x 2 = 0.5 ml 4 Answer Problems related to Dilution and concentration Q. How many grams of active ingredient are in 50 g of 10% ointment? 10 % means = 10 g active ingredient in 100g base ? = 50g base

= 10 x 50 = 5 g of AI 100 Answer Q. How many grams of a 5% ointment can be made from 5g of active ingredient? 5 % means = 100 g onit. contain 5 g of active ingredient ? 5 g of AI 5: 100 :: 5 : x ; = 100 x 5 = 100g of ointment 5 Answer The term trituration means a dilute powder mixture of a drug.

a) Determine the amount of drug in 10 g of trituration. Q. How much of a 1 in 10 trituration of a potent drug contains 200mg of the drug? 1 in 10 trituration means = 1g of drug in 10 g of mixture = 1 g of drug plus 9 g diluent (w/w) 1000mg 10g of mixture; 200 V V = 200 x 10 = 2 g of mixture 1000 Answer Q. How much diluent must be added to 10 g of a 1:100 trituration to make a mixture that contains 1 mg of drug in each 10 g of the final mixture a) Determine the amount of drug in 10 g of trituration. 1g drug 100g of trituration X g 10 g of trituration h

1 x 10 = 0.1g of drug 100 Answer b) Determine the amount of mixture that can be made from 0.1 g (100 mg) of drug. 0.1g of drug = 100 mg of drug 10 g mixture x 100 mg drug = 1000g of mixture 1mg drug Answer c) Determine the amount of diluents needed. 1000 g mixture - 10 g trituration = 990 g diluents Answer

Dilution of Alcohol When water and alcohol are mixed, there is a physical contraction such that the resultant volume is less than the total of the individual volumes of the two liquids. Thus, to prepare a volume-in-volume strength of an alcohol dilution, the alcohol ‘‘solute’’ may be determined and water used to ‘‘q.s.’’ to the appropriate volume. Because the contraction of the liquids does not affect the weights of the components, the weight of water needed to dilute alcohol to a desired weight-in-weight strength may be calculated.

Q. How much water should be mixed with 5000 ml of 85% v/v alcohol to make 50% v/v alcohol? N1 V1 = N2 V2 85% x 5000 = 50% x V2 V2 = 85% x 5000 ml = 8500ml 50 (%) Therefore, use 5000 ml of 85% v/v alcohol and enough water to make 8500 ml. Answer Q. How many milliliters of 95% v/v alcohol and how much water should be used in compounding the following prescription? Rx Xcaine 1 g Alcohol 70% 30 ml Sig. Ear drops.

N1 V1 = N2 V2 0.7 x 30 = 0.95 x V2 V2 = 0.7 x 30 ml = 22.1ml 0.95 Therefore, use 22.1 ml of 95% v/v alcohol and enough water to make 30 ml. Answer. Q. How much water should be added to 4000 g of 90% w/w alcohol to make 40% w/w alcohol? 0.90 x 4000g = 0.40 x V2 V2 = 0.90 x 4000 = 9000g 0. 40 V2 = 9000 g, weight of 40% w/w alcohol equivalent to 4000 g of 90% w/w alcohol So 9000 g - 4000 g = 5000 g or 5000 ml Answer.

I: 25ml alcohol x 5000ml (25%) = 1250ml alcohol Q. What is the percent of alcohol in a mixture made by mixing 5 L of 25%, I L of 50% and 1 L of 95% alcohol? a) Determine the total amount of alcohol in the three solutions and the total amount of solution (1 L = 1000mI). Assume additivity of volumes on mixing. I: 25ml alcohol x 5000ml (25%) = 1250ml alcohol 100 ml (25%) 2: 50 ml alcohol x 1000 ml(50%) = 500 mL aIcohol 100ml (50%) 3: 95 ml alcohol x 1000 ml (95%) = 950 ml alcohol 100 mL (95%) Total amount of alcohol = 2700ml Total amount of solution = 7000ml Answer

b) Determine the percent of alcohol in the mixture. There is a total of 2700 ml of alcohol in 7000 ml of total solution. 2700ml alcohol x 100 = 38.6% alcohol 7000ml mixture Answer Q. What is the strength of a mixture obtained by mixing 50 g of a 5%,100 g of a 7.5% and 40 g of a 10% ointment? 1. 5gdrug x 50 g oint (5%) = 2.5 g drug 100 g oint (5%) 2. 7.5 g drug x 100 g oint (7.5%) = 7.5 g drug 100 g oint (7.5%) 3. 10 g drug x 40 g oint (10%) = 4.0 g drug 100 g oint (10%)

There is a total of 14.0 g of active ingredient in 190 g of total mixture. 14.0 g drug x 100 g mixture = 7.37% 190 g mixture Answer ALLIGATION ALTERNATE Alligation is a rapid method of calculation that is useful to the pharmacist. The name is derived from the Latin alligatio, meaning the act of attaching, and it refers to lines drawn during calculation to bind quantities together.

This method is used to find the proportions in which substances of different strengths or concentrations must be mixed to yield a mixture of desired strength or concentration. When the proportion is found, a calculation may be performed to find the exact amounts of the substances required. 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.

Q. In what proportion must a preparation containing 10% of drug be mixed with one containing 15% of drug to produce a mixture of 12% drug strength? 15% 2 parts of 15% 12% 10% 3 parts of 10% 5 parts of 12% Answer Q. In what proportion must 30% alcohol and 95% alcohol be mixed to make 500 ml of 50% alcohol? 95% 20 parts of95% 50% 30% 45 parts of 30% 65 parts of 50%

In a total of 65 parts, 20 parts of 95% alcohol + 45 part of 30% alcohol are needed. Since the total is proportional to 500 ml, the following can be calculated; 20 parts of 95% x 500 (ml) 50% = 153.85 ml 95% 65 parts of 50% 45 parts (ml) 30% x 500 ml 50% = 346 ml 30% 65 parts (ml) 50% 500ml of 50 % Answer Since volumes are not additive, sufficient water may be needed to make 500 mI Q. How many grams of an ointment containing 0.18% of active ingredient must be mixed with 50 grams of an ointment containing 0.14% of active ingredient to make a product containing 0.15% of active ingredient?

18% 0. 01 parts of 0. 18% 0. 15% 0. 14% 0. 03 parts (g) 0. 14% 0 0.18% 0.01 parts of 0.18% 0.15% 0.14% 0.03 parts (g) 0.14% 0.04 parts of 0.15 % 0.03 : 50 g :: 0.01 : X X = 50 x 0.01 = 16.6 g 0.18% 0.03 Answer Q. A prescription calls for 50 g of a 10% ointment. The pharmacist only has a 5% ointment and the pure ingredient available. How much of the 5% ointment and the pure ingredient are needed to compound the prescription?

100% 5 parts of 100% 10% 5% 90 parts of 5% 95 parts of 10% 5 parts (g) 100% : 95% parts (g) 10% :: X: 50 g 10% X = 5 x 50 = 2.63 g 100% 95 90 parts (g) 5% : 95 parts (g) 10% :: X : 50 g 10% X = 90 x 50 = 47.4 g 5% Total = 2.63g + 47.4g = 50.03g of 10% ointment Answer

Home work (p: 121 -123; Remington) 1. How many grams of a drug are needed to make 240 ml of a solution of such strength that when 5 ml is diluted to 2 qt, a 1:2500 solution results? 2. The following prescription order was received in a pharmacy. If the only R cream available is a 10% concentration, how much of the 10% cream and how much diluent are required to compound the prescription? Rx R Cream 3% .... 30 g 3. How many grams of a 1: 100 trituration contain 100µg of the active ingredient? 4. How many gram of a 1: 1000 dilution can be made from 1 g of a 1:25 trituration.

Problems related to alligation 1. How much ointment containing 12% drug and how much ointment containing 16% drug must be used to make 1 kg of a product containing 12.5% drug? 2. In what proportion should 50% alcohol and purified water be mixed to make a 35% alcohol solution? (The purified water is 0% Alcohol) 3. How many grams of 28% w/w ammonia water should be added to 500 g of 5% w/w ammonia water to produce a 10% w/w ammonia concentration? 4. How many ml of 20% dextrose in water and how many ml of 50% dextrose in water are needed to make 1 L of 35% dextrose in water?