Dilution Concentration and Alligation Jan 2018 Dilution Concentration

Dilution, Concentration and Alligation Jan -2018

Dilution, Concentration and Alligation • There are occasions in which the dilution or concentration of a pharmaceutical preparation is either desirable or required. • 1 st quantity X 1 st concentration = 2 nd quantity X 2 nd concentration • Q 1 X C 1 = Q 2 X C 2 • Example(1): if 500 ml of 15% v/v solution are diluted to 1500 ml, what will be the percentage strength (v/v) • Q 1 X C 1 = Q 2 X C 2 • 500(ml) X 15(%) = 1500(ml) X x(%) • Example(2): if 50 ml of 1: 20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/v)? • 1 : 20 = 5% • • X = 5% Q 1 X C 1 = Q 2 X C 2 50(ml) X 5(%) = 1000(ml) X x(%) Dr. Osama A. A. Ahmed X = 0. 25% = 1: 400 2

• 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 concentrations. • Example(1): 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% and 1: 2000 = 0. 05% and 4 liters = 4000 ml • Q 1 X C 1 = Q 2 X C 2 • 4000(ml) X 0. 05(%) = X (ml) X 0. 25(%) X = 800 ml • Example(2): How many milliliters of a 1: 50 stock solution of phenylephrine hydrochloride should be used in compounding the following prescription. • R/ phenylephrine HCl 0. 25% • Rose water ad 30 ml • Sig. For the nose • 1 : 50 = 2 % • x X 2% = 30 X 0. 25% X = 3. 75 ml

• Some calculations used in pharmacy 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. • e. g. when the pharmacist need to prepare and dispense a concentrated solution and direct the patient to use small volume (5 ml, teaspoonful) and dilute it in a specified volume of water. • Example(1): How much drug should be used in preparing 50 ml of a solution such that 5 ml diluted to 500 ml will yield a 1: 1000 solution? • 1 : 1000 solution 0. 5 g : 500 ml • 0. 5 g drug in 500 ml is the same amount found in 5 ml • 0. 5 • X 5 50 X=5 g • Example(2): How many milliliters of water should be added to 300 ml of a 1: 750 w/v solution of benzalkonium chloride to make a 1: 2500 w/v solution? (Page 242) • 1 750 • X 300 • 1 2500 • 0. 4 X X = 0. 4 g X = 1000 ml 1000 – 300 = 700 ml

• Dilution of alcohol: when water and alcohol are mixed there is a physical contraction in the resultant volume. • Example: How much water should be mixed with 5000 ml 85% v/v alcohol to make 50% v/v alcohol? • Q 1 X C 1 = Q 2 X C 2 5000(ml) X 85(%) = X (ml) X 50(%) X = 8500 ml so enough water is added to make 8500 ml. • Dilution of acids: the strength of concentrated acids (official) are expressed as % w/w, while diluted acids are expressed as w/v. Consider the specific gravity of conc. Acids in calculations • Example: How many mls of 37% w/w HCl having a specific gravity of 1. 20 are required to make 1000 ml of dil HCl 10% w/v? • 1000 X 0. 1 = 100 gm of HCl 100% in 1000 ml of 10% w/v • Q 1 X C 1 = Q 2 X C 2 100 g X 100(%) = X (ml) X 37(%) X = 270 g of 37% acid 270/1. 2 = 225 ml

• Dilution and concentration of solids and semisolids: • Example: How many grams of opium containing 15 w/w of morphine and how many grams of lactose should be used to prepare 150 g of opium containing 10% w/w of morphine? • Q 1 X C 1 = Q 2 X C 2 X g X 15(%) = 150 g X 10(%) X = 100 g of 15% opium 150 g - 100 g = 50 g of lactose • Example: How many grams of 20% benzocaine ointment and how many grams of ointment base (diluent) should be used in preparing 5 lb. of 2. 5 % benzocaine ointment? • 5 lb = 454 g X 5 = 2270 g • Q 1 X C 1 = Q 2 X C 2 2270 g X 2. 5(%) = X g X 20(%) X = 283. 75 or 284 g 2270 – 284 = 1986 g of ointment base

• • Alligation Arithmatical method of solving problems that involves the mixing of solutions or mixtures of solids possessing different percentage strengths. • • Alligation medial 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. • Example: what is the percentage (v/v) of alcohol in a mixture of 300 ml 40% (v/v) alcohol and 100 ml 60% (v/v) alcohol ? Q 1 C 1 + Q 2 C 2 + Q 3 C 3 +……. . = Qt. Ct Qt = 300+100 = 400 ml 300 X 40 + 100 X 60 = 400 X Ct Ct = 18000/400 = 45 %

• • • Alligation alternate A method by which we may calculate the number of parts of two or more components of a given strength when they are mixed to prepare a mixture of desired strength. A final proportion permits us to translate relative parts to any specific denomination. • Example: in what proportion should alcohol 95% and 55% strengths be mixed to make 70% alcohol? • Available strength • 95% mi n • • 55% us f ro m desired strength e giv number of parts 15 parts (95%) s 70% giv es 25 parts ( 55%)