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Transcript
ISOTONICITY #2 www.monash.edu.au Questions and Feedback 2 Methods for Adjustment of Tonicity Two ‘Classes’ of methods First Class – add a material to a hypotonic solution to adjust the freezing point depression to -0.52° C a) Cryoscopic method Calculate how much sodium chloride required to further drop FPD by X°C b) Sodium chloride equivalence Calculate contribution of drug in terms of sodium chloride equivalent and make up to 0.9% with addition of NaCl Second ‘class’ – start with drug powder, make an isotonic drug solution, then make up to final volume with isotonic salt solution or isotonic buffer c) White-Vincent method 3 Cryoscopic method (using an example) Make 100 mL of a 1% solution of boric acid isotonic with blood, by adding NaCl The freezing point of 1% Boric Acid = -0.290 C Overall for isotonic we want -0.520 C. Need to adjust FD by further 0.230 C, the FPD of 1% NaCl = 0.580 C 1% 0.58 = X therefore X = 0.4%. 0.23 To 1.0 g of boric acid in 100 mL, add 0.4 g of NaCl 4 Sodium chloride equivalence method A 0.90% NaCl solution is isotonic. Therefore the total SCE for an isotonic formulation needs to be 0.9: % of A x SCE(A) + % of B x SCE(B) + … = 0.9 note: SCE may be written just as E in some places (eg, Martin) 5 example You are asked to prepare a formulation of a new amphetamine hydrochloride derivative, for IV injection at 1%. What quantity of sodium chloride would you need to add to ensure that the fluid in the IV bag is isotonic to blood serum? SCE amphetamine hydrochloride derivative = 0.313 SCE of NaCl = 1 Required % of drug: 1.0 % of NaCl x 1 = 0.9 - ( % of drug X SCE ) = 0.9 - ( 1 X 0.313 ) = 0.587 % You would need to add 0.587% NaCl to ensure that it is isotonic with blood serum. 6 what if we weren’t adding NaCl? Eye drops commonly use boric acid to adjust the tonicity. You are asked to prepare an isotonic formulation of 6.25% streptomycin sulfate (SCE = 0.06) and 0.5% chlorbutol (SCE = 0.24). What amount of boric acid (SCE = 0.5) is necessary? %A x SCE(A) + %B x SCE(B) + … = 0.9 You would need to add 0.81% boric acid to ensure that the eye drops are isotonic. 7 we can calculate SCE for a new drug… if the new drug has a MW of 187 and is a 1:1 electrolyte (with an Liso of 3.4), then we can calculate its SCE: MWNaCl Liso 3.4 58.45 SCE x x 0.313 MW Liso ( NaCl ) 187 3.4 8 example if the new drug has a MW of 265 and is a nonelectrolyte (Liso = 1.9), calculate its SCE: SCE 0.123 9 White-Vincent method Start with drug powder, make a isotonic solution, then make up to volume with isotonic NaCl Example: ‘make 30 mL of a 1% procaine hydrochloride solution isotonic with body fluids by adding NaCl’. Its SCE is 0.21. Amount of drug is 1% x 30 mL = 0.3 g Calculate what weight of NaCl this is osmotically equivalent to: Weight drug x SCE = 0.3 x 0.21 = 0.063g example taken from Martin 10 cont’d For an isotonic solution the volume, V, required for 0.063g of NaCl (or 0.3 g of drug) is: 0.9g/100 mL (isotonic saline is 0.9%) = 0.063/V → V = 7 mL So make up 0.3g of drug in 7 ml water Top up with 23 mL of 0.9% NaCl to 30 mL total 11 example Make 60 mL of a 1% phenobarbital sodium (MW=254) isotonic with body fluids by adding NaCl. Liso = 3.4 Top up with 44 mL of 0.9% NaCl to 60 mL total 12