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Www.monash.edu.au ISOTONICITY #2. 2 Questions and Feedback.

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Presentation on theme: "Www.monash.edu.au ISOTONICITY #2. 2 Questions and Feedback."— Presentation transcript:

1 ISOTONICITY #2

2 2 Questions and Feedback

3 3 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

4 4 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 = C Overall for isotonic we want C. Need to adjust FD by further C, the FPD of 1% NaCl = C therefore X = 0.4%. To 1.0 g of boric acid in 100 mL, add 0.4 g of NaCl 1% X =

5 5 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)

6 6 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 = SCE of NaCl = 1 Required % of drug: 1.0 % of NaCl x 1 = ( % of drug X SCE ) = ( 1 X ) = % You would need to add 0.587% NaCl to ensure that it is isotonic with blood serum.

7 7 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.

8 8 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 L iso of 3.4), then we can calculate its SCE:

9 9 example if the new drug has a MW of 265 and is a nonelectrolyte (L iso = 1.9), calculate its SCE:

10 10 White-Vincent method Example: ‘make 30 mL of a 1% procaine hydrochloride solution isotonic with body fluids by adding NaCl’. Its SCE is 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 Start with drug powder, make a isotonic solution, then make up to volume with isotonic NaCl example taken from Martin

11 11 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

12 12 example Make 60 mL of a 1% phenobarbital sodium (MW=254) isotonic with body fluids by adding NaCl. L iso = 3.4 Top up with 44 mL of 0.9% NaCl to 60 mL total

13 13 example if the new drug has a MW of 265 and is a nonelectrolyte (L iso = 1.9), calculate its SCE:

14 14 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 %SS x SCE(SS) + %Ch x SCE(Ch) + %BA x SCE(BA) = 0.9 %BA = (0.9- [%SS x SCE(SS) + %Ch x SCE(Ch)])/SCE(BA) = (0.9 – [6.25x x0.24])/0.5 = 0.81% You would need to add 0.81% boric acid to ensure that the eye drops are isotonic.

15 15 example Make 60 mL of a 1% phenobarbital sodium (MW=254) isotonic with body fluids by adding NaCl. L iso = 3.4 Amount of drug is 1% x 60 mL = 0.6 g Calculate what weight of NaCl this is osmotically equivalent to: Weight drug x SCE = 0.6 x 0.23 = 0.138g

16 16 cont’d For an isotonic solution the volume, V, required for g of NaCl (or 0.6 g of drug) is: 0.9g/100 mL (isotonic saline is 0.9%) = 0.138/V → V = 15.3 mL So make up 0.6g of drug in 15.3 ml water Top up with 44.7 mL of 0.9% NaCl to 60 mL total


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