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ISOTONICITY #2

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**Questions and Feedback**

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**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 Cryoscopic method Calculate how much sodium chloride required to further drop FPD by X°C 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 White-Vincent method

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**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 0.58 0.23 =

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**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)

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**example SCE of NaCl = 1 Required % of drug: 1.0**

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

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**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.

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**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:

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

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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 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

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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

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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

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

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**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.

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example Make 60 mL of a 1% phenobarbital sodium (MW=254) isotonic with body fluids by adding NaCl. Liso = 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

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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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