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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 methodsFirst Class – add a material to a hypotonic solution to adjust the freezing point

depression to -0.52° C

a) Cryoscopic methodCalculate how much sodium chloride required to further drop FPD by X°C

b) Sodium chloride equivalenceCalculate 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

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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 = -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

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

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.

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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…

313.04.3

45.58x

187

4.3

)(x

NaClL

MW

MW

LSCE

iso

NaCliso

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

123.0 SCE

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

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

Start with drug powder, make a isotonic solution, then make up to volume with isotonic NaCl

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