Transcript isotonics_2012_part2x

ISOTONICITY #2
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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
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
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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
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
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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…
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
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example
if the new drug has a MW of 265 and is a
nonelectrolyte (Liso = 1.9), calculate its SCE:
SCE  0.123
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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 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
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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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