Isotonicity: Pharmaceutical Calculations

Table of Contents

What is isotonicity?

For a solution to be termed isotonic (equal tone) it must have the same osmotic pressure as a specific bodily fluid.

In pharmacy, isotonicity calculations are most often performed for parenteral and ophthalmic solutions which must have a freezing point depression of 0.52^◦C for them to be isotonic with blood plasma and tears; i.e pure water freeze at 0^◦C degree centigrade & while blood/tears freeze at -0.52^◦C (minus 0.52^◦C) – means on addition/dissolving to anything to water which result into deress in to freezing point.

To calculate the osmotic pressure of a solution is to utilize the more easily measured property of the freezing point depression as they are proportional to one another.

Why is it important?

It is important for a solution to be isotonic with a bodily fluid to prevent irritation and cell damage, and to maximize drug efficacy.

If a hypotonic solution (with lower osmotic pressure than that of body fluid) is administered intravenously water will pass into the red blood cells, causing them to swell and possibly burst (haemolysis).

If a hypertonic solution (with higher osmotic pressure than that of body fluid) is administered intravenously then water is drawn from the cells in an attempt to dilute the solution, causing them to shrink (crenation).

Freezing Point Depression Method

A hypotonic solution can be made isotonic by adding an adjusting substance, usually sodium chloride.

The exact amount of substance to be added is to be calculated by following the formula:-

Isotonicity, Pharmaceutical calculation, pharmapedia — Freezing point depression

W=Weight of the added substance (g/100ml);
a=Freezing point depression of the unadjusted hypotonic solution;
b=Freezing point depression of a 1% w/v solution of the adjusting substance.

Example

The freezing point depression of a 1% w/v solution of morphine sulfate is 0.08◦C and that of 1% w/v sodium chloride solution is 0.576. How many grams each of morphine sulfate and sodium chloride are required to prepare 50 ml of a 1% w/v morphine sulfate solution isotonic with blood plasma?

Note:

Confusing ‘freezing point’ and ‘freezing point depression’ is common source of error in isotonicity.
A solution with freezing point depression 0.5◦C has freezing point −0.5◦C as ‘depression’ indicates a decrease in value.
All units must be in same manners for calculation.

Solution:-
The freezing point of (a 1% w/v solution of) morphine sulfate is −0.08^◦C (means 1 % w/v morphine will reduce 0.08^◦C in freezing point, but we require to reduce 0.52 ^◦C for Isotonicity. The required freezing point is 0.52◦C, therefore the freezing point need to be lowered by: 0.52◦C – 0.08◦C (produced by morphine sulfate)= 0.44^◦C; we can say that we will require to 0.44^◦C depress the freezing point using NaCl substance. A 1% w/v sodium chloride solution produces a freezing point depression of 0.576^◦C, therefore the weight of sodium chloride required to produce a depression of 0.44◦C is calculated by

W=Weight of the added substance (g/100ml);
a=Freezing point depression of the unadjusted hypotonic solution;
b=Freezing point depression of a 1% w/v solution of the adjusting substance

0.7638 g NaCl is required for 100 ml, but as per the question, we have to prepare 50 mL so

morphine sulfate required for 50 ml, 1% w/v solution(i.e 1 g in to 100 ml) of morphine sulfate so for 50 ml solution, we require morphine=(1g/100ml) x50 ml=0.5 g.
Like this, 1 % w/v NaCl required for 50 mL solution=( 0.7638 g/100 mL)x 50 mL= 0.3819g

Example

A pharmacist receives a prescription for 10 ml of isotonic 0.5% w/v chloramphenicol eye drops, what weight of sodium chloride is required to make the solution isotonic with tears? Given data:- A 1% w/v solution of chloramphenicol has a freezing point depression of 0.06◦C &1% w/v sodium chloride produces a freezing point depression of 0.576◦C./ 0.9% w/V NaCl produces a freezing point depression of 0.52^◦C.

Solutions:

1% w/v solution of chloramphenicol has a freezing
point depression of 0.06◦C. Therefore a 0.5% w/v solution has a freezing point depression of 0.06◦C×0.5 = 0.03◦C. The required freezing point is −0.52◦C, therefore the freezing point must be lowered by: 0.52◦C−0.03◦C = 0.49◦C. 1% w/v sodium chloride produces a freezing point depression of 0.576◦C, therefore the weight of sodium chloride required to produce a depression of 0.49◦C is: