## b) Isotonicity calculation using freezing point depression method

*You would like to prepare 30 mL of eye drops containing 1% (w/v) of pilocarpine and 2% (w/v) procaine. How many mL of 0.9% (w/v) sodium chloride should be used to make the above eye drops isotonic?*

*(freezing point of depression of 1% solution of pilocarpine = 0.14, procaine = 0.11, sodium chloride = 0.576) *

*4 marks*

Freezing point depression units are all in degrees Celsius

You need to know:

- The
**freezing point depression**(fpd) of a solution is proportional to its**concentration**. - Reference values of freezing point depression are for
**1%**solutions (unless stated otherwise) - Total freezing point depression of an isotonic solution (NaCl 0.9 %) is 0.52C (NaCl 1 % is 0.58)

fpd

Pilocarpine 1% 0.14

Procaine 2% 0.22 (1 % is 0.11)

Total fpd here is 0.14 + 0.22 = 0.36

You want the fpd to be 0.52 in order to be isotonic.

0.52 - 0.36 = 0.16 further depression needed.

What concentration of NaCl will produce 0.16 fdp:

x = **0.28 % NaCl** **will produce 0.16 fpd**

FInal concentrations:

Final volume of eye drops required: 30 ml

Amount of g NaCl in 30 ml of 0.28% solution:

0.28% = 0.28 g/100ml

0.28 g x 30/100 = **0.084g**

Amount of ml of 0.9% NaCl which contains 0.084 g:

0.9 % = 0.9 g / 100 ml

0.084/0.9 * 100 ml = ** 9.33 ml of 0.9% NaCl **contains 0.084 g NaCl