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




https://studylib.net/doc/8119214/isotonicity-calculations

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