Determination of solubility :
The solubility of salts is generally determined by gravimetric method. First of all a saturated solution is prepared. Some part of this saturated solution is weighed out in a porcelain dish. The solution is evaporated slowly to dryness on a sand bath. The dish is cooled and weighed again. The observations are recorded as follows:
1. Mass of empty dish = w g
2. Mass of dish + solution = w1 g
3. So Mass of solution = (w1 – w) g
4. Mass of dish + residue = w2 g
So Mass of residue = (w2 – w) g = x g
and Mass of solvent = (mass of solution – mass of residue)
= (w1 – w) – (w2 – w)
= (w1 – w2) = y g
Thus, the solubility of salt = x/y × 100 g per 100 g of solvent.
Example: 50 g of a saturated aqueous solution of potassium chloride at 30°C is evaporated to dryness, when 13.2 g of dry KCl was obtained. Calculate the solubility of KCl in water at 30°C.
Solution: Mass of water in solution = (50 – 13.2) = 36.8 g
Solubility of KCl = Mass of KCl/Mass of water × 100 = 13.2/36.8 × 100
Example: How much copper sulphate will be required to saturate 100 g of a dilute aqueous solution of CuSO4 at 25°C if 10 g of a dilute solution leave on evaporation and drying 1.2g of anhydrous CuSO4? The solubility of CuSO4 in water at 25°C is 25.
Solution: 100 g of dilute solution of CuSO4 contain
= 1.2 × 10 = 12.0 g CuSO4
Mass of water present in dilute solution
= (100 – 12) = 88 g
To saturate 100 g of water, CuSO4 required = 25 g
So, to saturate 88 g of water, CuSO4 required 25/100 × 88
= 22 g
Thus, the mass of CuSO4 to be added to 100 g of dilute solution to saturate it = (22 – 12) = 10 g