Question

The solubility of BaSO₄ in water is 2.33 x 10^-4 g/100 mL. Calculate the percentage loss in weight when 0.3 g of BaSO4 is washed with 1.5 L of 0.01 N₂H₄SO . (Molar mass (BaSO₄) = 233 g/mole).

Answer 1

Loss in weight of BaSO₄ = amount of BaSO₄ soluble 

0.01 N NH₂SO₄ = 0.01 N SO₄^2- ions 

= 0.005 M SO₄^2- ions

The presence of SO₄^2- ions in the solution will suppress the solubility of BaSO₄ (due to common ion effect). The extent of suppression is governed by solubility product of salt.

Let ‘x’ be the solubility in mol/L in H₂SO₄ 

[Ba²⁺] in solution = x 

[SO₄^2-] in solution = (x + 0.005) 

Ionic product = [Ba²⁺][SO₄^2-] 

= (x) (x + 0.005)

At equilibrium

4.66 x 10^-6 g of BaSO₄ is washed away with 1L of given solution

⇒ 6.99 x 10^-6 g BaSO₄ is washed away with 1.5 L of given solution