Question

The solubility product (K_sp) of BaSO₄ is 1.5 X 10^-9.Calculate the solubility of barium sulphate in pure water and in 0.1 M BaCl₂.

Answer 1

(i) Let the solubility of BaSO₄ in pure water be x mol/L

In pure water `K_"sp" = [Ba^(2+)][SO_4^(2-)]`

`= x xx x`
`1.5 xx 10^(-9) = x^2`
`x = 3.87 xx 10^(-5) mol//L`

(ii) Let y be the solubility of BaSO₄ in 0.1 BaCl₂ solution.
Therefore,

`[Ba^(2+)] = 0.1 + y ≈ 0.1 mol//L`

`[SO_4^(2-)] = y mol//L`

Now, 
`K_(sp) = [Ba^(2+)][SO_4^(2-)]`
`1.5 x 10^(-9) = 0.1 xx y`
`y = 1.5 x 10^(-8) mol//L`