Question

The solubility of CdSO₄ in water is 8.0 × 10^{-4} mol L^{-1}. Its solubility in 0.01 M H₂SO₄ solution is ________ × 10^{-6} mol L^{-1}. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)

Hint:

The Common Ion Effect always suppresses the solubility of a sparingly soluble salt.

Answer 1

Correct Answer: 64

Step 1: $K_{sp} = s^2 = (8 \times 10^{-4})^2 = 64 \times 10^{-8}$.
Step 2: In $0.01 \text{ M } H_2SO_4$, $[SO_4^{2-}] = 0.01 + s' \approx 0.01 \text{ M}$.
Step 3: $K_{sp} = [Cd^{2+}][SO_4^{2-}] \implies 64 \times 10^{-8} = s' \times 10^{-2}$. \[ s' = 64 \times 10^{-6} \text{ mol L}^{-1}. \]