JEE PYQ: Ionic Equilibrium Question 5
Question 5 - 2021 (18 Mar Shift 2)
Type: Numerical
The solubility of $\text{CdSO}_4$ in water is $8.0 \times 10^{-4}$ mol $\text{L}^{-1}$. Its solubility in 0.01 M $\text{H}_2\text{SO}_4$ solution is ____. (Round off to the nearest integer) (Assume that solubility is much less than 0.01 M)
Show Answer
Answer: 64
Solution
In pure water: $K_{sp} = S^2 = (8 \times 10^{-4})^2 = 64 \times 10^{-8}$
In 0.01 M $\text{H}_2\text{SO}_4$:
$\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}$ gives $[\text{SO}_4^{2-}] = 0.01$
$\text{CdSO}_4 \rightleftharpoons \text{Cd}^{2+} + \text{SO}_4^{2-}$
$K_{sp} = x(x + 0.01)$
$x + 0.01 \approx 0.01$
$x(0.01) = 64 \times 10^{-8}$
$x = 64 \times 10^{-6}$ M
BETA
