JEE PYQ: Ionic Equilibrium Question 26
Question 26 - 2019 (12 Apr Shift 2)
Type: MCQ
The molar solubility of $\text{Cd(OH)}_2$ is $1.84 \times 10^{-5}$ M in water. The expected solubility of $\text{Cd(OH)}_2$ in a buffer solution of pH $= 12$ is:
(a) $1.84 \times 10^{-9}$ M
(b) $\frac{2.49}{1.84} \times 10^{-9}$ M
(c) $6.23 \times 10^{-11}$ M
(d) $2.49 \times 10^{-10}$ M
Show Answer
Answer: (c)
Solution
$\text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2\text{OH}^-$
At equilibrium: $K_{sp} = s(2s)^2 = 4s^3$
$K_{sp} = 4 \times (1.84 \times 10^{-5})^3$
In buffer with pH $= 12$: $[\text{OH}^-] = 10^{-2}$
$\text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2\text{OH}^-$
$K_{sp} = s’ \times (2s’ + 10^{-2})^2 \approx s’ \times (10^{-2})^2$
$s’ = \frac{K_{sp}}{10^{-4}} = \frac{4 \times (1.84 \times 10^{-5})^3}{10^{-4}} = \frac{24.9 \times 10^{-15}}{10^{-4}} = 2.49 \times 10^{-10}$ M
BETA
