JEE PYQ: Ionic Equilibrium Question 25
Question 25 - 2019 (12 Apr Shift 1)
Type: MCQ
What is the molar solubility of $\text{Al(OH)}_3$ in 0.2 M NaOH solution? Given that, solubility product of $\text{Al(OH)}_3 = 2.4 \times 10^{-24}$:
(a) $3 \times 10^{-19}$
(b) $12 \times 10^{-21}$
(c) $3 \times 10^{-22}$
(d) $12 \times 10^{-23}$
Show Answer
Answer: (b)
Solution
Let the solubility of $\text{Al(OH)}_3$ in 0.2 M NaOH be $s$.
$\text{Al(OH)}_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^-$
$[\text{Al}^{3+}] = s$ and $[\text{OH}^-] = 3s + 0.2 \approx 0.2$
$K_{sp} = 2.4 \times 10^{-24} = s \times (0.2)^3$
$s = \frac{2.4 \times 10^{-24}}{8 \times 10^{-3}} = 3 \times 10^{-22}$ mol/L
BETA
