For a sparingly soluble salt AB 2 , the equilibrium concentrations of A 2+ ions and B - ions are 1.2 × 10 -4 M and 0.24 × 10 -3 M , respectively. The solubility product of AB 2 is :

Chemistry · Ionic Equilibrium

JEE Main 2024 — 9 April, Shift 2 — Question 64

For a sparingly soluble salt $\mathrm{AB}_{2}$ , the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $\mathrm{B}^{-}$ ions are $1.2 \times 10^{-4} \mathrm{M}$ and $0.24 \times 10^{-3} \mathrm{M}$ , respectively. The solubility product of $\mathrm{AB}_{2}$ is :

Option A:
$0.069 \times 10^{-12}$
Option B:
$6.91 \times 10^{-12}$
Correct
Option C:
$0.276 \times 10^{-12}$
Option D:
$27.65 \times 10^{-12}$

Answer: B

Step-by-step solution

$\quad \mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+2}+2 \mathrm{~B}_{(\mathrm{aq})}^{-}$

\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{A}^{+2}\right]\left[\mathrm{B}^{-}\right]^{2} & =1.2 \times 10^{-4} \times\left(2.4 \times 10^{-4}\right)^{2} & =6.91 \times 10^{-12} \mathrm{M}^{3} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Sparingly Soluble Salts, Solubility Product & Precipitation Conditions