JEE Main 2024 — Ionic Equilibrium Question with Solution

For a sparingly soluble salt ${\mathrm{AB}}_2$, the dissolution in water can be represented by the following equilibrium equation:

${\mathrm{AB}}_2\rightleftharpoons {\mathrm{A}}^{2+}+2{\mathrm{B}}^{-}$

When ${\mathrm{AB}}_2$ dissolves in water, it generates one ${\mathrm{A}}^{2+}$ ion and two ${\mathrm{B}}^{-}$ ions. The solubility product constant, $K_{sp}$, for this reaction can be expressed as:

$K_{sp}=[{\mathrm{A}}^{2+}][{\mathrm{B}}^{-}{]}^2$

Given in the problem, the equilibrium concentrations are:

$[{\mathrm{A}}^{2+}]=1.2\times 10^{-4}\mathrm{M}$

$[{\mathrm{B}}^{-}]=0.24\times 10^{-3}\mathrm{M}=2.4\times 10^{-4}\mathrm{M}$

Substituting these values into the $K_{sp}$ expression:

$K_{sp}=(1.2\times 10^{-4})\times (2.4\times 10^{-4}{)}^2$

Calculating $K_{sp}$:

$K_{sp}=1.2\times 10^{-4}\times (5.76\times 10^{-8})$

$K_{sp}=6.912\times 10^{-12}$

Therefore, the correct option is:

Option C: $6.91\times 10^{-12}$