QJEEM24Ch971c: JEE Main 2024 Chemistry - Class 11 JEEM24Ch971c Solution: For a sparingly soluble salt , the equilibrium concentrations…

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