For a sparingly soluble salt $\mathrm{AB}_2$, the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $B^{-}$ 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 :

<p>For a sparingly soluble salt $\mathrm{AB}_2$, the dissolution in water can be represented by the following equilibrium equation:</p> <p>$ \mathrm{AB}_2 \rightleftharpoons \mathrm{A}^{2+} + 2\mathrm{B}^{-} $</p> <p>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:</p> <p>$ K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^2 $</p> <p>Given in the problem, the equilibrium concentrations are:</p> <p>$ [\mathrm{A}^{2+}] = 1.2 \times 10^{-4} \mathrm{M} $</p> <p>$ [\mathrm{B}^{-}] = 0.24 \times 10^{-3} \mathrm{M} = 2.4 \times 10^{-4} \mathrm{M} $</p> <p>Substituting these values into the $K_{sp}$ expression:</p> <p>$ K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2 $</p> <p>Calculating $K_{sp}$:</p> <p>$ K_{sp} = 1.2 \times 10^{-4} \times (5.76 \times 10^{-8}) $</p> <p>$ K_{sp} = 6.912 \times 10^{-12} $</p> <p>Therefore, the correct option is:</p> <p>Option C: $6.91 \times 10^{-12}$</p>