MHTCET 17 Apr 2026 Shift 2 Paper

Concept:For a sparingly soluble salt $\mathrm{AB}_2$, the solubility product is calculated from its molar solubility using the stoichiometry of dissociation.Explanation:The salt dissociates as:$$\mathrm{AB}_2(\mathrm{s}) \rightleftharpoons \mathrm{A}^{2+}(\mathrm{aq}) + 2\mathrm{B}^-(\mathrm{aq})$$Convert the given solubility into molar solubility:$$s = \frac{18.78 \times 10^{-4}\ \mathrm{g\,dm}^{-3}}{187.8\ \mathrm{g\,mol}^{-1}} = 1.0 \times 10^{-5}\ \mathrm{mol\,dm}^{-3}$$From the dissociation equation:$$[\mathrm{A}^{2+}] = s \quad \text{and} \quad [\mathrm{B}^-] = 2s$$The solubility product expression is:$$K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^-]^2$$Substitute the concentrations:$$K_{sp} = s(2s)^2 = 4s^3$$Now substitute $s = 1.0 \times 10^{-5}$:$$K_{sp} = 4(1.0 \times 10^{-5})^3 = 4 \times 10^{-15}$$Answer:The solubility product is $4 \times 10^{-15}$, which matches option B.