MHTCET 18 Apr 2026 Shift 1 Paper

Concept:The solubility product is the equilibrium constant for the dissociation of a sparingly soluble salt.It relates the molar solubility $s$ to the concentrations of the ions produced.Explanation:The sparingly soluble salt $\mathrm{AB}_{2}$ dissociates in water as follows:$$\mathrm{AB}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{A}^{2+}(\mathrm{aq}) + 2\mathrm{B}^{-}(\mathrm{aq})$$Let the molar solubility of $\mathrm{AB}_{2}$ be $s = 1 \times 10^{-6} \ \mathrm{mol\ dm}^{-3}$.From the dissociation equation, one formula unit of $\mathrm{AB}_{2}$ gives one $\mathrm{A}^{2+}$ ion and two $\mathrm{B}^{-}$ ions.Therefore, the equilibrium concentrations are:$$[\mathrm{A}^{2+}] = s$$$$[\mathrm{B}^{-}] = 2s$$The solubility product expression for $\mathrm{AB}_{2}$ is:$$K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^{2}$$Substitute the concentrations into the expression:$$K_{sp} = s(2s)^{2} = s \times 4s^{2} = 4s^{3}$$Now substitute $s = 1 \times 10^{-6}$:$$K_{sp} = 4\left(1 \times 10^{-6}\right)^{3}$$$$K_{sp} = 4 \times 10^{-18}$$Answer:The solubility product of $\mathrm{AB}_{2}$ is $4 \times 10^{-18}$.Hence, the correct option is D: $4 \times 10^{-18}$.