MHTCET 11 May 2024 Shift 2 Paper

Concept:For a sparingly soluble salt, the solubility product is the product of the concentrations of its constituent ions in a saturated solution.Explanation:$\mathrm{CaCO}_{3}$ dissociates in water as follows:$$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)$$This dissociation produces one $\mathrm{Ca}^{2+}$ ion and one $\mathrm{CO}_{3}^{2-}$ ion per formula unit.Therefore, the concentration of each ion is equal to the given solubility:$$[\mathrm{Ca}^{2+}] = 7 \times 10^{-5} \, \mathrm{mol \, dm^{-3}}$$$$[\mathrm{CO}_{3}^{2-}] = 7 \times 10^{-5} \, \mathrm{mol \, dm^{-3}}$$The solubility product expression is:$$K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}]$$Substituting the ion concentrations:$$K_{sp} = (7 \times 10^{-5})(7 \times 10^{-5})$$$$K_{sp} = 49 \times 10^{-10}$$Simplifying:$$K_{sp} = 4.9 \times 10^{-9}$$Answer:The solubility product of $\mathrm{CaCO}_{3}$ at $25^{\circ}\mathrm{C}$ is $4.9 \times 10^{-9}$.Hence, the correct option is D.