Given, pH of the saturated solution = 9 We know that at 298K, $pH + pOH = 14$. $pOH = 14 - 9 = 5$ Therefore, the concentration of hydroxyl ions is $[OH^-] = 10^{-5} \text{ M}$. The dissociation equilibrium of $Ca(OH)_2$ is: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$ If the molar solubility of $Ca(OH)_2$ is $S$, then $[Ca^{2+}] = S$ and $[OH^-] = 2S$. Given $[OH^-] = 10^{-5} \text{ M}$, we have: $2S = 10^{-5} \Rightarrow S = 0.5 \times 10^{-5} \text{ M}$ Now, the solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ca^{2+}][OH^-]^2 = (S)(2S)^2$ $K_{sp} = (0.5 \times 10^{-5})(10^{-5})^2$ $K_{sp} = 0.5 \times 10^{-5} \times 10^{-10} = 0.5 \times 10^{-15}$