To solve this problem, we need to find the solubility product $(K_{sp})$ of $Ca(OH)_2. \\$Given:$\\$• The pH of the saturated solution is $9. \\$First, calculate the concentration of $OH^-$ ions in the solution:$\\$Since pH + pOH = 14, we have:$\\ pOH = 14 - 9 = 5 \\$The concentration of $OH^-$ ions is given by:$\\ [OH^-] = 10^{-pOH} = 10^{-5} \\$The dissociation of $Ca(OH)_2$ in water is represented by the equation:$\\ Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2OH^- (aq) \\$Let $s$ be the solubility of $Ca(OH)_2$ in mol/L. Then:$\\ [Ca^{2+}] = s \\ [OH^-] = 2s \\$Since we know $[OH^-] = 10^{-5},$ we can write:$\\ 2s = 10^{-5} \\ s = \frac{10^{-5}}{2} = 5 \times 10^{-6} \\$Now, calculate the solubility product $K_{sp}: \\ K_{sp} = [Ca^{2+}][OH^-]^2 \\ K_{sp} = (s)(2s)^2 \\ K_{sp} = s \cdot 4s^2 \\ K_{sp} = 4s^3 \\$Substitute the value of $s: \\ K_{sp} = 4 \times (5 \times 10^{-6})^3 \\ K_{sp} = 4 \times 125 \times 10^{-18} \\ K_{sp} = 500 \times 10^{-18} \\ K_{sp} = 0.5 \times 10^{-15} \\$Therefore, the solubility product $K_{sp}$ of $Ca(OH)_2$ is $0.5 \times 10^{-15}. \\$This corresponds to Option 1.