Given that the solubility product constant (K_sp) for Ca(OH)_2 at 25°C is 5.5 × 10^{-6}, what is the molar solubility of Ca(OH)_2 in pure water?

• A. 1.2 × 10^{-2} mol/L
• B. 1.2 × 10^{-2} mol/L
• C. 1.1 × 10^{-2} mol/L
• D. 1.1 × 10^{-2} mol/L

Answer: (B)

Verified step by step guidance

1. Write the dissociation equation for calcium hydroxide: \(\mathrm{Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 OH^- (aq)}\).
2. Define the molar solubility of \(\mathrm{Ca(OH)_2}\) as \(s\), which means the concentration of \(\mathrm{Ca^{2+}}\) ions is \(s\) and the concentration of \(\mathrm{OH^-}\) ions is \$2s$ because each formula unit produces two hydroxide ions.
3. Write the expression for the solubility product constant \(K_{sp}\) in terms of \(s\): \(K_{sp} = [Ca^{2+}][OH^-]^2 = s \times (2s)^2 = 4s^3\).
4. Set the expression equal to the given \(K_{sp}\) value and solve for \(s\): \$4s^3 = 5.5 \times 10^{-6}$.
5. Rearrange to isolate \(s\): \(s^3 = \frac{5.5 \times 10^{-6}}{4}\), then take the cube root of both sides to find the molar solubility \(s\).