Calculate the solubility (in grams per 1.00×10^2 mL of solution) of magnesium hydroxide in a solution buffered at pH = 12 given that the Ksp for Mg(OH)2 is 2.06×10^(-13).

• A. 1.52×10^(-4) g
• B. 2.06×10^(-5) g
• C. 7.60×10^(-5) g
• D. 3.04×10^(-4) g

Answer: (A)

Verified step by step guidance

1. Identify the dissolution equation for magnesium hydroxide: \( \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq) \).
2. Write the expression for the solubility product constant (Ksp): \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \). Given \( K_{sp} = 2.06 \times 10^{-13} \).
3. Since the solution is buffered at pH = 12, calculate the concentration of \( \text{OH}^- \) ions using the relation \( \text{pOH} = 14 - \text{pH} \). Therefore, \( \text{pOH} = 2 \) and \( [\text{OH}^-] = 10^{-2} \text{M} \).
4. Substitute \( [\text{OH}^-] = 10^{-2} \text{M} \) into the Ksp expression: \( 2.06 \times 10^{-13} = [\text{Mg}^{2+}](10^{-2})^2 \). Solve for \( [\text{Mg}^{2+}] \).
5. Convert the molar solubility of \( \text{Mg(OH)}_2 \) to grams per 100 mL of solution using the molar mass of \( \text{Mg(OH)}_2 \) and the calculated \( [\text{Mg}^{2+}] \).