Given the solubility product constant K_sp for Mg(OH)_2 is 1.8 × 10^{-11}, what is the molar solubility of Mg(OH)_2 in a 0.202 M solution of Mg(NO_3)_2?

• A. 8.9 × 10^{-4} M
• B. 0.202 M
• C. 2.1 × 10^{-6} M
• D. 1.8 × 10^{-11} M

Answer: (C)

Verified step by step guidance

1. Write the dissociation equation for magnesium hydroxide: \(\mathrm{Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2 OH^- (aq)}\).
2. Express the solubility product constant \(K_{sp}\) in terms of the ion concentrations: \(K_{sp} = [Mg^{2+}][OH^-]^2\).
3. Identify the initial concentration of \(Mg^{2+}\) from the \(Mg(NO_3)_2\) solution, which is 0.202 M, and let the molar solubility of \(Mg(OH)_2\) be \(s\). Then, \([Mg^{2+}] = 0.202 + s \approx 0.202\) (since \(s\) is very small) and \([OH^-] = 2s\).
4. Substitute these concentrations into the \(K_{sp}\) expression: \$1.8 \times 10^{-11} = (0.202)(2s)^2$.
5. Solve the equation for \(s\) (the molar solubility) by isolating \(s\) and calculating its value without performing the final arithmetic.