According to solubility rules and the solubility product constant (K_sp) for Fe(OH)_2, above what Fe^{2+} concentration will Fe(OH)_2 begin to precipitate in a solution with [OH^-] = 1.0 imes 10^{-4} M? (K_{sp} = 8.0 imes 10^{-16})

• A. 8.0 imes 10^{-8} M
• B. 8.0 imes 10^{-4} M
• C. 8.0 imes 10^{-8} M
• D. 8.0 imes 10^{-8} M

Answer: (A)

Verified step by step guidance

1. Write the dissociation equation for Fe(OH)\_2: \(\mathrm{Fe(OH)_2 (s) \rightleftharpoons Fe^{2+} (aq) + 2 OH^- (aq)}\).
2. Express the solubility product constant \(K_{sp}\) in terms of ion concentrations: \(K_{sp} = [Fe^{2+}][OH^-]^2\).
3. Rearrange the \(K_{sp}\) expression to solve for the concentration of \(Fe^{2+}\): \([Fe^{2+}] = \frac{K_{sp}}{[OH^-]^2}\).
4. Substitute the given values into the equation: \(K_{sp} = 8.0 \times 10^{-16}\) and \([OH^-] = 1.0 \times 10^{-4} \; M\).
5. Calculate \([Fe^{2+}]\) using the substituted values to find the concentration above which \(Fe(OH)_2\) will begin to precipitate.