Question

A solution contains 0.1 M $ \text{Mg}^{2+} $ ion. Indicate the optimum pH value so that $ \text{Mg(OH)}_2 $ can be precipitated from this solution ($ K_{\text{sp}}[\text{Mg(OH)}_2] = 1.0 \times 10^{-11} $).

• A. 9
• B. 5
• C. 8
• D. 6

Hint:

To precipitate a salt like \( \text{Mg(OH)}_2 \), the pH needs to be high enough to generate a sufficient concentration of hydroxide ions to exceed the solubility product \( K_{\text{sp}} \).

Answer 1

The Correct Option is A

The solubility product \( K_{\text{sp}} \) for \( \text{Mg(OH)}_2 \) is given as \( 1.0 \times 10^{-11} \). The dissolution equilibrium for \( \text{Mg(OH)}_2 \) is:
\[ \text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^- \] The concentration of \( \text{OH}^- \) ions required to precipitate \( \text{Mg(OH)}_2 \) can be found by solving for \( [\text{OH}^-] \) using the equation: \[ K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2 \] Substitute \( [\text{Mg}^{2+}] = 0.1 \, \text{M} \) and \( K_{\text{sp}} = 1.0 \times 10^{-11} \): \[ 1.0 \times 10^{-11} = (0.1)[\text{OH}^-]^2 \] Solving for \( [\text{OH}^-] \): \[ [\text{OH}^-] = \sqrt{\frac{1.0 \times 10^{-11}}{0.1}} = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, \text{M} \] The pOH is 
given by:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(1.0 \times 10^{-5}) = 5 \] Thus, the pH is: \[ \text{pH} = 14 - \text{pOH} = 14 - 5 = 9 \]