Question

Compare the molar solubility of Mg(OH)₂ in pure water with that in a solution buffered at pH 9.00.

Short answer

The solubility of Mg(OH)₂ is more in a solution that is buffered with pH 9.00.

Step-by-step solution

The concentration of hydroxide ion in solution buffered at pH 9.00

Given pH of the solution is 9.00.

Hence, $\left[O{H}^{-}\right]=1.0\times {10}^{-5}M$

Molar solubility of Mg(OH)₂ in this solution:

localid="1661941913777" $$\begin{gathered} Mg{\left(OH\right)}_2\left(s\right)\to M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right) \\ -s2s \\ {K}_{sp}=\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2 \\ 1.2\times {10}^{-11}=\left[M{g}^{2+}\right]\times {\left(1.0\times {10}^{-5}\right)}^2 \\ \left[M{g}^{2+}\right]=0.12mol/L \end{gathered}$$

Thus, the molar solubility of Magnesium (II) is 0.12 mol/L.----------(1)

Molar solubility of Mg(OH)2 in pure water

The molar solubility of can be calculated as shown below:

$$\begin{gathered} Mg{\left(OH\right)}_2\left(s\right)\square M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right) \\ -s2s \\ {K}_{sp}=4s^3 \\ 1.2\times {10}^{-11}=4s^3 \\ s=1.44\times {10}^{-4}mol/L \end{gathered}$$----------------(2)

Comparison

If the solubilities of (1) and (2) are compared, the solubility of Mg(OH)₂ is more in a solution that is buffered with pH 9.00.