Question

The $K_{sp}$ of $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$ is $1\times {10}^{-12}$. $0.01\mathrm{M}\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$ will precipitate at the limiting pH equal to________

Answer 1

Solubility product

• A substance's ability to dissolve in a solvent to form a solution is known as a substance's solubility.
• Temperature affects the value of the solubility product $\left(K_{sp}\right)$, which is a type of equilibrium constant.
• Due to increasing solubility, $K_{sp}$ often rises as temperature rises.

Solubility of $\mathbf{Mg}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}$

• The reaction is as follows:

$\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\to {\mathrm{Mg}}^{2+}+2{\mathrm{OH}}^{-}$

• Let $s_1$ is the solubility of ${\mathrm{Mg}}^{2+}$ and $s_2$ is the solubility of ${\mathrm{OH}}^{-}$.
• The concentration of ${\mathrm{Mg}}^{2+}$, $\left[{\mathrm{Mg}}^{2+}\right]=0.01\mathrm{M}$
• Therefore, $s_1=0.01$
• The solubility of ${\mathrm{OH}}^{-}$ ion is as follows:

$\begin{array}{rcl}{K}_{sp}& =& \left(s_1\right){\left(s_2\right)}^2\\ 1\times {10}^{-12}& =& \left(0.01\right){\left(s_2\right)}^2\\ s_2& =& \sqrt{\frac{1\times {10}^{-12}}{0.01}}\\ & =& 1\times {10}^{-5}\end{array}$

Calculation for pOH

• The concentration of ${\mathrm{OH}}^{-}$ ion, $\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-5}\mathrm{M}$.
• The pOH is as follows:
  :$\begin{array}{rcl}\mathrm{pOH}& =& -\log \left[{\mathrm{OH}}^{-}\right]\\ & =& -\log \left(1\times {10}^{-5}\right)\\ & =& 5\end{array}$

Calculation for pH

• The pH is as follows:

$\begin{array}{rcl}\mathrm{pH}& =& {\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH}\end{array}$

• The $\begin{array}{rcl}{\mathrm{pK}}_{\mathrm{w}}& =& 14\end{array}$.
• The pH of the solution is as follows:

$\begin{array}{rcl}\mathrm{pH}& =& {\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH}\\ & =& 14-5\\ & =& 9\end{array}$

• Therefore, the pH is equal to $\begin{array}{rcl}& & 9\end{array}$.