Question

Consider saturated solutions of the following compounds.

a.$\mathit{M}\mathit{g}{\mathbf{\left(}\mathbf{O}\mathbf{H}\mathbf{\right)}}_{\mathbf{2}}$

b.$\mathit{C}\mathit{d}{\mathbf{\left(}\mathbf{O}\mathbf{H}\mathbf{\right)}}_{\mathbf{2}}$

c.$\mathit{P}\mathit{b}{\mathbf{\left(}\mathbf{O}\mathbf{H}\mathbf{\right)}}_{\mathbf{2}}$

Calculate the pH of each saturated solution.

Short answer

ThepH of each given saturated solution is

a .$pH=10.52$

b.$pH=9.36$

c.$pH=9.13$

Step-by-step solution

The pH for Mg(OH)2.

The equation can be written as,

$Mg\left(O{H}_2\right)\left(s\right)\rightleftharpoons M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

And,

$\begin{array}{c}{K}_{sp}=\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ 1.8\times {10}^{-11}=\mathrm{S}\times {\left(2\mathrm{S}\right)}^2\end{array}$

Hence,

$\begin{array}{l}s={\left(\frac{1.8\times {10}^{-11}}{4}\right)}^{1/3}\\ s=0.000165\end{array}$

Also,

$\left[O{H}^{-}\right]=2s$

$\left[O{H}^{-}\right]=0.00033M$

And,

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log \left(0.00033\right)\\ =3.48\end{array}$

Therefore,

$\begin{array}{l}pH=14-pOH\\ pH=14-3.48\\ pH=10.52\end{array}$

The pH forCd(OH)2. 

The equation can be written as,

$Cd\left(O{H}_2\right)\left(s\right)\rightleftharpoons C{d}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

As,

$\begin{array}{c}{K}_{sp}=\left[C{d}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ 5.9\times {10}^{-15}=s\times {\left(2\mathrm{s}\right)}^2\end{array}$

And,

$\begin{array}{l}s={\left(\frac{5.9\times {10}^{-15}}{4}\right)}^{1/3}\\ s=0.000165\end{array}$

So,

$\left[O{H}^{-}\right]=2s$

$\left[O{H}^{-}\right]=0.000023M$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log \left(0.000023\right)\\ =4.63\end{array}$

Therefore,

$\begin{array}{l}pH=14-pOH\\ pH=14-4.63\\ pH=9.36\end{array}$

Calculating the pH for Pb(OH)2.

The equation can be written as,

$Pb\left(O{H}_2\right)\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

As,

$\begin{array}{c}{K}_{sp}=\left[P{b}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ 1.2\times {10}^{-15}=s\times {\left(2\mathrm{s}\right)}^2\end{array}$

And,

$\begin{array}{l}s={\left(\frac{1.2\times {10}^{-15}}{4}\right)}^{1/3}\\ s=0.000165\end{array}$

So,

$\left[O{H}^{-}\right]=2s$

$\left[O{H}^{-}\right]=0.000134M$

Also,

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ pOH=-\log \left(0.0000134\right)\\ =4.87\end{array}$

Therefore,

$\begin{array}{l}pH=14-pOH\\ pH=14-4.87\\ pH=9.13\end{array}$