Given the following equilibria, calculate the concentration of

each zinc species in a solution saturated with$\mathbf{Zn}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$and containing$\left[{\mathrm{OH}}^{-}\right]$

at a fixed concentration of$\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{27}}\mathbf{M}$.

$$\begin{gathered} \mathbf{Zn}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Ksp}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}} \\ \mathbf{Zn}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}^{\mathbf{+}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}} \\ \mathbf{Zn}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}} \\ \mathbf{Zn}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}} \\ \mathbf{Zn}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}} \end{gathered}$$

For a sparingly soluble salt, the solubility product is the mathematical product of the concentration of its dissolved constituent ions raised up to the power of their stoichiometric coefficient.

$$\begin{gathered} {\mathrm{M}}_{\mathrm{y}}{\mathrm{X}}_{\mathrm{z}\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{yM}}^{\mathrm{z}+}\left(\mathrm{aq}\right)+{\mathrm{zX}}^{\mathrm{y}-}\left(\mathrm{aq}\right) \\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{M}}^{\mathrm{z}+}\right]}^{\mathrm{y}}{\left[{\mathrm{X}}^{\mathrm{y}-}\right]}^{\mathrm{z}} \end{gathered}$$

The cumulative constant for the reaction

$\mathrm{M}+\mathrm{nX}\rightleftharpoons {\mathrm{MX}}_{\mathrm{n}}$

Is defined as:

${\mathrm{\beta }}_{\mathrm{n}}=\frac{\left[{\mathrm{MX}}_{\mathrm{n}}\right]}{\left[\mathrm{M}\right]{\left[\mathrm{X}\right]}^{\mathrm{n}}}$

Fixed concentration of${\mathbf{OH}}^{\mathbf{-}}$ in the solution$\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{M}$

Solubility product for $\mathbf{Zn}{\mathbf{(}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{OH}\mathbf{)}}_{\mathbf{2}\mathbf{(}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{s}\mathbf{)}}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}$

$$\begin{gathered} \mathrm{Zn}{\left(\mathrm{OH}\right)}_2\rightleftharpoons {\mathrm{Zn}}^{+2}+2{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Zn}}^{+2}\right]{\left[{\mathrm{OH}}^{-}\right]}^2 \\ \left[{\mathrm{Zn}}^{+2}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left[{\mathrm{OH}}^{-}\right]} \\ =\frac{3\times {10}^{-16}}{{\left(3.2\times {10}^{-7}\right)}^2} \\ =2.93\times {10}^{-3}\mathrm{M} \end{gathered}$$

$$\begin{gathered} {\mathrm{Zn}}^{+2}{\mathrm{OH}}^{-}\stackrel{\stackrel{{\mathrm{\beta }}_1}{\leftarrow }}{\leftarrow }{\mathrm{ZnOH}}^{+} \\ \left[{\mathrm{ZnOH}}^{+}\right]={\mathrm{\beta }}_1\left[{\mathrm{Zn}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right] \\ \left[{\mathrm{ZnOH}}^{+}\right]=1\times {10}^4\times 2.93\times {10}^{-3}\times 3.2\times {10}^{-7} \\ \left[{\mathrm{ZnOH}}^{+}\right]=9\times {10}^{-6} \end{gathered}$$

$$\begin{gathered} {\mathrm{Zn}}^{+2}+2{\mathrm{OH}}^{-}\stackrel{\stackrel{{\mathrm{\beta }}_2}{\leftarrow }}{\leftarrow }\mathrm{Zn}{\left(\mathrm{OH}\right)}_2 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_{2\left(\mathrm{aq}\right)}\right]={\mathrm{\beta }}_2\left[{\mathrm{Zn}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_2\right]=1\times {10}^4\times 2.93\times {10}^{-3}\times {\left(3.2\times {10}^{-7}\right)}^2 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_2\right]=6\times {10}^{-6} \end{gathered}$$

$$\begin{gathered} {\mathrm{Zn}}^{+2}+3{\mathrm{OH}}^{-}\stackrel{\stackrel{{\mathrm{\beta }}_3}{\leftarrow }}{\leftarrow }\mathrm{Zn}{\left(\mathrm{OH}\right)}_3^{-} \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_3^{-}\right]={\mathrm{\beta }}_3\left[{\mathrm{Zn}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^3 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_3^{-}\right]=1\times {10}^4\times 2.93\times {10}^{-3}\times {\left(3.2\times {10}^{-7}\right)}^3 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_3^{-}\right]=8\times {10}^{-9} \end{gathered}$$

$$\begin{gathered} {\mathrm{Zn}}^{+2}+4{\mathrm{OH}}^{-}\stackrel{\stackrel{{\mathrm{\beta }}_4}{\leftarrow }}{\leftarrow }\mathrm{Zn}{\left(\mathrm{OH}\right)}_4^{-2} \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_4^{-2}\right]={\mathrm{\beta }}_4\left[{\mathrm{Zn}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^4 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_4^2\right]=1\times {10}^4\times 2.93\times {10}^{-3}\times {\left(3.2\times {10}^{-7}\right)}^4 \\ \left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_4^{-2}\right]=9\times {10}^{14} \end{gathered}$$