Question

An acidified solution of $0.05M$ ${\mathrm{Zn}}^{2+}$ is saturated with$0.1M$ ${\mathrm{H}}_2\mathrm{S}$. What is the minimum molar concentration $\left(\mathrm{M}\right)$ of ${\mathrm{H}}^{+}$ required to prevent the precipitation of $\mathrm{ZnS}$?

Use$K_{sp}\left(ZnS\right)=1.25x{10}^{-22}$ and overall dissociation constant of ${\mathrm{H}}_2\mathrm{S}$, $K_{net}=K_1{K}_2=1\times {10}^{-21}$.

Answer 1

Step 1: It is given that $\left[{\mathrm{Zn}}^{2+}\right]$$=0.05M$, $\left[H_{2}S\right]$$=0.1M$, solubility product $K_{sp}\left(ZnS\right)$$=1.25\times {10}^{-22}$, overall dissociation constant of ${\mathrm{H}}_2\mathrm{S}$, $K_{NET}=K_1{K}_2=1\times {10}^{-21}$.

Step 2: We know that, $\left[{\mathrm{Zn}}^{2+}\right]\left[S^{2-}\right]=K_{\mathit{s}\mathit{p}}$

Step 3: So, $\left[{\mathrm{S}}^{2-}\right]=\frac{1.25\times {10}^{-22}}{0.05}=2.5\times {10}^{-21}M$

Now, dissociation of ${\mathrm{H}}_2\mathrm{S}$ is ${\mathrm{H}}_2\mathrm{S}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$

So, $K_{net}=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]}$

Putting the values in the equation,

we get, $$\begin{gathered} {10}^{-21}=\frac{{\left[H^{+}\right]}^2\times 2.5\times {10}^{-21}}{0.1} \\ {\left[H^{+}\right]}^2=\frac{0.1\times {10}^{-21}}{2.5\times {10}^{-21}}=\frac{1}{25}=0.2M \end{gathered}$$

Therefore, the minimum molar concentration $\left(\mathrm{M}\right)$ of ${\mathrm{H}}^{+}$ required to prevent the precipitation of $\mathrm{ZnS}$ is $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathit{M}$.