    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 ${\mathrm{H}}_{2}\mathrm{S}$, ${K}_{net}={K}_{1}{K}_{2}=1×{10}^{-21}$.

Open in App
Solution

## 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×{10}^{-22}$, overall dissociation constant of ${\mathrm{H}}_{2}\mathrm{S}$, ${K}_{NET}={K}_{1}{K}_{2}=1×{10}^{-21}$.Step 2: We know that, $\left[{\mathrm{Zn}}^{2+}\right]\left[{S}^{2-}\right]={K}_{\mathit{s}\mathit{p}}$Step 3: So, $\left[{\mathrm{S}}^{2-}\right]=\frac{1.25×{10}^{-22}}{0.05}=2.5×{10}^{-21}M$Now, dissociation of ${\mathrm{H}}_{2}\mathrm{S}$ is ${\mathrm{H}}_{2}\mathrm{S}⇌2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$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, ${10}^{-21}=\frac{{\left[{H}^{+}\right]}^{2}×2.5×{10}^{-21}}{0.1}\phantom{\rule{0ex}{0ex}}{\left[{H}^{+}\right]}^{2}=\frac{0.1×{10}^{-21}}{2.5×{10}^{-21}}=\frac{1}{25}=0.2M$Therefore, the minimum molar concentration $\left(\mathrm{M}\right)$ of ${\mathrm{H}}^{+}$ required to prevent the precipitation of $\mathrm{ZnS}$ is $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbit{M}$.  Suggest Corrections  1      Similar questions  Explore more