Question

What is the minimum pH required to prevent the precipitation of ZnS in a solution that is $0.01\mathrm{\text{M}}\mathrm{\text{Z}}\mathrm{\text{n}}\mathrm{\text{C}}{\mathrm{\text{l}}}_2$ and saturated with $0.10\mathrm{\text{M}}{\mathrm{\text{H}}}_2\mathrm{\text{S}}?$
$\left[\mathrm{\text{G}}\mathrm{\text{i}}\mathrm{\text{v}}\mathrm{\text{e}}\mathrm{\text{n}}{\mathrm{\text{K}}}_{\mathrm{\text{s}}\mathrm{\text{p}}}={10}^{-21},{\mathrm{\text{K}}}_{{\mathrm{\text{a}}}_1}\times {\mathrm{\text{K}}}_{{\mathrm{\text{a}}}_2}={10}^{-20}\right]$

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

$\text{Z}\text{n}\text{S}\stackrel{}{\rightleftharpoons }\text{Z}{\text{n}}^{2+}+{\text{S}}^{2-}$
${\text{K}}_{\text{s}\text{p}}=\left[\text{Z}{\text{n}}^{2+}\right]\left[{\text{S}}^{2-}\right]$
$\text{Z}\text{n}\text{C}{\text{l}}_2\to \text{Z}{\text{n}}^{2+}+2\text{C}{\text{l}}^{-}$


$\left[{\text{S}}^{2-}\right]=\frac{{\text{K}}_{\text{s}\text{p}}}{\left[\text{Z}{\text{n}}^{2+}\right]}=\frac{{10}^{-21}}{\left(0.01\right)}$
$\Rightarrow \left[{\text{S}}^{2-}\right]={10}^{-19}$
$\text{K}{\text{a}}_1.\text{K}{\text{a}}_2=\frac{{\left[{\text{H}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}$
Where, ${\text{K}}_{{\text{a}}_1}=\frac{\left[\text{H}{\text{S}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{H}}_2\text{S}\right]}$
${\text{K}}_{{\text{a}}_2}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{S}}^{2-}\right]}{\left[\text{H}{\text{S}}^{-}\right]}$
So, ${10}^{-20}=\frac{{\left[{\text{H}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}=\frac{{\left[{\text{H}}^{+}\right]}^2\left({10}^{-19}\right)}{\left(0.1\right)}$
$\Rightarrow \left[{\text{H}}^{+}\right]=0.1$
$\Rightarrow \text{p}\text{H}=-\mathrm{\text{log}}\left[{\text{H}}^{+}\right]=1$