Question

An aqueous solution of metal bromide,${\text{MBr}}_2\left(0.05\text{M}\right)$ is saturated with ${\text{H}}_2\text{S}$. What is the minimum pH at which MS will precipitate? $K_{sp}\text{for MS}=6.0\times {10}^{-21},$ concentration of saturated ${\text{H}}_2\text{S}=0.1\text{M}.{\text{K}}_1={10}^{-7}\text{and}{\text{K}}_2=1.3\times {10}^{-13}{\text{for H}}_2\text{S}$
log 1.04=0.017

• A. 0.98
• B. 0.45
• C. 1.2
• D. 0.02

Answer 1

The correct option is A 0.98

$\text{Minimum concentration of}{\text{S}}^{2-}\text{ions required to precipitate MS is}$

$\left[{\text{S}}^{2-}\right]=\frac{K_{sp}MS}{\left[{\text{M}}^{2+}\right]}$

$=\frac{6.0\times {10}^{-21}}{0.05}=1.2\times {10}^{-19}\text{M}$

${\text{H}}_2\text{S}\text{ionizes in solution in following steps}$

${\text{H}}_2\text{S}\rightleftharpoons {\text{H}}^{+}+{\text{HS}}^{-};{\text{K}}_1={10}^{-7}$

${\text{HS}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{S}}^{2-};{\text{K}}_2=1.3\times {10}^{-13}$

$\text{As}{\text{K}}_1=\frac{\left[H^{+}\right]\left[{\text{HS}}^{-}\right]}{\left[H_{2}S\right]}$

${\text{K}}_2=\frac{\left[{\text{H}}^{+}\right]\left[{\text{S}}^{2-}\right]}{\left[{\text{HS}}^{-}\right]}$

$\text{So}{\text{K}}_1{\text{K}}_2=\frac{{\left[{\text{H}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}$

${\left[{\text{H}}^{+}\right]}^2=\frac{{\text{K}}_1{\text{K}}_2\left[{\text{H}}_2\text{S}\right]}{\left[{\text{S}}^{2-}\right]}$

$=\frac{{10}^{-7}\times 1.3\times {10}^{-13}\times 0.1}{1.2\times {10}^{-19}}$

$=1.08\times {10}^{-2}$

$\left[{\text{H}}^{+}\right]=1.04\times {10}^{-1}$
$\text{pH}=-{\text{log}}_{10}\left[{\text{H}}^{+}\right]$
$=-{\text{log}}_{10}(1.04\times {10}^{-1})$
$=1-\text{log 1.04}$
$=0.98$