Chemistry - Ionic Equilibrium Question with Solution | TestHub

For H₂S,

${\text{H}}_2\text{S}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-}$

${\text{K}}_{\text{a}}={\text{K}}_{{\text{a}}_1}{\text{K}}_{{\text{a}}_2}=\frac{{\left[{\text{H}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}\text{;}{\left[{\text{H}}^{+}\right]}^2=\frac{10^{-21}\left[{\text{H}}_2\text{S}\right]}{\left[{\text{S}}^{2-}\right]}$

${\left[{\text{H}}^{+}\right]}^2=\frac{10^{-22}}{\left[{\text{S}}^{2-}\right]}$

$\Rightarrow$[S^{2 -}] for precipitation of Ag₂S  is

${\text{K}}_{\text{sp}}/{\left[{\text{Ag}}^{+}\right]}^2=\frac{6\text{.}3\times 10^{-50}}{{\left(10^{-2}\right)}^2}=6\text{.}3\times 10^{-46 }\text{M}$

i.e., too low value it will precipitate at any pH.

$\Rightarrow$[S^{2 -}] for precipitation of CoS$\Rightarrow$$\frac{4\times 10^{-21}}{10^{-2}}=4\times 10^{-19}$

$\therefore$$\left[{\text{H}}^{+}\right]=\sqrt{\frac{10^{-22}}{4\times 10^{-19}}}=1\text{.}58\times 10^{-2 }\text{M}$; pH = 1.8

$\Rightarrow$[S^{2 -}] for precipitation of MnS is

$\frac{2\text{.}5\times 10^{-10}}{10^{-2}}=2\text{.}5\times 10^{-8}$

$\left[{\text{H}}^{+}\right]=\sqrt{\frac{10^{-22}}{2\text{.}5\times 10^{-8}}}=6\text{.}3\times 10^{-8 }$;  pH = 7.2

pH < 1.8  Ag₂S will precipitate out.

pH - 1.8 to 7.2 only CoS willprecipitateout.

pH > 7.2  only MnS willprecipitateout.