Question

$H_{2}S$ gas is passed through a decimolar solution of $M{n}^{2+}$ containing $0.01M\left[H^{+}\right]$ ions. Neglect hydrolysis of $S^{2-}$ ions.
Which of the following statement is correct?
Given:
$K_{sp}\left(MnS\right)=1.4\times {10}^{-15}{K}_a\left(H_{2}S\right)=1.1\times {10}^{-21}\left[H_{2}S\right]=0.1M$

• A. $MnS$ precipitates.
• B. $MnS$ does not precipitate.
• C. $(Q_{sp}{)}_{MnS}=(K_{sp}{)}_{MnS}$
• D. Cannot tell about $MnS$ precipitation.

Answer 1

The correct option is B $MnS$ does not precipitate.

The dissociation of $H_{2}S$ is suppressed by presence of $H^{+}$ ions
$\therefore \left[S^{2-}\right]$ in presence of $H^{+}$ may be calculated as follows:
$H_{2}S\left(s\right)\rightleftharpoons H^{+}\left(aq\right)+S^{2-}\left(aq\right)$
$K_a=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H_{2}S\right]}1.1\times {10}^{-21}=\frac{(0.01{)}^2\times [S^{2-}]}{0.1}$
$\left[S^{2-}\right]=1.1\times {10}^{-18}M$

Ionic product $Q_{sp}$ of $MnS=\left[M{n}^{2+}\right]\left[S^{2-}\right]=0.1\times 1.1\times {10}^{-18}{Q}_{sp}=1.1\times {10}^{-19}$
Comparing it with $K_{sp}=1.4\times {10}^{-15}$
$Q_{sp}<K_{sp}$
$\therefore MnS$ does not precipitate.