Question

Consider an aqueous solution containing $0.001M{H}_{2}S$ and $0.25MHCl$. If the equilibrium constants for the formation $H{S}^{-}$ from $H_{2}S$ is $1.0\times {10}^{-7}$ and that of $S^{2-}$ from $H{S}^{-}$ is $1.2\times {10}^{-13}$. Then the concentration of $S^{2-}$ ions in aqueous solution is:

• A. $5\times {10}^{-16}M$
• B. $1.92\times {10}^{-22}M$
• C. $2.6\times {10}^{-24}M$
• D. $4.6\times {10}^{-18}M$

Answer 1

The correct option is B $1.92\times {10}^{-22}M$

$[H^{+}{]}_{total}\approx [H^{+}{]}_{HCl}$ (from strong acid only)
$\left[H_{2}S\right]=0.001M\left[HCl\right]=0.25M\Rightarrow \left[H^{+}\right]=0.25M{H}_{2}S\left(aq\right)\rightleftharpoons H{S}^{-}\left(aq\right)+H^{+}\left(aq\right);K_1=1.0\times {10}^{-7}$
$K_{a1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$

$H{S}^{-}\left(aq\right)\rightleftharpoons S^{2-}\left(aq\right)+H^{+}\left(aq\right);K_2=1.2\times {10}^{-13}$
$K_{a2}=\frac{\left[H^{+}\right]\left[S^{2^{-}}\right]}{\left[H{S}^{-}\right]}$
Multiplying $K_{a1}$ and $K_{a2}$

$K_{a1}\times K_{a2}=(1\times {10}^{-7})\times (1.2\times {10}^{-13})=1.2\times {10}^{-20}1.2\times {10}^{-20}=\frac{\left[S^{2-}\right]\times [H^{+}{]}^2}{\left[H_{2}S\right]}\left[S^{2-}\right]=\frac{1.2\times {10}^{-20}\times \left[H_{2}S\right]}{[H^{+}{]}^2}=\frac{1.2\times {10}^{-20}\times 0.001}{(0.25{)}^2}=1.92\times {10}^{-22}M$