Question

An aqueous solution contains $0.10M{H}_{2}S$ and $0.20MHCl$. 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 ions $H{S}^{-}$ is $1.2\times {10}^{-13}$ then the concentration of $S^{2-}$ ions in aqueous solution is:

• A. $3\times {10}^{-20}$
• B. $6\times {10}^{-21}$
• C. $5\times {10}^{-19}$
• D. $5\times {10}^{-8}$

Answer 1

The correct option is A $3\times {10}^{-20}$

$\left[H_{2}S\right]=0.10M\left[HCl\right]=0.20M\Rightarrow \left[H^{+}\right]=0.2M\left(1\right)H_{2}S\rightleftharpoons H{S}^{-}+H^{+};K_1=1.0\times {10}^{-7}\left(2\right)H{S}^{-}\rightleftharpoons S^{2-}+H^{+};K_2=1.2\times {10}^{-13}$
So,
$H_{2}S\rightleftharpoons S^{2-}+2H^{+}K=K_1\times K_{2}K=1.2\times {10}^{-20}K=\frac{\left[S^{2-}\right]\times [H^{+}{]}^2}{\left[H_{2}S\right]}\left[S^{2-}\right]=\frac{1.2\times {10}^{-20}\times \left[H_{2}S\right]}{[H^{+}{]}^2}=\frac{1.2\times {10}^{-20}\times {10}^{-1}}{4\times {10}^{-2}}=3\times {10}^{-20}M$
All the $\left[H^{+}\right]$ will come from strong acid $HCl$ only.