Question

An aqueous solution contains 0.10 M $H_{2}S$ and 0.20 M $HCl$. If the equilibrium constants for the formation of $H{S}^{-}$from $H_{2}S$ is $1.0\times {10}^{-7}$ and that of $S^{2-}$ from $H{S}^{-}$ ions is $1.2\times {10}^{-13}$, then the concentration of $S^{2-}$ ions in aqueous solution is:

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

Answer 1

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

According to the given data,
$H_{2}S\rightleftharpoons H{S}^{-}+H^{+};K_1=1\times {10}^{-7}...\left(1\right)$
$H{S}^{-}\rightleftharpoons S^{2-}+H^{+};K_2=1.2\times {10}^{-13}...\left(2\right)$

Adding (1) and (2)
we get, $H_{2}S\rightleftharpoons S^{2-}+2H^{+}$ $K_{eq}=K_1\times K_2=1.2\times {10}^{-20}$

Since $HCl$ is a strong acid, we can assume all the $H^{+}$ ions to come from $HCl$. Thus, $\left[H^{+}\right]$=0.2 M

$K_a=\frac{\left[S^{2-}\right][H^{+}{]}^2}{\left[H_{2}S\right]}$

$\left[S^{2-}\right]=\frac{1.2\times {10}^{-20}\times \left[H_{2}S\right]}{[H^{+}{]}^2}$
=$\frac{1.2\times {10}^{-20}\times {10}^{-1}}{4\times {10}^{-2}}$
= $3\times {10}^{-20}M$