Question

A solution contains $0.1M$ $H_{2}S$ and $0.3M$ $HCl$. Calculate the concentration of $S^{2-}$ and ${HS}^{-}$ ions is solution. Given $K_{a_1}$ and $K_{a_2}$ for $H_{2}S$ are ${10}^{-7}$ and $1.3\times {10}^{-13}$ respectively.

Answer 1

$H_{2}S\rightleftharpoons H^{+}+{HS}^{-}$
$K_{a_1}=\frac{\left[H^{+}\right]\left[{HS}^{-}\right]}{\left[H_{2}S\right]}$
Further ${HS}^{-}\rightleftharpoons H^{+}+S^{2-}$
$K_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[{HS}^{-}\right]}$
Multiplying both the equations
$K_{a_1}\times K_{a_2}=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]}$
Due to common ion, the ionisation of $H_{2}S$ is suppressed and the $\left[H^{+}\right]$ in solution is due to the presence of $0.3M$ $HCl$
$\left[S^{2-}\right]=\frac{K_{a_1}\times K_{a_2}\left[H_{2}S\right]}{{\left[H^{+}\right]}^2}=1.44\times {10}^{-20}M$
Putting the value of $\left[S^{2-}\right]$ in eq. (ii)
$1.3\times {10}^{-13}=\frac{0.3\times 1.44\times {10}^{-20}}{\left[{HS}^{-}\right]}$
or $\left[{HS}^{-}\right]=3.3\times {10}^{-8}$