Question

A solution contains $0.1$ M $H_{2}S$ and $0.3$ M $HCl$. Given $K_{a_1}$ and $K_{a_2}$ for $H_{2}S$ are ${10}^{-7}$ and $1.8\times {10}^{-13}$ respectively. If the concentration of $H{S}^{-}$ ions is $x\times {10}^{-8}M$, then what is the value of $x$? (Give your answer upto two decimal points)

Answer 1

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}{K}_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}......\left(i\right)\text{Further}H{S}^{-}\rightleftharpoons H^{+}+S^{2-}{K}_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}.......\left(ii\right)$
Multiplying both the equation
$K_{a_1}\times K_{a_2}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$
Due to common ion, the ionisation of $H_{2}S$ is suppresesed and the $\left[H^{+}\right]$ in solution is due to the presence of 0.3 M HCl.
$\left[S^{2-}\right]=\frac{K_{a_1}\times K_{a_2}\left[H_{2}S\right]}{[H^{+}{]}^2}=\frac{1.0\times {10}^{-7}\times 1.8\times {10}^{-13}\times \left(0.1\right)}{(0.3{)}^2}=2\times {10}^{-20}M$
Putting the value of $\left[S^{2-}\right]$ in eq. (ii)
$1.8\times {10}^{-13}=\frac{0.3\times 2\times {10}^{-20}}{\left[H{S}^{-}\right]}or\left[H{S}^{-}\right]=\frac{0.3\times 2\times {10}^{-20}}{1.8\times {10}^{-13}}=3.33\times {10}^{-8}M$