Question

The concentrations of $S^{2-}$ and $H{S}^{-}$ ions in the solution consisting of $0.1M{H}_{2}S$ and $0.3MHCl$, given that
$K_{a_1}=1.0\times {10}^{-7}$ and $K_{a_2}=1.3\times {10}^{-13}$ for $H_{2}S$

• A. $1.0\times {10}^{-7}$ and $1.3\times {10}^{-13}$
• B. $1.4\times {10}^{-20}$ and $3.3\times {10}^{-8}$
• C. $1.0\times {10}^{-13}$ and $1.3\times {10}^{-7}$
• D. None of the above

Answer 1

The correct option is B $1.4\times {10}^{-20}$ and $3.3\times {10}^{-8}$

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}{K}_{a_1}=1.0\times {10}^{-7}H{S}^{-}\rightleftharpoons H^{+}+S^{2-}{K}_{a_2}=1.3\times {10}^{-13}$
since $HCl$ superesses the ionization of $H_{2}S$ due to common ion effect, the concentration of $H^{+}$ is only due to HCl. $\left(0.3MHClgiven\right)$
$K_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}1.0\times {10}^{-7}=\frac{\left(0.3\right)\left[H{S}^{-}\right]}{0.1}\left[H{S}^{-}\right]=\frac{1.0\times {10}^{-7}\times 0.1}{0.3}=3.3\times {10}^{-8}M{K}_{a_2}=\frac{\left[H^{+}\right]\left[S^{-2}\right]}{\left[H{S}^{-}\right]}1.3\times {10}^{-13}=\frac{\left(0.3\right)\left[S^{2-}\right]}{3.3\times {10}^{-8}}\left[S^{2-}\right]=\frac{1.3\times {10}^{-13}\times 3.3\times {10}^{-8}}{0.3}\left[S^{2-}\right]=1.4\times {10}^{-20}M$