Question

Given that, $K_{a_1}=1.0\times {10}^{-7}$ and $K_{a_2}=1.3\times {10}^{-13}$ for $H_{2}S$.
Find the concentration of $S^{2-}$ and $H{S}^{-}$ ions (in $mol/L$) respectively in the solution consisting of $0.1M{H}_{2}S$ and $0.5MHCl$.

• A. $5.2\times {10}^{-21}$ and $2\times {10}^{-8}$
• B. $1.0\times {10}^{-18}$ and $2.3\times {10}^{-6}$
• C. $6.5\times {10}^{-19}$ and $1.3\times {10}^{-7}$
• D. $4.6\times {10}^{-20}$ and $1.3\times {10}^{-8}$

Answer 1

The correct option is A $5.2\times {10}^{-21}$ and $2\times {10}^{-8}$

$H_{2}S\rightleftharpoons S^{2^{-}}+2H^{+}HCl\to C{l}^{-}+H^{+}Commonion$

For $H_{2}S$ :

Step 1 :

$H_{2}S\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+H{S}^{-}\left(aq\right)$

$att=0C_{1}00$

$att=t_{eq}C(1-{\alpha }_1)(C{\alpha }_1{\alpha }_2+C{\alpha }_1)C{\alpha }_1(1-{\alpha }_2)$

Step 2 :

$H{S}^{-}\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+S^{2-}\left(aq\right)$

$att=t_{eq}C{\alpha }_1(1-{\alpha }_2)(C{\alpha }_1{\alpha }_2+C{\alpha }_1)C{\alpha }_1{\alpha }_2$

$[H^{+}{]}_{total}=0.5+C{\alpha }_1+C{\alpha }_1{\alpha }_2$
SInce ${\alpha }_1,{\alpha }_2$ are very less for weak acid $H_{2}S$ and ${\alpha }_1\times {\alpha }_2$ is even a lesser value than ${\alpha }_1$. Also, strong acid $HCl$ suppresses the ionization of $H_{2}S$ due to common ion effect. The concentration of $H^{+}$ can be taken only due to HCl.
$\therefore [H^{+}{]}_{total}\approx 0.5M$

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}{K}_{a_1}=1.0\times {10}^{-7}H{S}^{-}\rightleftharpoons H^{+}+S^{2-}{K}_{a_2}=1.3\times {10}^{-13}$

$K_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}1.0\times {10}^{-7}=\frac{\left(0.5\right)\left[H{S}^{-}\right]}{0.1}\left[H{S}^{-}\right]=\frac{1.0\times {10}^{-7}\times 0.1}{0.5}=2\times {10}^{-8}M{K}_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}1.3\times {10}^{-13}=\frac{\left(0.5\right)\left[S^{2-}\right]}{2\times {10}^{-8}}\left[S^{2-}\right]=\frac{1.3\times {10}^{-13}\times 2\times {10}^{-8}}{0.5}\left[S^{2-}\right]=5.2\times {10}^{-21}M$