Question

$K_1$ and $K_2$ for dissociation of $H_{2}S$ are $4\times {10}^{-3}$ and $1\times {10}^{-5}$. Calculate sulphide ion concentration in $0.1$ $M{H}_{2}S$ solution.

• A. ${10}^{-4}$
• B. ${10}^{-8}$
• C. ${10}^{-5}$
• D. ${10}^{-6}$

Answer 1

The correct option is C

${10}^{-5}$

$H_{2}S\rightleftharpoons H^{\oplus }+H{S}^{\ominus }$
$K_1=\frac{\left[H^{\oplus }\right]\left[H{S}^{\oplus }\right]}{\left[H_{2}S\right]}=4\times {10}^{-3}$
$\left[H^{\oplus }\right]=C\alpha ,\left[H{S}^{\ominus }\right]=C\alpha ,\left[H_{2}S\right]=C(1-\alpha )$
or $4\times {10}^{-3}=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{(1-\alpha )}$
$4\times {10}^{-3}=\frac{0.1\times {\alpha }^2}{(1-\alpha )}$ $(1-\alpha$ should not be neglected$)$
or $\alpha =0.18$,
$\therefore \left[H^{\oplus }\right]=C\alpha =0.1\times 0.18=0.018M$
$\left[H{S}^{\ominus }\right]=C\alpha =0.1\times 0.18=0.018M$
$\left[H_{2}S\right]=C(1-\alpha )=0.1(1-0.18)=0.082M$
Now, $H{S}^{\ominus }$ further dissociate to $H^{\ominus }$ and $S^{2-}$,
$C_1\left[H{S}^{\ominus }\right]=0.018M$
$H{S}^{\ominus }$ $H^{\oplus }+S^{2-}$ $\rightleftharpoons$
$1$ $0$ $0$
$\left(1{\alpha }_1\right){\alpha }_1{\alpha }_1$
$\therefore K_2=1\times {10}^{-5}=\frac{\left[H^{\oplus }\right]\left[S^{2-}\right]}{H{S}^{\ominus }}$
because $\left[H^{\oplus }\right]$ already in solution $=0.018$ and thus, dissociation of $\left[H{S}^{\ominus }\right]$ further suppresses due to common ion effect and $1-\alpha \approx 1.$
$\therefore 1\times {10}^{-5}=\frac{0.018\times C_1{\alpha }_1}{C_1(1-{\alpha }_1)=0.018\times {\alpha }_1}$
$\therefore {\alpha }_1=\frac{1\times {10}^{-5}}{0.018}=5.55\times {10}^{-4}$
$\therefore \left[S^{2-}\right]=C_1{\alpha }_1=0.018\times 5.55\times {10}^{-4}$
$={10}^{-5}$