Question

The dissociation constant for $H_{2}S$ and $H{S}^{-}$ are respectively ${10}^{-7}and{10}^{-13}$. The pH of $0.1M$ aqueous solution of $H_{2}S$ will be :

• A. 4
• B. 5
• C. 3
• D. 2.5

Answer 1

The correct option is A 4

Given, $K_{a_1}\left(H_{2}S\right)={10}^{-7},K_{a_2}\left(H{S}^{-}\right)={10}^{-13}$ $\left[H_{2}S\right]=0.1M$
Since, the second acid dissociation constant is very small in comparison to the first acid dissociation constant of $H_{2}S$ i.e. $K_{a_2}<<<<K_{a_1}$. Thus, $H^{+}$ ions are mainly obtained from the first step i.e.
$H_{2}S\rightleftharpoons H{S}^{-}+H^{+}\therefore \left[H^{+}\right]=\sqrt{K_{a_1}\times C}\left[H^{+}\right]=\sqrt{{10}^{-7}\times {10}^{-1}}\left[H^{+}\right]=\sqrt{{10}^{-8}}={10}^{-4}MpH=-\log \left[H^{+}\right]=-\log \left[{10}^{-4}\right]=4$