Question

Calculate the appoximate pH of $0.1M$ aqueous $H_{2}S$ solution. $K_1$ and $K_2$ for $H_{2}S$ are $1.0\times {10}^{-7}$ and $1.3\times {10}^{-13}$ respectively at ${25}^{o}C$.

Answer 1

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$; $K_1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}=1\times {10}^{-7}$

$H{S}^{-}\rightleftharpoons H^{+}+S^{-}$; $K_2=\frac{\left[H^{+}\right]\left[S^{-}\right]}{\left[H{S}^{-}\right]}=1.3\times {10}^{-13}$

Since $K_2<<K_1$, ionisation in the $2^{nd}$ step can be neglected.

Hence $\left[H{S}^{-}\right]=\left[H^{+}\right]$ and $\left[H_{2}S\right]=0.1-\left[H^{+}\right]$

$\therefore \frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{0.1-\left[H^{+}\right]}=1\times {10}^{-7}$

or, $\frac{[H^{+}{]}^2}{0.1}={10}^{-7}$ [Since, $\left[H^{+}\right]<<0.1$]

$\therefore \left[H^{+}\right]=\sqrt{{10}^{-7}\times 0.1}={10}^{-4}$

$\therefore pH=-\log {10}^{-4}=4$