Question

Calculate approximate $pH$ of a $0.1M$ aqueous solution of $H_{2}S,K_{1}and{K}_2$ for $H_{2}S$ are $1.00\times {10}^{-7}$ and $1.3\times {10}^{-13}$ respectively at ${25}^{o}C$.

Answer 1

Given,
$K_1=1\times {10}^{-7}{K}_2=1\times {10}^{-13}{K}_1>>K_2$
Thus, most of the $H^{+}$ ion will be produced by first ionization of $H_{2}S$. Thus, second dissociation may be ignored during calculation of $\left[H^{+}\right]$ ion concentration.
$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}att=0C00AtEquilibriumC(1-\alpha )C\alpha C\alpha$
$\therefore$
$K_1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$
$1\times {10}^{-7}=\frac{C{\alpha }^2}{1-\alpha }$
If $\alpha <<1$
Then, $(1-\alpha )\approx 1$
$1\times {10}^{-7}=\frac{0.1{\alpha }^2}{1}$
${\alpha }^2=1\times {10}^{-6}{\alpha }^2=1\times {10}^{-6}\alpha ={10}^{-3}M\left[H^{+}\right]={10}^{-4}M$
$pH=-log\left[H^{+}\right]=-log\left[H^{+}\right]=-log\left[{10}^{-4}\right]\Rightarrow pH=4$