Question

Approximate pH of 0.1 M aqueous $H_{2}S$ solution when $K_1$ and $K_2$ for $H_{2}S$ at ${25}^{\circ }Care1\times {10}^{-7}$ and $1.3\times {10}^{-13}$ respectively ?

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

The correct option is A 4

Since $K_2$ is much smaller than $K_1$, $H^{+}$ are largely formed from first dissociation
$H_{2}S\rightleftharpoons \left[H^{+}\right]+\left[H{S}^{-}\right];K_1$

$K_1=\frac{\left[H^{+}\right]\left[H{S}_2^{-}\right]}{\left[H_{2}S\right]}=\frac{[H^{+}{]}^2}{\left[H_{2}S\right]}$

$\left[H^{+}\right]=\sqrt{K_1\left[H_{2}S\right]}={10}^{-4}$

$pH=-log\left[H^{+}\right]=4$