Question

Calculate the $S^{2-}$ ion concentration in a saturated solution $\left(0.1M\right)$ of $H_{2}S$ whose pH was adjusted to $2$ by the addition of $HCl$. ($K_a=1.1\times {10}^{-21}$)

Answer 1

$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$
$K_a=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]}$
$\left[S^{2-}\right]=\frac{K_a\left[H_{2}S\right]}{{\left[H^{+}\right]}^2}$
$pH=2$; So, $\left[H^{+}\right]=1\times {10}^{-2}M$
$H_{2}S$ is a weak electrolyte. So, $\left[H_{2}S\right]=0.1M$
so, $\left[S^{2-}\right]=\frac{1.1\times {10}^{-21}\times 0.1}{1\times {10}^{-2}\times 1\times {10}^{-2}}=1.1\times {10}^{-18}M$