Question

An aqueous solution of metal bromide, $M{Br}_2\left(0.05M\right)$ is saturated with $H_{2}S$. What is the minimum pH at which $MS$ will precipitate? $K_{sp}$ for $MS=6.0\times {10}^{-21}$, conc of saturated $H_{2}S=0.1M$.
$K_1={10}^{-7};K_2=1.3\times {10}^{-13}$ for $H_{2}S$

Answer 1

The minimum concentration of $S^{2-}$ ions required to precipitate $MS$ is

$2MS\rightleftharpoons M^{2+}+S^{2-}$

$K_{sp}=\frac{\left[M^{2+}\right]\left[S^{2-}\right]}{{\left[MS\right]}^2}$
$\Rightarrow \left[S^{2-}\right]=\frac{K_{sp}MS}{\left[M^{2+}\right]}=1.2\times {10}^{-19}M$
$H_{2}S$ ionises in solution in two steps:
$H_{2}S\rightleftharpoons H^{+}+{HS}^{-};K_1={10}^{-7}$
${HS}^{-}\rightleftharpoons H^{+}+S^{2-};K_2=1.3\times {10}^{-13}$
$\frac{\left[H^{+}\right]\left[{HS}^{-}\right]}{\left[H_{2}S\right]}=K_1;\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[{HS}^{-}\right]}=K_2$
Hence, $K_1{K}_2=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[{HS}^{-}\right]}$
or ${\left[H^{+}\right]}^2=\frac{K_1{K}_2\left[H_{2}S\right]}{\left[S^{2-}\right]}=1.08\times {10}^{-2}$
so, $\left[H^{+}\right]=1.04\times {10}^{-1}$
$pH=-\log \left[H^{+}\right]=-\log (1.04\times {10}^{-1})=0.98$