Question

An aqueous solution of a metal bromide, $MB{r}_2\left(0.05M\right)$ is saturated with $H_{2}S$. What is the minimum pH at which $MS$ will precipitate out? Write the value to the nearest integer.

$(K_{sp}$ for $MS$ is $6.0\times {10}^{-21}$, concentration of saturated $H_{2}S$ is $0.1M$, $K_1={10}^{-7}$ and $K_2=1.3\times {10}^{-13}$ for $H_{2}S)$

Answer 1

$K_{sp}=\left[M^{2+}\right]\left[S^{2-}\right]$

$6.0\times {10}^{-21}=0.05\left[S^{2-}\right]$

$\left[S^{2-}\right]=1.2\times {10}^{-19}$

$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$

$K=K_1\times K_2={10}^{-7}\times 1.3\times {10}^{-13}$

$K=1.3\times {10}^{-20}$

$K=\frac{[H^{+}{]}^2\times 1.2\times {10}^{-19}}{0.1}=1.3\times {10}^{-20}$

$[H^{+}{]}^2=\frac{1.3\times {10}^{-20}\times 0.1}{1.2\times {10}^{-19}}=0.0108$

$\left[H^{+}\right]=0.1$

$pH=-log\left[H^{+}\right]=-log0.1=1$

The minimum pH at which $MS$ will precipitate out is $1$.