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? $K_{sp}$ for $MS=6.0\times 10^{-21}.$ Concentration of saturated $H_{2}S=0.1M;K_1=10^{-7}$ and $K_2=1.3\times 10^{-13}$ for $H_{2}S.$
[Report your answer by rounding off to one significant figure]

Answer 1

Answer: 1

For the precipitation of MS form a solution of $MB{r}_2$ by passing $H_{2}S$
${\left(\text{K}\right)}_{\text{sp}\left(\text{MS}\right)}=\left[{\left(\text{M}\right)}^{2+}\right]\left[{\left(\text{S}\right)}^{2-}\right]$
or $6\times 10^{-21}=\left[0.05\right]\left[S^{2-}\right]$
or $\left[S^{2-}\right]=1.2\times 10^{-19}M$
i.e. precipitation of MS will start when $H_{2}S$ provides
$1.2\times 10^{-19}M{\text{ion of S}}^{2-}$
For ${\text{H}}_2\text{S}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-}$
$K=K_1\times K_2=\frac{[H^{+}{]}^2[S^{2-}]}{[H_{2}S]}$
or $10^{-7}\times 1.3\times 10^{-13}=\frac{[H^{+}{]}^2[1.2\times 10^{-19}]}{0.1}$
or $\left[H^{+}\right]=1.04\times 10^{-1}$
$\therefore \text{pH=0.9826}$