Question

An aqueous solution of a metal bromide $MBr_{2}\left(0 . 05 M\right)$ is saturated with $H_{2}S.$ What is the minimum pH at which MS will precipitate? $K_{s p}$ 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

For the precipitation of MS form a solution of $MBr_{2}$ 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 $\text{H}_{2} \text{S} \rightleftharpoons 2 \text{H}^{+} + \text{S}^{2 -}$
$K=K_{1}\times K_{2}=\frac{\left[\right. H^{+} \left]\right.^{2} \left[\right. S^{2 -} \left]\right.}{\left[\right. H_{2} S \left]\right.}$
or $10^{- 7}\times 1.3\times 10^{- 13}=\frac{\left[\right. H^{+} \left]\right.^{2} \left[\right. 1 . 2 \times 10^{- 19} \left]\right.}{0 . 1}$
or $\left[H^{+}\right]=1.04\times 10^{- 1}$
$\therefore \text{pH=0.9826}$