Question

An aqueous solution of a metal bromide $M{Br}_2$ $\left(0.05M\right)$ is saturated with $H_{2}S$. The minimum $pH$ at which $MS$ will precipitate is $X$? $K_{sp}$ for $MS=6.0\times {10}^{-21}$.

X is:
Concentration of saturated $H_{2}S=0.1M$; $K_1={10}^{-7}$ and $K_2=1.3\times {10}^{-13}$ for $H_{2}S$.

• A. $4$
• B. $5$
• C. $1$
• D. $7$

Answer 1

The correct option is C $1$

For the precipitation of $MS$ from a solution of $M{Br}_2$ by passing $H_{2}S$

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

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 $1.2\times {10}^{-19}M$ ion of $S^{2-}$

For $H_{2}S$

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

$K=K_1\times K_2=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H_{2}S\right]}$

or ${10}^{-7}\times 1.3\times {10}^{-13}=\frac{\left[{}^{+}\right][1.2\times {10}^{-19}]}{0.1}$

or $\left[H^{+}\right]=1.04\times {10}^{-1}$

$\therefore$ $pH=0.9826$