Question

An aqueous solution of a metal bromide $MB{r}_2$ of $0.05M$ is saturated with $H_{2}S$. What is the minimum $pH$ at which $MS$ will precipitate? Given $K_{sp}$ of $MS$ = $6.0\times {10}^{-21}$. Concentration of saturated $H_{2}S$ = $0.1M$ ($K_1$ of $H_{2}S$ = ${10}^{-7}$ and $K_2$ = $1.3\times {10}^{-1}C$)

• A. 0.5
• B. 0.26
• C. 0.983
• D. 0.38

Answer 1

The correct option is C

0.983

$K_{SP}$ of MS = $6.0\times {10}^{-21}$

Conc. of $\left[M^{2+}\right]$ = $0.05M$

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

$\left[S^{2-}\right]$ = $\frac{6.0\times {10}^{-21}}{5\times {10}^{-2}}$ = $1.2\times {10}^{-19}$

$K_1$ of $H_{2}S$ = $1\times {10}^{-7}$ & $K_2$ = $1.3\times {10}^{23}$

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

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

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

$[H^{+}{]}^2$ = $\frac{K_1.K_2.\left[H_{2}S\right]}{\left[S^{2-}\right]}$ = $\frac{1\times {10}^{-7}\times 1.3\times {10}^{-13}\times 0.1}{1.2\times {10}^{-19}}$ = $1.08\times {10}^{-2}$

$\left[H^{+}\right]$ = $\sqrt{1.08\times {10}^{-2}}$ = $1.04\times {10}^{-1}$

$pH$ = $-log1.04\times {10}^{-1}$ = $0.983$