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? Ksp for M S = $6.0\times {10}^{-21}$ : concentration of saturated $H_{2}S$ = $0.1M$ : $K_1$ = ${10}^{-7}$ and $K_2$ = $1.3\times {10}^{-13}$ for $H_{2}S$.

• A. 0.850
• B. 0.301
• C. 0.463
• D. 0.983

Answer 1

The correct option is D

0.983

The first rate constant for step 1 ionization is :

$H_{2}S\left(g\right)\rightleftharpoons H^{+}\left(aq\right)+H{S}^{-1}\left(aq\right)$

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

$K_1$ = ${10}^{-7}$

$H{S}^{-1}\rightleftharpoons H^{+}+S^{2-}$ = $1.3\times {10}^{-13}$

The overall ionization constant for $H_{2}S$ is $K_1.K_2$ = ${10}^{-7}\times 1.3\times {10}^{-13}$

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

$MS\left(s\right)\rightleftharpoons M^{2+}\left(aq\right)+S^{2-}\left(aq\right)$

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

From here, we can calculate $\left[S^{2-}\right]$ = $Ksp\frac{\left(MS\right)}{[M2+]}$ = $1.2\times {10}^{-19}$

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

Substituting the known values in the above equation, we have

$[H^{+}{]}^2$ = $1.0833\times {10}^{-2}$

Taking -log both sides, we have

$pH$ = $0.983$