Chemical and Ionic Equilibrium 2 Question 64
64. An aqueous solution of a metal bromide $M \mathrm{Br}{2}(0.05 \mathrm{M})$ is saturated with $\mathrm{H}{2} \mathrm{~S}$. What is the minimum $\mathrm{pH}$ at which $M \mathrm{~S}$ will precipitate? $K_{\text {sp }}$ for $M \mathrm{~S}=6.0 \times 10^{-21}$, concentration of saturated $\mathrm{H}{2} \mathrm{~S}=0.1 \mathrm{M}, K{1}=10^{-7}$ and $K_{2}=1.3 \times 10^{-13}$, for $\mathrm{H}_{2} \mathrm{~S}$.
(1993, 3M)
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Solution:
- For $\mathrm{H}{2} \mathrm{~S}, \mathrm{H}{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}$
$$ K=K_{1} \times K_{2}=1.3 \times 10^{-20} $$
Minimum $\left[\mathrm{S}^{2-}\right]$ required to begin precipitation of
$$ \begin{aligned} M \mathrm{~S}= & \frac{6 \times 10^{-21}}{0.05}=1.2 \times 10^{-19} \ K= & 1.3 \times 10^{-20}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}=\left[\mathrm{H}^{+}\right]^{2} \frac{\left(1.2 \times 10^{-19}\right)}{0.10} \ & \quad\left[\mathrm{H}^{+}\right]=0.10 \mathrm{M} \Rightarrow \mathrm{pH}=1 \end{aligned} $$
