Chemical and Ionic Equilibrium - Result Question 120
64. An aqueous solution of a metal bromide $M Br _2(0.05 M)$ is saturated with $H _2 S$. What is the minimum $pH$ at which $M S$ will precipitate? $K _{\text {sp }}$ for $M S=6.0 \times 10^{-21}$, concentration of saturated $H _2 S=0.1 M, K _1=10^{-7}$ and $K _2=1.3 \times 10^{-13}$, for $H _2 S$.
(1993, 3M)
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Answer:
Correct Answer: 64. (1)
Solution:
For $H _2 S, H _2 S \rightleftharpoons 2 H^{+}+S^{2-}$
$ K=K _1 \times K _2=1.3 \times 10^{-20} $
Minimum $\left[S^{2-}\right]$ required to begin precipitation of
$ \begin{aligned} M S= & \frac{6 \times 10^{-21}}{0.05}=1.2 \times 10^{-19} \\ K= & 1.3 \times 10^{-20}=\frac{\left[H^{+}\right]^{2}\left[S^{2-}\right]}{\left[H _2 S\right]}=\left[H^{+}\right]^{2} \frac{\left(1.2 \times 10^{-19}\right)}{0.10} \\ & \quad\left[H^{+}\right]=0.10 M \Rightarrow pH=1 \end{aligned} $
BETA
