Chemical and Ionic Equilibrium - Result Question 120

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} $