Chemical And Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise

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 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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Answer:

Correct Answer: 64. (1)

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

$\begin{aligned} M \mathrm{~S}= & \frac{6 \times 10^{-21}}{0.05}=1.2 \times 10^{-19} \\ K= & 1.3 \times 10^{-20}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}=\left[\mathrm{H}^{+}\right]^2 \frac{\left(1.2 \times 10^{-19}\right)}{0.10} \\ & {\left[\mathrm{H}^{+}\right]=0.10 \mathrm{M} \Rightarrow \mathrm{pH}=1 } \end{aligned}$