Chemical And Ionic Equilibrium Result Question 16
16. A solution which is $10^{-3} \mathrm{M}$ each in $\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Zn}^{2+}$ and $\mathrm{Hg}^{2+}$ is treated with $10^{-16} \mathrm{M}$ sulphide ion. If $K_{\text {sp }}$ of $\mathrm{MnS}, \mathrm{FeS}, \mathrm{ZnS}$ and HgS are $10^{-15}, 10^{-23}, 10^{-20}$ and $10^{-54}$ respectively, which one will precipitate first?
(2003, 1M)
FeS
MgS
HgS
ZnS
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Answer:
Correct Answer: 16. ( a )
Solution:
$\begin{aligned} K_h\left(X^{-}\right)=\frac{K_w}{K_a} & =\frac{10^{-14}}{10^{-5}}=10^{-9} \Rightarrow \alpha=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{10^{-9}}{0.10}}=10^{-4} \\ \text { % hydrolysis } & =100 \alpha \times 0.01 \end{aligned}$
BETA
