Chemical and Ionic Equilibrium - Result Question 56
75. A solution contains a mixture of $\mathrm{Ag}^{+}(0.10 \mathrm{M})$ and $\mathrm{Hg}^{2+}(0.10 \mathrm{M})$ which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?
(1984, 4M)
$K_{\text {sp }}: \mathrm{AgI}=8.5 \times 10^{-17}, \mathrm{HgI}_2=2.5 \times 10^{-26}$
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Answer:
Correct Answer: 75. $(99.83)$
Solution:
- $K_{\text {sp }}(\mathrm{AgI})=8.5 \times 10^{-17}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]$
[ $\mathrm{I}^{-}$] required to start precipitation of AgI
$\begin{aligned} & =\frac{8.5 \times 10^{-17}}{0.10}=8.5 \times 10^{-16} \mathrm{M} \\ K_{\text {sp }}\left(\mathrm{HgI}_2\right) & =2.5 \times 10^{-26}=\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^2 \end{aligned}$
[ $\mathrm{I}^{-}$] required to start precipitation of $\mathrm{HgI}_2$
$=\sqrt{\frac{2.5 \times 10^{-26}}{0.10}}=5 \times 10^{-13} \mathrm{M}$
The above calculation indicates that lower $\left[\mathrm{I}^{-}\right]$ is required for precipitation of AgI. When $\left[\mathrm{I}^{-}\right]$ reaches to $5 \times 10^{-13}, \mathrm{AgI}$ gets precipitated almost completely. When $\mathrm{HgI}_2$ starts precipitating,
$\begin{aligned} {\left[\mathrm{Ag}^{+}\right] } & =\frac{8.5 \times 10^{-17}}{5 \times 10^{-13}}=1.70 \times 10^{-4} \mathrm{M} \\ % \mathrm{Ag}^{+} \text {remaining } & =\frac{1.70 \times 10^{-4} \times 100}{0.10}=0.17 \\ % \mathrm{Ag}^{+} \text {precipitated } & =100-0.17=99.83 \end{aligned}$
BETA
