Question

A solution contains a mixture of ${Ag}^{+}$ $\left(0.01M\right)$ and ${Hg}_2^{2+}$ $\left(0.10M\right)$ which are to be separated by selective precipitation using iodide. What percentage of silver ion is precipitated?
[$K_{{sp}_{\left(AgI\right)}}=8.5\times {10}^{-17};K_{{sp}_{\left({Hg}_2{I}_2\right)}}=2.5\times {10}^{-26}$].

• A. $99.84$%
• B. $99.83$%
• C. $99.85$%
• D. None of these

Answer 1

The correct option is B $99.83$%

$\left[I^{-}\right]$ required to precipitate ${Ag}^{+}$ and ${Hg}_2^{2+}$ are derived as:
For $AgI$:

$\left[{Ag}^{+}\right]\left[I^{-}\right]=K_{{sp}_{AgI}};\left[0.1\right]\left[I^{-}\right]=8.5\times {10}^{-17};\left[I^{-}\right]=8.5\times {10}^{-16}M.......\left(i\right)$

For ${Hg}_2{I}_2$:

$\left[{Hg}_2^{2+}\right]{\left[I^{-}\right]}^2=K_{{sp}_{{Hg}_2{I}_2}};\left[0.1\right]{\left[I^{-}\right]}^2=2.5\times {10}^{-26};\left[I^{-}\right]=5\times {10}^{-13}M........\left(ii\right)$

$\left[I^{-}\right]$ required to precipitate $AgI$ are lesser than that required to precipitate ${Hg}_2{I}_2$ and thus, precipitation of $AgI$ will take place first. It will continue till the $\left[I^{-}\right]$ becomes $5\times {10}^{-13}$ when ${Hg}_2{I}_2$ begins to precipitate and thus,

Maximum $\left[I^{-}\right]$ for $AgI$ preicipitation $=5\times {10}^{-13}M$. Also ${\left[{Ag}^{+}\right]}_{left}$ at this concentration of $\left[I^{-}\right]$ can be evaluated as:

${\left[{Ag}^{+}\right]}_{left}\left[I^{-}\right]=K_{{sp}_{AgI}}$

$\left[{Ag}^{+}\right]=\frac{K_{{sp}_{AgI}}}{\left[I^{-}\right]}=\frac{8.5\times {10}^{-17}}{5\times {10}^{-13}}=1.7\times {10}^{-4}M$

$\because$ $0.1M$ ${Ag}^{+}$ will left $=1.7\times {10}^{-4}M$ ${Ag}^{+}$ in solution

$\therefore$ % of ${Ag}^{+}$ left $=\frac{1.7\times {10}^{-4}\times 100}{0.1}=0.17$%$M$ ${Ag}^{+}$

$\therefore$ % of $Ag$ precipitated $=100-0.17=99.83$%