Question

A solution contains a mixture of $A{g}^{+}\left(0.10M\right)$ and $H{g}_2^{2+}\left(0.10M\right)$ which are to be separated by selective precipitation using iodide .Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of silver ion is precipitated at that concentration? ($K_{sp}$ of $AgI=8.5\times {10}^{-17}$ and $K_{sp}$ of $H{g}_2{I}_2=2.5\times {10}^{-26}$.)

• A. $\left[I^{-}\right]=5\times {10}^{-13}$ ; $99.87\%$
• B. $\left[I^{-}\right]=7\times {10}^{-13}$ ; $95.87\%$
• C. $\left[I^{-}\right]=5\times {10}^{13}$ ; $93.87\%$
• D. $\left[I^{-}\right]=5\times {10}^{-13}$ ; $91.87\%$

Answer 1

The correct option is A $\left[I^{-}\right]=5\times {10}^{-13}$ ; $99.87\%$

The $\left[I^{-}\right]$ needed for precipitation of $A{g}^{+}$ and are derived as:

For $AgI$:
$\left[A{g}^{+}\right]\left[I^{-}\right]=K_{sp}\left(AgI\right)\left(0.1\right)\left[1^{-}\right]=8.5\times {10}^{-17}\therefore \left[I^{-}\right]=8.5\times {10}^{-16}M$


For $H{g}_2{I}_2:\left[H{g}_2^{2+}\right][I^{-}{]}^2=2.5\times {10}^{-26}(0.1\left)\right[I^{-}]=2.5\times {10}^{-25}\therefore [I^{-}]=5\times {10}^{-13}M$

Since $\left[I^{-}\right]$ required for precipitation of $AgI$ is less and thus $AgI$ begins to precipitate first. Also, it will continue up to addition of $\left[I^{-}\right]=5\times {10}^{-13}$ when $H{g}_2{I}_2$ begins to precipitate.

Now at this concentration of $I^{-}the\left[A{g}^{+}\right]$ left in solution is $\left[A{g}^{+}{]}_{left}\right[I^{-}]=(K_{sp}\left)\right(AgI)$

$\therefore [A{g}^{+}{]}_{left}=\frac{8.5\times {10}^{-17}}{5.0\times {10}^{-13}}=1.7\times {10}^{-4}M$

$0.1MA{g}^{+}$ was initial concentration.

And now,
$=1.7\times {10}^{-4}MA{g}^{+}$ left in solution.

Amount of $A{g}^{+}$ precipitated=$0.1-1.7\times {10}^{-4}$

$\therefore \%A{g}^{+}$ precipitated $=\frac{0.1-1.7\times {10}^{-4}}{0.1}\times 10=99.87\%$