Question

A solution contains $\left[I^{-}\right]=\left[B{r}^{-}\right]=0.2M$ each.
Then find the percentage of $\left[I^{-}\right]$ that has reacted when the $AgBr$ first precipitates:
Given :
$Salt{K}_{sp}(mol{L}^{-1}{)}^2$
$AgBr5\times {10}^{-13}$
$AgI8.3\times {10}^{-17}$

• A. $85\%$
• B. $0.01\%$
• C. $90\%$
• D. $99.98\%$

Answer 1

The correct option is D $99.98\%$

The minimum concentration of $A{g}^{+}$ required for the precipitation of $AgBr$ is:
$(K_{sp}{)}_{AgBr}=[A{g}^{+}\left]\right[B{r}^{-}]$
$\left[A{g}^{+}\right]=\frac{(K_{sp}{)}_{AgBr}}{\left[B{r}^{-}\right]}$ $=\frac{5\times {10}^{-13}}{0.2}$

For ,$\left[A{g}^{+}\right]>2.5\times {10}^{-12}M$ precipitation of $AgBr$ occurs.

When AgBr first precipitates, $\left[A{g}^{+}\right]>2.5\times {10}^{-12}M$. Therefore, $\left[I^{-}\right]$ unreacted at this point can be calculated as:
$(K_{sp}{)}_{AgI}=[A{g}^{+}\left]\right[I^{-}{]}_{unreacted}$
$8.3\times {10}^{-17}=[2.5\times {10}^{-12}][I^{-}{]}_{unreacted}$
$[I^{-}{]}_{unreacted}=3.32\times {10}^{-5}M$

$\%$ of $\left[I^{-}\right]$ unreacted:

$\%[I^{-}{]}_{unreacted}=\frac{3.32\times {10}^{-5}M}{0.2M}\times 100=1.66\times {10}^{-2}$

$\%[I^{-}{]}_{reacted}=100-1.66\times {10}^{-2}\%[I^{-}{]}_{reacted}=99.9834\%$