On addition of increasing amount of $AgN{O}_3$ to 0.1M each of $NaCl$  and $NaBr$ in a solution, What % of Br ion gets precipitated when $C{l}^{-}$ ion starts precipitating? $K_{sp}\left(AgCl\right)$  $=1.0\times {10}^{-10};K_{sp}\left(AgBr\right)=1\times {10}^{-13}$

$\left[A{g}^{+}\right]$  required for commencement of precipitation of $\left[A{g}^{+}\right]=\frac{K_{sp}\left(AgCl\right)}{\left[C{l}^{-}\right]}=\frac{1.0\times {10}^{-10}}{0.1}=1.0\times {10}^{-9}M$

$\left[B{r}^{-}\right]$  remaining at this stage

$=\frac{K_{sp}\left(AgBr\right)}{\left[A{g}^{+}\right]}=\frac{1.0\times {10}^{-13}}{1.0\times {10}^{-9}}=1.0\times {10}^{-4}M$

$\%ofB{r}^{-}$ remaining unprecipitated

$=\frac{1.0\times {10}^{-4}}{0.1}\times 100=0.1$

$\%ofB{r}^{-}$  precipitated $=\text{1}00-0.\text{1}=\text{99}.\text{9}$