An amount of 0.10 moles of AgCl is added to one liter of water. Next the crystals of NaBr are added until 80% of AgCl is converted to AgBr(s). If the  $B{r}^{-}$ at this point is  $Y\times {10}^{-8}$ then the value of Y is ____
(Given  $K_{SP}$ of  $AgCl={10}^{-10}$ and $K_{SP}$ of  $AgBr=4\times {10}^{-13}$)

Concept of simultaneous solubility 
80%  $AgCl\to AgBr$
 $AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
(0.01-9)   (a+b)    a     …….(1)
$AgBr\rightleftharpoons A{g}^{+}+B{r}^{-}$
0.08       (a+b)     b    ………(2)
 $\frac{K_{sp}}{K_{sp}}=\frac{AgCl}{AgBr}=\frac{\left(a+b\right)a}{\left(a+b\right)b}=\frac{{10}^{-10}}{4\times {10}^{-13}}$
 $\frac{a}{b}=\frac{1000}{4}=250 \therefore a=250b$   
Substitute a in 2 eq., for  $B{r}^{-}$   conc.,
 $$\begin{gathered} \left(a+b\right)b=4\times {10}^{-13} \\ \left(250b+b\right)b=4\times {10}^{-13} \\ 250b^2+b^2=4\times {10}^{-13} \\ 251b^2=4\times {10}^{-13} \\ {b}^2=15.936\times {10}^{-16} \\ \therefore b=3.99\times {10}^{-8} \end{gathered}$$
   So  $b=4\times {10}^{-8}$