Question

The solubility of $AgBr$ in the presence of $AgSCN$ in water is $x\times {10}^{-7}mol$ $L^{-1}$. The $x$ is ______
($K_{sp}AgSCN=0.96\times {10}^{-12}{M}^2,K_{sp}AgBr=4.8\times {10}^{-13}{M}^2$)

Answer 1

$AgCNS----->A{g}^{+}+CN{S}^{-}\left(X\right)\left(X\right)$

$AgBr------>A{g}^{+}+B{r}^{-}\left(Y\right)\left(Y\right)$

$K_{sp}\left(AgCNS\right)=0.96\times {10}^{-12}=(X+Y)X$ ---(1)

$K_{sp}\left(AgBr\right)=4.8\times {10}^{-13}=(X+Y)Y$ ---(2)

Solving (1) and (2)

$\left[AgCNS\right]=X=8\times {10}^{-7}$

$\left[AgBr\right]=Y=4\times {10}^{-7}$ Which is the required answer