The solubility of A g B r ( s ) , having solubility product 5 × 10 - 10 in 0 . 2   M   NaBr…

Here, the solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$ of $\mathrm{AgBr}$ is $5\times {10}^{-10} \mathrm{M}$.

We know that for the reaction,

$$\begin{gathered} \mathrm{AgBr} \to {\mathrm{Ag}}^{+}+{\mathrm{Br}}^{-} \\ {\mathrm{K}}_{\mathrm{sp}} = \left[{\mathrm{Ag}}^{+}\right]\times \left[{\mathrm{Br}}^{-}\right] \end{gathered}$$

Also, it is given that the concentration of ${\mathrm{Br}}^{-}$ is $0.2 \mathrm{M}$. Therefore,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}} = {\mathrm{s}}_{\mathrm{Ag}}\times {\mathrm{s}}_{\mathrm{Br}} \\ 5\times {10}^{-10} = {\mathrm{s}}_{\mathrm{Ag}}\times 0.2 \\ \therefore {\mathrm{s}}_{\mathrm{Ag}} = 25\times {10}^{-10} \end{gathered}$$