Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{CuBr}\to {\mathrm{Cu}}^{+}+{\mathrm{Br}}^{-}$ Solubility of $\mathrm{CuBr}$ is $2\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ Therefore, solubility of ${\mathrm{Cu}}^{+}=2\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ $\Rightarrow$ Solubility product $=2\times 10^{-4}\times 2\times 10^{-4}$ $=4\times 10^{-8}{\mathrm{mol}}^2{\mathrm{L}}^{-2}$