Solubility of `Ag_(2)CrO_(4)` is `8 xx 10^(-5)` mol `L^(-1)`. Calculate the solubility product.

Explanation

We are given that
Solubility product of `PbBr_(2) = 8.0 xx 10^(-5)`
We are to calculate the solubility of `PbBr_(2)`.
`{:(PbBr_(2) ,hArr, Pb^(2+), + ,2Br^(-)),(,,S,,2S):}`
`K_(sp) = [Pb^(2+)][Br^(-)]^(2) = 4S^(3)`
`S = [K_(sp))/(4)]^(1//3) = (0.94 xx 10^(-2))/(1.41) = 0.66 xx 10^(-2) hArr S = 6.6 xx 10^(-1)` mol `L^(-1)`.