If the solubility product of `CuS` is `6xx10^(-16)`, calculate the maximum molarity of `CuS` in aqueous solution.

Answer

`S=2.45xx10^(-8)mol^(-1)`

Explanation

Maximum molarity of `CuS` in aqueous solution is the solubility of `CuS` in `mol L^(-1)`. Let `S` is the solubility of `CuS` in `mol L^((-1)`, then
`CuS hArr Cu^(2+)+S^(2-)`
`K_(sp)=[Cu^(2+)][S^(2-)]`
`=SxxS` ltbr. `=S^(2) mol^(2)L^(-2)`
`:. S=sqrt(K_(sp))=sqrt(6xx10^(-16))`
`2.45xx10^(-8)mol L^(-1)`