The solubility of `CuBr` is `2xx10^(-4)` at `25^(@)C`. The `K_(sp)` value for `CuBr` is

Options

A. `4xx10^(-8) mol^(2) l^(-2)`
B. `4xx10^(-11) mol^(2) L^(-1)`
C. `4xx10^(-4) mol^(2) l^(-2)`
D. `4xx10^(-15) mol^(2) l^(-2)`

Answer

A

Explanation

`underset(K_(sp))(CuBr)hArrunderset((s))(Cu^(+))+ underset((S))(Br^(-))`
`K_(sp)=S^(2)=(2xx10^(-4))^(2)=4xx10^(-8)(mol^(2))/(l^(2))`