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

Options

A. ` 4 xx 10^(-8) "mol"^(2)L^(-2)`
B. ` 4 xx 10^(-4) "mol"^(2)L^(-2)`
C. ` 4 xx 10^(-11) "mol"^(2)L^(-2)`
D. ` 4 xx 10^(-15) "mol"^(2)L^(-2)`

Answer

A

Explanation

`CuBr to Cu^(+)+Br^(-)`
Solubility of CuBr is ` 2 xx 10^(-4) ` mol `L^(-1)`
Therefore , solubility of `Cu^(+)= 2 xx 10^(-4) ` mol `L^(-1)`
`implies ` Solubility product `=2 xx 10^(-4) xx 2 xx 10^(-4)`
`= 4 xx 10^(-8) mol^(2)L^(-2)`