The solubility product of a sparingly soluble salt `AX_(2)` is `3.2xx10^(-11)`. Its solubility (in `mo//L`) is

Options

A. `5.6xx10^(-6)`
B. `3.1xx10^(-4)`
C. `2xx10^(-4)`
D. `4xx10^(-4)`

Answer

C

Explanation

`AX_(2)` is ionised as follows:
`underset(S mol L^(-1))(AX_(2))hArrunderset(S)(A_(2)^(+)+underset(2S)(2X^(-))`
Solubility product of `AX_(2)`,
`K_(sp)=[A^(2+)][X^(-)]^(2)=Sxx(2S)=4S^(3)`
`:' K_(sp)` of `AX_(2)=3.2xx10^(-11)`
`:. 3.2xx10^(-11)=4S^(3)`
`S^(3)=0.8xx10^(-11)=8xx10^(-12)`
Solubility `=2xx10^(-4) mol//L`