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 as follows
`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 `=2x10^(-4)` mol/L