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
`{:(A X_(2),hArr,A^(2+),+2X^(-)),("S mol L"^(-),,S," 2S"):}`
Solubility product of `AX_(2)`,
`K_(sp)=[A^(2+)][X^(-)]^(2)=Sxx(2S)^(2)=4S^(3)`
`because K_(sp)` of `A X_(2)=32xx10^(-11)`
`therefore 32xx10^(-11)=4S^(3)`
`S^(3)=0.8xx10^(-11)=8xx10^(-12)`
Solubility `=2xx10^(-4)` mol/L