The solubility of a saturated solution of calcium fluoride is `2xx10^(-4)` mol/L. Its solubility product is

Options

A. `12xx10^(-2)`
B. `14xx10^(-4)`
C. `22xx10^(-11)`
D. `32xx10^(-12)`

Answer

D

Explanation

`underset(2xx10^(-4))(CaF_(2)hArr underset(2xx10^(-4)M)(Ca^(2+))+underset(2xx2xx10^(-4)M)(2F^(-))`
`K_(sp)` of `CaF_(2)=[Ca^(2+)][F^(-)]^(2)`
`=[2xx10^(-4)][4xx10^(-4)]^(2)=32xx10^(-12)(mol//L)^(2)`