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

Options

A. `12 xx 10^(-2)`
B. `14 xx 10^(-4)`
C. `22 xx 10^(-11)`
D. `32 xx 10^(-12)`

Answer

D

Explanation

`CaF_(2) hArr underset(2 xx 10^(-4))(Ca^(2+))+underset(2 xx 2 xx 10^(-4))(2F^(-))`
Solubility Product `=[Ca^(2+)][F^(-)]^(2)`
`=[2 xx 10^(-4)][2 xx 2 xx 10^(-4)]^(2)`
`=32 xx 10^(-12)`