The solubility of calcium fluoride in saturated solution, it its solubility product is `3.2 xx 10^(-11)` is :

Options

A. `2.0xx 10^(-4) mol //L`
B. `12.0 xx 10^(-3) mol //L`
C. `0.2 xx 10^(-4) mol //L`
D. ` 2 xx 10^(-3) mol //L`

Answer

A

Explanation

`CaF_(2) hArr Ca^(2+)+2F^(-)`
Let solubility of `CaF_(2)=x mol L^(-1)`
`[Ca^(2+)]=x M, [F^(-)]=2xM`
`K_(sp)=[Ca^(2+)][F^(-)]^(2)`
`3.2 xx 10^(-11)=x.(2x)^(2)`
`4x^(3)=3.2 xx 10^(-11)`
`x^(3) =8 xx 10^(-12)`
`x=2xx10^(-4)M`.