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The solubility of CaF(2) is 2xx10^(-4)m...

The solubility of `CaF_(2)` is `2xx10^(-4)mol L^(-1)` . Its solubility product is :

A

`2.0xx10^(-8)`

B

`4.0xx10^(-12)`

C

`8.0xx10^(-12)`

D

`3.2xx10^(-11)`

Text Solution

Verified by Experts

The correct Answer is:
D
`K_(sp)=[Ca^(2+)][F^(-)]^(2)`
`[Ca^(2+)]=2xx10^(-4), [F^(-)]=2xx2xx10^(-4)`
`K_(sp)=(2xx10^(-4))(4xx10^(-4))^(2)`
`3.2xx10^(-11)`
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