1.

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

Answer»

`2.0xx10^(-8)`
`4.0xx10^(-12)`
`8.0xx10^(-12)`
`3.2xx10^(-11)`

Solution :`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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