1.

The solubility of CaF_(2) is 2 xx 10^(-4) moles/litre. Its solubility product (K_(sp)) is

Answer»

`2.0 XX 10^(-4)`
`4.0 xx 10^(-3)`
`8.0 xx 10^(-12)`
`3.2 xx 10^(-11)`

Solution :`K_(sp)` for `CaF_(2) = 4S^(3) = 4 xx [2 xx 10^(-4)]^(3) = 3.2 xx 10^(-11)`.


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