The solubility of `CaF_(2) " in water at " 298 K is 1.7 xx 10^(-3)` grams per 100 mL of the solution. Calculate solubility product of `CaF_(2).`

Answer

`4.14xx10^(-11)`

Explanation

Solubility of `CaF_2`
`=(1.7xx10^(-3))/100xx1000=1.7xx10^(-2) gL^(-1)`
Mol. Wt. of `CaF_2` = 40+2 x 19 = 78
`therefore` Solubility =`(1.7xx10^(-2))/78`
`=2.18xx10^(-4) "mol L"^(-1)`
`CaF_2 hArr Ca^(2+) + 2F^(-)`
`[Ca^(2+)]=2.18xx10^(-4) "mol L"^(-1), [F^-]=2xx2.18xx10^(-4)`
`=4.36xx10^(-4) "mol L"^(-1)`
`K_(sp)=[Ca^(2+)][F^-]^2`
`=(2.18xx10^(-4))(4.36xx10^(-4))^2`
`=4.14xx10^(-11)`