At `25^(@)C`, solublity product of `CaF_(2)` is `4xx10^(-11)`. Determine the solubility of `CaF_(2)` in its saturated solution and also the molar concentrations of the constituent ions.

Explanation

In saturated product, `CaF_(2)` exists in the following equilibrium: `CaF_(2)(s)hArr Ca^(2+)(aq)+2F^(-)(aq)`
`therefore`Solubility of `CaF_(2)(s),(K_(sp))=[Ca^(2+)]xx[F^(-)]^(2)`
If in the solution, the molar concentration of `CaF_(2)(s)` is S `mol*L^(-1)`, then `[Ca^(2+)]=S" "mol*L^(-1)& [F^(-)]=2"S "mol*L^(-1)`
`therefore K_(sp)=[Ca^(2+)]xx[F^(-)]^(2)=Sxx(2S)^(2)=4S^(3)`
As given, `K_(sp)=4xx10^(-11)" "therefore 4S^(3)=4xx10^(-11)`
or, `S^(3)=10^(-11)" "therefore S=2.15xx10^(-4)mol*L^(-1)`
`therefore `At `25^(@)C` solubility of `CaF_(2)=2.15xx10^(-4)mol*L^(-1)`
In the solution, `[Ca^(2+)]=2.15xx10^(-4)mol*L^(-1) and [F^(-)]=2xx2.15xx10^(-4)=4.3xx10^(-4)mol*L^(-1)`.