At a certain temperature, solubility of `CaF_(2) ` in its saturated aqueous solution is `2xx10^(-4)mol*L^(-1)`. What is its solubility produced at that temperature?

Explanation

In the saturated aqueous solution of `CaF_(2)`, the following equilibrium is established: `CaF_(2)(s)hArrCa^(2+)(aq)+2F^(-)(aq)`
Hence, solubility product of `CaF_(2),K_(sp)=[Ca^(2+)]xx[F^(-)]^(2)`
As given, solubility of `CaF_(2)=2xx10^(-4)mol*L^(-1)`
`therefore`In the solution, `[Ca^(2+)]=2xx10^(-4)mol*L^(-1)` and
`[F^(-)]=2xx2xx10^(-4)=4xx10^(-4)mol*L^(-1)" "[because"1 molecule of "CaF_(2)" dissociates into one "Ca^(2+)" and two "F^(-)" ions"]`
`therefore` Solubility product of `CaF_(2),K_(sp)=[Ca^(2+)]xx[F^(-)]^(2)`
`=(2xx10^(-4))xx(4xx10^(-4))^(2)=3.2xx10^(-11)`.