Calculate the `[F^(Θ)]` in a solution saturated with respect ot `MgF_(2)` and `SrF_(2)`.
`K_(sp) (MgF_(2)) = 6.0 xx 10^(-9), K_(sp) (SrF_(2)) = 3.0 xx 10^(-9)`

Explanation

Note the `K_(sp)` values of two slats are very similar. So the concentration of `F^(Θ)` (ions the common ion) can not be calculated fro a single salt alone.
Let the simuitaneous solubilities of `MgF_(2)` and `SrF_(2)` be `x` and `y` respectively in `mol L^(-1)`.
`{:(MgF_(2)(s)hArr,Mg^(2+)+,2F^(Θ),,),(,x,2x,,),(SrF_(2)(s)hArr,Sr^(2+)+,2F^(Θ),,),(,y,2y,,):}`
At equilibrium `[F^(Θ)] = 2x + 2y, [Mg^(2+)] = x, [Sr^(2+)] = y`
`rArr [Mg^(2+)] [F^(Θ)] = K_(sp) MgF_(2)`
and `[Sr^(2+)] [F^(Θ)]^(2) = K_(sp) SrF_(2)`
Note that: `[F^(Θ)] = 2 [(Mg^(2+)) + (Sr^(2+))]`
[This is an Electrical charge neutrality equation)
`[F^(Θ)] = 2 [(K_(sp_(MgF_(2))))/([F^(Θ)]^(2))+(K_(spSrF_(2)))/([F^(Θ)]^(2))]`
`rArr [F^(Θ)] = sqrt(2(K_(sp(MgF_(2)))+K_(sp(SrF_(2))))) = 2.62 xx 10^(-3)M`