Question

`2M` solution of `Na_(2)CO_(3)` is boiled in a closed container with excess of `CaF_(2)`. Very small amount of `CaCO_(3)` and `NaF` are formed. If `K_(sp)` of `CaCO_(3)` is `x` and molar solubility of `CaF_(2)` is `y`, find the molar after cocentration of `F^(Theta)` in the resulting solution after equilibrium is attained.

Answer 1

`{:(,Na_(2)CO_(3)+,CaF_(2)(s)rarr,CaCO_(3)+,2NaF),("Mole taken",2,,0,0),("Mole left",(2-a),,a,2a):}`
where `a` is very-very small and thus assume that `CaCO_(3)` is in soluble form
Now, `K_(sp)` of `CaCO_(3) = x = [Ca^(2+)] [CO_(3)^(2-)]`
Also `[CO_(3)^(2-)] =2 -a+a =2 :. [Ca^(2+)] = (x)/(2)`
For `{:(CaF_(2) hArr,Ca^(2+)+,2F^(Theta),,),(,y,2y,,):}`
`K_(sp(CaF_(2))) = [Ca^(2+)] [F^(Theta)]^(2) = (y) (2y)^(2) = 4y^(3)`
Further for `[F^(Theta)]`, we can have
`[F^(Theta)] = [F^(Theta)]` frm `CaF_(2) + [F^(Theta)]` from `NaF`
`[F^(Theta)] = sqrt((K_(sp(CaF_(2))))/([Ca^(2+)])) +` Negligible value
`[F^(Theta) = sqrt((4y^(3))/(x//2))sqrt((8y^(3))/(x))`