InterviewSolution
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InterviewSolution
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`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. |
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Answer» `{:(,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))` |
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