Question

17.) An aqueous solution of barium nitrate is added dropwise to a solution containing 0.10 M sulfate ions and 0.10 M fluoride ions. The Ksp value for barium sulfate = 1.1 x 10-10 and the Ksp value for barium fluoride = 1.7 x 10-6
a.) Which salt will precipitate from solution first?

b.) What is the minimum [Ba2+] concentration necessary to precipitate the first salt? c.) What is the minimum [Ba2+] concentration necessary to precipitate the second salt?

Answer 1

Solution:- (a) A precipitate formed if ionic product is greater than solubilitiy product. Since we have equal concentrations of F^- and SO4^2- ions, the precipitate would form first for the salt to which the Ksp is lower as the ionic product would easily exceed the solubility product. Ksp is lower for barium sulfate(BaSO₄) so it's precipitate would form first.

(b)for barium sulfate..

BaSO₄(s) <-------> Ba²⁺(aq) + SO₄^2-(aq)

Ksp= [Ba²⁺] [SO₄^2-]

[Ba²⁺] = Ksp/[SO₄^2-]

[Ba²⁺] = 1.1 x 10^-10/0.10 = 1.1 x 10^-9 M

for barium fluoride..

BaF₂(s) <-----> Ba²⁺(aq) + 2F^-(aq)

Ksp = [Ba²⁺] [F^-]²

[Ba²⁺] = Ksp/[F^-]² = 1.7 x 10^-6/(0.10)² = 1.7 x 10^-4 M

From the minimum required concentrations of Ba²⁺ ion it is also clear that a precipitate of barium sulfate would form first as it needs less amount of barium ion.