Calculate the molar solubility of barium fluoride in each liquid or solution.

Question

Calculate the molar solubility of barium fluoride in each liquid or solution.

BaF2 Ksp = 2.45 x 10-5

Ksp(BaF2)=2.45×10−5

a) Pure water

b) 0.15 M NaF

Answer 1

BaF₂(s) <==> Ba²⁺(aq) + 2F^-(aq) ... dissolution of BaF₂

Ksp = [Ba²⁺][F^-]²

2.45x10^-5 = (x)(2x)² = 4x³

x³ = 6.125x10^-6

x = 0.0183 M = molar solubility of BaF₂ in pure water

When the BaF₂ is in 0.15 M NaF, we have a situation where we have a common ion, which is F^-. The concentration of F^- is mainly supplied by the NaF (0.15 M), so we can use that as the final [F^-]. According to Le Chatelier's principle, the presence of F^- will shift equilibrium to the reactant side, thus decreasing the solubility relative to that in pure water.

Ksp = 2.45x10^-5 = [Ba²⁺][F^-]²

2.45x10^-5 = (x)(0.15)² = 0.0225x

x = 0.00109 M = molar solubility of BaF₂ in 0.15 M NaF