Question

Question 4.

The K_sp for BaF₂ is 1.7 x 10^-6 .

What is the molar solubility of barium fluoride?

Will PbSO₄ be more or less soluble in 0.1 M H₂SO₄ compared to in pure water? Why? Explain in terms of LeChatelier’s Principle.

Answer 1

1: Let 'S' be the molar solubility of the BaF2   

BaF2 + H2O --- > Ba²⁺(aq) + 2F^-(aq), Ksp = 1.7 x 10^-6

eqm.conc Na ---> S M 2S M

Ksp = 1.7 x 10^-6 = [Ba²⁺(aq)]x[F^-(aq)]² = S x (2S)² = 4S³

=>  1.7 x 10^-6 = 4S³

=> S = cuberoot ( 1.7 x 10^-6 / 4) = 7.52 x 10^-3  M (answer)

2: PbSO4 is less soluble in 0.1M H2SO4 due to the following reason

The dissociation of PbSO4 and H2SO4 is

PbSO4(s) --- > Pb²⁺(aq) + SO₄^2-(aq)

H2SO4 --- > H⁺(aq) +  SO₄^2-(aq)

Hence when PbSO4 is dissolved in 0.1M H2SO4, the concentration of SO₄^2-(aq) increases. Now according to LeChatelier's principele, in order to decrease the concentration of SO₄^2-(aq), more and more SO₄^2-(aq) combines with Pb²⁺(aq) to form PbSO4 and hence solubility decreases.

But this doesn't happen when PbSO4 is dissolved in pure water due to absence of Pb²⁺(aq) in pure water.