equilibrium summary

Question

Calculate the molar solubility of BaSO₄ (K_sp = 1.1 x 10^-10) in water AND in a solution containing 1.0 M barium ions.

Answer 1

BaSO4(s) <==> Ba²⁺(aq) + SO₄^2-(aq)

Ksp = [Ba²⁺][SO₄^2-]

In water:

1.1x10-¹⁰ = (x)(x)

x = 1.05x10^-5 M solubility

In 1.0 M Ba²⁺

This is a common ion, and since it is present at such a high concentration (1.0 M), we can ignore the very small contribution of Ba²⁺ from BaSO₄. Thus, we have ...

Ksp = [Ba²⁺][SO₄^2-]

1.1x10^-10 = (1.0)(x)

x = 1.1x10^-10 M solubility

NOTE: In the presence of a common ion, the solubility will always be less than in water because of Le Chatelier's principle.