Unit 7: Topic 12 - Common-Ion Effect

Question 1

What is the common-ion effect?

Answer 1

The common-ion effect says that if a salt is dissolved into a solution already containing ions that make up the salt, then solubility will decrease.

For example : If Ksp of AB(s) <-> A⁻(aq) + B⁺(aq) is some value “x” in pure water and the Ksp of the same dissolution is some other value “y” in water that has some concentration of A⁻ or B⁺ :

Then, y must be less than x.

Question 2

Explain why the common-ion effect exists using Le Châtelier’s principle.

Answer 2

We can once again use AB(s) <-> A⁻(aq) + B⁺(aq)

A⁻(aq) and B⁺(aq) are both products so if we already have either of them in solution, then the system will shift to favor the reactant side. This decreases Ksp and therefore also decreases solubility.

Question 3

How much more soluble is AB₂ (Ksp = 1.72*10⁻⁷) in distilled water compared to water with :

1. [A²⁺] = 0.140 M
2. [B⁻] = 0.095 M

Answer 3

AB₂(s) <-> A²⁺(aq) + 2B⁻(aq)

Ksp = [A²⁺][B⁻]² = 1.7210⁻⁷

Since the Ksp is a lot less than 1, we know that the amount of A²⁻ and B⁺ from the dissociation is very tiny and negligible.

Therefore, for (1), we can just directly use that [A²⁺] = 0.14 M

So now, let [B⁺] = x = solubility
We get the equation : (0.14)(2x)² = 1.72*10⁻⁷
Solving for x gives that solubility = 5.54 * 10⁻⁴ moles/liter

For (2), we can directly use [B⁻] = 0.095 M
Let [A²⁺] = y = solubility
We get that y(0.095) = 1.72*10⁻⁷, solving for y gives that solubility = 1.81 * 10⁻⁶ moles/liter

In distilled water you get the equation :
z(2z)²=1.7210⁻⁷ , solving for z gives that solubility = 3.5*10⁻³ moles/liter

z/x = 6.3, so AB₂ is 6.3 times more soluble in distilled water compared to condition (1)
z/y = 1934 so AB₂ is roughly 2000 times more soluble in distilled water compared to condition (2)