Common-Ion Effect

A saturated solution is a solution that contains the maximum amount of dissolved salt
If a second compound, which has an ion in common with the dissolved salt, is added to the saturated solution, the solubility of the salt reduces and a solid precipitate will be formed
This is also known as the common ion effect
For example, if a solution of potassium chloride (KCl) is added to a saturated solution of silver chloride (AgCl) a precipitate of silver chloride will be formed
- The chloride ion is the common ion
The solubility product can be used to predict whether a precipitate will form or not
- A precipitate will form if the product of the ion concentrations is greater than the solubility product (K_sp)

Common ion effect in silver chloride

When a KCl solution is added to a saturated solution of AgCl, an AgCl precipitate forms
In a saturated AgCl solution, the silver chloride is in equilibrium with its ions

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq)

When a solution of potassium chloride is added:
- Both KCl and AgCl have the common Cl^- ion
- There is an increased Cl^- concentration so the equilibrium position shifts to the left
- The increase in Cl^- concentration also means that [Ag⁺ (aq)] [Cl^-(aq)] is greater than the K_sp for AgCl
- As a result, the AgCl is precipitated

The common ion effect with KCl (aq) and AgCl (aq)

Equilibria - Common Ion Effect, downloadable AS & A Level Chemistry revision notes

The addition of potassium chloride to a saturated solution of silver chloride results in the precipitate of silver chloride

Worked example

Calculations using the K_sp values and the concentration of the common ion

Predict whether a precipitate of CaSO₄ will form if a saturated solution of 1.0 x 10^-3 M CaSO₄ is mixed with an equal volume of 1.0 x 10^-3 M Na₂SO₄.

K_sp CaSO₄ = 2.0 x 10^-5

Answer:

Step 1: Determine the equilibrium reaction of CaSO₄:
- CaSO₄ (s) ⇌ Ca²⁺ (aq) + SO₄^2- (aq)
Step 2: Write down the equilibrium expression for K_sp:
- K_sp = [Ca²⁺ (aq)] [SO₄^2- (aq)]

CaSO₄ (s)	$⇌$	Ca²⁺ (aq)	+	SO₄^2- (aq)
x		0		0
-x		+ x		+ x
		5.0 x 10^-4		x + 1 x 10^-3

Step 3: Determine the concentrations of the ions:
- There are equal volumes of each solution
- This means that the total solution was diluted by a factor of 2
- The new concentration of the Ca²⁺ ion is halved:
  - [Ca²⁺] = $\frac{1.0 \times 10^{- 3}}{2}$
  - [Ca²⁺] = 5.0 x 10^-4 M
- The sulfate ion concentration remains the same as it is a common ion and its concentration is the same in both solutions

Step 4: Substitute the values into the expression:
- Product of the ion concentrations = [Ca²⁺ (aq)] x [SO₄^2- (aq)]
- Product of the ion concentrations = (5.0 x 10^-4) x (1.0 x 10^-3)
- Product of the ion concentrations = 5.0 x 10^-7 M

Step 5: Determine if a precipitate will form:
- As Q (5.0 x 10^-7 M) is smaller than the K_sp value (2.0 x 10^-5 M), the CaSO₄ precipitate will not be formed

Worked example

The K_sp for lead chloride, PCl₂ is 1.7 x 10^-5at 25 °C.

Calculate the molar solubility of PCl₂ at 25 °C which is in a solution of 0.10 NaCl.

Answer:

Step 1: Write equation and expression

PbCl₂ (s) $⇌$ Pb²⁺ (aq) + 2Cl^– (aq)

K_sp = [Pb²⁺ (aq)][Cl^– (aq)]²

Step 2: Calculate concentrations of ions

PbCl₂ (s)	$⇌$	Pb²⁺ (aq)	+	2Cl^– (aq)
x		0		0
-x		+ x		+ 2x
		5.0 x 10^-4		2x + 0.10

Step 3: Place values into expression
- K_sp = [Pb²⁺ (aq)][Cl^– (aq)]²
- 1.7 x 10^-5 = (x)(2x + 0.10)²

The value for x is very small, so we can approximate and use 0.10 M for 2x + 0.10

- 1.7 x 10^-5 = (x)(0.10)²
- x = 1.7 x 10^-3M

Additional Note

We can compare this value of to molar solubility of PbCl₂ when it is NOT in a solution of 0.10 M NaCl

K_sp = [Pb²⁺ (aq)][Cl^– (aq)]²
1.7 x 10^-5 = (x)(2x)²
1.7 x 10^-5 = 4x³
x = $\sqrt[3]{\frac{1.75 \times 10^{- 5}}{4}}$
x = 0.016 M

So we can see that the addition of a common ion (Cl^–) decreased the solubility of PbCl₂ by roughly a factor of 10

Common-Ion Effect (College Board AP Chemistry)

Revision Note