The Common Ion Effect (Cambridge International A-Level Chemistry)

The Solubility Product & the 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)

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 mol dm^-3 CaSO₄ is mixed with an equal volume of 1.0 x 10^-3 mol dm^-3 Na₂SO₄.

K_sp* *CaSO₄ = 2.0 x 10^-5 mol² dm^-6.

Answer

• **Step 1: **Determine the equilibrium reaction of CaSO₄:

CaSO₄** (s) ⇌ Ca**²⁺** (aq) + SO ₄^2- (aq)**

• **Step 2: **Write down the equilibrium expression for K_sp:

K_sp** = [Ca**²⁺** (aq)] [SO ₄^2- (aq)]**

-** 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⁺] = $\frac{1.0\times10^{-3}}{2}$
  • [Ca²⁺] = 5.0 x 10^-4 mol dm^-3

• 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 mol² dm^-6

• Step 5: Determine if a precipitate will form:

  • As the product of the ion concentration (5.0 x 10^-7 mol dm^-3) is smaller than the K_sp value (2.0 x 10^-5 mol² dm^-6), the CaSO₄ precipitate will not be formed