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Common Ion Effect and Complex Ions

Learn and apply Common Ion Effect and Complex Ions in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Common Ion Effect and Complex Ions: Orientation

Solubility shifts in aqueous chemistry often involve two linked equilibria: dissolution (with Kₛₚ) plus either a common ion or complex formation. This lesson shows how to explain the direction of change correctly and how to do quick common-ion calculations without claiming Kₛₚ changes.

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. Common ion effect

The common ion effect is the decrease in solubility (or dissociation) when an ion already present in solution is added.

B. Complex ion formation

A complex ion forms when a metal ion binds to ligands (e.g. NH₃), often reducing the free metal-ion concentration in solution.

C. Free (uncomplexed) ion concentration

The free ion concentration is the concentration of the ion not tied up in a complex (e.g. “free [Ag⁺]” means Ag⁺ not in [Ag(NH₃)₂]⁺).

Detailed Explanations

A. Common ion effect on solubility (linked to Kₛₚ)

For: MX(s) ⇌ M + (aq) + X-(aq) Kₛₚ = [M⁺][X⁻]

If [X⁻] is increased by adding a soluble salt containing X⁻, then because Kₛₚ must stay constant at that temperature, therefore the free [M⁺] at equilibrium must decrease. This is achieved by shifting the dissolution equilibrium left, so solubility decreases.

Workflow (explain questions):

  1. Write the dissolution equilibrium.
  2. State the added ion is a product of the dissolution equilibrium (common ion).
  3. State the shift direction (left) to oppose the increase.
  4. Conclude solubility decreases (less solid dissolves).

Mini example: Adding NaCl(aq) increases [Cl⁻] in AgCl(s) ⇌ Ag⁺ + Cl⁻, so the equilibrium shifts left and AgCl becomes less soluble.

B. Complex ions can increase solubility (removing free metal ions)

When a ligand forms a stable complex with the metal ion, the free metal-ion concentration drops.

Example system: AgCl(s) ⇌ Ag + (aq) + Cl-(aq) (Kₛₚ) Ag + (aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂] + (aq)

Because complex formation removes free Ag⁺ from solution, therefore [Ag⁺][Cl⁻] becomes smaller than Kₛₚ. More AgCl(s) dissolves to restore Qₛₚ = Kₛₚ for the free ions, so overall solubility increases.

C. A useful calculation shortcut (common ion present in large excess)

If the added common ion concentration is much larger than the solubility, you can often assume it stays approximately constant.

Example: If AgCl is in 0.100 mol dm⁻³ Cl⁻, then:

Kₛₚ = [Ag⁺][Cl⁻] ≈ [Ag⁺](0.100)

so [Ag⁺] ≈ Kₛₚ/0.100 (this gives the solubility for a 1:1 salt).

Worked Examples

Modelled example 1

Explain the common-ion effect on AgCl

Core

Problem

Explain why adding NaCl(aq) to a saturated AgCl suspension reduces the solubility of AgCl.
Study the worked solution

  1. Write the relevant equilibrium

    Method

    Represent dissolution using free aqueous ions.

    Reason

    The added ion must be located in the equilibrium before a shift is predicted.

    Working

    AgCl(s) ⇌ Ag + (aq) + Cl-(aq).

  2. Identify the disturbance

    Method

    Recognise chloride from sodium chloride as a common ion.

    Reason

    Added Cl⁻ directly increases the concentration of one dissolution product.

    Working

    [Cl⁻] increases.

  3. Predict the response

    Method

    Shift the dissolution equilibrium to the left.

    Reason

    Silver and chloride ions combine to oppose the increased product concentration while maintaining Kₛₚ at fixed temperature.

    Working

    Less AgCl remains dissolved, so its solubility decreases.

Guided practice 2

Calculate solubility with excess chloride

About 7 min

Problem

Calculate the solubility of AgCl in 0.100 mol dm⁻³ NaCl(aq) at 25°C. Given Kₛₚ(AgCl) = 1.8 × 10⁻¹⁰.

Try this before viewing the solution

Hints

Hint 1: use the excess-ion approximation
The chloride supplied by dissolving AgCl is negligible beside 0.100 mol dm⁻³, so use [Cl⁻] ≈ 0.100 mol dm⁻³.
Hint 2: solve for free silver
Rearrange Kₛₚ = [Ag⁺][Cl⁻] for [Ag⁺].
View solution step by step

  1. Set the common-ion concentration

    Method

    Approximate free chloride by the sodium chloride concentration.

    Reason

    The common ion is present in very large excess compared with the expected AgCl solubility.

    Working

    [Cl⁻] ≈ 0.100 mol dm⁻³.

  2. Use the fixed Ksp

    Method

    Calculate the equilibrium free silver concentration.

    Reason

    At saturation, the free-ion product equals Kₛₚ.

    Working

    [Ag⁺] = (1.8 × 10⁻¹⁰)/0.100 = 1.8 × 10⁻⁹ mol dm⁻³.

  3. Convert to molar solubility

    Method

    Equate AgCl solubility with the silver concentration produced.

    Reason

    One mole of dissolved AgCl releases one mole of Ag⁺.

    Working

    s = 1.8 × 10⁻⁹ mol dm⁻³.

Common misconception 3

Separate Ksp from solubility

Find and correct the mistake

Learner claim

A learner says adding sodium chloride makes AgCl less soluble because the added common ion decreases Kₛₚ. Diagnose the explanation at fixed temperature.

Choose what changes

At fixed temperature, adding chloride changes

View solution step by step

  1. Keep the constant fixed

    Method

    Reject the claimed change in Kₛₚ.

    Reason

    Kₛₚ for a specified dissolution equilibrium depends on temperature, which is unchanged here.

    Working

    The equilibrium must still satisfy [Ag⁺][Cl⁻] = Kₛₚ.

  2. Change the composition instead

    Method

    Increase chloride and decrease the equilibrium free silver concentration and solubility.

    Reason

    The dissolution equilibrium shifts left until the free-ion product again equals the same Kₛₚ.

    Working

    Solubility decreases even though the equilibrium constant does not.

Examiner practice 4

Explain dissolution in aqueous ammonia

4 marks

Problem

Explain why AgCl(s) dissolves in aqueous ammonia. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Show complex formation

    1 mark

    Method

    Write silver-ion coordination by ammonia.

    Reason

    The ligand binds dissolved silver ions.

    Working

    Ag + (aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂] + (aq).

  2. Identify the free-ion change

    1 mark

    Reason

    Complexed silver is not free Ag⁺ in the Kₛₚ expression.

    Working

    Free [Ag⁺] decreases.

  3. Shift dissolution

    1 mark

    Reason

    The AgCl equilibrium responds by replacing removed free silver ions.

    Working

    AgCl(s) ⇌ Ag + (aq) + Cl-(aq) shifts to the right.

  4. Conclude about solubility

    1 mark

    Method

    State the observable equilibrium consequence.

    Reason

    Rightward dissolution consumes more solid.

    Working

    More AgCl dissolves, so its total solubility increases.

Challenge 5

Compare two opposing solubility interventions

Minimal support

Problem

Two identical saturated AgCl suspensions contain excess solid. Sodium chloride is added to sample A; excess aqueous ammonia is added to sample B. Compare the change in AgCl solubility in the two samples and explain each change using free ions.

Try this before viewing the solution

Hints

Hint 1: track chloride in sample A
Added sodium chloride supplies a product ion of the dissolution equilibrium.
Hint 2: track free silver in sample B
Ammonia binds free Ag⁺; distinguish free silver concentration from total dissolved silver.
View solution step by step

  1. Analyse sample A

    Method

    Treat added chloride as a common ion.

    Reason

    Increasing a dissolution product shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left.

    Working

    Less AgCl is dissolved in sample A, so its solubility decreases.

  2. Analyse sample B

    Method

    Use ammonia to form [Ag(NH₃)₂]⁺ from free Ag⁺.

    Reason

    Removing free silver shifts AgCl dissolution to the right to replace it.

    Working

    More AgCl dissolves in sample B, so total dissolved silver increases.

  3. Compare the directions

    Method

    State the opposing outcomes using the controlled variable.

    Reason

    The common ion adds a free dissolution product, whereas complexation removes one.

    Working

    NaCl decreases AgCl solubility; ammonia increases it, even though both treatments can lower free [Ag⁺] at their new equilibria.

Mind Stretchers

Mind stretcher 1Extension

Why does complex formation affect solubility but does not “change Kₛₚ” at a fixed temperature?

Show Answer

Mark scheme:

  • Kₛₚ is defined for the dissolution equilibrium and depends only on temperature.
  • Complex formation reduces free [Ag⁺], so more solid dissolves until the free-ion product [Ag⁺][Cl⁻] equals Kₛₚ again.
  • Therefore the equilibrium composition (how much dissolves) changes, but the value of Kₛₚ does not.