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A LevelChemistryCIERevision Notes25. Equilibria (AL)25.1 Acids & BasesThe Common Ion Effect

The Common Ion Effect (CIE A Level Chemistry)

Revision Note

Philippa Platt

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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 (Ksp)

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 Ksp 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 Ksp 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.

Ksp CaSO4 = 2.0 x 10-5 mol2 dm-6.

Answer

  • Step 1: Determine the equilibrium reaction of CaSO4:
    • CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
  • Step 2: Write down the equilibrium expression for Ksp:
    • Ksp = [Ca2+ (aq)] [SO42- (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+] =  fraction numerator 1.0 cross times 10 to the power of negative 3 end exponent over denominator 2 end fraction
      • [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 [SO42- (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 Ksp value (2.0 x 10-5 mol2 dm-6), the CaSO4 precipitate will not be formed

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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.