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The Common Ion Effect (CIE A Level Chemistry)

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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 actually 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

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

Equilibria - Worked Example - Calculations using the Ksp values and the concentration of the common ion, downloadable AS & A Level Chemistry revision notes

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 concentrations of the Ca2+ ion is halved 

      begin mathsize 14px style Ca to the power of 2 plus end exponent space equals space fraction numerator 1 space cross times 10 blank to the power of negative 3 end exponent over denominator 2 end fraction end style

= 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)]

= (5.0 x 10-4) x (1.0 x 10-3)

= 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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Francesca

Author: Francesca

Expertise: Head of Science

Fran studied for a BSc in Chemistry with Forensic Science, and since graduating taught A level Chemistry in the UK for over 11 years. She studied for an MBA in Senior Leadership, and has held a number of roles during her time in Education, including Head of Chemistry, Head of Science and most recently as an Assistant Headteacher. In this role, she used her passion for education to drive improvement and success for staff and students across a number of subjects in addition to Science, supporting them to achieve their full potential. Fran has co-written Science textbooks, delivered CPD for teachers, and worked as an examiner for a number of UK exam boards.