Common Ion Effect on Solubility Equilibrium

Introduction

• The presence of a common ion can impact solubility equilibrium.
• Example: Saturated solution of Lead (II) Chloride (PbCl₂).
  • PbCl₂ is a white solid at equilibrium with its ions in solution.
  • Ions: Pb²⁺ and Cl⁻ with a mole ratio of 1:2.

Equilibrium and Addition of Common Ion

• Equilibrium State:
  • Rate of dissolution = Rate of precipitation.
  • Ion concentration remains constant.
• Adding Potassium Chloride (KCl):
  • KCl dissociates into K⁺ and Cl⁻.
  • Addition disrupts equilibrium (Le Chatelier's Principle).
  • Increase in Cl⁻ concentration.
  • System shifts left to reduce stress by forming more PbCl₂.
  • Result: Increased PbCl₂ precipitate until new equilibrium.

Common Ion Effect Explanation

• Common Ion: Cl⁻ from both PbCl₂ and added KCl.
• Le Chatelier's Principle: System shifts to reduce Cl⁻ concentration.
• Result: Solubility of PbCl₂ decreased.

Quantitative Analysis of Common Ion Effect

• Reaction Quotient (Q):
  • At equilibrium, Qₛₚ = Kₛₚ.
  • Adding Cl⁻ increases Qₛₚ, causing shift left until Qₛₚ = Kₛₚ again.
  • More PbCl₂ solid forms, decreasing solubility.
• Kₛₚ remains constant at constant temperature.

Calculation of Molar Solubility

• Objective: Calculate molar solubility of PbCl₂ in 0.10M KCl at 25°C.
• Kₛₚ for PbCl₂: 1.7 x 10⁻⁵.
• Use ICE table:
  • Initial: Pb²⁺ = 0, Cl⁻ = 0 from PbCl₂, Cl⁻ = 0.10M from KCl.
  • Change: PbCl₂ dissolves (-x), Pb²⁺ (+x), Cl⁻ (+2x).
  • Equilibrium: Pb²⁺ = x, Cl⁻ = 0.10 + 2x.
• Approximation:
  • Assuming x is small, 0.10 + 2x ≈ 0.10.
  • Simplifies calculation.
• Result: Molar solubility of PbCl₂ = 1.7 x 10⁻³ M in 0.10M KCl.

Effect without Common Ion

• Without KCl:
  • Molar solubility of PbCl₂ = 0.016 M.
• Comparison:
  • Common ion decreases solubility by ~factor of 10.

Conclusion

• Common ion effect decreases solubility of slightly soluble salts.
• Quantitative analysis confirms significant reduction in solubility due to common ion presence.