Unit 7 · Topic 7.12 Beta

Common-Ion Effect

The common-ion effect: a salt is less soluble in a solution that already contains one of its ions.

Practice 5: Mathematical RoutinesPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

Lead(II) chloride is fairly soluble in pure water for a "slightly soluble" salt, about four and a half grams per liter. Pour in some table salt, and almost all of that lead chloride falls out as a white solid. Nothing about lead chloride has changed. The solution simply already had one of its ions, and that changes the balance.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For PbCl2, what is the Ksp expression?

  1. [Pb2+][Cl−]2
  2. [Pb2+][Cl−]
  3. [Pb2+]2[Cl−]
  4. [PbCl2] / ([Pb2+][Cl−]2)
Show the answer

Ion concentrations, each to its coefficient; the solid is left out.

  • Correct: [Pb2+][Cl−]2:
  • [Pb2+][Cl−]:
  • [Pb2+]2[Cl−]:
  • [PbCl2] / ([Pb2+][Cl−]2):

2. Adding a product to a system at equilibrium makes it shift

  1. toward reactants
  2. toward products
  3. nowhere
  4. only if the temperature changes
Show the answer

Le Châtelier: it partly uses up what was added.

  • Correct: toward reactants:
  • toward products:
  • nowhere:
  • only if the temperature changes:

3. When should you check the small-x approximation?

  1. Every time you use it
  2. Only when K is large
  3. Never; it always works
  4. Only for gases
Show the answer

The 5% check confirms the dropped term really is small.

  • Correct: Every time you use it:
  • Only when K is large:
  • Never; it always works:
  • Only for gases:

Part 4 · See it

See it first

A bar chart of the molar solubility of PbCl2 at 25 °C: 1.6 × 10−2 M in pure water, falling to 1.3 × 10−2 M in 0.010 M NaCl, 4.8 × 10−3 M in 0.050 M NaCl and 2.7 × 10−4 M in 0.25 M NaCl. A side panel explains: Q = [Pb2+][Cl−] squared; added chloride raises Q above Ksp, so solid forms until Q = Ksp; Ksp is unchanged but the solubility is lower.
More Cl− from NaCl means less PbCl2 dissolves. Ksp does not change; Q does. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A soluble salt adds an ion that is also in a solubility equilibriumQ rises above Ksp
  2. Q is greater than Kspthe slightly soluble salt precipitates until Q = Ksp
  3. The equilibrium is restored with the common ion presentless of the salt is dissolved: its molar solubility is lower
  4. The temperature has not changedKsp is the same; only the solubility changed

Part 6 · Key ideas

Key ideas

  • The common-ion effect: an ion already in solution lowers the solubility of a salt that contains it.
  • Justify it with Q: the common ion raises Q above Ksp, so solid forms until Q = Ksp. Ksp does not change.
  • In calculations, the shared ion is (common ion + coefficient × s) ≈ common ion. Count ions per formula unit; check the 5% rule.
  • An ion that is squared in Ksp has a larger effect.

Part 7 · Misconception

A common mistake

The wrong idea: Adding NaCl lowers the Ksp of PbCl₂, so less dissolves.

What actually happens: Ksp depends only on temperature. Added Cl⁻ raises Q above Ksp, so PbCl₂ precipitates until Q = Ksp; the solubility falls, the constant does not.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Lead(II) chloride in salt solutions

A student measures the molar solubility of PbCl2 (Ksp = 1.7 × 10−5 at 25 °C) in water and in three NaCl solutions at 25 °C. NaCl is a soluble salt that dissociates completely. PbCl2(s) ⇌ Pb2+(aq) + 2 Cl−(aq).

Molar solubility of PbCl2 at 25 °C
Solvent[NaCl] (M)Molar solubility of PbCl2 (M)
pure water01.6 × 10−2
NaCl(aq)0.0101.3 × 10−2
NaCl(aq)0.0504.8 × 10−3
NaCl(aq)0.252.7 × 10−4

1. What trend does the table show?

  1. PbCl2 is less soluble when more Cl− is already in solution.
  2. PbCl2 is more soluble in salt solutions than in water.
  3. The solubility of PbCl2 does not depend on [NaCl].
  4. PbCl2 is less soluble because Ksp decreases in salt solutions.
Show the answer

This is the common-ion effect: an ion already present (Cl⁻ from NaCl) lowers the solubility of a salt that contains it.

  • Correct: PbCl2 is less soluble when more Cl− is already in solution.: Right: the solubility falls from 1.6 × 10−2 M in water to 2.7 × 10−4 M in 0.25 M NaCl.
  • PbCl2 is more soluble in salt solutions than in water.: The numbers fall, not rise, as [NaCl] increases.
  • The solubility of PbCl2 does not depend on [NaCl].: The solubility changes by a factor of about 60 across the table.
  • PbCl2 is less soluble because Ksp decreases in salt solutions.: The trend is right, but the cause is wrong: Ksp is the same in every solution at 25 °C.

2. Which explanation of the trend would earn credit?

  1. Cl− from NaCl raises Q, so less PbCl2 dissolves before Q = Ksp.
  2. NaCl lowers Ksp for PbCl2.
  3. Na+ ions react with Pb2+ and remove it.
  4. NaCl takes up the water, leaving less for PbCl2 to dissolve in.
Show the answer

Reason with Q and Ksp: Cl⁻ from NaCl raises Q, so the equilibrium sits further toward the solid and less PbCl₂ dissolves.

  • Correct: Cl− from NaCl raises Q, so less PbCl2 dissolves before Q = Ksp.: Right: with Cl⁻ already present, a smaller [Pb²⁺] is enough to make Q reach Ksp.
  • NaCl lowers Ksp for PbCl2.: Ksp depends only on temperature; the common ion changes Q.
  • Na+ ions react with Pb2+ and remove it.: Na⁺ is a spectator ion here. The shared ion, Cl⁻, is what matters.
  • NaCl takes up the water, leaving less for PbCl2 to dissolve in.: The solutions are dilute; the effect is an equilibrium effect of Cl⁻, not a shortage of water.

3. Using the small-x approximation ([Cl−] ≈ 0.25 M), calculate the molar solubility of PbCl2 in 0.25 M NaCl.

Type a number and its unit.

Show the answer

PbCl₂ ⇌ Pb²⁺ + 2 Cl⁻: [Pb²⁺] = s, [Cl⁻] = 0.25 + 2s ≈ 0.25 M. Ksp = s(0.25)² → s = 1.7 × 10⁻⁵ / 0.0625 = 2.7 × 10−4 M. Check: 2s = 5.4 × 10−4 M is 0.2% of 0.25 M, well under 5%.

  • Answer: 2.7 × 10-4 M

4. Ksp of AgCl is 1.8 × 10−10. Calculate its molar solubility in 0.020 M CaCl2.

Type a number and its unit.

Show the answer

[Cl⁻] from CaCl₂ = 2 × 0.020 = 0.040 M. Ksp = s(0.040) → s = 1.8 × 10⁻¹⁰ / 0.040 = 4.5 × 10−9 M.

  • Answer: 4.5 × 10-9 M

5. Which added solutes would lower the solubility of CaF2 in water? Select all that apply.

  1. NaF
  2. NaCl
  3. KNO3
  4. Sugar (C12H22O11)
Show the answer

Only an ion that appears in the solubility equilibrium (Ca²⁺ or F⁻) raises Q and lowers the solubility. Of these, only NaF supplies one.

  • Correct: NaF: Right: it supplies F⁻, a common ion.
  • NaCl: Neither Na⁺ nor Cl⁻ is in the CaF₂ equilibrium, so Q is unchanged.
  • KNO3: Neither K⁺ nor NO₃⁻ is in the CaF₂ equilibrium.
  • Sugar (C12H22O11): A nonelectrolyte adds no ions, so it does not change Q.

6. A saturated solution of BaSO4 is at equilibrium with the solid. Concentrated Na2SO4 solution is added. Which statement gives the best justification of what happens?

  1. Q rises above Ksp, so BaSO4 precipitates.
  2. Ksp rises because more sulfate is present, so more BaSO4 dissolves.
  3. Na+ replaces Ba2+ in the solid.
  4. The reverse reaction speeds up, so the system shifts.
Show the answer

Adding a common ion raises Q above Ksp; the net reaction is precipitation, which lowers [Ba²⁺] until Q = Ksp.

  • Correct: Q rises above Ksp, so BaSO4 precipitates.: Right: claim (precipitate forms), evidence (sulfate added), reasoning (Q > Ksp).
  • Ksp rises because more sulfate is present, so more BaSO4 dissolves.: Ksp does not change with concentration, and adding a common ion lowers solubility.
  • Na+ replaces Ba2+ in the solid.: Na⁺ is a spectator; the solid stays BaSO₄.
  • The reverse reaction speeds up, so the system shifts.: Rate language alone does not earn the point; compare Q with Ksp.

7. Why does adding a common ion lower solubility, at the particle level?

  1. Ions rejoin the solid more often, so less salt dissolves before the rates match.
  2. The added ions coat the surface of the solid so water does not reach it.
  3. The added ions make the water less polar.
  4. The added ions react with the solid to make a new compound.
Show the answer

Precipitation happens when dissolved ions meet the solid. A higher concentration of one of the ions speeds precipitation, so the dissolving and precipitation rates become equal with less salt dissolved.

  • Correct: Ions rejoin the solid more often, so less salt dissolves before the rates match.: Right: the precipitation rate rises with the ion concentration, so less salt needs to dissolve before the rates match.
  • The added ions coat the surface of the solid so water does not reach it.: Dissolved ions do not coat the solid; the effect comes from the equilibrium.
  • The added ions make the water less polar.: Dilute ions barely change water as a solvent.
  • The added ions react with the solid to make a new compound.: The solid is the same compound; only the balance between dissolving and precipitation changes.

Part 9 · Summary

Summary

The common-ion effect: a salt is less soluble in a solution that already contains one of its ions. The common ion raises Q above Ksp, so the salt precipitates until Q = Ksp again; Ksp is unchanged and only the solubility drops. To calculate the new solubility, write the shared ion as common ion plus a small term, drop the small term, solve Ksp for s and check the 5% rule.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections