AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-12)
Unit 7 · Topic 7.12
7.12 Common-Ion Effect
A salt is less soluble in a solution that already contains one of its ions, which is called the common-ion effect. You can explain it with Le Châtelier's principle and calculate the lower solubility from Ksp by including the ion that's already there.
Key terms
- common-ion effect
- Le Châtelier's principle
- molar solubility
- Ksp
What the common-ion effect is
A common ion is an ion that the solution and the salt share. AgCl dissolves less in a NaCl solution than in pure water, because the NaCl already supplies Cl⁻.
Le Châtelier explanation: for AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the extra Cl⁻ is an added product, so the equilibrium shifts left, toward the solid. Less AgCl can dissolve.
Q explanation: extra Cl⁻ makes Q = [Ag⁺][Cl⁻] larger. To keep Q = Ksp, [Ag⁺] must be much smaller, so less AgCl dissolves.
Ksp stays the same
The common ion lowers the solubility, but Ksp doesn't change, since only temperature changes an equilibrium constant. The ion concentrations just rearrange so that their product still equals Ksp.
If you calculated Ksp from solubility data measured in a solution that already contained a common ion, and you forgot to include that ion, your Ksp would come out far too small.
Calculating solubility with a common ion
- Write the dissolving equation and the Ksp expression.
- Set up an ICE table: the common ion starts at its concentration from the other salt, not at zero.
- Add s (or 2s, depending on coefficients) for each ion from the dissolving salt.
- Because the salt is now even less soluble, s is tiny compared with the common-ion concentration, so 0.10 + s ≈ 0.10. Check afterward.
- Solve for s.
Why it matters in the lab
Chemists use the common-ion effect to make a precipitation more complete. If you want to collect as much of an ion as possible as a solid, add more than enough of the precipitating ion. Washing a precipitate with a dilute solution containing a common ion, instead of pure water, also dissolves less of it.
Which common ion you add matters. For CaF₂, adding F⁻ lowers solubility much more than adding the same concentration of Ca²⁺, because [F⁻] is squared in the Ksp expression.
The same idea in acid–base chemistry
The common-ion effect isn't limited to solubility. Adding sodium acetate to a solution of acetic acid adds CH₃COO⁻, a product of CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻. The ionization shifts left, [H₃O⁺] drops and the pH rises. A weak acid mixed with its conjugate base is a buffer, which you'll study in Unit 8 (topics 8.8 and 8.9).
In every case the reasoning is the same: an ion that is already present pushes the equilibrium away from the side that produces it.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
AgCl in sodium chloride solution
Ksp of AgCl is 1.8 × 10⁻¹⁰. Find the molar solubility of AgCl in 0.10 M NaCl, and compare it with its solubility in pure water (1.3 × 10⁻⁵ M).
Show the solutionHide the solution
- Step 1: NaCl dissolves completely, so [Cl⁻] starts at 0.10 M.
- Step 2: ICE: [Ag⁺] = s; [Cl⁻] = 0.10 + s.
- Step 3: Ksp = s(0.10 + s) = 1.8 × 10⁻¹⁰. Since s will be tiny, 0.10 + s ≈ 0.10.
- Step 4: s = 1.8 × 10⁻¹⁰ ÷ 0.10 = 1.8 × 10⁻⁹ M. Check: s is far less than 5% of 0.10.
- Step 5: Comparison: 1.3 × 10⁻⁵ ÷ 1.8 × 10⁻⁹ ≈ 7 × 10³, so AgCl is about 7000 times less soluble here.
Answer: s = 1.8 × 10⁻⁹ M, roughly 7000 times lower than in pure water
- Example 2Calculator allowed
Trap: squaring the common ion
Ksp of CaF₂ is 3.9 × 10⁻¹¹. Find the molar solubility of CaF₂ in 0.010 M NaF.
Show the solutionHide the solution
- Step 1: [F⁻] starts at 0.010 M. ICE: [Ca²⁺] = s; [F⁻] = 0.010 + 2s ≈ 0.010.
- Step 2: Ksp = [Ca²⁺][F⁻]² = s(0.010)².
- Step 3: s = 3.9 × 10⁻¹¹ ÷ (1.0 × 10⁻⁴) = 3.9 × 10⁻⁷ M.
- Step 4: Two traps: forgetting to square 0.010 (which gives 3.9 × 10⁻⁹ M), and doubling the 0.010 to 0.020. The 0.010 M F⁻ comes from NaF, which gives one F⁻ per formula unit, so it isn't doubled; only the F⁻ from CaF₂ is written as 2s.
Answer: s = 3.9 × 10⁻⁷ M (compared with 2.1 × 10⁻⁴ M in pure water)
- Example 3Calculator allowed
Comparing two common ions
Find the molar solubility of CaF₂ (Ksp = 3.9 × 10⁻¹¹) in 0.010 M Ca(NO₃)₂, and compare it with the answer for 0.010 M NaF.
Show the solutionHide the solution
- Step 1: Now Ca²⁺ is the common ion: [Ca²⁺] = 0.010 + s ≈ 0.010 and [F⁻] = 2s.
- Step 2: Ksp = (0.010)(2s)² = 0.040s² = 3.9 × 10⁻¹¹.
- Step 3: s² = 9.75 × 10⁻¹⁰, so s = 3.1 × 10⁻⁵ M.
- Step 4: In 0.010 M NaF, s was 3.9 × 10⁻⁷ M. Adding F⁻ lowers solubility far more than adding the same concentration of Ca²⁺, because [F⁻] is squared in Ksp.
Answer: s = 3.1 × 10⁻⁵ M; the F⁻ solution suppresses solubility much more.
Common mistakes
- Starting the common ion at zero in the ICE table.
- Saying the common ion lowers Ksp. It lowers solubility; Ksp stays the same at a given temperature.
- Forgetting to square the common-ion concentration when its coefficient is 2.
- Doubling the common-ion concentration from a salt that only provides one of that ion per formula unit.
On the exam
- You may be asked to explain qualitatively (with Le Châtelier or Q versus Ksp) why a salt is less soluble in a given solution, or to calculate the new solubility. Mention the shift toward the solid.
- Experimental-design questions may ask how adding a common ion changes the mass of precipitate collected or the measured solubility.
Connected topics
Videos
Check yourself
4 questions on 7.12 Common-Ion Effect. Pick an answer to see if you got it, and why.
Ksp for AgCl is 1.8 × 10⁻¹⁰ at 25 °C. What is the approximate molar solubility of AgCl in 0.10 M NaCl at 25 °C?
Solid NaF is dissolved in a saturated solution of CaF₂ that is in contact with undissolved CaF₂ at constant temperature. Which of the following describes the result once equilibrium is reestablished?
Ksp for Ag₂CrO₄ is 1.1 × 10⁻¹² at 25 °C. What is the approximate molar solubility of Ag₂CrO₄ in 0.10 M K₂CrO₄?
In which of the following would AgCl have the lowest molar solubility at 25 °C?
0 of 4 answered