Calculate solubility with a common ion
Use Ksp and the pre-existing common-ion concentration to calculate a sparingly soluble salt's solubility.
Section 25 · Equilibria
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 25. Use the equilibrium expressions and data stated in the question; written practice is self-checked and human subject-teacher review remains pending.
Before you begin
- Use AS equilibrium expressions, mole calculations, logarithms, and Brønsted–Lowry acid–base ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For AgCl dissolving in 0.10 mol dm⁻³ NaCl, what is [Cl⁻] if solubility is s?
What you will learn
- Set equilibrium concentrations in a common-ion solution
- Substitute into Ksp and solve for the small solubility
- State and check any concentration approximation
Include the common ion already present
For AgCl(s) ⇌ Ag⁺ + Cl⁻ in 0.10 mol dm⁻³ NaCl, let the AgCl solubility be s. Then [Ag⁺]=s and [Cl⁻]=0.10+s.
If s is much smaller than 0.10, use [Cl⁻] ≈ 0.10 mol dm⁻³, giving Ksp ≈ s(0.10). Check the resulting s against the approximation before accepting it.
Use the correct powers and temperature data
For a salt with a 2:1 ion ratio, the common-ion contribution may be squared in Ksp. Write each equilibrium concentration from the dissolution stoichiometry and starting solution before substitution.
Ksp must correspond to the same temperature. A common-ion calculation finds the new equilibrium solubility; it does not calculate a new Ksp.
Pause and explain
If s = 1.8 × 10⁻⁹ mol dm⁻³ in 0.10 mol dm⁻³ chloride, is [Cl⁻]≈0.10 justified?
Worked example
AgCl has Ksp = 1.8 × 10⁻¹⁰ in 0.10 mol dm⁻³ NaCl. Estimate its molar solubility.
Show the worked solution
- Let [Ag⁺]=s and [Cl⁻]=0.10+s.
- Assume s ≪ 0.10, so Ksp ≈ s(0.10).
- Solve and check the approximation.
Answer s ≈ 1.8 × 10⁻⁹ mol dm⁻³. Since this is far below 0.10 mol dm⁻³, the approximation is valid.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find AgCl solubility in 0.20 mol dm⁻³ NaCl if Ksp = 1.8 × 10⁻¹⁰.
Give me a hint
Use [Ag⁺]=s and [Cl⁻]≈0.20 if s is small.
Compare my reasoning
- Write Ksp ≈ s(0.20).
- Solve for s.
- Check s ≪ 0.20 mol dm⁻³.
s ≈ 9.0 × 10⁻¹⁰ mol dm⁻³; the approximation is valid.
Look for these in your work
- Common concentration included
- Equation solved
- Approximation checked
For PbI₂ in 0.10 mol dm⁻³ KI, let solubility be s and Ksp = 8.0 × 10⁻⁹. Estimate s using [I⁻]≈0.10 mol dm⁻³.
Give me a hint
Use [Pb²⁺]=s and Ksp=[Pb²⁺][I⁻]².
Compare my reasoning
- Ksp ≈ s(0.10)².
- Rearrange for s.
- Check that 2s is small relative to 0.10.
s ≈ 8.0 × 10⁻⁷ mol dm⁻³. The extra 2s iodide is negligible beside 0.10 mol dm⁻³.
Look for these in your work
- Iodide power retained
- Common-ion value substituted
- Approximation checked
A student uses only s for chloride when calculating AgCl solubility in 0.050 mol dm⁻³ NaCl. Explain and correct the setup.
Give me a hint
Write total equilibrium chloride as initial chloride plus chloride from dissolved AgCl.
Compare my reasoning
- NaCl supplies 0.050 mol dm⁻³ Cl⁻ before AgCl dissolves.
- If AgCl solubility is s, total [Cl⁻]=0.050+s.
- Use Ksp=s(0.050+s), or s(0.050) if justified.
The setup omitted the common ion. Use Ksp = s(0.050+s); when s ≪ 0.050, approximate s ≈ Ksp/0.050 and check it afterward.
Look for these in your work
- Initial concentration included
- Equilibrium concentration written
- Approximation qualified
Common mistakes
- Using [Cl⁻]=s and forgetting the salt solution
- Adding the common ion concentration without checking stoichiometry
- Failing to verify s is small compared with the initial concentration
- Changing Ksp instead of solving for solubility
What can you explain now?
For AgCl dissolving in 0.10 mol dm⁻³ NaCl, what is [Cl⁻] if solubility is s?
Compare with the explanation
0.10 + s
The original chloride remains and dissolution adds s more.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do you combine the starting common-ion concentration with the ions added by dissolution?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.