Calculate solubility from Ksp
Relate the molar solubility of a salt to its ion concentrations and calculate Ksp or solubility.
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
If PbI₂ dissolves to give solubility s, what is [I⁻]?
What you will learn
- Set ion concentrations from molar solubility and stoichiometry
- Substitute into Ksp and solve for solubility
- Check powers and physical units
Translate solubility into equilibrium ion concentrations
For AgCl(s) ⇌ Ag⁺ + Cl⁻, if its molar solubility in pure water is s, then [Ag⁺] = [Cl⁻] = s and Ksp = s².
For PbI₂(s) ⇌ Pb²⁺ + 2I⁻, [Pb²⁺] = s and [I⁻] = 2s, so Ksp = s(2s)² = 4s³. Include the stoichiometric factors before solving.
Solve and check the result
Use the supplied Ksp and isolate s algebraically. For a 1:1 salt, take a square root; for the PbI₂ example, take the cube root of Ksp/4.
Verify by substituting the result into the expression. State mol dm⁻³ for molar solubility and make sure the concentration is plausible for a sparingly soluble solid.
Pause and explain
For AgCl in pure water, if Ksp = 1.8 × 10⁻¹⁰, which equation gives s?
Worked example
AgCl has Ksp = 1.8 × 10⁻¹⁰ mol² dm⁻⁶ at a given temperature. Calculate its molar solubility in pure water.
Show the worked solution
- Let solubility be s; then [Ag⁺]=[Cl⁻]=s.
- Ksp = s², so s = √(1.8 × 10⁻¹⁰).
- State the positive concentration root.
Answer s = 1.34 × 10⁻⁵ mol dm⁻³.
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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.
A salt MX has Ksp = 4.0 × 10⁻¹² and dissolves as MX(s) ⇌ M⁺ + X⁻. Find its molar solubility in pure water.
Give me a hint
For a 1:1 salt, both ion concentrations equal s.
Compare my reasoning
- Write Ksp = s².
- Take the square root of 4.0 × 10⁻¹².
- Give mol dm⁻³.
s = 2.0 × 10⁻⁶ mol dm⁻³.
Look for these in your work
- Stoichiometry translated
- Square root evaluated
- Unit stated
PbI₂ has Ksp = 8.0 × 10⁻⁹ mol³ dm⁻⁹. Calculate s in pure water.
Give me a hint
Use [Pb²⁺]=s and [I⁻]=2s.
Compare my reasoning
- Ksp = s(2s)² = 4s³.
- s³ = 8.0 × 10⁻⁹/4.
- Take the cube root.
s = 1.26 × 10⁻³ mol dm⁻³.
Look for these in your work
- Iodide factor included
- Expression rearranged
- Positive solubility and unit stated
A learner sets [I⁻]=s for PbI₂ and obtains Ksp=s³. Explain and correct the missing factor.
Give me a hint
Read the coefficient of iodide in PbI₂ dissolution.
Compare my reasoning
- One PbI₂ unit releases two I⁻ ions.
- Therefore [I⁻]=2s.
- Substitute into Ksp=[Pb²⁺][I⁻]².
The correct expression is Ksp = s(2s)² = 4s³. The learner omitted the two iodide ions per formula unit.
Look for these in your work
- Equation coefficient identified
- Concentration corrected
- Squared factor applied
Common mistakes
- Using ion concentrations equal to s when coefficients are not 1:1
- Forgetting that a coefficient also scales the ion concentration
- Taking a negative root for a concentration
- Dropping concentration units
What can you explain now?
If PbI₂ dissolves to give solubility s, what is [I⁻]?
Compare with the explanation
2s
Two moles of iodide are released per mole of PbI₂.
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Make it stick
How do dissolution coefficients connect a salt's molar solubility to the powers in Ksp?
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