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II-12: Solubility products and precipitation

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Module II-12 · Recommended before: Before Ksp calculations

Solubility products and precipitation

Solubility equilibrium expressions reflect formula coefficients. For a 1:1 solid MX, Ksp=s²; for MX₂, Ksp=4s³.

Learn the method

Ksp = [M⁺][X⁻] = s² for ideal 1:1 MX

  1. Write the dissolution equation.
  2. Convert molar solubility s into each ion concentration.
  3. Substitute into Ksp.
  4. Solve the power equation.
  5. Check the stoichiometric factors.

Common mistake: The relation Ksp=s² only applies to 1:1 dissolution in this simple case.

Worked example

Problem: For MX(s) ⇌ M⁺ + X⁻ with Ksp = 1.0×10⁻⁸, find s in pure water.

Step-by-step solution: Ksp = s² so s = √(1.0×10⁻⁸) = 1.0×10⁻⁴ M.

Check the units and whether the answer is reasonable before continuing.

More applications of this mathematics

Try each extension before expanding its solution.

Different dissolution stoichiometry

For MX₂(s) ⇌ M²⁺+2X⁻ with molar solubility s, Ksp=s(2s)²=4s³. If s=1.0×10⁻³ M, Ksp=4.0×10⁻⁹.

Precipitation threshold

If ion product Qsp exceeds Ksp, precipitation is thermodynamically favored toward lower dissolved ion concentrations until equilibrium is approached.

Complex ion formation

For M²⁺+4L ⇌ ML₄²⁺, a simplified formation constant is Kf=[ML₄²⁺]/([M²⁺][L]⁴). Complexation can change free metal concentration and apparent solubility.

Independent practice — five questions

Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.

Choose one response for each question, then select “Check my answers.” No timer. Hints are optional. You can retry freely. Questions are formative, not proctored.

Question 1. For MX(s) ⇌ M⁺ + X⁻, Ksp=2.50e-07. What is molar solubility s in pure water?
Hint for question 1

For 1:1 dissolution, both ion concentrations equal s.

Question 2. For MX(s) ⇌ M⁺ + X⁻, Ksp=3.60e-07. What is molar solubility s in pure water?
Hint for question 2

For 1:1 dissolution, both ion concentrations equal s.

Question 3. For MX(s) ⇌ M⁺ + X⁻, Ksp=4.90e-09. What is molar solubility s in pure water?
Hint for question 3

For 1:1 dissolution, both ion concentrations equal s.

Question 4. For MX(s) ⇌ M⁺ + X⁻, Ksp=6.40e-09. What is molar solubility s in pure water?
Hint for question 4

For 1:1 dissolution, both ion concentrations equal s.

Question 5. For MX(s) ⇌ M⁺ + X⁻, Ksp=8.10e-09. What is molar solubility s in pure water?
Hint for question 5

For 1:1 dissolution, both ion concentrations equal s.

Supporting reference

OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.

W3C WCAG 2.2 accessibility guidance.

Applied practice: Solubility with stoichiometry and a common ion

These ungraded questions extend the scored five-question check. Work each problem before opening its solution; progress records are not changed.

  1. Challenge 1. For M(OH)₂(s) ⇌ M²⁺ + 2OH⁻, let Ksp = 4.0 × 10⁻¹². Estimate molar solubility in pure water, neglecting water autoionization.

    Show worked solution for challenge 1

    If s dissolves, [M²⁺] = s and [OH⁻] = 2s. Ksp = s(2s)² = 4s³, so s = (Ksp/4)^(1/3) = 1.0 × 10⁻⁴ M.

  2. Challenge 2. For MX(s) ⇌ M⁺ + X⁻ with Ksp = 1.0 × 10⁻¹⁰, estimate [M⁺] at equilibrium when the solution already contains [X⁻] ≈ 0.010 M.

    Show worked solution for challenge 2

    Ksp = [M⁺][X⁻], so [M⁺] ≈ (1.0 × 10⁻¹⁰)/(0.010) = 1.0 × 10⁻⁸ M, assuming the common-ion concentration remains nearly constant.

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