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Solubility Equilibria — Ksp, Common Ion & Precipitation

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Stoichiometry drives the Ksp–solubility link

For a salt that releases more than one of an ion, the solubility s enters Ksp with both a coefficient and an exponent. Take Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq): dissolving s mol/L gives [Ca²⁺] = s but [OH⁻] = 2s, so Ksp = (s)(2s)² = 4s³. The factor of 4 comes from the coefficient (the 2 in 2s) and the exponent (squaring). Always rebuild this from the balanced equation — never assume Ksp = s².

Solubility product and molar solubility
AₘBₙ(s) ⇌ m Aⁿ⁺ + n Bᵐ⁻: Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ. MX: Ksp = s². MX₂ or M₂X: Ksp = 4s³. MX₃: Ksp = 27s⁴.
The numerical prefix (4, 27, …) is (coefficient)^(coefficient) summed over the ions. To get s from Ksp for a 1:2 salt, take the cube root of Ksp/4.
Worked example

Calcium hydroxide, Ca(OH)₂, has Ksp = 5.0×10⁻⁶. Calculate its molar solubility in pure water.

  1. 1.Dissolution: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq). Let s = molar solubility, so [Ca²⁺] = s and [OH⁻] = 2s.
  2. 2.Ksp = [Ca²⁺][OH⁻]² = (s)(2s)² = (s)(4s²) = 4s³.
  3. 3.Set equal: 4s³ = 5.0×10⁻⁶, so s³ = 1.25×10⁻⁶.
  4. 4.Take the cube root: s = (1.25×10⁻⁶)^(1/3). Rewrite as (1250×10⁻⁹)^(1/3); cube root of 10⁻⁹ is 10⁻³ and cube root of 1250 ≈ 10.8.
  5. 5.s ≈ 10.8×10⁻³ = 1.1×10⁻² M.
  6. 6.Check: 4(1.1×10⁻²)³ ≈ 4(1.3×10⁻⁶) ≈ 5.0×10⁻⁶ ✓. Note [OH⁻] = 2s ≈ 2.2×10⁻² M.
Answer: s ≈ 1.1×10⁻² M (0.011 M). The 4s³ relationship — not s² — is essential; treating it as s² would give a very different, wrong answer.

The common-ion effect, quantitatively

When a salt dissolves in a solution that already contains one of its ions, Le Châtelier pushes the dissolution equilibrium back toward the solid, sharply lowering solubility. Quantitatively, you plug the pre-existing ion concentration straight into Ksp. If the common ion carries a coefficient, it is still raised to that power — a subtlety that trips people up for salts like PbI₂, where the common ion I⁻ appears squared.

Worked example

Lead(II) iodide, PbI₂, has Ksp = 7.1×10⁻⁹. Calculate its molar solubility in 0.10 M NaI, and compare to pure water.

  1. 1.Dissolution: PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq), so Ksp = [Pb²⁺][I⁻]².
  2. 2.Let s = molar solubility in the NaI solution. Then [Pb²⁺] = s. The iodide is dominated by the NaI: [I⁻] ≈ 0.10 M (the little 2s added by PbI₂ is negligible).
  3. 3.Substitute: Ksp = (s)(0.10)² = s(0.010) = 7.1×10⁻⁹.
  4. 4.Solve: s = 7.1×10⁻⁹ / 0.010 = 7.1×10⁻⁷ M.
  5. 5.Compare to pure water: 4s³ = 7.1×10⁻⁹ → s³ = 1.78×10⁻⁹ → s ≈ 1.2×10⁻³ M.
Answer: In 0.10 M NaI, s ≈ 7.1×10⁻⁷ M — roughly 1700× less soluble than in pure water (1.2×10⁻³ M). Note the common ion enters as (0.10)², not 0.10, because I⁻ has coefficient 2.

Predicting precipitation: Q versus Ksp

Mixing two solutions may or may not form a precipitate. Compute the ion product Q — the same form as Ksp but using the actual mixed concentrations — and compare: Q > Ksp means supersaturated, so a precipitate forms until Q falls to Ksp; Q < Ksp means unsaturated, everything stays dissolved; Q = Ksp is exactly saturated. The critical trap is dilution: combining volumes lowers every concentration, so recompute each ion in the combined volume before finding Q.

Precipitation criterion
Q > Ksp → precipitate forms; Q = Ksp → just saturated; Q < Ksp → no precipitate
Q uses concentrations in the combined solution (after mixing dilutes them). Same Q-vs-K logic as reaction equilibria, applied to dissolving.
Worked example

Equal volumes of 0.0020 M Pb(NO₃)₂ and 0.020 M KI are mixed. Does PbI₂ precipitate? Ksp(PbI₂) = 7.1×10⁻⁹.

  1. 1.Account for dilution first. Mixing equal volumes halves each concentration: [Pb²⁺] = 0.0020/2 = 0.0010 M and [I⁻] = 0.020/2 = 0.010 M.
  2. 2.Write the ion product with the correct exponent: Q = [Pb²⁺][I⁻]² (I⁻ has coefficient 2).
  3. 3.Substitute: Q = (0.0010)(0.010)² = (1.0×10⁻³)(1.0×10⁻⁴) = 1.0×10⁻⁷.
  4. 4.Compare to Ksp: Q = 1.0×10⁻⁷ versus Ksp = 7.1×10⁻⁹. Q > Ksp.
Answer: Q (1.0×10⁻⁷) > Ksp (7.1×10⁻⁹), so PbI₂ precipitates. Forgetting to dilute, or forgetting to square [I⁻], would each change Q by orders of magnitude and can flip the conclusion.
Checkpoint

A salt MX₂ dissolves as MX₂(s) ⇌ M²⁺(aq) + 2X⁻(aq) with Ksp = 3.2×10⁻¹¹. What is its molar solubility in pure water?

Checkpoint

Equal volumes of 0.0020 M AgNO₃ and 0.0020 M NaCl are mixed. Ksp(AgCl) = 1.8×10⁻¹⁰. What happens?

Checkpoint

CaF₂ (Ksp = 3.9×10⁻¹¹) is dissolved in 0.10 M NaF. What is its molar solubility s? (CaF₂(s) ⇌ Ca²⁺ + 2F⁻.)

On the exam

Two exam reflexes for solubility problems: (1) after mixing solutions, recompute every concentration in the combined volume before finding Q — dilution is the most common oversight; (2) always carry the ion coefficient as an exponent, so a common ion or trial ion with coefficient 2 is squared. Both slips move the answer by orders of magnitude.

Watch out

Do not rank solubility by comparing Ksp values unless the salts share the same ion-count stoichiometry. A 1:2 salt (Ksp = 4s³) with a larger Ksp can be less soluble than a 1:1 salt (Ksp = s²). Convert each Ksp to molar solubility s first, then compare the s values.

Answer the 3 checkpoints as you read.

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