7.11 Ksp

Ksp & Precipitation

A saturated beaker solved from Ksp itself: watch molar solubility built up step by step, then push Qsp above or below Ksp and see the crystal bed grow or dissolve. Ranks 6 salts by Ksp and by s.

KspMolar SolubilityQsp vs Ksp6 Salts
Topic 7.11

Introduction to Solubility Equilibria

Calculate the solubility of a salt based on the value of Ksp for the salt.

The dissolution of a salt is a reversible process whose extent is described by Ksp, the solubility-product constant. For a saturated solution:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = [Ag⁺][Cl⁻]

The solid is excluded, so Ksp is just the product of the ion concentrations, each raised to its coefficient.

Molar solubility (s) is the moles of salt that dissolve per litre to give a saturated solution. Converting between s and Ksp is the core skill, and the relationship depends on the ion ratio:

  • AB (e.g. AgCl): [A⁺] = s, [B⁻] = s → Ksp = s²
  • AB₂ (e.g. PbI₂): [A²⁺] = s, [B⁻] = 2s → Ksp = s(2s)² = 4s³
  • A₂B (e.g. Ag₂CrO₄): [A⁺] = 2s, [B²⁻] = s → Ksp = (2s)²(s) = 4s³
  • A₂B₃: Ksp = (2s)²(3s)³ = 108s⁵

EK 7.11.A.3 connects back to Unit 4: the solubility rules can be quantitatively related to Ksp, with Ksp values greater than 1 corresponding to soluble salts.

Qsp versus Ksp predicts precipitation:

  • Qsp < Ksp → unsaturated, all solid dissolves.
  • Qsp = Ksp → exactly saturated.
  • Qsp > Ksp → supersaturated; a precipitate forms until Qsp falls back to Ksp.

A critical comparison rule: you may compare Ksp values directly to rank solubility only when the salts have the same ion ratio. Otherwise convert both to molar solubility first.

Key points

  • Ksp is the product of ion concentrations at saturation, each raised to its coefficient.
  • The Ksp–to–s relationship depends on the ion ratio: s², 4s³, 108s⁵, and so on.
  • Compare Ksp directly only for salts with the same ion ratio; otherwise convert to molar solubility.
  • Qsp > Ksp means a precipitate forms.

Equations

  • not on the sheetFor AnB_m(s) ⇌ n A^{m+} + m B^{n-}. A special case of the Kc on the equation sheet.
  • not on the sheetFor an AB₂ or A₂B salt, where s is molar solubility.

Common mistakes

  • Forgetting the coefficient appears twice — once as a multiplier inside the bracket and once as an exponent. For PbI₂, [I⁻] = 2s and it is squared.
  • Comparing Ksp across different ion ratios. A 1:2 salt with a smaller Ksp can be more soluble than a 1:1 salt with a larger one.
  • Including the solid in Ksp.
  • Confusing molar solubility with solubility in g/L. Convert with molar mass if the question asks for grams.

Worked example

Ksp for PbI₂ is 7.1 × 10⁻⁹ at 25 °C. (a) Calculate its molar solubility. (b) Calculate its solubility in g/L. (c) Ag₂CrO₄ has Ksp = 1.1 × 10⁻¹², and AgCl has Ksp = 1.8 × 10⁻¹⁰. Which is more soluble?

(a) Molar solubility of PbI₂.
PbI₂(s) ⇌ Pb²⁺(aq) + 2 I⁻(aq)
Let s = molar solubility. Then [Pb²⁺] = s and [I⁻] = 2s.

Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = 4s³
7.1 × 10⁻⁹ = 4s³
s³ = 1.775 × 10⁻⁹
s = 1.2 × 10⁻³ M

(b) In g/L. M(PbI₂) = 207.2 + 2(126.9) = 461.0 g/mol
(1.2 × 10⁻³ mol/L)(461.0 g/mol) = 0.55 g/L

(c) Comparing across different ion ratios — convert both to s.

AgCl (1:1): Ksp = s² → s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M

Ag₂CrO₄ (2:1): Ksp = (2s)²(s) = 4s³ → s³ = 1.1 × 10⁻¹²/4 = 2.75 × 10⁻¹³ → s = 6.5 × 10⁻⁵ M

Ag₂CrO₄ is about five times more soluble, even though its Ksp is nearly 200 times smaller. This is exactly why Ksp values cannot be compared directly across different ion ratios.

Full notes for topic 7.11 →