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Solubility Equilibria (Ksp)

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Even "insoluble" salts are in equilibrium

A slightly soluble salt in water sits at equilibrium between the undissolved solid and its dissolved ions. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), dissolution and re-precipitation balance. The equilibrium constant for this dissolving is the solubility product, Ksp — the product of the dissolved ion concentrations, each raised to its coefficient. The solid is a pure solid, so (as always) it is left out of the expression.

Solubility product
For AₘBₙ(s) ⇌ m Aⁿ⁺(aq) + n Bᵐ⁻(aq): Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Only the dissolved ions appear, each raised to its coefficient. A larger Ksp (for the same ion-count stoichiometry) means a more soluble salt.

Molar solubility: how much actually dissolves

The molar solubility, s, is how many moles of the salt dissolve per liter to reach a saturated solution. Set s equal to the amount of solid that dissolves, express each ion concentration in terms of s (using the coefficients), and substitute into Ksp. The stoichiometry matters: a 1:1 salt gives Ksp = s², a 1:2 salt gives Ksp = (s)(2s)² = 4s³.

Ksp ↔ molar solubility
MX type: Ksp = s². MX₂ type: Ksp = 4s³. M₂X type: Ksp = 4s³.
Always rebuild the relationship from the balanced dissolution equation; do not assume Ksp = s² for every salt.
Worked example

Silver chloride has Ksp = 1.8×10⁻¹⁰. Calculate the molar solubility of AgCl in pure water.

  1. 1.Write the dissolution equilibrium: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
  2. 2.Let s = molar solubility. Each formula unit gives one Ag⁺ and one Cl⁻, so [Ag⁺] = s and [Cl⁻] = s.
  3. 3.Substitute into Ksp: Ksp = [Ag⁺][Cl⁻] = (s)(s) = s².
  4. 4.s² = 1.8×10⁻¹⁰, so s = √(1.8×10⁻¹⁰).
  5. 5.s = 1.3×10⁻⁵ (since √1.8 ≈ 1.34 and √10⁻¹⁰ = 10⁻⁵).
Answer: s ≈ 1.3×10⁻⁵ M. That is how many moles of AgCl dissolve per liter — and [Ag⁺] = [Cl⁻] = 1.3×10⁻⁵ M in the saturated solution.

The common-ion effect

Dissolving a salt in a solution that already contains one of its ions suppresses its solubility. By Le Châtelier, the extra common ion pushes the dissolution equilibrium back toward the solid. So AgCl is far less soluble in 0.10 M NaCl than in pure water: the added Cl⁻ shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left.

Checkpoint

What is the correct Ksp expression for Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq)?

Checkpoint

Barium sulfate, BaSO₄, has Ksp = 1.1×10⁻¹⁰. What is its molar solubility in pure water?

Checkpoint

AgCl (Ksp = 1.8×10⁻¹⁰) dissolves in 0.10 M NaCl instead of pure water. Its molar solubility is closest to:

On the exam

To decide whether a precipitate forms, compute the ion product Q for the trial concentrations and compare to Ksp: Q > Ksp means the solution is supersaturated and a precipitate forms; Q < Ksp means it stays dissolved; Q = Ksp is exactly saturated. It is the same Q-versus-K logic from Lesson 1, applied to dissolving.

Watch out

Do not compare Ksp values directly to rank solubility unless the salts share the same ion-count stoichiometry. A 1:2 salt with a larger Ksp can still be less soluble than a 1:1 salt, because Ksp relates to s differently (4s³ vs. s²). Convert to molar solubility before comparing.

Answer the 3 checkpoints as you read.

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