This AP Chemistry lesson on solubility product Ksp notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on solubility product Ksp and check yourself with the quiz.
Quick answer: The solubility product Ksp is the equilibrium constant for a sparingly soluble ionic solid dissolving, such as AgCl(s) ⇌ Ag⁺ + Cl⁻, with Ksp = [Ag⁺][Cl⁻]. It lets you calculate molar solubility, predict precipitation by comparing Q with Ksp, and see how common ions and pH change solubility.
Level: AP Chemistry · Curriculum: AP Chem Unit 7
Solubility product Ksp notes: the key ideas
What is the solubility product, Ksp?
For a sparingly soluble ionic solid, Ksp is the equilibrium constant for dissolving. For MₐXᵦ(s) ⇌ aM^(n+)(aq) + bX^(m−)(aq), Ksp = [M^(n+)]ᵃ[X^(m−)]ᵇ. The solid does not appear in the expression. Molar solubility s is the amount in mol that dissolves per dm³ of saturated solution.
A saturated solution is a dynamic equilibrium: ions leave the crystal and return at the same rate. A larger Ksp, for salts with the same formula type, means a more soluble salt.
| Salt | Dissolving equation | Ksp expression | Ksp in terms of s |
|---|---|---|---|
| AgCl | AgCl ⇌ Ag⁺ + Cl⁻ | [Ag⁺][Cl⁻] | s² |
| CaF₂ | CaF₂ ⇌ Ca²⁺ + 2F⁻ | [Ca²⁺][F⁻]² | 4s³ |
| Ag₂CrO₄ | Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻ | [Ag⁺]²[CrO₄²⁻] | 4s³ |
| Ca₃(PO₄)₂ | Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻ | [Ca²⁺]³[PO₄³⁻]² | 108s⁵ |
How do you find solubility from Ksp?
Find the molar solubility and mass solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰, M = 143.3 g/mol). Ksp = s², so s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/dm³. Mass = 1.34 × 10⁻⁵ × 143.3 = 1.92 mg/dm³.
Find s for CaF₂, Ksp = 3.5 × 10⁻¹¹. Each unit gives one Ca²⁺ and two F⁻, so [Ca²⁺] = s and [F⁻] = 2s. Ksp = s(2s)² = 4s³, so s = ³√(3.5 × 10⁻¹¹ / 4) = ³√(8.75 × 10⁻¹²) = 2.06 × 10⁻⁴ mol/dm³.
To compare Q with Ksp: if Q < Ksp the solution is unsaturated, if Q = Ksp it is saturated, and if Q > Ksp a precipitate forms. Q has the same form as Ksp but uses the concentrations present right now, after mixing.
How does the common ion effect change solubility?
Adding a solution that contains one of the salt’s ions pushes the equilibrium to the left, in line with Le Chatelier’s principle. The salt becomes less soluble. When the added ion concentration is large, it is almost equal to the total concentration of that ion, so the other ion’s concentration equals s.
Find s for AgCl in 0.10 mol/dm³ NaCl. Here [Cl⁻] ≈ 0.10 mol/dm³, so [Ag⁺] = s = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ mol/dm³. That is 7454 times smaller than in pure water.
Why does pH change the solubility?
If the anion is basic, H₃O⁺ removes it from solution and pulls the equilibrium to the right. Hydroxides, carbonates, phosphates and fluorides therefore dissolve better in acid, while chlorides, bromides and iodides of silver do not. For a hydroxide, a lower pH means a lower [OH⁻] and a higher s.
Mg(OH)₂ has Ksp = 5.6 × 10⁻¹². In pure water s = ³√(Ksp/4) = 1.12 × 10⁻⁴ mol/dm³, so [OH⁻] = 2.24 × 10⁻⁴ mol/dm³ and pH = 10.35. In a buffer at pH 10.00, [OH⁻] = 1.0 × 10⁻⁴ mol/dm³, so s = Ksp/[OH⁻]² = 5.6 × 10⁻⁴ mol/dm³, five times larger.
Students write s = √Ksp for every salt. That works only for salts with a 1:1 ratio such as AgCl or BaSO₄. For CaF₂ or Ag₂CrO₄ use 4s³, and always write the Ksp expression first. Also do not compare Ksp values of salts with different formulas to rank solubility.
Use a three-line ICE table for every Ksp problem: write the dissolving equation, put s (or 2s, 3s) under each ion and substitute into the Ksp expression. Check at the end by multiplying the ion concentrations again.
- Ksp = product of ion concentrations (each raised to its coefficient) at saturation; the solid is omitted.
- Molar solubility: AB gives s = √Ksp; AB₂ and A₂B give s = ³√(Ksp/4).
- Precipitation: Q > Ksp precipitate forms; Q = Ksp saturated; Q < Ksp unsaturated.
- A common ion lowers solubility, as in s = Ksp/[Cl⁻] for AgCl in NaCl.
- Salts of basic anions (OH⁻, CO₃²⁻, F⁻, PO₄³⁻) dissolve better in acid.
Ksp, precipitation and the common ion effect: frequently asked questions
What is the difference between solubility and Ksp?
Solubility is the amount that dissolves in a given volume, often written as molar solubility s in mol/dm³. Ksp is the equilibrium constant that links the ion concentrations at saturation. You can convert between them with the dissolving equation.
Why is the solid left out of the Ksp expression?
The concentration of a pure solid is constant, so it is absorbed into the constant. Only dissolved ions appear in Ksp.
How do you know whether a precipitate forms?
Calculate Q from the concentrations after mixing, remembering that volumes add. If Q is greater than Ksp, a precipitate forms until Q falls to Ksp.
Why does a common ion reduce solubility?
The extra ion shifts the dissolving equilibrium to the left, so more solid stays undissolved. The shift is a direct Le Chatelier effect and Ksp itself does not change.
Why does CaCO₃ dissolve in acid?
The carbonate ion reacts with H₃O⁺ to form CO₂ and water, which removes CO₃²⁻ from the equilibrium. The solid then dissolves to replace it.
