The Solubility-Product Constant, Ksp

Core Concept

Even "insoluble" salts dissolve to a tiny extent, establishing an equilibrium between the solid and its dissolved ions. The solubility-product constant Ksp describes that equilibrium: for a salt MₓAᵧ, Ksp = [M]ˣ[A]ʸ, with each ion raised to its stoichiometric coefficient and the solid itself omitted (a pure solid has constant concentration). A smaller Ksp means a less soluble salt, and Ksp is the gateway to every quantitative solubility calculation — molar solubility, precipitation, and selective precipitation.

Key Ideas

• Dissolution equilibrium: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
• Ksp = [Ag⁺][Cl⁻]; the solid is excluded because pure solids have constant activity.
• Stoichiometric exponents: for PbCl₂, Ksp = [Pb²⁺][Cl⁻]²; for Ag₂CrO₄, Ksp = [Ag⁺]²[CrO₄²⁻].
• Saturated solution: the ion product exactly equals Ksp.
• Ksp is temperature-dependent (most salts are more soluble when hot).
• Comparing Ksp values directly is valid only for salts of the same ion ratio (e.g., both 1:1).

Equations and Variables

• General: MₓAᵧ(s) ⇌ x Mⁿ⁺(aq) + y Aᵐ⁻(aq)
• Ksp = [Mⁿ⁺]ˣ [Aᵐ⁻]ʸ
• Examples: AgCl: Ksp = [Ag⁺][Cl⁻]; PbCl₂: Ksp = [Pb²⁺][Cl⁻]²; Ag₂CrO₄: Ksp = [Ag⁺]²[CrO₄²⁻]
• Qsp (ion product under any conditions) vs Ksp (at saturation)

How It Works

1. Write the balanced dissolution equation, being careful with coefficients.
2. Write Ksp as the product of the ion concentrations, each raised to its coefficient in the balanced equation.
3. Omit the solid — its "concentration" is constant and is folded into the constant.
4. At equilibrium (a saturated solution), the ion product equals Ksp; this is the maximum amount of ions the solution can hold before solid remains.
5. The magnitude of Ksp reflects solubility: for same-type salts, smaller Ksp = less soluble.

Worked Example

Write the Ksp expression for (a) AgCl, (b) CaF₂, and (c) Ag₃PO₄, and order the 1:1 salts AgCl (Ksp = 1.8 × 10⁻¹⁰) and AgBr (Ksp = 5.0 × 10⁻¹³) by solubility.

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

(b) CaF₂(s) ⇌ Ca²⁺ + 2 F⁻ → Ksp = [Ca²⁺][F⁻]²

(c) Ag₃PO₄(s) ⇌ 3 Ag⁺ + PO₄³⁻ → Ksp = [Ag⁺]³[PO₄³⁻]

Because AgCl and AgBr are both 1:1 salts, we may compare Ksp directly: AgBr (5.0 × 10⁻¹³) is less soluble than AgCl (1.8 × 10⁻¹⁰), since its Ksp is smaller.

How it works

1. Write the balanced dissolution equation, being careful with coefficients.
2. Write Ksp as the product of the ion concentrations, each raised to its coefficient in the balanced equation.
3. Omit the solid — its "concentration" is constant and is folded into the constant.
4. At equilibrium (a saturated solution), the ion product equals Ksp; this is the maximum amount of ions the solution can hold before solid remains.
5. The magnitude of Ksp reflects solubility: for same-type salts, smaller Ksp = less soluble.

Common confusions

• "Ksp includes the solid." — Solids (and pure liquids) are always omitted from equilibrium expressions.
• "Ksp = [M][A] for every salt." — Only for 1:1 salts; you must raise each ion to its coefficient (e.g., [F⁻]² in CaF₂).
• "Smaller Ksp always means less soluble." — Only when comparing salts with the same stoichiometry (both 1:1, or both 1:2); otherwise convert Ksp to molar solubility first.
• "Ksp is the molar solubility." — Ksp is a constant; molar solubility (mol/L) is derived from it and depends on stoichiometry.
• "Ksp is a fixed universal number." — It changes with temperature.

Quick review

• MₓAᵧ(s) ⇌ x Mⁿ⁺ + y Aᵐ⁻; Ksp = [Mⁿ⁺]ˣ[Aᵐ⁻]ʸ.
• Omit solids; include stoichiometric exponents.
• Smaller Ksp = less soluble (same stoichiometry only).
• Saturated ⇔ ion product = Ksp.
• Ksp varies with temperature.