Solubility Product Constant Ksp

Ksp is the equilibrium constant for a sparingly soluble salt dissolving into its ions, and it quantifies how much of the salt can dissolve.

K_{sp} = [M^{n+}]^x[A^{m-}]^y

Ksp = [cation]^x [anion]^y

What it means

For the dissolution equilibrium MxAy(s) ⇌ x Mⁿ⁺(aq) + y Aᵐ⁻(aq), the solubility product is Ksp = [Mⁿ⁺]^x[Aᵐ⁻]^y. It works exactly like Kc, except the reactant is a pure solid that drops out of the ratio. From Ksp you can compute the molar solubility s — for a 1:1 salt like AgCl, Ksp = s², so s = √Ksp. Salts with tiny Ksp values are the ones chemists call insoluble.

Variables

Symbol	Name	Unit	Meaning
$[Mⁿ⁺]$	Cation concentration at saturation	mol/L (M)	The molar concentration of the dissolved metal cation in a saturated solution at equilibrium with the solid.
$[Aᵐ⁻]$	Anion concentration at saturation	mol/L (M)	The molar concentration of the dissolved anion in a saturated solution at equilibrium with the solid.
$x, y$	Ion counts in the formula unit	dimensionless	The subscripts from the salt's formula (for example, x = 2 and y = 1 for Ag2S); each ion concentration is raised to its count.
$Ksp$	Solubility product constant	dimensionless	Measures the salt's solubility: a larger Ksp means more of the solid dissolves before the solution saturates.

Assumptions

The solid salt sits in equilibrium with a saturated solution — no more can dissolve.
Concentrations approximate activities, which holds for these very dilute solutions.
Temperature is constant, since Ksp is temperature-dependent.
The undissolved solid is omitted from the expression, as with any pure solid.

Derivation

Ksp is the equilibrium constant Kc applied to a dissolution reaction. The reactant is a pure solid, and pure solids are always omitted from equilibrium expressions, so only the product ion concentrations remain: Ksp = [cation]^x[anion]^y. That is why Ksp depends only on the dissolved ions.

Worked example

Problem

Silver chloride dissolves as AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) with Ksp = 1.8 × 10⁻¹⁰. Calculate the molar solubility of AgCl in pure water.

Given:

Ksp(AgCl) = 1.8 × 10⁻¹⁰
Dissolution: AgCl(s) ⇌ Ag⁺ + Cl⁻ (1:1 salt)
Let s = moles of AgCl that dissolve per liter

Each mole dissolved gives one mole of each ion, so [Ag⁺] = [Cl⁻] = s.
Write Ksp = [Ag⁺][Cl⁻] = s × s = s².
So s² = 1.8 × 10⁻¹⁰.
Take the square root: s = √(1.8 × 10⁻¹⁰) = √1.8 × 10⁻⁵.
√1.8 = 1.34, so s = 1.34 × 10⁻⁵ M.

Molar solubility = 1.34 × 10⁻⁵ mol/L

Only about 13 micromoles of AgCl dissolve per liter of water — an extremely low solubility, which is why AgCl is classified as insoluble.