Solubility Product as an Equilibrium Constant

Introduction

Silver chloride is called "insoluble", yet a tiny amount does dissolve. Add solid silver chloride to water and, after a while, the solution contains about 1.3 × 10⁻⁵ mol dm⁻³ of silver ions and the same concentration of chloride ions. At that point dissolving and precipitating continue at equal rates. This is an equilibrium like any other, and it has an equilibrium constant, the solubility product, which lets us treat precipitation and dissolution with the same quantitative tools used for acids and bases.

Core explanation

The dissolution equilibrium. For a solid of general formula MₓAᵧ:

MₓAᵧ(s) ⇌ x Mᵐ⁺(aq) + y Aⁿ⁻(aq)

where charge balance requires x × m = y × n.

The Ksp expression. Following the usual rules for equilibrium constants, each aqueous ion appears raised to the power of its stoichiometric coefficient:

Ksp = [Mᵐ⁺]ˣ [Aⁿ⁻]ʸ

The solid is omitted. Its activity is 1 as long as some pure solid is present, whatever its amount. This is why a saturated solution has the same ion concentrations whether it sits over a speck of solid or a large heap.

Examples.

Solid Dissolution Ksp expression Ksp at 25 °C --- --- --- --- AgCl AgCl ⇌ Ag⁺ + Cl⁻ [Ag⁺][Cl⁻] 1.8 × 10⁻¹⁰ BaSO₄ BaSO₄ ⇌ Ba²⁺ + SO₄²⁻ [Ba²⁺][SO₄²⁻] 1.1 × 10⁻¹⁰ CaF₂ CaF₂ ⇌ Ca²⁺ + 2F⁻ [Ca²⁺][F⁻]² 3.9 × 10⁻¹¹ Ag₂CrO₄ Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻ [Ag⁺]²[CrO₄²⁻] 1.1 × 10⁻¹² Mg(OH)₂ Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻ [Mg²⁺][OH⁻]² 5.6 × 10⁻¹² Ca₃(PO₄)₂ Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻ [Ca²⁺]³[PO₄³⁻]² very small

Units. Ksp has units of (mol dm⁻³) raised to the total number of ions: mol² dm⁻⁶ for AgCl, mol³ dm⁻⁹ for CaF₂ and mol⁵ dm⁻¹⁵ for Ca₃(PO₄)₂. Strictly, a thermodynamic Ksp is dimensionless because it is built from activities, but units help you track the powers.

Scope. Ksp is used for sparingly soluble salts. For very soluble salts such as sodium chloride, ion concentrations are high, activities differ greatly from concentrations, and simple Ksp calculations are unreliable. Even for sparingly soluble salts, Ksp values are only strictly constant at a fixed temperature and in solutions of low ionic strength.

Comparing Ksp values. Ksp values can be compared directly to rank solubility only when the salts have the same ion ratio, such as AgCl and BaSO₄ (both 1:1). Comparing AgCl with Ag₂CrO₄ this way would be misleading, because the powers in the expressions differ.

Step-by-step reasoning

1. Write the formula of the solid and identify its ions, with charges. 2. Write the balanced dissolution equation, with (s) and (aq) state symbols. 3. Write Ksp as the product of ion concentrations, each raised to its coefficient. 4. Leave out the solid, and check that the powers match the formula subscripts. 5. Deduce units from the total power.

Visual explanation

Imagine a beaker with a layer of white solid at the bottom. Arrows leave the surface as ions break free, and arrows return as ions rejoin the lattice. In a saturated solution the arrows in each direction are equally numerous. The solid layer can be thick or thin; the ion concentrations above it stay the same.

Real-world analogy

A saturated solution is like a busy shop with a fixed number of fitting rooms. People constantly go in and come out, but the number inside at any moment stays at capacity. Adding more people to the queue outside (more solid) does not increase the number inside.

Real-world example

Barium sulfate is given to patients as a "barium meal" before X-ray imaging of the digestive tract. Free barium ions are toxic, but BaSO₄ has such a small Ksp that the dissolved Ba²⁺ concentration stays extremely low, while the dense barium compound blocks X-rays and outlines the gut.

Why?

Why is the concentration of the solid left out? A pure solid's composition does not change as it dissolves: the number of formula units per unit volume of solid is fixed. Its activity is therefore constant and is absorbed into the equilibrium constant itself.

Common misconception

"Ksp for CaF₂ is [Ca²⁺][2F⁻]." The coefficient 2 becomes a power, not a multiplier inside the bracket: Ksp = [Ca²⁺][F⁻]². The factor of 2 only enters later, when concentrations are expressed in terms of molar solubility.

Worked example

Question: Write the Ksp expression and its units for aluminium hydroxide, Al(OH)₃.

Reasoning: Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH⁻(aq). Each Al³⁺ is accompanied by three OH⁻, so the hydroxide concentration is cubed. Total power is 1 + 3 = 4.

Answer: Ksp = [Al³⁺][OH⁻]³, units mol⁴ dm⁻¹².

Quick check

1. Write the Ksp expression for lead(II) iodide, PbI₂, including its units. Answer: Ksp = [Pb²⁺][I⁻]², with units mol³ dm⁻⁹.

Exam focus

Always include state symbols in the dissolution equation. Examiners penalise including the solid in the expression and writing coefficients inside brackets. Remember that direct comparison of Ksp values to rank solubility works only for salts of the same formula type.

Advanced insight

A rigorous Ksp uses activities: Ksp = a(M) ˣ a(A) ʸ. In solutions containing other dissolved salts, activity coefficients fall below 1, so more solid must dissolve to reach the same activity product. This "salt effect" means solubility rises slightly in inert electrolytes such as KNO₃, the opposite of the common-ion effect.

Summary

The solubility product is the equilibrium constant for a sparingly soluble solid dissolving into its ions. It is written as the product of ion concentrations raised to their stoichiometric coefficients, with the solid omitted because its activity is constant. Units depend on the total power. Ksp values rank solubility directly only for salts with the same ion ratio.

Practice questions

1. Write the dissolution equation and Ksp expression for silver phosphate, Ag₃PO₄. Answer: Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq); Ksp = [Ag⁺]³[PO₄³⁻]. 2. Give the units of Ksp for magnesium hydroxide. Answer: Ksp = [Mg²⁺][OH⁻]², so mol³ dm⁻⁹. 3. Explain why adding more solid AgCl to a saturated solution does not change [Ag⁺]. Answer: The solid's activity is constant and does not appear in Ksp, so the equilibrium ion concentrations are unchanged. 4. Can Ksp values of AgCl and BaSO₄ be compared directly to decide which is more soluble? Explain. Answer: Yes, because both are 1:1 salts, so Ksp equals the square of molar solubility in each case.