Solubility Equilibrium and Ksp

Introduction

A sparingly soluble salt still dissolves to a measurable extent. When undissolved solid coexists with dissolved ions at equilibrium, the ion activities satisfy a solubility product, Ksp. This constant describes an equilibrium, whereas the amount dissolved depends on the salt's stoichiometry and on other species already present in solution.

Core explanation

For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the thermodynamic equilibrium expression is Ksp = a(Ag⁺)a(Cl⁻). The pure solid's activity is one, so it does not appear. In dilute textbook calculations, ion activities are approximated by concentrations relative to a standard concentration, giving the familiar Ksp ≈ [Ag⁺][Cl⁻] numerical form. A saturated solution containing solid maintains this relation at fixed temperature. Dissolution and recrystallization continue microscopically even though measured concentrations are steady.

For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, Ksp ≈ [Ca²⁺][F⁻]². The coefficient two becomes an exponent in the equilibrium expression. For Al(OH)₃(s) ⇌ Al³⁺ + 3OH⁻, the hydroxide activity appears cubed. The exact formula of the dissolving solid must therefore be written before any Ksp equation; guessing an expression from the compound name can omit stoichiometric powers.

Ksp is not itself a molar solubility. A 1:1 salt and a 1:2 salt produce different ion combinations for the same amount of dissolved formula units. Numerical Ksp values cannot always be ranked to compare solubilities across salts with different stoichiometries. Even for a given salt, solubility changes when a common ion is supplied or when ions undergo acid-base or complex-formation reactions. Ksp remains temperature-specific while those other equilibria alter free-ion concentrations.

Adding more pure solid to an already saturated solution does not raise free-ion concentrations under the same conditions, as long as solid remains and equilibrium is reached. A solid must be present for the saturated-equilibrium relation to be enforced by dissolution/precipitation. In an unsaturated solution with no solid, the ion product may be below Ksp. Temperature changes can change Ksp, but the direction is not universally predictable without dissolution enthalpy and temperature data.

Step-by-step reasoning

1. Balance dissolution into constituent aqueous ions. 2. Omit the pure solid from the activity expression. 3. Raise each free-ion activity to its stoichiometric coefficient. 4. Relate Ksp to solubility only after specifying solution composition.

Visual explanation

Draw a solid crystal at the bottom of a beaker, with ions leaving and returning at equal rates. Beside it show the product of free-ion concentrations fixed at saturation.

Real-world analogy

A station can have passengers entering and leaving at equal rates, keeping the platform count steady. Likewise a saturated solid can keep exchanging ions with water without a net concentration change.

Real-world example

Calcium fluoride dissolves only slightly in water. Its saturated solution contains calcium and fluoride ions, and the calcium-to-fluoride dissolution ratio is one to two even though Ksp uses their concentrations with powers.

Why?

Why omit pure solid from Ksp? Its thermodynamic activity is fixed at one when that phase is present, so changing the amount of solid does not alter the equilibrium expression.

Common misconception

“Ksp equals the concentration of a dissolved salt.” Ksp is a product of free-ion activities with stoichiometric exponents; molar solubility must be derived separately.

Worked example

Write Ksp for PbI₂(s) ⇌ Pb²⁺ + 2I⁻. The expression is Ksp ≈ [Pb²⁺][I⁻]² in the dilute model. If molar solubility in pure water is s and no other source supplies ions, [Pb²⁺] ≈ s and [I⁻] ≈ 2s, giving Ksp ≈ s(2s)² = 4s³. The factor four follows from ion stoichiometry, not from the solid's activity.

Quick check

1. Does the mass of remaining pure solid appear in Ksp? Answer: No. Pure solid activity is one while the phase is present.

Exam focus

Write the balanced dissolution equation first. Use concentrations of free ions, not total analytical element concentrations when complexation or acid-base reactions are significant.

Advanced insight

Thermodynamic Ksp is dimensionless because activities are relative to standard states. Textbook concentration products carry apparent units unless the standard-concentration normalization is understood implicitly.

Summary

Ksp gives the equilibrium product of free dissolved ion activities for a solid's dissolution. Pure solid is omitted; aqueous stoichiometric coefficients become exponents. Solubility requires an additional mass-balance calculation.

Practice questions

1. Write the dilute Ksp expression for CaF₂. Answer: Ksp ≈ [Ca²⁺][F⁻]². 2. Why does a saturated solution remain dynamic? Answer: Dissolution and precipitation continue at equal rates. 3. Can two different salt formulas be ranked by numerical Ksp alone for molar solubility? Answer: Not reliably; their ion stoichiometries differ.