Ksp: the solubility product constant
Learn what the solubility product Ksp means, how to compute solubility from Ksp, and how comparing Qsp to Ksp predicts precipitation.
Quick answer
The solubility product Ksp is the equilibrium constant for a solid dissolving into its ions. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰ at 25 °C.
Example: Molar solubility of AgCl: s² = 1.8 × 10⁻¹⁰, so s = 1.3 × 10⁻⁵ M — only about a milligram of AgCl dissolves per liter of water.
What you will learn
- Write the Ksp expression for a salt from its dissolution equation.
- Calculate molar solubility from Ksp for 1:1 and other stoichiometries.
- Predict precipitation by comparing Qsp with Ksp.
- Explain the common-ion effect quantitatively with Ksp.
What Ksp is
The solubility product, Ksp, is the equilibrium constant for a solid salt dissolving:
The pure solid is omitted (as in all K expressions), and each ion concentration is raised to its coefficient. For lead(II) iodide:
Ksp measures how far dissolution proceeds: a tiny Ksp means an 'insoluble' salt, but every salt dissolves slightly — Ksp quantifies exactly how slightly.
From Ksp to molar solubility
For a 1:1 salt like AgCl, if s mol/L dissolves, then [Ag⁺] = [Cl⁻] = s:
That is about 1.9 mg of AgCl per liter — 'insoluble' in practice, nonzero in principle.
For other stoichiometries, track the coefficients: for PbI₂ (Ksp = 7.9 × 10⁻⁹), dissolving s mol/L gives [Pb²⁺] = s and [I⁻] = 2s, so Ksp = (s)(2s)² = 4s³, and s = ∛(Ksp/4) = 1.3 × 10⁻³ M. The takeaway: convert Ksp to solubility s before comparing salts — raw Ksp values mislead across different stoichiometries.
Predicting precipitation: Qsp versus Ksp
Mix two solutions and a precipitate may form. Compute the ion product Qsp from the current (mixed, diluted!) concentrations, then compare:
- Qsp > Ksp: supersaturated — precipitate forms until Qsp falls to Ksp
- Qsp = Ksp: saturated, at equilibrium
- Qsp < Ksp: unsaturated — no precipitate
Example: mixing equal volumes of 0.0020 M AgNO₃ and 0.0020 M NaCl halves each concentration to 0.0010 M. Qsp = (0.0010)(0.0010) = 1.0 × 10⁻⁶ > 1.8 × 10⁻¹⁰, so AgCl precipitates. Forgetting the dilution-on-mixing step is the classic error here.
Ksp and the common-ion effect
Dissolving AgCl in 0.10 M NaCl instead of pure water: the common ion Cl⁻ is already at 0.10 M, so Ksp = (s)(0.10 + s) ≈ 0.10s, giving s = 1.8 × 10⁻⁹ M — roughly ten-thousand times less soluble than in pure water.
This quantitative common-ion effect is Le Châtelier's principle with numbers: the extra product ion pushes the dissolution equilibrium left, and Ksp tells you exactly how far.
Variables and units
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| Molar solubility | mol/L | Moles of salt that dissolve per liter of saturated solution | |
| Ion product | varies | Ion product computed from current concentrations (compare with Ksp) |
Common mistakes
Only for salts with the same stoichiometry. Comparing AgCl (Ksp = 1.8 × 10⁻¹⁰, s = 1.3 × 10⁻⁵ M) with PbI₂ (Ksp = 7.9 × 10⁻⁹, s = 1.3 × 10⁻³ M) is fine — but Ag₂CrO₄ has Ksp = 1.1 × 10⁻¹² yet s = 6.5 × 10⁻⁵ M, more soluble than AgCl despite the smaller Ksp. Always convert to s before comparing.
Pure solids are omitted from all equilibrium expressions, including Ksp. For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻] — no [AgCl] term. Including the solid is one of the most frequent Ksp errors.
No precipitate forms, but the ions stay dissolved at their current concentrations — the solution is simply unsaturated. 'Nothing happens' is correct only in the sense that no solid appears; the ions remain in solution.
Real-world applications
Water treatment: Ksp predicts whether treatment will precipitate contaminants or scale pipes.
Medicine: understanding kidney-stone formation as a Ksp problem guides prevention.
Analytical chemistry: selective precipitation schemes separate ions by their Ksp values.
Geology: predicting which minerals precipitate from groundwater and seawater.
Frequently asked questions
How do I write the Ksp expression for any salt?
Write the dissolution equation, then Ksp = [cation]^coeff × [anion]^coeff — pure solid omitted, coefficients as exponents. For PbI₂: PbI₂(s) ⇌ Pb²⁺ + 2I⁻, so Ksp = [Pb²⁺][I⁻]². The exponents follow the equation, not the charges.
Is Ksp the same as the ion product?
No. Ksp is the equilibrium constant — a property of the salt at a temperature. The ion product Qsp is what you compute from current concentrations. Precipitation is predicted by comparing them: Qsp > Ksp means precipitate forms.
Does Ksp depend on temperature or pH?
Temperature changes Ksp (like any K), and it is defined for the dissolution as written. Ksp values are also medium-dependent in reality (activity effects), but textbook problems treat them as constants in water at 25 °C.
What is the difference between Ksp and Ka?
They share the math but describe different equilibria: Ksp for solid ⇌ ions, Ka for acid ⇌ H⁺ + base. Do not mix them — a 'Ksp of an acid' is a category error. Some problems chain them (e.g., dissolving hydroxides in acid), which is legitimate and common.