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Solubility Product Ksp: Predicting When a Salt Precipitates

Multiply the ion concentrations of a sparingly-soluble salt together and compare that product to a fixed constant — bigger, it precipitates; smaller, more can dissolve.

mysimulator teamUpdated June 2026≈ 6 min read▶ Open the simulation

An equilibrium like any other

A sparingly-soluble ionic salt sitting at the bottom of a saturated solution is in dynamic equilibrium with its dissolved ions: ions are dissolving off the solid surface and re-depositing onto it at exactly equal rates. Because a pure solid's activity is defined as 1, it drops out of the equilibrium expression, leaving only the dissolved ion concentrations, each raised to the power of its stoichiometric coefficient — the solubility product Ksp.

Molar solubility vs Ksp

Ksp itself is a fixed number at a given temperature, but the salt's molar solubility s — the actual concentration that dissolves — depends on how the stoichiometry converts s into ion concentrations. For a simple 1:1 salt like AgCl, each mole that dissolves gives one Ag+ and one Cl-, so Ksp = s². For a salt like PbI2, each mole gives one Pb2+ but two I-, so Ksp = [Pb2+][I-]² = s · (2s)² = 4s³. Two salts can share numerically similar Ksp values yet have very different molar solubilities once the exponents diverge.

AB(s) ⇌ A⁺ + B⁻           Ksp = [A⁺][B⁻] = s²
AB₂(s) ⇌ A²⁺ + 2B⁻         Ksp = [A²⁺][B⁻]² = 4s³

if Q < Ksp  →  solution undersaturated, more solid can dissolve
if Q > Ksp  →  solution supersaturated, precipitation occurs
if Q = Ksp  →  saturated, at equilibrium

Q vs Ksp: the precipitation test

The reaction quotient Q is calculated the same way as Ksp, but using whatever ion concentrations actually exist right now — not necessarily at equilibrium. Comparing Q to Ksp instantly predicts which direction the system will move: if Q exceeds Ksp, the solution holds more dissolved ions than equilibrium allows and solid precipitates out until Q drops back to Ksp; if Q is below Ksp, the solution can still dissolve more solid.

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The common-ion effect

Adding a soluble source of one of the salt's own ions — say, dissolving NaCl into a saturated AgCl solution to add extra Cl- — pushes Q above Ksp immediately, by Le Chatelier's principle the equilibrium shifts back toward the solid, and AgCl precipitates until Ksp is restored. Because Ksp itself never changes, raising one ion's concentration forces the other ion's equilibrium concentration, and hence the salt's overall molar solubility, to fall. This common-ion effect is used deliberately in analytical chemistry to suppress unwanted dissolution.

Limits of the simple model

The straightforward Ksp expression assumes ideal, dilute behaviour. At higher ionic strength, activity coefficients deviate from 1 and the true thermodynamic Ksp diverges from a naive concentration-based calculation. Complexation with other dissolved species, and pH-dependence for salts whose anion is the conjugate base of a weak acid, can also shift the effective solubility well beyond what the bare Ksp expression predicts — the classical model is an excellent starting approximation, not the final word.

Frequently asked questions

What's the difference between Ksp and molar solubility?

Ksp is a fixed equilibrium constant for a given salt at a given temperature. Molar solubility s is the actual concentration of dissolved salt at equilibrium, and it is derived from Ksp using the salt's specific stoichiometry — for a simple 1:1 salt like AgCl, Ksp = s^2, but for a salt like PbI2 that releases two iodide ions per formula unit, Ksp = 4s^3. Two salts can have the same Ksp value and very different molar solubilities if their stoichiometries differ.

How do you predict whether a precipitate will form when two solutions are mixed?

Calculate the reaction quotient Q using the actual (not necessarily equilibrium) ion concentrations right after mixing, using the same expression as Ksp. If Q > Ksp, the solution is supersaturated and precipitation occurs until Q falls back to Ksp. If Q < Ksp, the solution is undersaturated and no precipitate forms — more solid could still dissolve. If Q equals Ksp, the solution is exactly saturated and at equilibrium.

Why does adding a common ion decrease a salt's solubility?

By Le Chatelier's principle, increasing the concentration of one of the product ions shifts the dissolution equilibrium back toward the solid, undissolved form. Because Ksp is a fixed constant, raising one ion's concentration forces the other ion's equilibrium concentration to drop, which in turn lowers the salt's overall molar solubility — the classic common-ion effect, used deliberately to suppress unwanted solubility in analytical chemistry.

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