🧂 Solubility Product Ksp — Precipitation & the Common-Ion Effect

Watch a sparingly-soluble salt dissolve and precipitate as you add a common ion. Compare the ion product Q to Ksp live and see molar solubility drop with the classic common-ion effect.

ChemistryInteractive
Left: beaker with solid precipitate and dissolved ions · Right: molar solubility s vs added common-ion concentration

How it Works

A sparingly-soluble ionic solid MX sits at the bottom of a beaker, releasing M⁺ and X⁻ ions into solution until the ion product Q = [M⁺][X⁻] equals the solubility product Ksp. For the simple 1:1 salt modeled here, the intrinsic molar solubility with no other source of ions is s₀ = √Ksp. When you add extra X⁻ from an external soluble salt (the common-ion slider), the system can no longer support as much dissolved M⁺: the new equilibrium solubility s solves Ksp = s·(s + c), where c is the added common-ion concentration.

Whenever Q is less than Ksp, the solution is unsaturated and solid keeps dissolving; whenever Q exceeds Ksp, the solution is supersaturated and ions precipitate back onto the solid pile until Q returns to Ksp. Watch the beaker: as you raise the common-ion slider, dissolved particles rejoin the solid and the pile visibly grows, while the plot on the right traces the falling solubility curve — the signature of the common-ion effect.

Dissolution: MX(s) ⇌ M⁺(aq) + X⁻(aq)
Solubility product: Ksp = [M⁺][X⁻]
No common ion (1:1 salt): s₀ = √Ksp
With common ion c: Ksp = s·(s + c) → s = (−c + √(c² + 4Ksp)) / 2
1:2 salt (e.g. PbI₂): Ksp = 4s³ → s = ³√(Ksp / 4)

Frequently Asked Questions

What is the solubility product Ksp?

The solubility product Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic solid, such as MX(s) ⇌ M⁺(aq) + X⁻(aq). It equals the product of the equilibrium molar concentrations of the dissolved ions, each raised to its stoichiometric coefficient, and tells you exactly how far dissolution proceeds before the solution saturates.

How do you calculate molar solubility from Ksp?

For a simple 1:1 salt MX, molar solubility s satisfies Ksp = s × s = s², so s = √Ksp. For salts with different stoichiometry the exponents change: for a 1:2 salt like PbI₂ (Pb²⁺ + 2I⁻), Ksp = s × (2s)² = 4s³, so s = ³√(Ksp/4). Stoichiometry must always be built into the Ksp expression before solving for s.

What is the common-ion effect?

The common-ion effect is the decrease in solubility of a sparingly soluble salt when a soluble salt containing one of its ions is added. By Le Chatelier's principle, adding extra of one ion pushes the dissolution equilibrium back toward the solid side, so less of the original salt stays dissolved.

Why does adding a common ion reduce solubility?

Ksp is fixed at a given temperature, so [M⁺][X⁻] cannot exceed it at equilibrium. If extra X⁻ is added from another source, [M⁺] must fall to keep the product equal to Ksp, meaning less solid MX can remain dissolved — the same logic Le Chatelier's principle predicts for any equilibrium under an added-product stress.

What do saturation, unsaturation, and supersaturation mean?

The ion product Q = [M⁺][X⁻] is compared directly to Ksp. If Q < Ksp the solution is unsaturated and more solid can dissolve. If Q = Ksp the solution is saturated and at equilibrium. If Q > Ksp the solution is supersaturated and ions precipitate out until Q falls back to Ksp.

Why is the solubility product important in real life?

Ksp chemistry governs water hardness and limescale formation from CaCO₃ in pipes and kettles, crystallization of minerals in kidney stones, selective precipitation used to separate metal ions in qualitative analysis, and the chemistry of antacids and mineral supplements dissolving or precipitating in the body.

Why does temperature usually increase Ksp and solubility?

For most ionic solids, dissolution is endothermic — breaking the crystal lattice absorbs more energy than is released hydrating the ions. By Le Chatelier's principle, adding heat to an endothermic equilibrium shifts it toward more dissolution, so Ksp and molar solubility typically increase as temperature rises.

How does Ksp relate to the general equilibrium constant K?

Ksp is a special case of the general equilibrium constant K applied to a heterogeneous dissolution reaction, where the solid's activity is defined as 1 and dropped from the expression. The same mass-action reasoning used for Ka, Kb, and Kc applies directly to Ksp, and the common-ion effect is the same Le Chatelier logic used in acid-base buffer chemistry.

About this simulation

Written by MySimulator Team · Reviewed by MySimulator Editorial Review

Last updated: 11 July 2026

This simulator turns the solubility product Ksp into a live tug-of-war between a solid salt and its dissolved ions. A beaker holds a sparingly-soluble MX solid with cations and anions diffusing above it; as you add a common ion, the ion product Q = [M⁺][X⁻] is forced above Ksp, and ions visibly precipitate back onto the pile until equilibrium is restored at a lower dissolved concentration. The chart on the right traces that falling molar solubility curve — the classic signature of the common-ion effect.

🔬 What it shows

Two synchronized views: a beaker where dissolved M⁺ and X⁻ particles float above a solid pile that grows or shrinks as equilibrium shifts, and a solubility-vs-common-ion curve that always bends downward — more common ion always means less of the original salt stays dissolved.

🎮 How to use

Pick a real salt preset (AgCl, BaSO₄, PbI₂, CaCO₃) or drag the pKsp slider directly, then raise the common-ion slider to watch precipitation happen in real time. The temperature slider mimics how Ksp typically rises with heat, and the Reset button returns to the default AgCl scenario.

💡 Did you know?

The common-ion effect is why adding table salt to a silver nitrate solution makes AgCl crash out of solution almost instantly, and it is the same chemistry that keeps calcium carbonate scale building up faster in hard water once other calcium salts are already present.

Frequently asked questions

What does the Q vs Ksp comparison in the stats box mean?

Q = [M⁺][X⁻] is the instantaneous ion product using whatever concentrations are currently present, while Ksp is the fixed equilibrium value at the current temperature. Comparing them tells you which way the system is moving: Q < Ksp means dissolving, Q > Ksp means precipitating, and Q = Ksp means the beaker has settled at equilibrium.

Why does the solid pile grow when I raise the common-ion slider?

Adding common ion X⁻ from an external source raises [X⁻] instantly, pushing Q above Ksp. To bring the product back down to Ksp, some dissolved M⁺ must leave solution and rejoin the solid — so the pile visibly grows while the floating particle count drops, exactly the common-ion effect in action.

Why do the salt presets have different Ksp values?

Each real salt has its own crystal lattice energy and hydration energetics, which set how favorable dissolution is. AgCl and BaSO₄ are extremely insoluble (Ksp near 10⁻¹⁰), while CaCO₃ and PbI₂ are only slightly more soluble (Ksp near 10⁻⁹) — small differences in Ksp translate into large differences in how much solid you would need to fully dissolve a given volume of water.

Why does raising the temperature slider usually increase solubility?

The simulator scales Ksp upward as temperature rises, reflecting that dissolution of most ionic solids is endothermic. By Le Chatelier's principle, supplying heat favors the side of the equilibrium that absorbs it, which is usually the dissolved-ion side, so both Ksp and equilibrium solubility increase with temperature.

Why does the sim use a 1:1 salt model even though PbI₂ is a preset?

To keep the visual particle-and-pile model simple and directly comparable across presets, the simulator applies the 1:1 relationships s₀ = √Ksp and Ksp = s(s + c) to every preset's Ksp value. Real PbI₂ actually follows Ksp = 4s³ because of its 1:2 stoichiometry — the FAQ and formula block above spell out that distinction explicitly.

How is this related to acid-base equilibrium concepts?

Ksp is just the equilibrium constant K applied to a dissolution reaction, and the common-ion effect here is the identical Le Chatelier reasoning used to explain acid-base buffers: adding a conjugate or common ion suppresses further reaction in the direction that produces more of it, whether that reaction is a salt dissolving or a weak acid ionizing.