The solubility-product expression can be evaluated for ion concentrations that are not at equilibrium. The result is the ion product, usually written $Q_{sp}$.

For

$$\mathrm{M_pX_q(s)\rightleftharpoons p,M^{m+}(aq)+q,X^{n-}(aq)},$$

$$Q_{sp}=[\mathrm{M^{m+}}]^p[\mathrm{X^{n-}}]^q$$

uses the current ion concentrations, while $K_{sp}$ is the value of that product at equilibrium with the solid.

The comparison determines the saturation state:

• $Q_{sp}<K_{sp}$: the solution is unsaturated with respect to the solid; more solid can dissolve if it is available;
• $Q_{sp}=K_{sp}$: the solution is saturated and at solubility equilibrium;
• $Q_{sp}>K_{sp}$: the ion concentrations are supersaturated; net precipitation is favored until the ion product falls back toward $K_{sp}$.

Example

For silver chloride,

$$\mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)},$$

suppose

$$K_{sp}=1.8\times10^{-10}.$$

A solution currently has

$$[\mathrm{Ag^+}]=2.0\times10^{-5},\mathrm M,$$

$$[\mathrm{Cl^-}]=3.0\times10^{-5},\mathrm M.$$

Then

$$Q_{sp}=(2.0\times10^{-5})(3.0\times10^{-5}) =6.0\times10^{-10}.$$

Because

$$Q_{sp}>K_{sp},$$

the mixture is supersaturated with respect to $\mathrm{AgCl}$, so net formation of solid is favored.

This gives a quantitative replacement for qualitative solubility rules. A pair of ions can form a sparingly soluble compound without necessarily precipitating: precipitation begins only when their actual concentrations make $Q_{sp}$ exceed $K_{sp}$.

At the precipitation threshold, $Q_{sp}=K_{sp}$. Solving this equality for one ion concentration gives the concentration at which precipitation first becomes favored. This idea underlies controlled and selective precipitation in analytical chemistry.