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A precipitate is the solid formed by precipitation. Fine particles can initially remain dispersed as a particle suspension, rather than immediately settling to the vessel bottom.

Consider an original toy AB salt model with $K_{\mathrm{sp},c}=10^{-8}{\mathrm{M}}^2$ stipulated at fixed temperature and fixed activity coefficient product. This is a conditional concentration constant, not measured data for calcium carbonate or any identified material. A stock contains A cations and X anions; another contains Y cations and B anions. All four ions have unit electric charge magnitude. Only AB can form a solid; X and Y remain spectator ions. Assume full dissociation, no other chemical reactions, and a mixed volume equal to the sum of the initial volumes.

Mix two stocks, each of volume $V$,, each at molar concentration $C$ of its indicated salt. Before any precipitation, the mixed volume is $2V$, soMultiplying the two stock concentrations would overestimate the mixed ion product by a factor of four. The saturation threshold is $C=2\sqrt{K_{\mathrm{sp},c}}=2\times 10^{-4}\mathrm{M}$.

If $C=10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=2.5\times 10^{-9}{\mathrm{M}}^2<K_{\mathrm{sp},c}$: no AB precipitate is favored. If $C=4\times 10^{-4}\mathrm{M}$, mixing gives $Q_{c,0}=4\times 10^{-8}{\mathrm{M}}^2>K_{\mathrm{sp},c}$. Suppose nucleation occurs and bulk AB solid remains at chemical equilibrium. Equal starting inventories and one-to-one stoichiometry require equal remaining concentrations, henceFor two $10\mathrm{mL}$ portions in the second case, each reacting ion starts with $4\mu \mathrm{mol}$; the final $20\mathrm{mL}$ chemical solution retains $2\mu \mathrm{mol}$ of each, and $2\mu \mathrm{mol}$ of AB enters the solid. The respective X and Y inventories stay at $4\mu \mathrm{mol}$. Exactly half of each reacting inventory precipitates in this toy model.

The chemical equilibrium calculation adds a phase constraint to the inventory balance. Write $x=n_{\mathrm{AB},s}/(2V)$ and $c=C/2-x$. In the equilibrated bulk model,When $x>0$, the last condition requires saturation; when $x=0$, the inventory can leave an undersaturated chemical solution. Together these give $x=\max \{0,C/2-\sqrt{K_{\mathrm{sp},c}}\}$. This excludes a persistent supersaturated solution by its explicit chemical equilibrium assumption; it does not model a nucleation delay.

Precipitation is formation of a solid from a chemical solution; the formed solid is a precipitate. Dissolved chemical species enter a separate thermodynamic phase. A newly formed solid can create a surface for heterogeneous catalysis; observing it does not establish the composition or catalytic role of that surface.

For example, mixing suitable chemical solutions of calcium chloride and sodium carbonate can form calcium carbonate:Here $aq$ means dissolved in water, and $s$ marks the solid. Both sides contain one calcium chemical atom, one carbon chemical atom, and three oxygen chemical atoms; net electric charge is $+2-2=0$ on the left and zero on the right. The arrow describes the net change when precipitation occurs, rather than promising complete removal of either ion. The calcium carbonate precipitation experiment makes this phase separation visible.^[ref]

The attraction is that an invisible dissolved inventory becomes a recoverable solid. Net ionic equations track that transfer; the saturation criterion for precipitation determines whether it is favored under specified conditions. This combination supports separations and chemical analysis.^[ref]