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A net ionic equation retains the chemical species changed by a chemical reaction and cancels spectator ions present unchanged on both sides. Dissolved strong electrolytes are represented by their separated ions, while a solid product retains its chemical formula. Cancellation preserves both elemental inventories and electric charge conservation.^[ref]

For the calcium carbonate example, the complete ionic account isCanceling the sodium cations and chloride anions yields the net ionic equation above. Neither cancellation nor the conventional sodium chloride formula implies that the remaining dissolved ions have become paired molecules.

For this schematic, let $\xi$ be the amount of substance of AB formed. With no competing chemical reactions or losses,Adding each dissolved inventory to its contribution in the solid recovers its initial total. Each dissolved pair has net electric charge zero, so transferring a pair into neutral AB also preserves electric charge conservation. This is a short stoichiometry argument; choosing the actual $\xi$ additionally needs chemical equilibrium or chemical kinetics information.

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.

A spectator ion cancels from the chosen net ionic equation because its net inventory is unchanged. This label does not make it physically irrelevant: dissolved spectator ions contribute to ionic strength, which affects activity coefficients. Their concentrations can also change during dilution even though their amount of substance remains fixed.