Net ionic equation

Precipitation (chemistry)

Precipitation

Spectator ion

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>.<Chemistry 2e by Paul Flowers and others>[Chemistry 2e, §4.2, Precipitation Reactions and Solubility Rules: molecular and net ionic equations for silver chloride]{ref}

For the <calcium carbonate> example, the complete ionic account is
$$
\begin{aligned}
&\mathrm{Ca}^{2+}(aq)+2\mathrm{Cl}^-(aq)+2\mathrm{Na}^+(aq)+\mathrm{CO}_3^{2-}(aq)\\
&\qquad\longrightarrow\mathrm{CaCO}_3(s)+2\mathrm{Na}^+(aq)+2\mathrm{Cl}^-(aq).
\end{aligned}
$$
Canceling 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>.

\Image[/precipitation-species-and-phase-balance.png]
{title=<Precipitation (chemistry)> transfers selected <ions> into a <solid> while conserving inventories}
{id=image-precipitation-species-and-phase-balance}
{description=Original one-to-one model: four A <cations> and four B <anions> initially dissolved; three of each enter AB <solid>, leaving one of each dissolved. Four X <anions> and four Y <cations> are <spectator ions> throughout. Counts are illustrative, not measured, and do not represent the <calcium carbonate> experiment or determine a <solubility product>.}

For this schematic, let $\xi$ be the <amount of substance> of AB formed. With no competing <chemical reactions> or losses,
$$
n_{\mathrm A,aq}=n_{\mathrm A,0}-\xi,\qquad n_{\mathrm B,aq}=n_{\mathrm B,0}-\xi,\qquad n_{\mathrm{AB},s}=\xi,
\qquad 0\le\xi\le\min(n_{\mathrm A,0},n_{\mathrm B,0}).
$$
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.


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