Total & Net Ionic Equations

The balanced equation written with complete, neutral compound formulas — exactly what you've been writing so far. It must be balanced and must carry correct state symbols, because the states decide what gets split in the next stage. Use the solubility rules to decide whether each product is (aq) or (s).

Rewrite every strong soluble ionic compound marked (aq) as its separate, dissociated ions. Solids (s), liquids (l), and gases (g) are NOT split apart — they stay written as full formulas, since they are not actually dissolved and free-moving. Strong acids in water (HCl, HBr, HI, ${\text{HNO}}_3$ and ${\text{H}}_2{\text{SO}}_4$) also exist as ions, so they are split too. Molecular substances such as water and sugar are never split. Two details trip people up: the coefficient multiplies every ion that comes out of a formula, and a subscript outside a bracket tells you how many polyatomic ions are released, but the polyatomic ion itself stays in one piece.

Cross out any ion that appears unchanged, in identical form, on both the reactant side and the product side — these are called spectator ions, because they don't actually participate in the reaction; they're just present in solution before and after. What's left is the net ionic equation: only the substances that actually react. An ion counts as a spectator only if it has the same formula, the same charge, the same state (aq) and the same number on both sides. If ${\text{Ag}}^{+}\text{(aq)}$ is on the left but ends up inside $\text{AgCl(s)}$ on the right, it has changed and must stay. When a spectator has the same coefficient on both sides (say $2{\text{Na}}^{+}$ on each side), cancel all of it; if the numbers differ, recheck your balancing, because a correct total ionic equation always gives matching spectator counts.

Net ionic equations are not only for precipitates. Every neutralization of a strong acid by a strong base, whether it is HCl with NaOH or ${\text{HNO}}_3$ with KOH, reduces to the same net ionic equation, ${\text{H}}^{+}\text{(aq)}+{\text{OH}}^{-}\text{(aq)}\to {\text{H}}_2\text{O(l)}$, which is why these reactions all release about the same heat per mole of water formed. Single displacements shrink too: in $\text{Zn(s)}+{\text{CuSO}}_4\text{(aq)}\to {\text{ZnSO}}_4\text{(aq)}+\text{Cu(s)}$ the sulfate is a spectator, leaving $\text{Zn(s)}+{\text{Cu}}^{2+}\text{(aq)}\to {\text{Zn}}^{2+}\text{(aq)}+\text{Cu(s)}\text{,}$ which shows the electron transfer from zinc to copper directly. If every ion cancels, as when NaCl(aq) and KNO₃(aq) are mixed, there is no net ionic equation: no reaction.