Mix silver nitrate and sodium chloride solutions and a white solid appears (Precipitation and Solubility Rules). But Electrolytes and Ions in Solution says both solutions were really ions all along. Which ions actually reacted? A net ionic equation shows only those.
Key idea
Write every soluble strong electrolyte in water (aq) as its separate ions. Keep everything else as written: solids (s), liquids such as $\mathrm{H_2O}$, gases (g), and weak acids and bases. Ions that appear unchanged on both sides are spectator ions; cross them out. What remains is the net ionic equation.
Why it works
In a solution of $\mathrm{AgNO_3}$ there are no $\mathrm{AgNO_3}$ units, only $\mathrm{Ag^+}$ and $\mathrm{NO_3^-}$ ions moving independently, and the same goes for NaCl. When the solutions mix, $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$ lock together into solid AgCl; the $\mathrm{Na^+}$ and $\mathrm{NO_3^-}$ ions just float around before and after. Writing the spectators in suggests they take part, which they don't. A weak acid such as acetic acid, though, is mostly whole molecules in solution, so it is written whole: that is what actually meets the base.
Worked example: A precipitation
Write the net ionic equation for mixing $\mathrm{Pb(NO_3)_2(aq)}$ and $\mathrm{KI(aq)}$, which forms solid $\mathrm{PbI_2}$.
- Molecular: $$\mathrm{Pb(NO_3)_2(aq)} + 2\,\mathrm{KI(aq)} \to \mathrm{PbI_2(s)} + 2\,\mathrm{KNO_3(aq)}$$
- Complete ionic: $$\mathrm{Pb^{2+}} + 2\,\mathrm{NO_3^-} + 2\,\mathrm{K^+} + 2\,\mathrm{I^-} \to \mathrm{PbI_2(s)} + 2\,\mathrm{K^+} + 2\,\mathrm{NO_3^-}$$
- Cancel the spectators, $\mathrm{K^+}$ and $\mathrm{NO_3^-}$.
Answer: $\mathrm{Pb^{2+}(aq)} + 2\,\mathrm{I^-(aq)} \to \mathrm{PbI_2(s)}$
Worked example: A strong acid and a strong base
Write the net ionic equation for $\mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \to \mathrm{NaCl(aq)} + \mathrm{H_2O(l)}$.
- Complete ionic: $\mathrm{H^+} + \mathrm{Cl^-} + \mathrm{Na^+} + \mathrm{OH^-} \to \mathrm{Na^+} + \mathrm{Cl^-} + \mathrm{H_2O(l)}$.
- Water is a liquid and stays whole; $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ are spectators.
Answer: $\mathrm{H^+(aq)} + \mathrm{OH^-(aq)} \to \mathrm{H_2O(l)}$, the same for every strong acid with every strong base.
Worked example: A weak acid
Write the net ionic equation for acetic acid reacting with sodium hydroxide.
- Acetic acid, $\mathrm{CH_3COOH}$, is weak: it stays mostly as molecules, so it is written whole.
- $\mathrm{NaOH}$ splits; the product sodium acetate is a soluble salt and splits too, leaving $\mathrm{Na^+}$ as a spectator.
Answer: $$\mathrm{CH_3COOH(aq)} + \mathrm{OH^-(aq)} \to \mathrm{CH_3COO^-(aq)} + \mathrm{H_2O(l)}$$
Common mistake
Splitting things that don't exist as free ions (a solid precipitate, water, a gas such as $\mathrm{CO_2}$, or a weak acid), or leaving spectator ions in the "net" equation. Check: everything in a net ionic equation should change; if an ion appears identically on both sides, cancel it. And the charges must balance on both sides.