Writing Net Ionic Equations

The molecular equation shows compounds as intact formulas. The complete ionic equation shows every dissolved strong electrolyte as separate ions. The net ionic equation shows only the species that change. All three describe the same reaction, but the net ionic form is the one that generalises across different starting salts.

Each representation conserves the same matter. For 3 CaCl₂(aq) + 2 Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6 NaCl(aq), the complete ionic line is 3 Ca²⁺ + 6 Cl⁻ + 6 Na⁺ + 2 PO₄³⁻ → Ca₃(PO₄)₂(s) + 6 Na⁺ + 6 Cl⁻. Each soluble formula contributes its coefficient times its ion subscripts. Phosphate remains PO₄³⁻; dissolving does not break its P–O bonds into separate phosphorus and oxygen ions.

Split only strong electrolytes that are dissolved: soluble ionic compounds, strong acids and strong bases, all marked (aq). Keep intact anything marked (s), (l) or (g), plus weak acids, weak bases and water. A solid never dissociates on paper even if its ions are the whole point of the reaction.

An (aq) label is necessary but insufficient for splitting. Sugar(aq) and ethanol(aq) are dissolved molecules, not strong electrolytes. Conversely, soluble ionic salts dissociate even if their solution is dilute. A sparingly soluble hydroxide written as a solid stays whole; a stated dissolved strong-base portion is represented by its ions. Classify the species and its physical state before rewriting it.

For a molecular strong acid, “ionization” means reaction with water rather than release of pre-existing lattice ions. H⁺(aq) is convenient shorthand for hydrated protons; H₃O⁺(aq) is also acceptable when water coefficients are balanced consistently.

A correct net ionic equation balances atoms and total charge. In the silver chloride example the left side carries 1+ and 1-, a net of zero, and the right side is a neutral solid, also zero. If the charges do not match, a coefficient or a species has been dropped.

After cancelling the six Na⁺ and six Cl⁻ spectators in the calcium phosphate reaction, write 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s). The left charge is 3(+2) + 2(−3) = 0, matching the neutral solid. Calcium, phosphorus and oxygen counts are 3, 2 and 8 on each side. Count whole polyatomic units when they are unchanged, then check the individual elements if needed.

Cancel only the same chemical species in the same state. Cu²⁺(aq) cannot cancel against Cu(s), and CH₃COOH cannot cancel against CH₃COO⁻: those are the substances changing. If every species cancels, report no net reaction instead of inventing a product.

For a strong acid and strong base, H⁺ + OH⁻ → H₂O captures the change. For acetic acid, keep the weak reactant intact: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. For ammonia with strong acid, NH₃ + H⁺ → NH₄⁺. In the first weak-electrolyte example OH⁻ removes a proton; in the second NH₃ accepts one. Neither requires pretending the weak reactant was fully ionized before reacting.

The aqueous acetate product is still an ion; the instruction to keep a weak acid molecular does not mean keeping all of its salts molecular. Sodium acetate is a soluble ionic compound and splits into Na⁺ and CH₃COO⁻. This distinction is essential when writing buffer and titration reactions later.