Master AP Chemistry Topic 4.2 with this free practice quiz. Test your skills on spectator ions, complete ionic equations, precipitation reactions, and solubility rules to prepare for your AP exam.
Report a question
Three Ways to Write a Reaction in Solution
A molecular equation shows every substance as a complete formula, including spectator ions and insoluble products: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
A complete ionic equation splits every soluble ionic compound and strong acid or base into its separate ions, showing what is actually present in the beaker: Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq).
A net ionic equation deletes every ion that appears unchanged on both sides. Those ions are spectator ions: present in the solution, but chemically uninvolved. What survives is the actual chemical change: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Net ionic equations are useful because they strip a reaction down to the particles that react, which is why dozens of different molecular equations can share one net ionic equation.
What Splits and What Stays Whole
Soluble ionic compounds dissociate completely in water and are written as ions. Strong acids and strong bases ionize completely and are also written as ions; they are strong electrolytes. Pure solids, pure liquids, and gases are never split: a precipitate stays as its formula with (s), water stays as H₂O(l), and an evolved gas stays as its formula with (g). Weak acids and weak bases stay whole too, because they ionize only slightly.
The charge must balance in a net ionic equation just as atoms do. The total charge on the left must equal the total charge on the right; an imbalance means an ion was dropped or a coefficient is wrong.
Solubility Rules That Decide Precipitates
Every nitrate, and every compound of sodium, potassium, or ammonium, is soluble; these ions essentially never form precipitates. Chlorides are soluble except those of Ag⁺, Pb²⁺, and Hg₂²⁺. Sulfates are soluble except those of Ba²⁺, Pb²⁺, Ca²⁺, and Sr²⁺. Carbonates, phosphates, and sulfides are insoluble except with the always-soluble cations above. Hydroxides are insoluble except with those cations and, to a degree, Ca²⁺, Sr²⁺, and Ba²⁺.
A precipitation reaction occurs only if swapping the partners in a mixture produces one of the insoluble combinations. If every possible product is soluble, all ions remain in solution and there is no reaction, even though a molecular equation can be written on paper.
Worked Examples
Example 1: precipitation, full derivation
Write the net ionic equation for mixing CaCl₂(aq) and Na₂CO₃(aq).
Molecular: CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq). Calcium carbonate is insoluble, so a precipitate forms.
Complete ionic: Ca²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → CaCO₃(s) + 2Na⁺(aq) + 2Cl⁻(aq).
Cancel Na⁺ and Cl⁻, which appear unchanged on both sides. Net ionic:
Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
Check: atoms balance, and charge balances, +2 − 2 = 0 on the left and a neutral solid on the right.
Example 2: metal reacting with acid
Magnesium ribbon is dropped into dilute hydrochloric acid. Write the net ionic equation.
Molecular: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). HCl is a strong acid, written as H⁺ and Cl⁻; the solid metal and the gas stay whole.
Complete ionic: Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g).
Cancel Cl⁻. Net ionic:
Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
Charge check: +2 on the left, +2 on the right.
Traps Specific to This Topic
Do not split a precipitate, a gas, or water into ions. Writing CaCO₃(s) as Ca²⁺ + CO₃²⁻ in the complete ionic equation destroys the point of the reaction: forming the solid is the reaction.
Sulfuric acid questions often hide the electrolyte rules: H₂SO₄ is strong for its first proton, and in these equations it is written as ions, while a weak acid such as CH₃COOH must be left as whole molecules on the reactant side.
Spectator ions must be removed completely, including their coefficients. If 2Cl⁻ appears on both sides, both copies cancel; leaving one behind unbalances the charge.
A mixture of two soluble salts whose four ions form no insoluble pair has no net ionic equation at all. The correct answer is “no reaction,” not an equation with every ion cancelled and nothing left.