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How do we write a net ionic equation that shows only the species that actually take part in a reaction?

Topic 4.2 Net Ionic Equations: write balanced molecular, complete ionic and net ionic equations for reactions in aqueous solution, removing spectator ions.

A focused answer to AP Chemistry Topic 4.2, covering molecular, complete ionic and net ionic equations, how to identify and cancel spectator ions, and how solubility rules guide which species are written as ions, with full worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The three forms of equation
  3. Which species are written as ions
  4. Finding the spectator ions
  5. Try this

What this topic is asking

The College Board (Topic 4.2) wants you to write three forms of a reaction in aqueous solution, the molecular, complete ionic and net ionic equations, and to identify and remove the spectator ions that take no part in the reaction. The net ionic equation captures the actual chemical change, which is the goal. Doing this correctly requires knowing which species are strong electrolytes (and so written as separated ions) and which are not.

The three forms of equation

Each form serves a purpose. The molecular equation is the bookkeeping form that balances atoms. The complete ionic equation shows what actually exists in solution. The net ionic equation strips out the bystanders to reveal the genuine chemical change. The AP exam most often asks for the net ionic equation, because it is the most honest representation of the reaction.

Which species are written as ions

Useful solubility guidelines: nitrates (NO3−\text{NO}_3^-), and the salts of group 1 metals and ammonium (NH4+\text{NH}_4^+), are always soluble; most chlorides are soluble except those of silver, lead and mercury(I); most sulfates are soluble except those of barium, lead, calcium and strontium; most carbonates, phosphates, hydroxides and sulfides are insoluble. These rules decide which product is the precipitate (written as a solid) and which ions stay in solution as spectators.

Finding the spectator ions

A spectator ion is one that is present as an aqueous ion on both sides of the complete ionic equation, unchanged. Because it neither forms a new substance nor disappears, it plays no role and is cancelled. What is left, the net ionic equation, contains only the ions that combine to form the precipitate (or the molecules involved in the reaction). The net ionic equation must still balance in both atoms and charge, a useful final check.

Try this

Q1. Identify the spectator ions when aqueous sodium hydroxide is mixed with aqueous copper(II) sulfate to form a copper(II) hydroxide precipitate. [2 points]

  • Cue. Na+\text{Na}^+ and SO42−\text{SO}_4^{2-} (both stay as aqueous ions); the net ionic equation is Cu2+(aq)+2OH−(aq)→Cu(OH)2(s)\text{Cu}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cu(OH)}_2(s).

Q2. State which species in a complete ionic equation are NOT split into ions. [1 point]

  • Cue. Solids (precipitates), gases, pure liquids and water, and weak electrolytes.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (short FRQ). Aqueous silver nitrate (AgNO3\text{AgNO}_3) is mixed with aqueous sodium chloride (NaCl\text{NaCl}), forming a white precipitate. (a) Write the balanced molecular equation. (b) Write the complete ionic equation. (c) Write the net ionic equation. (d) Identify the spectator ions and justify why they are removed.
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A 4-point FRQ on the three forms of equation.

(a) Molecular (1 point): AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq).
(b) Complete ionic (1 point): Ag+(aq)+NO3−(aq)+Na+(aq)+Cl−(aq)→AgCl(s)+Na+(aq)+NO3−(aq)\text{Ag}^+(aq) + \text{NO}_3^-(aq) + \text{Na}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) + \text{Na}^+(aq) + \text{NO}_3^-(aq).
(c) Net ionic (1 point): Ag+(aq)+Cl−(aq)→AgCl(s)\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s).
(d) Spectators (1 point): Na+\text{Na}^+ and NO3−\text{NO}_3^- are spectator ions; they appear unchanged on both sides (still aqueous ions), take no part in forming the precipitate, and so are cancelled.

Markers reward a balanced molecular equation, dissociating all soluble strong electrolytes in the ionic equation, cancelling spectators correctly, and identifying the spectators by reasoning.

AP 2021 (style)1 marksSection I (multiple choice). In the reaction of aqueous barium chloride with aqueous sodium sulfate, which is the correct net ionic equation? (A) Ba2++SO42−→BaSO4(s)\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(s) (B) Ba2++2Cl−→BaCl2\text{Ba}^{2+} + 2\text{Cl}^- \rightarrow \text{BaCl}_2 (C) Na++Cl−→NaCl\text{Na}^+ + \text{Cl}^- \rightarrow \text{NaCl} (D) 2Na++SO42−→Na2SO42\text{Na}^+ + \text{SO}_4^{2-} \rightarrow \text{Na}_2\text{SO}_4. Justify your choice.
Show worked answer →

A 1-point conceptual MCQ. The answer is (A).

The only insoluble product is barium sulfate, so the net ionic equation shows just the ions that combine to form it: Ba2+(aq)+SO42−(aq)→BaSO4(s)\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s). Sodium and chloride are spectator ions (both their compounds are soluble), so they do not appear.

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