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Writing ionic equations

Ionic equations: split aqueous ions, cancel spectator ions, and balance atoms and charge for common precipitation, neutralisation, and gas reactions.

  • SEC G3 Pure Chemistry 2027

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A full equation lists all the substances involved. A net ionic equation highlights the particles that change, leaving out ions that remain unchanged in solution.

Reacting particles and spectator ions

An ionic equation (net ionic equation) shows only the ions/particles that actually react. Spectator ions are ions that remain unchanged and are cancelled out.

Keep atoms and charge balanced

  • Write dissolved ionic compounds as their separate ions, and represent strong acids such as hydrochloric acid as ions. Do not split every substance labelled (aq).
  • Cancel ions that appear identical on both sides (including the same charge and state).
  • The net ionic equation must still be balanced (atoms and total charge).
  • Many standard net ionic equations repeat in exams (neutralisation, precipitation, carbonate + acid).

The total charges on the two sides must be equal, but they need not be zero. For example, zinc displacing copper(II) ions has total charge + 2 on each side.

Derive the net ionic equation

Aqueous does not always mean separate ions
  • Dissolved salts and ionic alkalis are represented as their aqueous ions.
  • Strong acids such as hydrochloric acid are represented as ions; weak acids such as ethanoic acid are mainly molecular and are kept intact in the net equation.
  • Keep solid precipitates, liquid water and gases intact.
  • Cancel only unchanged species with the same formula, charge and state, in equal amounts on both sides.

Ions in solution

  • Dissociation: when an ionic compound dissolves in water, its ions separate and move freely.
  • Spectator ions: ions present in solution that do not change; they cancel out.

Split, cancel and check

  1. Write the balanced full equation with state symbols.
  2. Represent dissolved ionic compounds and strong acids as ions. Keep other species intact.
  3. Cancel equal amounts of identical spectator ions on both sides.
  4. Write the net ionic equation and check atoms and total charge.
Keep the reacting solid or molecule visible

Split the aqueous species shown as ions. Never split a precipitate (s), water (l), or a gas (g).

Which species stay intact?

Substance typeExampleSplit into ions?
Aqueous ionic compoundNaCl(aq)Yes
Strong aqueous acidHCl(aq)Yes
Dissolved ionic alkaliNaOH(aq)Yes
Solid ionic compoundAgCl(s)No
WaterH₂O(l)No
GasCO₂(g)No

A weak acid such as CH₃COOH(aq) is only partly ionised in water, so it is kept as a molecule when deriving its net ionic reaction with an alkali. A dissolved molecular substance such as sugar is also not split into ions. The symbol (aq) tells you that a substance is dissolved in water; its bonding and ionisation determine which particles are present.

Deriving a net ionic equationThe balanced equation between silver nitrate and sodium chloride is split into aqueous ions. Sodium and nitrate ions are unchanged spectator ions and are cancelled. The net equation shows aqueous silver ions with charge one plus reacting with aqueous chloride ions with charge one minus to form solid silver chloride.1 · Balanced equation with state symbolsAgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)2 · Write dissolved salts as ions; keep the solid intactAg⁺ + NO₃⁻ + Na⁺ + Cl⁻ → AgCl(s) + Na⁺ + NO₃⁻All ions here are (aq); Na⁺ and NO₃⁻ are unchanged spectators.3 · Cancel spectators and check atoms + total chargeAg⁺(aq) + Cl⁻(aq) → AgCl(s)left charge: +1 + (−1) = 0 · right charge: 0
Represent dissolved salts as aqueous ions, keep the solid precipitate intact, cancel unchanged spectator ions, then check atoms and total charge.

Check the derivation

  • Splitting a precipitate into aqueous ions, which incorrectly removes the solid formed in the reaction.
  • Cancelling ions that are not identical (different charge or different state).
  • Forgetting to check charge balance at the end.
  • Writing H⁺ + OH⁻ → H₂O without state symbols when asked.
  • Splitting every (aq) formula, including weak acids and dissolved molecules. An ionic equation can correctly contain both ions and intact molecules.

A final balance check

Final line checklist
  1. Count each element on both sides.
  2. Add the charges on each side, including coefficients.
  3. Check that no equal amounts of unchanged spectator ions remain.
Where ionic equations show up

Neutralisation, precipitation, displacement, and gas-forming reactions (e.g., carbonate + acid).

Worked examples

Modelled example 1

Neutralisation

Core

Problem

Derive the net ionic equation for NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
Study the worked solution

  1. Split aqueous electrolytes

    Method

    Represent sodium hydroxide, hydrochloric acid and sodium chloride as aqueous ions.

    Reason

    Sodium hydroxide and sodium chloride dissociate in water; hydrochloric acid is a strong acid that ionises. Liquid water remains intact.

    Working

    Na + (aq) + OH-(aq) + H + (aq) + Cl-(aq) → Na + (aq) + Cl-(aq) + H₂O(l)

  2. Cancel spectator ions

    Method

    Remove identical Na⁺ and Cl⁻ ions from both sides.

    Reason

    Their formula, charge and aqueous state are unchanged.

    Working

    Spectators: Na + (aq), Cl-(aq).

  3. Write and check the net equation

    Method

    Retain only reacting particles.

    Reason

    Atoms and total charge must remain balanced.

    Working

    H + (aq) + OH-(aq) → H₂O(l)

Guided practice 2

Forming a silver chloride precipitate

About 7 min

Problem

Write the net ionic equation for aqueous silver nitrate reacting with aqueous sodium chloride to form silver chloride precipitate.

Choose what splits and what remains intact

AgCl(s) treatment
Spectator ions

Hints

Hint 1: write the full equation
Start with AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
Hint 2: apply state rules
Split every aqueous ionic compound, but not AgCl(s).
View solution step by step

  1. Write and split the equation

    Method

    Split the three aqueous compounds into ions.

    Reason

    The solid silver chloride precipitate remains intact.

    Working

    Ag + (aq) + NO₃-(aq) + Na + (aq) + Cl-(aq) → AgCl(s) + Na + (aq) + NO₃-(aq)

  2. Cancel spectators

    Method

    Cancel sodium and nitrate ions.

    Reason

    They appear in identical aqueous forms on both sides.

    Working

    Cancel Na + (aq) and NO₃-(aq).

  3. Write the net equation

    Working

    Ag + (aq) + Cl-(aq) → AgCl(s)

Common misconception 3

Keep the solid precipitate intact

Find and correct the mistake

Learner response

A student writes AgCl(s) → Ag + (aq) + Cl-(aq) while deriving a precipitation equation. Locate the state-rule error and correct the representation.

Use the state symbol as the decision

Correct treatment
Species normally split

View solution step by step

  1. Locate the state-rule error

    Method

    Read the (s) state symbol.

    Reason

    It identifies an insoluble solid precipitate, not freely separated aqueous ions.

    Working

    The product is solid silver chloride, not separated aqueous silver and chloride ions.

  2. Apply the correct split rule

    Method

    Write dissolved ionic compounds as their separate aqueous ions.

    Reason

    Solids, liquids and gases remain as complete species in net ionic equations.

    Working

    Correct representation: AgCl(s).

Examiner practice 4

Zinc displaces copper from solution

3 marks

Examination question

Derive the net ionic equation for Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). [3 marks]

Split, cancel and check charge

View solution step by step

  1. Split aqueous sulfates

    1 mark

    Method

    Separate CuSO₄(aq) and ZnSO₄(aq) into ions.

    Reason

    Solid zinc and copper remain intact.

    Working

    Zn(s) + Cu²⁺(aq) + SO₄²⁻(aq) → Zn²⁺(aq) + SO₄²⁻(aq) + Cu(s)

  2. Cancel sulfate

    1 mark

    Method

    Remove SO₄²⁻(aq) from both sides.

    Reason

    It is unchanged in charge and state and is therefore a spectator.

    Working

    Spectator: SO₄²⁻(aq).

  3. Write the net equation

    1 mark

    Method

    Retain zinc, copper(II) ions, zinc ions and copper.

    Reason

    These are the particles whose oxidation states change.

    Working

    Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Challenge 5

Carbonate ions react with acid

Minimal support

Words-to-ions transfer

Carbonate ions in solution react with dilute strong acid, producing bubbles of carbon dioxide and liquid water. Construct the balanced net ionic equation with state symbols.

Balance atoms and total charge

Product states

Hints

Hint 1: balance charge first

The carbonate ion contributes total charge 2-; products are neutral.

Hint 2: then check hydrogen

Two hydrogen ions supply the two H atoms needed for one water molecule.

View solution step by step

  1. Write the reacting species

    Method

    Use carbonate ions and hydrogen ions.

    Reason

    Spectator cations and acid anions are not part of the stated net change.

    Working

    CO₃²⁻(aq) + H + (aq) → CO₂(g) + H₂O(l).

  2. Balance charge and hydrogen

    Method

    Place coefficient 2 before H⁺.

    Reason

    -2 + 2(+1) = 0, matching neutral products, and two H atoms form water.

    Working

    CO₃²⁻(aq) + 2H + (aq) → CO₂(g) + H₂O(l).

  3. Check the final equation

    Method

    Count C, O, H and total charge.

    Reason

    A valid net ionic equation conserves both atoms and charge.

    Working

    CO₃²⁻(aq) + 2H + (aq) → CO₂(g) + H₂O(l)

Try it yourself

Mind stretcher 1: Find the spectator ionsExtension

Question: Find the net ionic equation for: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq).

Show answer

Split the dissolved ionic compounds:

Ba²⁺(aq) + 2Cl-(aq) + 2Na + (aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na + (aq) + 2Cl-(aq)

Cancel Na⁺ and Cl⁻:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Mind stretcher 2: Check both atoms and chargeExtension

Question: A student writes H + (aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l). Explain what is wrong and fix it.

Show answer

Charge is not balanced: left side total charge is + 1 + (-2) = -1, right side is 0.

Hydrogen is also unbalanced: there is one H atom on the left and two on the right.

You need 2 hydrogen ions:

CO₃²⁻(aq) + 2H + (aq) → CO₂(g) + H₂O(l)

Mind stretcher 3: An aqueous substance that stays molecularExtension

Ethanoic acid is a weak acid. Sodium hydroxide and sodium ethanoate are dissolved ionic compounds. Derive the net ionic equation from:

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

Show answer

Keep the weak acid as CH₃COOH(aq). Write sodium hydroxide and sodium ethanoate as separate aqueous ions, then cancel Na + (aq) from both sides:

CH₃COOH(aq) + OH-(aq) → CH₃COO-(aq) + H₂O(l)

Each side has two C atoms, five H atoms and three O atoms. The total charge is -1 on each side. An ionic equation can include an intact molecule as well as ions.

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