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Chemistry Lab · Free lesson · 22 minutes

Neutralisation: follow the reacting ions

Represent a dilute strong-acid/strong-base neutralisation, identify spectator ions and calculate reacting amounts with correct volume units.

SS2 · acid–base reactions

01 · Think first

Start with a question.

An acid and an alkali are mixed in a teacher-supervised laboratory. The ions that actually form water are more informative than saying simply 'the acid disappeared'. Here we analyse supplied data only.

What you already know
  • Balance atoms and charge
  • Recognise aqueous ions
  • Convert cm³ to dm³ and calculate amounts
Which net ionic equation describes a dilute strong-acid/strong-base neutralisation?

02 · See it

Keep reacting and spectator ions separate

HCl and NaOH are used as dilute aqueous model reagents. This is symbolic analysis, not a home experiment.

Molecular equation
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Reacting ions
H⁺ + OH⁻ → H₂O
Spectators in this reaction
Na⁺ and Cl⁻
Amount ratio
1 mol H⁺ : 1 mol OH⁻

Use the justified worked steps below to explore this idea.

03 · Show steps · explain why

One justified step at a time.

  1. 1

    HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

    This equation describes the stated dilute aqueous strong-acid/strong-base system. It does not describe every acid/base combination.

04 · Mistake clinic

Find the first wrong idea.

Tempting wrong line

n = 0.100 × 25.0 = 2.50 mol for a 25.0 cm³ sample.

Compare the correction and its reason

n = 0.100 × 0.0250 = 0.00250 mol.

The concentration is per dm³, not per cm³. A missing factor of 1,000 turns a small dilute sample into an implausibly large amount.

Check: Write the volume conversion and units before multiplying. Keep stoichiometric equivalence separate from a universal pH claim.

A mistake is evidence about the next idea to repair. It is not a label for the learner.

05 · Your turn

Try first. Then use a hint.

1. Convert 50.0 cm³ to dm³. Enter the number in dm³.guided practice
2. Find moles of H⁺ in 50.0 cm³ of 0.200 mol dm⁻³ HCl. Enter mol.independent practice
3. A dilute suitable acid reacts with a carbonate. Which product pattern is expected?far practice

06 · Check & reflect → next step

Understanding includes explaining.

Which ions actually react, which are spectators, and which assumption made the calculation valid?

Choose your next learning need

Practise distinguishing observations from conclusions using chromatography data.

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Source and review status

JEM-CH-C08-003 · Phase 5C Chemical Change Acids Redox Electrolysis Energetics Kinetics Equilibrium · Lab 03 · printed pp. 16–17

This published free lesson contains selected explanations and authored practice. Complete canonical volumes and tutor solutions remain in the protected Content Vault.