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6.1.7Reactivity 1 — What drives chemical reactions?

Neutralisation Reactions — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 7 of 8 Unit quiz

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Part 1 of 6Acid Reactions

6.1.7 Neutralisation Reactions

In every neutralisation, an acid donates protons to a base to give a salt, and water or carbon dioxide may also form.

Acid + alkali
  • HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l); the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).
Acid + metal oxide
  • CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); the oxide ion accepts protons.
Acid + carbonate
  • Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g), with effervescence.
Acid + hydrogencarbonate
  • NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g).
Acid + ammonia
  • NH₃(aq) + HCl(aq) → NH₄Cl(aq); no water forms because ammonia has no oxygen.

Part 2 of 6Energy and Observations

Energy and Observations
Exothermic
  • Neutralisation releases heat; strong acid with strong base gives about −57 kJ mol⁻¹ of water formed.
Effervescence
  • Carbonates and hydrogencarbonates produce bubbles of carbon dioxide, which turns limewater cloudy.
Disappearing solid
  • Insoluble metal oxides dissolve as the acid reacts and a coloured salt solution forms.

Part 3 of 6Titration Calculation

1
Write the balanced equation to find the mole ratio between acid and base.
2
Find the moles of the standard solution using n = c × V, with volume converted to dm³.
3
Use the ratio to calculate the moles of the unknown.
4
Divide by the volume of the unknown to find its concentration in mol dm⁻³.
5
Choose a suitable indicator and use concordant titres (within 0.10 cm³) for the mean.

Part 4 of 6Practical Points and Applications

Practical Points and Applications
Indicator choice
  • Methyl orange or phenolphthalein can be used for a strong acid with a strong base titration.
Antacids
  • Indigestion tablets contain bases such as magnesium hydroxide or calcium carbonate that neutralise excess stomach acid.
Soil treatment
  • Farmers add lime, which is calcium oxide or hydroxide, to neutralise acidic soils.
Spillages
  • Acid spillages are cleared using sodium hydrogencarbonate, which effervesces as it neutralises the acid.

Part 5 of 6Worked Example

Question

  • A 25.00 cm³ sample of NaOH requires 20.50 cm³ of 0.100 mol dm⁻³ HCl for neutralisation.
  • Calculate the concentration of the NaOH.

  1. EquationHCl + NaOH → NaCl + H₂O, a 1:1 ratio.
  2. n(HCl) = 0.100 × 0.02050 = 2.05 × 10⁻³ mol, so n(NaOH) = 2.05 × 10⁻³ mol.
  3. c(NaOH) = 2.05 × 10⁻³ / 0.02500 = 0.0820 mol dm⁻³.

Answer [NaOH] = 0.0820 mol dm⁻³.

Part 6 of 6Exam Checklist & Tips

6.1.7 – Key Points for the Exam

  • Neutralisation A reaction between an acid and a base that produces a salt.
  • Salt An ionic compound formed when the H⁺ of an acid is replaced by a metal or ammonium ion.
  • Titration A technique for finding an unknown concentration by reacting measured volumes of solutions.
  • Concordant titres Results that agree within 0.10 cm³.

Common Mistakes

  • Forgetting that carbonates also give carbon dioxide and water.
  • Writing H₂O as a product of ammonia and acid; only a salt forms.
  • Not converting volume from cm³ to dm³ in titration calculations.
  • Using the molar ratio 1:1 when the equation shows otherwise.

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