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
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 6Acid 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
- Acid + alkali HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l); the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).
Acid + metal oxide
- Acid + metal oxide CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); the oxide ion accepts protons.
Acid + carbonate
- Acid + carbonate Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g), with effervescence.
Acid + hydrogencarbonate
- Acid + hydrogencarbonate NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g).
Acid + ammonia
- Acid + ammonia NH₃(aq) + HCl(aq) → NH₄Cl(aq); no water forms because ammonia has no oxygen.
Section 2 of 6Energy and Observations
Energy and Observations
Exothermic
- Exothermic Neutralisation releases heat;
- strong acid with strong base gives about −57 kJ mol⁻¹ of water formed.
Effervescence
- Effervescence Carbonates and hydrogencarbonates produce bubbles of carbon dioxide, which turns limewater cloudy.
Disappearing solid
- Disappearing solid Insoluble metal oxides dissolve as the acid reacts and a coloured salt solution forms.
Section 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.
Section 4 of 6Practical Points and Applications
Practical Points and Applications
Indicator choice
- Indicator choice Methyl orange or phenolphthalein can be used for a strong acid with a strong base titration.
Antacids
- Antacids Indigestion tablets contain bases such as magnesium hydroxide or calcium carbonate that neutralise excess stomach acid.
Soil treatment
- Soil treatment Farmers add lime, which is calcium oxide or hydroxide, to neutralise acidic soils.
Spillages
- Spillages Acid spillages are cleared using sodium hydrogencarbonate, which effervesces as it neutralises the acid.
Section 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.
- EquationHCl + NaOH → NaCl + H₂O, a 1:1 ratio.
- n(HCl) = 0.100 × 0.02050 = 2.05 × 10⁻³ mol, so n(NaOH) = 2.05 × 10⁻³ mol.
- c(NaOH) = 2.05 × 10⁻³ / 0.02500 = 0.0820 mol dm⁻³.
Answer [NaOH] = 0.0820 mol dm⁻³.
Section 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.
Last updated:
\n