6.1.7Reactivity 1 — What drives chemical reactions?
Neutralisation Reactions — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 9 of 12 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 5Typical Reactions
Neutralisation is the reaction of an acid with a base to give a salt.
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), a salt and water form.
Acid + carbonate
- Acid + carbonate CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g); effervescence shows CO₂ being released.
Acid + metal
- Acid + metal Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g); this is a redox reaction, not a neutralisation.
Section 2 of 5Energy and Evidence
Energy and Evidence
Exothermic
- Exothermic Neutralisation releases heat, so the temperature of the mixture rises.
Enthalpy
- Enthalpy For a strong acid with a strong base the enthalpy change is about −57 kJ mol⁻¹ of water formed.
Effervescence
- Effervescence Bubbles of carbon dioxide indicate a reaction with a carbonate or a hydrogencarbonate.
Indicator
- Indicator A colour change in an indicator marks the point at which an acid and a base are in equal amounts.
| Reactants | Products |
|---|---|
| Acid + base | Salt + water |
| Acid + metal oxide | Salt + water |
| Acid + carbonate | Salt + water + carbon dioxide |
| Acid + reactive metal | Salt + hydrogen |
Section 3 of 5Titration Method
1
Measure an exact volume of one solution with a pipette into a conical flask and add a few drops of indicator.
2
Fill a burette with the other solution and record the initial reading.
3
Add the solution slowly while swirling, until the indicator changes colour at the end point.
4
Record the final reading, repeat until two results agree within 0.10 cm³, and calculate the mean titre.
5
Use the balanced equation to find the moles of the unknown, and then divide by its volume to find the concentration.
Section 4 of 5Worked Example
Question
- 25.00 cm³ of NaOH solution is neutralised by 21.40 cm³ of 0.100 mol dm⁻³ HCl.
- Calculate the concentration of the NaOH.
- n(HCl) = 0.100 × 21.40 / 1000 = 2.140 × 10⁻³ mol.
- HClNaOH is 1 : 1, so n(NaOH) = 2.140 × 10⁻³ mol.
- c(NaOH) = 2.140 × 10⁻³ / 0.02500 = 0.0856 mol dm⁻³.
Answer c(NaOH) = 0.0856 mol dm⁻³
Section 5 of 5Exam Checklist & Tips
6.1.7 – Key Points for the Exam
- Neutralisation The reaction of an acid with a base to produce a salt and water.
- Salt An ionic compound formed when the H⁺ of an acid is replaced by a metal or ammonium ion.
- Titration A technique to find the concentration of a solution by reacting it with a standard solution.
- End point The point at which the indicator changes colour.
Common Mistakes
- Writing hydrogen as a product of an acid with a carbonate.
- Using the volume of the burette solution in the wrong place in n = cV.
- Forgetting to convert cm³ to dm³ when calculating moles.
- Averaging the rough trial titre with the concordant results, which gives a wrong mean.
Last updated:
\n