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2.6.3Inorganic Chemistry

Acid-Alkali Titrations — Edexcel International GCSE Science Double Award 4SD0 - Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 30 of 52

📚 Edexcel International GCSE Science Double Award 4SD0 - Chemistry Hub / Topical Revision
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Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
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Part 1 of 5Carrying Out a Titration

2.6.3 Acid-Alkali Titrations

A titration measures the exact volume of one solution needed to react completely with a measured volume of another.

1
Use a pipette and pipette filler to place exactly 25.0 cm³ of the alkali into a clean conical flask.
2
Fill a burette with the acid, run some through the tap to remove air bubbles, and record the starting volume to the nearest 0.05 cm³.
3
Add a few drops of a suitable indicator to the flask and place it on a white tile so that the colour change is easy to see.
4
Run in the acid while swirling, first in a rough titration, then slowly near the end-point, and record the final volume once the colour just changes.

Part 2 of 5Getting Reliable Results

Accuracy comes from repeating the titration until two results agree closely.

Rough titration
  • A quick first run shows roughly where the end-point is, and it is not included in the mean.
Concordant results
  • Two titres within 0.10 cm³ of each other are described as concordant and are used for the mean.
Mean titre
  • The mean is calculated from the concordant titres only, giving a value to two decimal places.
Reading the burette
  • The meniscus is read at eye level to avoid parallax error in each volume reading.
RunVolume of acid (cm³)
Rough15.50
Titre 114.90
Titre 215.00
Mean (concordant)14.95

Part 3 of 5Analysing the Data

Analysing the Data
Moles of alkali
  • Moles = concentration × volume in dm³, so 0.100 mol/dm³ × 0.0250 dm³ gives 0.00250 mol.
Mole ratio
  • The balanced equation shows how many moles of acid react with each mole of alkali, such as 1:1 for HCl and NaOH.
Unknown concentration
  • Divide the moles of acid by the volume of acid in dm³ to find its concentration.
Units
  • Always convert cm³ to dm³ by dividing by 1000 before calculating concentrations in mol/dm³.

Part 4 of 5Worked Example

Question

  • 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is neutralised by 20.0 cm³ of hydrochloric acid.
  • Calculate the concentration of the acid.
  • (NaOH + HCl → NaCl + H₂O)

  1. Moles of NaOH = 0.100 mol/dm³ × 0.0250 dm³ = 0.00250 mol.
  2. The mole ratio NaOHHCl is 1 : 1, so moles of HCl = 0.00250 mol.
  3. Concentration of HCl = moles ÷ volume = 0.00250 mol ÷ 0.0200 dm³ = 0.125 mol/dm³.

Answer 0.125 mol/dm³ hydrochloric acid.

Part 5 of 5Exam Checklist & Tips

2.6.3 – Key Points for the Exam

  • Titration A method of finding the exact volume of one solution that reacts with a known volume of another.
  • End-point The point at which the indicator changes colour, showing neutralisation is complete.
  • Titre The volume of solution delivered from the burette in one titration.
  • Concordant Describes two titres that agree to within 0.10 cm³ of each other.

Common Mistakes

  • Including the rough titre in the mean; only concordant results should be used.
  • Failing to convert cm³ to dm³ before calculating moles and concentration.
  • Leaving an air bubble in the jet of the burette, which gives a false final reading.
  • Leaving a funnel in the burette during the titration, which can drip liquid into it and change the start reading.

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