Titration – IGCSE Chemistry Definition and Key Facts

Titration is a quantitative technique tested in the Cambridge 0620 syllabus under acids and bases and stoichiometry. Paper 4 regularly sets titration calculations worth 4-5 marks, and Paper 6 (Alternative to Practical) tests the procedural steps. This is a topic where procedural detail matters — vague descriptions lose marks.

The 0620 definition

A titration is a technique for determining the exact volume of one solution needed to react completely with a known volume of another solution. The endpoint is detected using an indicator.

Apparatus

Equipment	Purpose
Burette (0–50.0 cm³)	Delivers the solution dropwise; reads volume to 0.05 cm³
Pipette (25.0 cm³)	Measures a fixed, accurate volume of the other solution
Conical flask	Holds the pipetted solution and indicator
Indicator (methyl orange or phenolphthalein)	Shows the endpoint by changing colour
White tile	Placed under the flask to see the colour change clearly

Standard procedure

1. Rinse the pipette with the solution it will contain (e.g. alkali)
2. Pipette 25.0 cm³ of alkali into a clean conical flask
3. Add 2-3 drops of a suitable indicator
4. Rinse the burette with the acid solution
5. Fill the burette with acid, recording the initial reading
6. Add acid from the burette to the flask, swirling constantly
7. Near the expected endpoint, add acid dropwise until the indicator just changes colour permanently
8. Record the final burette reading
9. Calculate the titre (volume of acid used = final reading - initial reading)
10. Repeat until concordant results are obtained (titres within 0.10 cm³ of each other)

Choosing an indicator

Indicator	Colour in acid	Colour in alkali	Suitable for
Methyl orange	Red	Yellow	Strong acid + strong/weak alkali
Phenolphthalein	Colourless	Pink	Strong/weak acid + strong alkali
Litmus	Not recommended — colour change is not sharp
Universal indicator	Not recommended — gradual colour change across many colours

An indicator must change colour over a narrow pH range to give a sharp endpoint.

Titration calculations (Supplement)

The calculation follows a standard pattern:

1. Calculate moles of the known solution: n = c x V (where V is in dm³)
2. Use the balanced equation to find the mole ratio
3. Calculate moles of the unknown solution
4. Calculate the unknown concentration: c = n / V

Worked example

25.0 cm³ of NaOH of unknown concentration is neutralised by 20.0 cm³ of 0.10 mol/dm³ HCl.

NaOH + HCl → NaCl + H₂O (ratio 1:1)

Step 1: Moles of HCl = 0.10 x (20.0/1000) = 0.0020 mol

Step 2: Ratio is 1:1, so moles of NaOH = 0.0020 mol

Step 3: Concentration of NaOH = 0.0020 / (25.0/1000) = 0.080 mol/dm³

Using titration to prepare a pure soluble salt

Titration can determine the exact volumes needed for complete reaction. The experiment is then repeated without indicator, using those exact volumes. The resulting neutral solution is evaporated to crystallise the pure salt. This avoids indicator contamination and ensures no excess acid or alkali remains.

Worked exam question

A student titrates 25.0 cm³ of sodium hydroxide solution with 0.20 mol/dm³ sulfuric acid. The mean titre is 12.5 cm³. Calculate the concentration of the sodium hydroxide solution. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O [3]

n(H₂SO₄) = 0.20 x 12.5/1000 = 0.0025 mol [1]. Mole ratio 2:1, so n(NaOH) = 2 x 0.0025 = 0.005 mol [1]. c(NaOH) = 0.005 / (25.0/1000) = 0.20 mol/dm³ [1].

Common exam mistakes

1. Forgetting to convert cm³ to dm³ (divide by 1000) before using c = n/V.
2. Using the wrong mole ratio — always check the balanced equation. NaOH + HCl is 1:1 but 2NaOH + H₂SO₄ is 2:1.
3. Saying “universal indicator” is used in titrations — universal indicator gives a gradual colour change and is not suitable. Use methyl orange or phenolphthalein.
4. Recording only one titre. Concordant results (at least two titres within 0.10 cm³) are needed for reliability.