IGCSE · Practice questions · Acid-base titrations

Find the end-point.

Six original Cambridge-style questions on carrying out and processing an acid-base titration.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Pipette, indicator, burette, repeat.

Pipette a fixed volume of alkali; burette adds the acid.
Methyl orange: red in acid, yellow in alkali. Thymolphthalein: colourless in acid, blue in alkali.
Titre = final − initial reading; concordant = within 0.10 cm³.
Extended: moles = concentration × volume in dm³, then use the mole ratio.

Recall

[4 marks]

Describe how to carry out a titration to find the volume of dilute hydrochloric acid that neutralises 25.0 cm³ of sodium hydroxide solution.

Use a volumetric pipette to put 25.0 cm³ of sodium hydroxide into a conical flask and add a few drops of methyl orange (or thymolphthalein). ✓
Fill a burette with the acid and record the initial reading. ✓
Add acid while swirling, then drop by drop near the end-point, until the indicator just changes colour; record the final reading. ✓
Repeat until concordant titres are obtained and find their mean. ✓

Recall

[2 marks]

State the colour of methyl orange (a) in the sodium hydroxide at the start and (b) at the end-point when the acid is in slight excess.

(a) yellow. ✓
(b) red (orange at the exact end-point). ✓

Calculation

[2 marks]

The burette reading is 0.50 cm³ before a titration and 24.70 cm³ after. Calculate the titre.

Titre = final reading − initial reading ✓
= 24.70 − 0.50 = 24.20 cm³ ✓

Analysis

[2 marks]

A student obtains titres of 23.90 (rough), 23.80, 23.85 and 23.75 cm³. Identify the concordant results and calculate the mean titre.

Concordant titres: 23.80, 23.85 and 23.75 cm³ (all within 0.10 cm³); the rough titre is ignored. ✓
Mean = (23.80 + 23.85 + 23.75) ÷ 3 = 23.80 cm³ ✓

Calculation

[3 marks]

(Extended) 25.0 cm³ of sodium hydroxide solution is neutralised by 22.40 cm³ of 0.150 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide. Give your answer to 3 significant figures.

Moles HCl = 0.150 × 22.40 ÷ 1000 = 0.00336 mol ✓
NaOH + HCl → NaCl + H₂O, so moles NaOH = 0.00336 mol ✓
Concentration = 0.00336 ÷ 0.0250 = 0.134 mol/dm³ ✓

Calculation

[3 marks]

(Extended) 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide is neutralised by 20.0 cm³ of dilute sulfuric acid. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Calculate the concentration of the sulfuric acid.

Moles NaOH = 0.200 × 25.0 ÷ 1000 = 0.00500 mol ✓
Moles H₂SO₄ = 0.00500 ÷ 2 = 0.00250 mol ✓
Concentration = 0.00250 ÷ 0.0200 = 0.125 mol/dm³ ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.