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Titration lab answers for GCSE chemistry (free PDF download)

The method and the results are on the titration lab. Do the analysis yourself first, then check the answers below.

Quick answer: A titration finds an unknown concentration by measuring the exact volume of one solution that reacts with a known volume of another. These answers go with the Chemistry 9-10 lab on titration: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.

Titration lab answers for GCSE chemistry (free PDF download)

The answers use the sample results. Your own titres will differ slightly, but your method for choosing concordant titres and for calculating the concentration should match.

Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4 RP

Question 1 Select the concordant titres

Runs 1, 2 and 3 (20.10, 20.05 and 20.00 cm³) are within 0.10 cm³ of each other. The rough titre is excluded.

Question 2 Mean titre

(20.10 + 20.05 + 20.00) ÷ 3 = 20.05 cm³.

Question 3 Moles of acid

n = 0.100 × 0.02005 = 0.002005 mol.

Question 4 Concentration of NaOH

The ratio is 1 : 1, so n(NaOH) = 0.002005 mol and c = 0.002005 ÷ 0.0250 = 0.0802 mol/dm³.

Question 5 Concentration in g/dm³

M(NaOH) = 40 g/mol; 0.0802 × 40 = 3.21 g/dm³.

Question 6 Uncertainty

A titre uses two readings, so the uncertainty is ±0.10 cm³. 0.10 ÷ 20.05 × 100 = 0.50%.

Question 7 Why swirl and go slowly

One drop of acid is about 0.05 cm³. Slow addition lets you stop when one drop causes the permanent colour change, giving a precise titre.

Question 8 Improving the lab

Use a white tile and a wash bottle to rinse the walls of the flask, and repeat the titration until three titres agree within 0.10 cm³.

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