Question 1Warm-upHigher
Titres of 24.10, 24.20 and 24.15 cm³ were recorded. Calculate the mean titre.
Add them and divide by 3.
Worked solution
- cm³
Answer: cm³
GCSE & IGCSE chemistry practice
Titration calculations questions and answers
In a titration you add one solution from a burette until it exactly reacts with a measured volume of another. From the volumes and the known concentration you can work out the unknown concentration. The three steps: moles of the known solution, use the equation's ratio, then concentration = moles ÷ volume.
15 questions · Higher tier · Every answer is checked as you type it, with a full worked solution. Updated .
Where titration calculations comes up
| GCSE (AQA) | GCSE Chemistry only, Higher only: calculate an unknown concentration (or volume) from titration results. Titration is a required practical. |
|---|---|
| Cambridge IGCSE (0620) | Supplement (Extended): use titration data to calculate the moles of solute, or the concentration or volume of a solution. |
The method
- Mean titre: average the concordant results (within 0.10 cm³ of each other).
- Moles of the known solution = concentration × volume (dm³).
- Use the balanced equation's ratio to get the moles of the unknown.
- Unknown concentration = moles ÷ its volume (dm³).
Titration calculations questions
Try each one before looking at the hint. If you're stuck, the AI tutor can talk you through it step by step.
Warm-up
Question 2Warm-upHigher
Titres of 25.60 (rough), 24.30, 24.40 and 24.35 cm³ were recorded. Using concordant results only, what is the mean titre?
Leave out the rough titre.
Worked solution
- cm³
Answer: cm³
Question 9Warm-upHigher
Titres of 22.50, 22.60 and 22.55 cm³ were recorded. Calculate the mean titre.
Add and divide by 3.
Worked solution
- cm³
Answer: cm³
Question 10Warm-upHigher
How many moles of hydrochloric acid are in 18.0 cm³ of a 0.25 mol/dm³ solution?
Moles = concentration × volume in dm³.
Worked solution
- cm³ dm³
- mol
Answer: mol
Exam-style
Question 3Exam-styleHigher
25.0 cm³ of sodium hydroxide is neutralised by 20.0 cm³ of 0.10 mol/dm³ hydrochloric acid. . Calculate the concentration of the sodium hydroxide.
1 : 1 ratio.
Worked solution
- Moles HCl
- Moles NaOH
- Concentration mol/dm³
Answer: mol/dm³
Question 4Exam-styleHigher
25.0 cm³ of 0.20 mol/dm³ sodium hydroxide is neutralised by 12.5 cm³ of hydrochloric acid. Calculate the concentration of the acid.
Moles of NaOH first; ratio 1 : 1.
Worked solution
- Moles NaOH
- Concentration HCl mol/dm³
Answer: mol/dm³
Question 5Exam-styleHigher
. 20.0 cm³ of 0.10 mol/dm³ sulfuric acid neutralises 25.0 cm³ of sodium hydroxide. Calculate the concentration of the sodium hydroxide.
1 mol of acid reacts with 2 mol of NaOH.
Worked solution
- Moles
- Moles NaOH
- Concentration mol/dm³
Answer: mol/dm³
Question 6Exam-styleHigher
What volume of 0.50 mol/dm³ hydrochloric acid neutralises 25.0 cm³ of 0.40 mol/dm³ sodium hydroxide? Give your answer in cm³.
Volume = moles ÷ concentration.
Worked solution
- Moles NaOH
- Volume HCl dm³ cm³
Answer: cm³
Question 11Exam-styleHigher
25.0 cm³ of sodium hydroxide is neutralised by 15.0 cm³ of 0.20 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide. ()
Moles of acid = moles of alkali (1 : 1).
Worked solution
- Moles HCl mol
- Moles NaOH mol
- mol/dm³
Answer: mol/dm³
Question 12Exam-styleHigher
20.0 cm³ of 0.15 mol/dm³ sodium hydroxide is neutralised by 24.0 cm³ of hydrochloric acid. Calculate the concentration of the acid.
Find the moles of NaOH first (1 : 1 ratio).
Worked solution
- Moles NaOH mol
- mol/dm³
Answer: mol/dm³
Question 13Exam-styleHigher
. 25.0 cm³ of 0.10 mol/dm³ sodium hydroxide is neutralised by 12.5 cm³ of sulfuric acid. Calculate the concentration of the acid.
The ratio is 1 acid : 2 alkali.
Worked solution
- Moles NaOH mol
- Moles mol
- mol/dm³
Answer: mol/dm³
Challenge
Question 7ChallengeHigher
In question 3, the sodium hydroxide's concentration was 0.08 mol/dm³. What is that in g/dm³? (NaOH: = 40)
Multiply by Mr.
Worked solution
- g/dm³
Answer: g/dm³
Question 8ChallengeHigher
Why is a few drops of indicator used rather than a large amount, and why is the burette read at the bottom of the meniscus?
Think about accuracy.
Worked solution
- A large amount of indicator could react and change the titre.
- Reading at the bottom of the meniscus, at eye level, avoids parallax errors.
Answer: Indicators are themselves weak acids or bases, so too much affects the titre; reading at the bottom of the meniscus (at eye level) gives an accurate reading
Question 14ChallengeHigher
Sulfuric acid has a concentration of 0.25 mol/dm³. What is this in g/dm³? ( of = 98)
g/dm³ = mol/dm³ × Mr.
Worked solution
- g/dm³
Answer: g/dm³
Question 15ChallengeHigher
25.0 cm³ of potassium hydroxide is neutralised by 20.0 cm³ of 0.125 mol/dm³ nitric acid (). Calculate the concentration of the potassium hydroxide in g/dm³. ( of KOH = 56)
Find mol/dm³ first, then × Mr.
Worked solution
- Moles mol = moles KOH
- mol/dm³
- g/dm³
Answer: g/dm³
Common mistakes
- Using volumes in cm³ in the moles calculation.
- Ignoring a 1 : 2 ratio (e.g. H₂SO₄ with NaOH).
- Including a rough or anomalous titre in the mean.
Stuck on one of these?
The TutorAura AI tutor works through it with you step by step, with the reason for each step, then gives you one like it to try. It checks your working too.
Try the AI tutor freeMore chemistry questions
Sources
The questions are our own. The “where it comes up” notes are from each board's published specification, checked on 27 October 2026:
Practise chemistry with your own AI tutor
Questions at your level, help when you're stuck, and a plan for your exams. Free for 3 days, no card needed.