✓ Revise Right course
Find moles of the known solution with moles = concentration × volume, use the equation's mole ratio, then divide by the unknown's volume. Free notes.
Titration Calculations is part of Reactions of acids in GCSE Edexcel Higher Chemistry.
3 min
Reading time
22
Lessons in this topic
GCSE Edexcel Higher
Pathway
Reviewed by Revise Right
Content review
Find moles of the known solution with moles = concentration × volume, use the equation's mole ratio, then divide by the unknown's volume. Free notes.
Edexcel GCSE Chemistry mock draft, 26–31 Oct. Draft coverage: Chemistry Paper 1 (Higher tier). Reviewed for an original-practice sitting; check student availability before registering. Marking and results will follow later, after separate approval. Eligible new accounts receive a five-day Pro trial when signup email verification completes. Previous trials keep their original expiry. After expiry, new papers need Pro unless you have an explicitly confirmed free sitting offer for this event. That offer adds no Pro time or AI learning. Saved attempts and released results remain readable. No card or automatic charge.
Use this lesson to stay inside Reactions of acids while connecting the detail back to the wider Topic 3: Chemical changes topic.
Read the core explanations from Titration Calculations before testing yourself from memory.
You've done titrations in the lab - carefully adding acid to alkali until the indicator changes colour. But here's the real power: those volume readings can tell you the exact concentration of unknown solutions. This is how chemists analyse everything from soil samples to medicines.
The key insight is that at the end point, the moles of acid and alkali react in the ratio shown by the balanced equation, for example 1:1 for HCl and NaOH. Once you know the moles, you can calculate concentration in two different ways - and examiners love to test both.
All titration calculations use the same core relationship between moles, concentration and volume. Master this triangle and you'll crack any titration question.
Every titration calculation follows the same logical sequence. Let's work through a complete example to see how the pieces fit together.
Worked example: Finding Unknown Concentration (mol/dm³)
25.0 cm³ of sodium hydroxide solution is neutralised by 23.2 cm³ of 0.150 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide in mol/dm³. [4 marks]
Answer: 0.139 mol/dm³
Notice how we used the balanced equation to find the mole ratio. This is always step 1 - without it, you can't connect the moles of one substance to the other.
Examiners often ask you to express concentration in g/dm³ instead of mol/dm³. The trick is to use relative atomic masses to convert between moles and grams.
Worked example: Converting mol/dm³ to g/dm³
A solution of sodium chloride has a concentration of 0.200 mol/dm³. Calculate its concentration in g/dm³. (Relative atomic masses: Na = 23, Cl = 35.5) [3 marks]
Answer: 11.7 g/dm³
Higher-tier questions often involve acids or bases that don't have a 1:1 ratio. The method is identical, but you must use the balanced equation carefully to find the correct mole ratio.
Worked example: Diprotic Acid Titration
20.0 cm³ of sulfuric acid is neutralised by 35.6 cm³ of 0.250 mol/dm³ potassium hydroxide. Calculate the concentration of sulfuric acid in mol/dm³. [5 marks]
Answer: 0.223 mol/dm³
Titration calculations are mark-heavy questions that follow predictable patterns. Here's how to maximise your marks and avoid the common traps that catch students out.
Common Exam Mistakes: Students often forget to convert volumes or use the wrong mole ratio. The equation H₂SO₄ + 2NaOH shows that 1 mole of acid neutralises 2 moles of base - not 1:1.
Mark Scheme Language: Use phrases like "from the balanced equation" and "using n = cv" to show your method clearly. Examiners give method marks even if your final answer is wrong.
Exam precision note: In Titration Calculations - Titrations (chemistry only) - Reactions of acids, always identify the key chemical idea first, then link it to evidence, data, or an equation. A high-scoring GCSE answer names the particles involved, states what changes, and justifies the conclusion using correct scientific language. Where calculations are used, show each step with units and keep significant figures sensible. Finally, check that terminology matches AQA wording so your answer is unambiguous and mark-scheme friendly.
Titration calculations are most secure when students separate the stages of the method. First, identify the mean titre from concordant results. Second, use concentration and volume to calculate the number of moles in the known solution. Third, apply the balanced equation to find the number of moles in the unknown solution, which may or may not be a 1:1 ratio. Finally, use the flask volume to calculate the unknown concentration if required. This staged approach prevents the common mistake of assuming equal moles in every titration.
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-22.
Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
Converting cm³ to dm³ works like this: 23.2 cm³ = 0.0232 dm³ and 35.6 cm³ = 0.0356 dm³. Students often forget to convert volumes in titration calculations.
The mean titre is the mean of the concordant titres only. It is the first stage of a titration calculation, before working out moles of the known solution.
The relative formula mass (Mr) of sodium chloride, NaCl, is 23 + 35.5 = 58.5. A 0.200 mol/dm³ solution therefore has a concentration of 0.200 × 58.5 = 11.7 g/dm³.
Calculate moles of the known solution with moles = concentration × volume, use the balanced equation to find moles of the unknown, then divide by its volume. For 23.2 cm³ of 0.150 mol/dm³ HCl neutralising 25.0 cm³ of NaOH, moles = 0.00348, so NaOH = 0.00348 ÷ 0.0250 = 0.139 mol/dm³.
To convert mol/dm³ to g/dm³, use mass = moles × Mr for 1 dm³ of solution. For 0.200 mol/dm³ sodium chloride, Mr = 23 + 35.5 = 58.5, so the concentration is 0.200 × 58.5 = 11.7 g/dm³.
Students often forget to convert volumes from cm³ to dm³ or use the wrong mole ratio, or assume equal moles in every titration. For example, H₂SO₄ + 2NaOH shows that 1 mole of acid neutralises 2 moles of base, not 1:1.
First find the mean titre from concordant results. Second, use concentration and volume to calculate moles in the known solution. Third, use the balanced equation to find moles in the unknown solution, which may not be a 1:1 ratio. Finally, use the flask volume to calculate the unknown concentration.
Acids and alkalis do not always react in a 1:1 ratio. HCl and NaOH react 1:1, but H₂SO₄ + 2NaOH shows 1 mole of acid neutralising 2 moles of base, so assuming equal moles is a common mistake.