Chemical changes · GCSE Chemistry
Titration
GCSE Chemistry revision on titration: burette, pipette, indicator, concordant titres within 0.10 cm³, the mean, and calculating concentration from a worked titre.
Pipette the alkali, burette the acid, indicator for the end-point. Rough titre first, then concordant titres within 0.10 cm³, then mean of those concordant results — never include the rough.
The important bits
What you need to know
- 1
A titration finds the exact volume of one solution that reacts completely with a known volume of another. Typical reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).
- 2
Rinse the pipette with the solution it will transfer (often 25.0 cm³ of alkali), rinse the burette with the acid, and rinse the conical flask with distilled water.
- 3
Add a few drops of indicator: phenolphthalein (colourless in acid, pink in alkali) for strong acid–strong alkali; methyl orange (yellow in alkali, red in acid) is an alternative.
- 4
Perform a rough titration to the end-point, then repeat dropwise near the colour change. Record burette start and end to 0.05 cm³. Titre = end − start.
- 5
Concordant titres agree within 0.10 cm³. Calculate the mean of the concordant titres only. Do not include the rough titre or any result that is not concordant.
- 6
Moles of the known solution = concentration × volume in dm³ (divide cm³ by 1000). Use the balanced equation ratio, then concentration of the unknown = moles / its volume in dm³.
- 7
White tile under the flask makes the colour change clearer. Swirl throughout. Stop at the first permanent colour change (pale pink with phenolphthalein).
- 8
To make a soluble salt, repeat the mean volumes without indicator so crystals are not contaminated, then evaporate and crystallise.
Quotations worth analysing
Short evidence. Real method.
“HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)”
Mole ratio 1:1, so moles of acid equal moles of alkali at the end-point. Sulfuric acid is 1:2 with NaOH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
“Concordant titres agree within 0.10 cm³. Mean = sum of concordant titres / number of concordant titres.”
If titres are 24.40, 24.50 and 25.20 cm³, mean the first two only: 24.45 cm³. 25.20 is not concordant. The rough titre is never averaged.
“concentration (mol/dm³) = moles / volume (dm³)”
Volume must be in dm³. 25.0 cm³ is 0.0250 dm³. Forgetting the /1000 is the most common numerical error in the whole practical.
Go deeper
The method marks are in the glassware and the indicator
A pipette measures the fixed volume more precisely than a measuring cylinder — that is why 25.0 cm³ is written to one decimal place. The burette measures the variable volume; read the bottom of the meniscus at eye level. Indicator choice: phenolphthalein is colourless to pink, easy to judge as a single drop of extra alkali (or the last drop of acid that just removes the pink, depending which way you titrate). Universal indicator is wrong: the colour slides through green and you cannot define an end-point. Two or three drops is enough. A white tile is not decoration; it is how you see pale pink against a lab bench.
Go deeper
Concordant, then mean, then moles — in that order
Examiners award a mark for selecting concordant titres and another for the mean. Show the subtraction for each titre. Then convert the mean to dm³. Worked pattern: moles of HCl = 0.100 × (24.45/1000) = 0.002445 mol. Ratio 1:1, so moles of NaOH = 0.002445. Volume of NaOH was 25.0 cm³ = 0.0250 dm³, so concentration = 0.002445 / 0.0250 = 0.0978 mol/dm³. Three significant figures if the data support it. If the acid is H₂SO₄, double the alkali moles or halve, matching the 1:2 ratio — check which reagent you pipetted.
Go deeper
Uncertainty and why we repeat
One titration could be a splash or a missed end-point. Concordant repeats show the end-point is reproducible. Burette uncertainty is often ±0.05 cm³ on each reading, so ±0.10 cm³ on a titre; that is why 0.10 cm³ is the concordant window. A 24.00 cm³ titre then has a percentage uncertainty of (0.10/24.00) × 100 ≈ 0.4%. Smaller titres have larger percentage uncertainty, which is why we do not chase tiny volumes. Rinsing the flask with alkali would add extra moles and make the titre too large; rinse with water and the extra water does not change the moles of alkali, only the volume of the mixture.
See the idea in action
25.0 cm³ of NaOH is titrated with 0.100 mol/dm³ HCl. Phenolphthalein is pink in the alkali and becomes colourless at the end-point. Rough titre: 25.80 cm³. Accurate titres: 24.40 cm³ and 24.50 cm³ (concordant within 0.10 cm³). Mean titre = (24.40 + 24.50) / 2 = 24.45 cm³. Equation HCl + NaOH → NaCl + H₂O, ratio 1:1. Moles HCl = 0.100 × (24.45/1000) = 0.002445 mol. Concentration of NaOH = 0.002445 / 0.0250 = 0.0978 mol/dm³.
Exam technique
Turn knowledge into marks
Write indicator, rough titre, concordant titres within 0.10 cm³, mean of concordant only, then moles of the known, ratio, concentration of the unknown. Convert cm³ to dm³. State the colour change.
Common mistakes
Do not give these marks away
- 01
Averaging the rough titre, or including a titre that is more than 0.10 cm³ from the others.
- 02
Using universal indicator, or forgetting to divide volume by 1000.
- 03
Using the wrong mole ratio for sulfuric acid (1:2 with NaOH).
Titres of 24.30, 24.40 and 25.10 cm³ are recorded after a rough titre of 26 cm³. What mean should be used?
A24.90 cm³
B24.35 cm³
C25.10 cm³
D26.00 cm³
Show the answer
24.35 cm³. Concordant titres agree within 0.10 cm³: 24.30 and 24.40. Mean = 24.35 cm³. Discard 25.10 and the rough titre.
Quick questions
If this is the bit you searched
What are concordant titres GCSE Chemistry?
Repeat titres that agree within 0.10 cm³. You calculate the mean from those concordant results only, and you never include the rough titre.
Which indicator is used in a strong acid–strong alkali titration?
Phenolphthalein (colourless in acid, pink in alkali) or methyl orange (red in acid, yellow in alkali). Do not use universal indicator for the end-point.
How do you calculate concentration from a titration?
Find the mean concordant titre, convert cm³ to dm³, calculate moles of the known solution, use the equation ratio, then divide moles of the unknown by its volume in dm³.
Why is a pipette used instead of a measuring cylinder?
A pipette transfers a fixed volume such as 25.0 cm³ more precisely, which reduces uncertainty in the known moles of that solution.