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CIE 0620 Chemistry · IGCSE · Topic 12.2

Acid-base titrations

Clear, syllabus-mapped CIE 0620 Chemistry revision notes on acid-base titrations: explanations, worked examples and exam technique, then a free targeted practice drill.

CIE 0620 ChemistryIGCSEFree revision notes
Contents: 6 sections

Cambridge IGCSE Chemistry 0620 · Core and Extended

Syllabus points

Both objectives in 12.2 are Core, so everything above is required of every candidate. The calculation below is not a 12.2 objective at all: it sits in 3.3, "use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution", which is Supplement.

The apparatus and why each piece is there

ApparatusJob
Volumetric pipette with a fillerDelivers exactly 25.0 cm³ of one solution into the flask
BuretteAdds the other solution in any amount, read to 0.05 cm³
Conical flaskIts sloping sides let the mixture be swirled hard without splashing out
White tilePlaced under the flask so the first hint of colour change is visible
Small funnelUsed to fill the burette, then removed, so drips do not fall in afterwards

A beaker cannot replace the conical flask, because a beaker cannot be swirled without loss, and a measuring cylinder cannot replace the pipette, because it is not accurate enough.

The method

  1. Rinse the pipette with the solution it will hold, and the burette with the solution it will hold. Rinsing either with distilled water would leave water behind that dilutes the solution, and the titre would be wrong.
  2. Rinse the conical flask with distilled water only. Water left in the flask does not matter: it dilutes the solution but does not change the number of moles of alkali in it, and the number of moles is what the acid reacts with.
  3. Use the pipette to transfer 25.0 cm³ of the alkali into the conical flask. Let it drain under gravity and touch the tip on the liquid surface; do not blow out the last drop, since the pipette is calibrated to leave it behind.
  4. Add 2 or 3 drops of indicator. More than that and the indicator itself starts to react with the acid.
  5. Fill the burette above the zero mark, then run liquid through the tap until the jet below the tap is full and the air bubble has gone. A bubble that escapes during the titration adds its volume to the titre.
  6. Record the initial reading to the nearest 0.05 cm³, reading the bottom of the meniscus with your eye level with it.
  7. Run the acid in while swirling the flask constantly, standing it on the white tile.
  8. Near the end-point, add the acid dropwise, and rinse the inside walls of the flask with distilled water so that no acid is left clinging above the liquid.
  9. Stop at the first permanent colour change and record the final reading.
  10. Subtract the initial reading from the final reading to get the titre. An initial reading of 0.05 cm³ and a final reading of 22.45 cm³ give titre = 22.45 - 0.05 = 22.40 cm³.

Do a rough titration first to find roughly where the end-point lies, then repeat accurately. Continue until two titres are concordant, meaning they agree within 0.10 cm³, and average only those. Averaging the rough titration with the accurate ones is a standard way of losing a mark.

The end-point

The end-point is the point at which the indicator changes colour, showing that the acid has exactly neutralised the alkali. One further drop past it and the colour change is complete, which is why the last additions are made a drop at a time.

IndicatorColour in acidColour in alkali
Methyl orangeRedYellow
PhenolphthaleinColourlessPink
LitmusRedBlue
ThymolphthaleinColourlessBlue

Each of these changes over a narrow pH range, so the colour flips within a single drop and the end-point is sharp.

Universal indicator is not suitable. It passes through a whole sequence of colours over a wide pH range, so there is no single sharp change to stop at. Its job is measuring pH, not finding an end-point. That distinction is examined regularly.

The calculation (Extended)

This calculation is syllabus point 3.3, not 12.2, and it is Supplement content, so a Core candidate can stop at the practical technique above.

Worked example. 25.0 cm³ of aqueous sodium hydroxide of unknown concentration is exactly neutralised by 22.40 cm³ of hydrochloric acid of concentration 0.100 mol/dm³. Find the concentration of the sodium hydroxide.

Step 1. Moles of the known solution. Convert the volume to dm³ by dividing by 1000, then multiply by the concentration.

moles of HCl = 22.40 / 1000 × 0.100 = 0.00224 mol

Step 2. Use the equation for the ratio.

NaOH + HCl → NaCl + H₂O

One mole of acid reacts with one mole of alkali, so moles of NaOH = 0.00224 mol.

Step 3. Divide by the volume of the unknown, in dm³. The pipette delivered 25.0 cm³, which is 0.0250 dm³.

concentration = 0.00224 / 0.0250 = 0.0896 mol/dm³

Step 4, if grams are wanted. The relative formula mass of NaOH is

Mr = 23 + 16 + 1 = 40

0.0896 × 40 = 3.584 g/dm³

The ratio in step 2 is the step that separates a right answer from a wrong one. With sulfuric acid the equation is 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, so the moles of alkali are twice the moles of acid, and skipping that doubling halves the answer.

Common mistakes

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