Titrations

In a titration, a pipette measures an exact volume of one solution into a conical flask and a burette adds the other until they just react, shown by a sudden colour change of an indicator: the end point. Repeat until the titres are concordant and take their mean. From the known solution’s moles and the ratio in the equation, you can find the unknown concentration.

Key facts

  • In a titration a pipette measures one solution (often 25.0 cm³) into a flask, and a burette adds the other until the indicator changes colour: the end point.
  • Repeat until the titres are concordant (within 0.10 cm³ of each other) and use their mean.
  • 1 dm³ = 1 litre = 1000 cm³: divide a volume in cm³ by 1000 to get dm³.
  • concentration (mol/dm³) = moles ÷ volume (dm³), so moles = concentration × volume.

Worked examples

Example 1

Which piece of apparatus measures exactly 25.0 cm³ of alkali into the flask for a titration?

  1. A pipette measures one fixed volume very accurately.
  2. A pipette filler is used to draw the liquid up safely.
  3. The burette is for adding the other solution.

Answer: a pipette (with a pipette filler)

Example 2

Which indicator would you use for a titration of a strong acid with a strong alkali?

  1. Universal indicator changes gradually through many colours.
  2. A titration needs one sudden colour change at the end point.
  3. So use phenolphthalein or methyl orange.

Answer: phenolphthalein or methyl orange (a sharp single colour change)

Example 3

In a titration, why is the conical flask swirled while the acid is added?

  1. The acid must mix fully with the alkali.
  2. Otherwise the colour may change in one place too early.
  3. Swirling makes the end point accurate.

Answer: to mix the solutions so the end point is seen accurately

Example 4

Why are titrations repeated until concordant results are obtained?

  1. A single titre could contain a mistake.
  2. Concordant results (within 0.10 cm³) show the results are repeatable.
  3. Their mean is a reliable value.

Answer: to check the results are repeatable and take a reliable mean

Example 5

Titres (cm³): rough 27.10, then 25.95, 26.00, 26.40, 26.05. Concordant results are within 0.10 cm³ of each other. What is the mean titre, to 2 decimal places?

  1. Ignore the rough titre.
  2. 26.40 is not within 0.10 cm³ of the others, so leave it out.
  3. Mean = (25.95 + 26.00 + 26.05) ÷ 3 = 26, which is 26.00 cm³ to 2 decimal places.

Answer: 26.00 cm³

Example 6

25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is neutralised by 20.0 cm³ of hydrochloric acid. NaOH + HCl → NaCl + H2O. Find the concentration of the acid.

  1. Moles of NaOH in the flask. 0.100 × 25.0 ÷ 1000 = 0.00250 mol
  2. The ratio NaOH : HCl is 1 : 1, so the acid contains 0.00250 mol of HCl.
  3. Divide by the acid’s volume in dm³. 0.00250 ÷ 0.0200 = 0.125 mol/dm³

Answer: 0.125 mol/dm³

Example 7

25.0 cm³ of 0.200 mol/dm³ potassium hydroxide is neutralised by 12.5 cm³ of sulfuric acid. 2KOH + H2SO4 → K2SO4 + 2H2O. Find the concentration of the sulfuric acid.

  1. Moles of KOH. 0.200 × 25.0 ÷ 1000 = 0.00500 mol
  2. 2 mol of KOH react with 1 mol of H2SO4, so halve it. 0.00500 ÷ 2 = 0.00250 mol
  3. Divide by the volume in dm³. 0.00250 ÷ 0.0125 = 0.200 mol/dm³

Answer: 0.200 mol/dm³

Example 8

25 cm³ of 0.1 mol/dm³ chem KOH is exactly neutralised by 23.1 cm³ of chem H2SO4. 2KOH + H2SO4 → K2SO4 + 2H2O. What is the concentration of the chem H2SO4, in mol/dm³, to 3 significant figures?

  1. Moles of chem KOH: 0.1 × 25 ÷ 1,000 = 0.0025 mol
  2. The ratio is 2 chem KOH : 1 chem H2SO4, so moles of chem H2SO4 = 0.0025 ÷ 2 = 0.00125 mol.
  3. Concentration = 0.00125 ÷ 0.0231 = 0.0541 mol/dm³

Answer: 0.0541 mol/dm³