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3.2.5Stoichiometry

Titration Calculations — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 6 parts Lesson 27 of 28 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Part 1 of 6From Burette Readings to a Reliable Titre

3.2.5 Titration Calculations

A titration measures the volume of one solution that exactly reacts with a fixed volume of another, so the raw readings must be processed before any calculation.

Titre
  • The volume of solution delivered from the burette at the end point, found by subtracting the start reading from the final reading.
Rough run
  • A first quick titration that shows roughly where the end point lies, so later runs can add the solution drop by drop near it.
Concordant results
  • Repeat titres that agree within about 0.10 cm³;
    • only these are used because they show the technique was consistent.
Mean titre
  • The average of the concordant titres only; any rough or outlying result is left out of the mean.
End point
  • The point where the indicator changes colour permanently, showing that the acid and alkali have reacted in exactly the amounts given by the equation.

Part 2 of 6The Calculation Method

Every titration calculation follows the same logical route, whichever quantity is unknown.

1
Write the balanced symbol equation, because the mole ratio between acid and alkali comes from its coefficients.
2
Convert the volume of the solution whose concentration is known into dm³ by dividing the cm³ value by 1000.
3
Find the moles of that known solution using moles = concentration (mol/dm³) × volume (dm³).
4
Use the mole ratio in the equation to convert to moles of the unknown solution, multiplying or dividing as the coefficients require.
5
Divide the moles of the unknown by its volume in dm³ to give its concentration, then multiply by the molar mass if g/dm³ is wanted.

Part 3 of 6Using the Mole Ratio

The ratio step is where most marks are lost, so check it against the equation every time.

Using the Mole Ratio
1 : 1 reactions
  • HCl + NaOH → NaCl + H₂O, so moles of acid equal moles of alkali and no scaling is needed.
1 : 2 reactions
  • H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so the moles of NaOH are double the moles of H₂SO₄ that reacted.
2 : 1 reactions
  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O, so the moles of HCl are double the moles of barium hydroxide.
Rearranging
  • Concentration = moles ÷ volume and volume = moles ÷ concentration, so the same relationship finds any one missing quantity.
ReactionMole ratio acid : alkali
HCl + NaOH1 : 1
H₂SO₄ + 2NaOH1 : 2
2HCl + Ba(OH)₂2 : 1

Part 4 of 6Units and Presentation

Volume units
  • Divide cm³ by 1000 to get dm³ before using concentration in mol/dm³;
    • forgetting this gives answers a thousand times wrong.
Concentration in g/dm³
  • Multiply the concentration in mol/dm³ by the molar mass of the solute to convert it to a mass concentration.
Significant figures
  • Give the final answer to the same number of significant figures as the least precise data, usually three.
Show working
  • Marks are awarded for the equation, the moles of each substance and the ratio step, even if the final number is wrong.

Part 5 of 6Worked Example

Question

  • 25.0 cm³ of sodium hydroxide solution of unknown concentration was neutralised by a mean titre of 20.0 cm³ of 0.100 mol/dm³ sulfuric acid.
  • Calculate the concentration of the sodium hydroxide in mol/dm³ and in g/dm³.
  • (Ar: Na = 23, O = 16, H = 1)

  1. EquationH₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
  2. Moles of H₂SO₄ = 0.100 × (20.0 ÷ 1000) = 2.00 × 10⁻³ mol.
  3. Mole ratio H₂SO₄NaOH is 1 : 2, so moles of NaOH = 2 × 2.00 × 10⁻³ = 4.00 × 10⁻³ mol.
  4. Concentration of NaOH = 4.00 × 10⁻³ ÷ (25.0 ÷ 1000) = 0.160 mol/dm³.
  5. Mr of NaOH = 23 + 16 + 1 = 40, so concentration = 0.160 × 40 = 6.4 g/dm³.

Answer 0.160 mol/dm³, which is 6.4 g/dm³.

Part 6 of 6Exam Checklist & Tips

3.2.5 – Key Points for the Exam

  • Titration A method that finds the volume of one solution needed to react exactly with a known volume of another.
  • Titre The volume added from the burette to reach the end point.
  • Concordant Titration results that agree closely, normally within 0.10 cm³ of each other.
  • Mole ratio The ratio of the coefficients in a balanced equation, used to convert between substances.
  • Concentration The amount of solute per unit volume of solution, in mol/dm³ or g/dm³.

Common Mistakes

  • Using the volume in cm³ directly in moles = concentration × volume without dividing by 1000.
  • Ignoring the mole ratio and assuming one mole of acid always reacts with one mole of alkali.
  • Including the rough titre when calculating the mean titre.
  • Dividing the moles of the known solution, instead of the unknown, by the volume of the unknown at the last step.

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