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Titration Calculations

Learn and apply Titration Calculations in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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Titration Calculations: Orientation

Titration calculations are method-mark questions: if you write the same 5 lines every time (equation → n = cV → ratio → n → c), you stop losing marks to “small” slips like cm³ vs dm³ or flipped coefficients.

Build this on Mole and Avogadro Constant and keep the Stoichiometry hub open so unit conversions and mole logic stay coherent.

Definitions (Must Know)

A. Titration calculation

A titration calculation uses a balanced equation to connect the moles in the burette solution to the moles in the pipetted solution.

B. Mean titre

The mean titre is the average of the concordant titres (excluding the rough titre and any clear outlier, if instructed).

Detailed Explanations

A. Standard workflow (always works)

Because the balanced equation fixes the mole ratio at the endpoint, the known moles from n = cV let you calculate the unknown moles, then its concentration.

  1. Write a balanced equation.
  2. Convert volume(s) to dm³.
  3. Calculate moles of the known solution: n = cV.
  4. Use the mole ratio to find moles of the unknown.
  5. Convert back to concentration (if needed): c = n/V.

Mini example (Step 2 → Step 3):

  • 20.0 cm³ = 0.0200 dm³, so n = cV = 0.100 × 0.0200 = 0.00200 mol.

Worked Examples

Modelled example 1

Find Sodium Hydroxide Concentration

Core

Problem

25.0 cm³ NaOH is titrated with 0.100 mol dm⁻³ H₂SO₄. The mean titre is 20.0 cm³. Find the sodium hydroxide concentration.

Study the worked solution

  1. Write the equation

    Method

    Use the balanced neutralisation equation.

    Reason

    The coefficients supply the amount ratio at the endpoint.

    Working

    H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

  2. Find sulfuric acid amount

    Reason

    The titrant concentration and mean titre are known.

    Working

    n(H₂SO₄) = (0.100)(20.0/1000) = 0.00200 mol

  3. Apply the mole ratio

    Reason

    One mole of sulfuric acid neutralises two moles of sodium hydroxide.

    Working

    n(NaOH) = 2(0.00200) = 0.00400 mol

  4. Calculate sodium hydroxide concentration

    Method

    Divide the aliquot amount by its volume in cubic decimetres.

    Reason

    Concentration is amount per solution volume.

    Working

    c(NaOH) = 0.00400/0.0250 = 0.160 mol dm⁻³

Guided practice 2

Use a Carbonate–Acid Ratio

About 7 min

Problem

25.0 cm³ of 0.0500 mol dm⁻³ Na₂CO₃ is titrated with HCl. The mean titre is 18.0 cm³. Find the hydrochloric acid concentration.

Try this before viewing the solution

Hints

Hint 1: write the balanced equation
One carbonate ion reacts with two hydrogen ions.
Hint 2: start from the known solution
Calculate the amount in the sodium carbonate aliquot before using the ratio.
View solution step by step

  1. Calculate carbonate amount

    Method

    Convert 25.0 cm³ to 0.0250 dm³ and use n = cV.

    Reason

    The sodium carbonate solution has the known concentration.

    Working

    n(Na₂CO₃) = (0.0500)(0.0250) = 0.00125 mol

  2. Apply the equation ratio

    Reason

    Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ gives a 1:2 ratio.

    Working

    n(HCl) = 2(0.00125) = 0.00250 mol

  3. Calculate acid concentration

    Method

    Divide by the 0.0180 dm³ titre.

    Reason

    The calculated acid amount occupied the measured burette volume.

    Working

    c(HCl) = 0.00250/0.0180 = 0.139 mol dm⁻³

Common misconception 3

Correct an Inverted Mole Ratio

Find and correct the mistake

Learner attempt

In the carbonate titration above, a learner obtains 0.00125 mol Na₂CO₃ but then calculates n(HCl) = 0.00125/2. Identify the first error and correct the hydrochloric acid amount and concentration.

Diagnose the coefficient direction

From carbonate amount to HCl amount

View solution step by step

  1. Read the ratio in the requested direction

    Method

    Multiply carbonate amount by two.

    Reason

    Two moles of HCl react per one mole of Na₂CO₃.

    Working

    n(HCl) = 2(0.00125) = 0.00250 mol

  2. Correct the concentration

    Method

    Divide by the 0.0180 dm³ acid titre.

    Reason

    Concentration uses the volume of the solution whose amount was calculated.

    Working

    c(HCl) = 0.00250/0.0180 = 0.139 mol dm⁻³

Examiner practice 4

Determine Nitric Acid Concentration

4 marks

Problem

20.0 cm³ of 0.150 mol dm⁻³ Ba(OH)₂ requires 24.0 cm³ HNO₃ for complete neutralisation. Calculate the nitric acid concentration. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate base amount

    1 mark

    Method

    Use n = cV with 0.0200 dm³.

    Reason

    The barium hydroxide concentration is known.

    Working

    n(Ba(OH)₂) = (0.150)(0.0200) = 0.00300 mol

  2. Apply the neutralisation ratio

    1 mark

    Reason

    Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O gives two acid moles per base mole.

    Working

    n(HNO₃) = 2(0.00300) = 0.00600 mol

  3. Convert the titre

    1 mark

    Reason

    The acid concentration is required per cubic decimetre.

    Working

    V(HNO₃) = 24.0/1000 = 0.0240 dm³

  4. Calculate concentration

    1 mark

    Method

    Divide acid amount by acid volume.

    Reason

    Both quantities now refer to the titrant.

    Working

    c(HNO₃) = 0.00600/0.0240 = 0.250 mol dm⁻³

Challenge 5

Infer Acid Basicity from Titration Data

Minimal support

Problem

25.0 cm³ of 0.100 mol dm⁻³ acid HₓA is neutralised by 20.0 cm³ of 0.250 mol dm⁻³ NaOH. Assuming complete neutralisation of all acidic hydrogen atoms, determine x.

Try this before viewing the solution

Hints

Hint 1: calculate both amounts
Use n = cV separately for the acid and sodium hydroxide.
Hint 2: interpret hydroxide per acid
At complete neutralisation, x equals moles of hydroxide consumed per mole of acid.
View solution step by step

  1. Calculate the acid amount

    Method

    Use the acid concentration and aliquot volume.

    Reason

    This gives moles of acid molecules rather than acidic hydrogen atoms.

    Working

    n(HₓA) = (0.100)(0.0250) = 0.00250 mol

  2. Calculate hydroxide amount

    Reason

    Sodium hydroxide supplies one mole of hydroxide per mole.

    Working

    n(OH⁻) = (0.250)(0.0200) = 0.00500 mol

  3. Infer the stoichiometric count

    Method

    Divide hydroxide amount by acid amount.

    Reason

    Each acid molecule consumes x hydroxide ions when all acidic hydrogens are neutralised.

    Working

    x = 0.00500/0.00250 = 2; the acid is diprotic.

Mind Stretchers

Mind stretcher 1Extension

25.0 cm³ of a NaOH solution is diluted to 250 cm³. Then 25.0 cm³ of the diluted solution is titrated with 0.100 mol dm⁻³ H₂SO₄ and the mean titre is 20.0 cm³. Find the concentration of the original NaOH solution.

Show Hint

Use the mean concordant titre, calculate the titrant amount, then apply the balanced-equation ratio.

Show Answer

Mark scheme:

  • Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
  • n(H₂SO₄) = 0.100 × 0.0200 = 0.00200 mol
  • In the aliquot: n(NaOH) = 2(0.00200) = 0.00400 mol
  • c(diluted NaOH) = 0.00400/0.0250 = 0.160 mol dm⁻³
  • Dilution factor is 250/25.0 = 10, so c(original) = 10(0.160) = 1.60 mol dm⁻³

Mind stretcher 2: Inferring purity from a titreExtension

Question. A 0.250 g impure Na₂CO₃ sample requires 20.0 cm³ of 0.100 mol dm⁻³ HCl for complete reaction. Calculate its percentage purity. Use M(Na₂CO₃) = 106 g mol⁻¹.

Show Hint

Use Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.

Show Answer

n(HCl) = 0.100(0.0200) = 0.00200 mol, so n(Na₂CO₃) = 0.00100 mol and its mass is 0.106 g. Purity = (0.106/0.250) × 100 = 42.4%.