Titration Calculations
Calculate amounts and concentrations from titration data using the equation's mole ratio.
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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 on the mole and Avogadro constant, using your course’s Stoichiometry topic navigation to review them when needed. Keep unit conversions and mole ratios together in your working.
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).
Key Ideas (What Earns Marks)
- Convert titres to dm³ before using n = cV.
- The mole ratio comes from the balanced equation coefficients.
- State whether the concentration required is the acid or the base.
- n(known) = cV (with V in dm³). - Use coefficients to convert to n(unknown). - c = n/V for the solution you are asked about.
Data table
| Titration run | Titre |
|---|---|
| Rough | 24.6 |
| 1 | 24.1 |
| 2 | 24.15 |
| 3 | 24.12 |
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.
- Write a balanced equation.
- Convert volume(s) to dm³.
- Calculate moles of the known solution: n = cV.
- Use the mole ratio to find moles of the unknown.
- 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
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
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)Find sulfuric acid amount
Reason
The titrant concentration and mean titre are known.Working
n(H₂SO₄) = (0.100)(20.0/1000) = 0.00200 molApply the mole ratio
Reason
One mole of sulfuric acid neutralises two moles of sodium hydroxide.Working
n(NaOH) = 2(0.00200) = 0.00400 molCalculate 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
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
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 molApply the equation ratio
Reason
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ gives a 1:2 ratio.Working
n(HCl) = 2(0.00125) = 0.00250 molCalculate 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
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
View solution step by step
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 molCorrect 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
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
Calculate base amount
1 markMethod
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 molApply the neutralisation ratio
1 markReason
Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O gives two acid moles per base mole.Working
n(HNO₃) = 2(0.00300) = 0.00600 molConvert the titre
1 markReason
The acid concentration is required per cubic decimetre.Working
V(HNO₃) = 24.0/1000 = 0.0240 dm³Calculate concentration
1 markMethod
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
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit the known amount, ratio, volume conversion and concentration separately.
Challenge 5
Infer Acid Basicity from Titration Data
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
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 molCalculate hydroxide amount
Reason
Sodium hydroxide supplies one mole of hydroxide per mole.Working
n(OH⁻) = (0.250)(0.0200) = 0.00500 molInfer 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.
Common Mistakes
- Using V in cm³ in n = cV.
- Flipping the mole ratio (using coefficients the wrong way round).
- Using the pipetted volume as the burette volume (or vice versa).
Use the topic check in Practise and check below to practise and check your understanding.
Exam Tips
- Write the ratio explicitly: “1 mol H₂SO₄ : 2 mol NaOH”.
- Check units in the final concentration: mol dm⁻³.
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%.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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