Titration Calculations (Higher)

This lesson covers titration calculations — a Higher tier topic required by the Edexcel GCSE Chemistry specification (1CH0). Titrations combine practical technique with quantitative chemistry and are a common source of extended calculation questions in the exam.

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What Is a Titration?

A titration is an experimental technique used to find the exact volume of one solution needed to react completely with a known volume of another solution. It is most commonly used for acid-base reactions.

By knowing the concentration of one solution and measuring the volumes carefully, you can calculate the unknown concentration of the other solution.

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Core Practical: Acid-Base Titration

Equipment

Item	Purpose
Burette	Measures the volume of solution added (to ±0.05 cm³)
Pipette (usually 25.00 cm³)	Measures a precise, fixed volume of solution
Conical flask	Where the reaction takes place
Indicator (e.g. phenolphthalein or methyl orange)	Shows the end point (colour change)
White tile	Placed under the conical flask to see the colour change clearly

Method

1. Use a pipette to measure exactly 25.00 cm³ of alkali (e.g. NaOH) into a clean conical flask.
2. Add 2–3 drops of a suitable indicator.
3. Fill a burette with the acid (e.g. HCl). Record the initial burette reading.
4. Add acid from the burette slowly to the alkali in the conical flask, swirling the flask constantly.
5. When the indicator changes colour, record the final burette reading.
6. The titre is the volume of acid added: titre = final reading − initial reading.
7. Repeat until you obtain at least two concordant results (results that agree to within 0.10 cm³).

Exam Tip: In the exam, always use the term "concordant results" — this means results within 0.10 cm³ of each other. The first (rough) result is usually ignored.

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Processing Titration Results

Concordant Results

A typical set of titration results might look like this:

Trial	Rough	1	2	3
Initial reading (cm³)	0.00	0.50	0.30	0.20
Final reading (cm³)	24.80	24.70	24.55	24.45
Titre (cm³)	24.80	24.20	24.25	24.25

Identifying concordant results:

• Rough titre = 24.80 cm³ (ignore — it is a rough trial)
• Trial 1 = 24.20 cm³
• Trial 2 = 24.25 cm³
• Trial 3 = 24.25 cm³

Trials 2 and 3 are concordant (within 0.10 cm³). Trial 1 is also close, but we should use the best agreement.

Mean titre = (24.20 + 24.25 + 24.25) ÷ 3 = 24.23 cm³

Or, using only the two concordant results: (24.25 + 24.25) ÷ 2 = 24.25 cm³

Exam Tip: Exclude the rough result when calculating the mean. Use concordant results only. If asked to justify which results to include, explain that you choose those within 0.10 cm³ of each other.

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Titration Calculation Method

Once you have the mean titre, you can calculate the unknown concentration. The method follows these steps:

1. Write the balanced equation for the reaction.
2. Calculate moles of the substance whose concentration you know, using: moles = concentration × volume (in dm³).
3. Use the molar ratio from the equation to find moles of the other substance.
4. Calculate the unknown concentration using: concentration = moles ÷ volume (in dm³).

flowchart TD
    A[Balanced equation] --> B[Known: concentration and volume]
    B --> C["Moles known = c × V (in dm³)"]
    C --> D[Apply molar ratio from equation]
    D --> E[Moles of unknown]
    E --> F["Concentration unknown = moles ÷ V (in dm³)"]
    F --> G["Optional: convert mol/dm³ → g/dm³ using Mr"]

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Worked Examples

Example 1: Finding the Concentration of an Acid

Question: 25.00 cm³ of 0.10 mol/dm³ sodium hydroxide was neutralised by 20.00 cm³ of hydrochloric acid. Calculate the concentration of the acid.

Step 1: Balanced equation:

NaOH + HCl → NaCl + H₂O

Step 2: Calculate moles of NaOH:

• Volume = 25.00 cm³ = 0.02500 dm³
• moles = 0.10 × 0.02500 = 0.00250 mol

Step 3: Use the molar ratio:

• 1 NaOH : 1 HCl (ratio is 1:1)
• moles of HCl = 0.00250 mol

Step 4: Calculate concentration of HCl:

• Volume of HCl = 20.00 cm³ = 0.02000 dm³
• concentration = 0.00250 ÷ 0.02000 = 0.125 mol/dm³

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Example 2: Finding the Concentration of an Alkali

Question: 25.00 cm³ of potassium hydroxide was titrated against 0.20 mol/dm³ sulfuric acid. The mean titre was 15.00 cm³. Calculate the concentration of the potassium hydroxide in mol/dm³ and g/dm³.

Step 1: Balanced equation:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Step 2: Calculate moles of H₂SO₄:

• Volume = 15.00 cm³ = 0.01500 dm³
• moles = 0.20 × 0.01500 = 0.00300 mol

Step 3: Use the molar ratio:

• 2KOH : 1H₂SO₄ (ratio is 2:1)
• moles of KOH = 0.00300 × 2 = 0.00600 mol

Step 4: Calculate concentration of KOH:

• Volume of KOH = 25.00 cm³ = 0.02500 dm³
• concentration = 0.00600 ÷ 0.02500 = 0.24 mol/dm³

Convert to g/dm³:

• Mr of KOH = 39 + 16 + 1 = 56
• concentration (g/dm³) = 0.24 × 56 = 13.44 g/dm³

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Example 3: Full Titration Problem

Question: A student titrated 25.00 cm³ of sodium carbonate solution against 0.50 mol/dm³ hydrochloric acid. The results are shown below.

Trial	1	2	3
Titre (cm³)	23.40	22.50	22.55

Calculate the concentration of the sodium carbonate solution.

Processing results:

• Trial 1 = 23.40 cm³ (not concordant — discard or treat as rough)
• Trial 2 = 22.50 cm³
• Trial 3 = 22.55 cm³
• These are concordant (within 0.10 cm³)
• Mean titre = (22.50 + 22.55) ÷ 2 = 22.525 cm³

Step 1: Balanced equation:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step 2: Calculate moles of HCl:

• Volume = 22.525 cm³ = 0.022525 dm³
• moles = 0.50 × 0.022525 = 0.011263 mol

Step 3: Use the molar ratio:

• 1 Na₂CO₃ : 2 HCl
• moles of Na₂CO₃ = 0.011263 ÷ 2 = 0.005631 mol

Step 4: Calculate concentration of Na₂CO₃:

• Volume = 25.00 cm³ = 0.02500 dm³
• concentration = 0.005631 ÷ 0.02500 = 0.225 mol/dm³

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Example 4: Finding Mass from Titration

Question: 25.00 cm³ of sodium hydroxide solution was neutralised by 18.50 cm³ of 0.10 mol/dm³ hydrochloric acid. Calculate the mass of NaOH in the 25.00 cm³ sample.

NaOH + HCl → NaCl + H₂O