Titration Calculations [H]

This lesson covers titration calculations as required by the AQA GCSE Chemistry specification (4.3.3). This is Higher Tier only content. Titrations are used to find the concentration of an unknown solution by reacting it with a solution of known concentration. You need to be able to process titration results, identify concordant values, and carry out calculations involving moles, concentration, and volume.

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What Is a Titration?

A titration is an experimental technique used to determine the exact volume of one solution needed to react completely with a known volume of another solution. Typically, an acid is neutralised by a base (or vice versa), and an indicator is used to show the end point — the moment the reaction is complete.

Term	Definition
Titration	An experiment to find the volume of solution needed to react exactly with another solution
Burette	A graduated tube used to accurately measure the volume of solution added
Pipette	Used to measure a precise, fixed volume of solution
End point	The point at which the indicator changes colour, showing the reaction is complete
Concordant results	Titration results that are within 0.10 cm3 of each other
Mean titre	The average of the concordant results

Exam Tip: When processing titration data, always discard any anomalous results (results that are not concordant with the others) before calculating the mean. The mean should only be calculated from concordant values.

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Processing Titration Results

In a titration, you typically perform several trial runs and then calculate a mean titre from the concordant results.

Worked Example: Finding the Mean Titre

Run	Initial Reading (cm3)	Final Reading (cm3)	Titre (cm3)
1 (rough)	0.00	26.50	26.50
2	0.00	25.20	25.20
3	0.15	25.40	25.25
4	0.00	25.30	25.30

Step 1: Identify concordant results (within 0.10 cm3 of each other).

• Run 1 (rough) = 26.50 — this is the rough run and is too far from the others. Discard.
• Runs 2, 3, and 4 give 25.20, 25.25, and 25.30 — these are concordant.

Step 2: Calculate mean titre.

• Mean = (25.20 + 25.25 + 25.30) / 3 = 25.25 cm3

Exam Tip: Always discard the rough run and any anomalous results. Only average concordant values. The rough run is used to get an approximate end point and is never included in the mean.

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Titration Calculation Method

Once you have the mean titre, you can calculate the concentration of the unknown solution. The general method is:

graph TD
    A["Step 1: Calculate mean titre from concordant results"] --> B["Step 2: Calculate moles of the known solution"]
    B --> C["Step 3: Use mole ratio from balanced equation"]
    C --> D["Step 4: Calculate moles of the unknown solution"]
    D --> E["Step 5: Calculate concentration of the unknown solution"]

    style A fill:#3498db,color:#fff
    style B fill:#2980b9,color:#fff
    style C fill:#e67e22,color:#fff
    style D fill:#27ae60,color:#fff
    style E fill:#9b59b6,color:#fff

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Worked Example: Full Titration Calculation

25.0 cm3 of sodium hydroxide (NaOH) of unknown concentration was neutralised by 20.0 cm3 of 0.10 mol/dm3 hydrochloric acid (HCl). Calculate the concentration of the NaOH solution in mol/dm3.

Balanced equation: NaOH + HCl --> NaCl + H2O

Step 1: Calculate moles of HCl (the known solution).

• Volume of HCl = 20.0 cm3 = 20.0 / 1000 = 0.020 dm3
• moles of HCl = concentration x volume = 0.10 x 0.020 = 0.002 mol

Step 2: Use the mole ratio.

• From the equation, NaOH : HCl = 1 : 1
• So moles of NaOH = 0.002 mol

Step 3: Calculate the concentration of NaOH.

• Volume of NaOH = 25.0 cm3 = 25.0 / 1000 = 0.025 dm3
• concentration = moles / volume = 0.002 / 0.025 = 0.08 mol/dm3

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Worked Example: Different Mole Ratio

25.0 cm3 of sulfuric acid (H2SO4) of unknown concentration was neutralised by 30.0 cm3 of 0.20 mol/dm3 sodium hydroxide (NaOH). Calculate the concentration of the H2SO4 in mol/dm3.

Balanced equation: H2SO4 + 2NaOH --> Na2SO4 + 2H2O

Step 1: moles of NaOH = 0.20 x (30.0 / 1000) = 0.20 x 0.030 = 0.006 mol

Step 2: From the equation, H2SO4 : NaOH = 1 : 2 So moles of H2SO4 = 0.006 / 2 = 0.003 mol

Step 3: concentration of H2SO4 = 0.003 / (25.0 / 1000) = 0.003 / 0.025 = 0.12 mol/dm3

Exam Tip: Pay close attention to the mole ratio. For sulfuric acid and sodium hydroxide, the ratio is 1:2 (one acid to two base), so you must divide the moles of NaOH by 2 to find the moles of H2SO4. This is a very common source of error.

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Converting to g/dm3

If the question asks for the concentration in g/dm3, multiply the mol/dm3 concentration by the Mr of the solute.

Worked Example

From the previous example, convert 0.08 mol/dm3 NaOH to g/dm3.

• Mr of NaOH = 23 + 16 + 1 = 40
• concentration in g/dm3 = 0.08 x 40 = 3.2 g/dm3

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Finding Mass of Solute from Titration

You may be asked to find the mass of solute in the original solution.

Worked Example

In a titration, 25.0 cm3 of NaOH was neutralised by 18.0 cm3 of 0.50 mol/dm3 HCl. What mass of NaOH was in the 25.0 cm3 sample?

NaOH + HCl --> NaCl + H2O

Step 1: moles of HCl = 0.50 x (18.0 / 1000) = 0.009 mol

Step 2: moles of NaOH = 0.009 mol (1:1 ratio)

Step 3: mass of NaOH = moles x Mr = 0.009 x 40 = 0.36 g

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Common Titration Indicators

Indicator	Colour in Acid	Colour in Alkali	End Point Colour
Methyl orange	Red	Yellow	Orange (at end point)
Phenolphthalein	Colourless	Pink	Colourless (acid added to alkali) or pink (alkali added to acid)
Litmus	Red	Blue	Purple (at end point)

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Practical Titration Tips

1. Use a pipette (not a measuring cylinder) to measure the fixed volume — it is more accurate.
2. Use a white tile under the flask to see the colour change more clearly.
3. Add the solution from the burette drop by drop near the end point.
4. Record all burette readings to 2 decimal places (the second decimal is always 0 or 5).
5. Rinse the conical flask with distilled water during the titration — this does not affect the result because the same number of moles is present.