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Titration errors: trace the cause and direction

Key idea: Trace a titration error through sample amount, titrant concentration and recorded volume before predicting its effect. Concordant repeats can still share the same dilution or end-point bias.

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  • GCE A-Level H2 Chemistry 2027
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A titre can be repeatable and still be wrong. To explain an error, follow what changes: the amount in the sample aliquot, the titrant concentration, or the volume recorded by the burette. Then follow that change through the calculation.

If readings and concordance are unfamiliar, start with titration technique.

Trace the error before naming its direction

For an acid aliquot titrated with alkali of known concentration, the calculation uses the recorded alkali volume to find its amount, applies the reaction ratio, then divides by the acid aliquot volume.

n(alkali) = c(alkali)V(alkali)

Ask three questions:

  1. Did the error change the actual amount of acid in the pipetted aliquot?
  2. Did it change the actual concentration of the alkali delivered?
  3. Does the recorded volume equal the volume that reached the flask?

State the known and unknown solutions before predicting a calculated concentration. A high titre alone cannot show the direction for every possible analysis.

Where did the recorded liquid go?

Guided practice 1

Trace the effect of an air bubble in the burette tip

About 6 min

Problem

An air bubble is present in the burette tip at the initial reading. It clears during the titration as liquid fills the tip. Explain the effect on the recorded titre.

Try this before viewing the solution

Hints

Hint 1

follow the liquid

Compare liquid leaving the reservoir with liquid reaching the flask.
Hint 2

compare volumes

The reading change includes both tip filling and flask delivery.
Show solution step by step

  1. Account for tip filling

    Method

    State that some titrant displaces the air and fills the jet.

    Reason

    That portion changes the burette reading but does not reach the flask.

    Working

    Answer

    V_reading = Vₜᵢₚ + V_flask.

  2. Give the direction

    Method

    Conclude that the recorded titre is too high.

    Reason

    The indicated volume is larger than the volume that actually reacted in the flask.

    Working

    Answer

    Improvement: run titrant through the jet and remove bubbles before the initial reading.

Check your understanding 1 Check the bubble’s liquid account

A bubble is present in the burette tip at the initial reading and clears during titration. Check where some of the indicated liquid goes first and how that affects the recorded titre.

Part of indicated volume first
Recorded titre
Show answer

Fills the burette tip; Too high. The burette change includes liquid used to fill the tip, so it exceeds the volume delivered to the flask.

If a bubble remains unchanged throughout the titration, it does not consume the extra volume described above. The high-titre explanation requires the tip to fill after the initial reading.

Has the endpoint been passed?

Exam-style question 1

Evaluate endpoint overshoot

3 marks

Examination question

A student adds titrant rapidly near the endpoint and the indicator colour passes well beyond the first permanent change. Explain the effect on the titre and state a specific prevention. [3 marks]

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Show solution step by step

  1. Identify overshoot

    1 mark

    Method

    State that excess titrant is delivered after the endpoint is reached.

    Reason

    Rapid addition does not allow the first permanent colour change to be judged promptly.

    Working

    Answer

    Delivered volume exceeds the volume needed for the chosen endpoint.

  2. Give the direction

    1 mark

    Method

    State that the recorded titre is too high.

    Reason

    The final burette reading includes excess titrant.

    Working

    Answer

    V_recorded > V_endpoint.

  3. Prevent the error

    1 mark

    Method

    Add titrant dropwise with swirling near the endpoint and use a white tile to see the first permanent change.

    Reason

    Small additions and better contrast reduce the chance of passing the endpoint.

    Working

    Answer

    Dropwise + swirl + first permanent colour change.

Water in a pipette or water in the flask?

The pipette fixes the sample aliquot. Water mixed with the sample before that aliquot is delivered reduces its concentration. A small deionised-water wash after the aliquot is in the conical flask changes the total volume but leaves the sample amount unchanged. Use enough water to wash splashes down, while keeping the specified endpoint detectable.

Try it yourself 2

Trace water left in a pipette

Minimal support

Problem

A pipette contains residual deionised water before it is filled to the mark with an acid sample. The acid is titrated with alkali of known concentration. Predict the effect on the mean titre and on the calculated acid concentration.

Try this before viewing the solution

Mean titre
Calculated acid concentration

Hints

Hint 1

locate dilution

Water in the pipette mixes with the acid before the aliquot is delivered.

Hint 2

follow moles

Fewer acid moles require fewer alkali moles at equivalence.

Show solution step by step

  1. Trace the aliquot

    Method

    State that residual water dilutes the acid inside the pipette, so the delivered aliquot contains fewer acid moles.

    Reason

    The pipette still delivers its nominal total volume, but not the intended acid concentration.

    Working

    Answer

    n(acid in aliquot) is too low.

  2. Predict the titre

    Method

    Conclude that less alkali is required and the mean titre is too low.

    Reason

    Equivalence moles follow the reaction ratio.

    Working

    Answer

    V(alkali) is too low.

  3. Predict the reported concentration

    Method

    Conclude that the calculated concentration of the original acid is too low.

    Reason

    The calculation assigns the underestimated moles to the nominal pipette volume.

    Working

    Answer

    Rinse the pipette with the acid sample before transfer.

Water left in the burette

Check your understanding 2 Trace dilution of the titrant

A burette should contain 0.100 mol dm⁻³ sodium hydroxide. A student leaves deionised water in it, then fills it with this solution. Assume the water mixes with the titrant. The titration is used to find the concentration of a fixed acid aliquot. Explain the effect on the titre and the reported acid concentration.

Show hint

Compare the actual alkali concentration with the concentration used in the calculation.

Show answer

The actual alkali concentration is lower. The unchanged acid amount therefore needs a larger alkali volume at equivalence. Using 0.100 mol dm⁻³ with this larger volume overestimates the alkali amount, and hence the acid amount and concentration. Rinse the burette with the titrant and fill the jet before starting.

Check your understanding 3 Can close agreement hide a concentration error?

A 25.0 cm³ aliquot of a monoprotic acid reacts 1:1 with sodium hydroxide. Concordant titres give a mean of 20.0 cm³. The student assumes the alkali is 0.100 mol dm⁻³, but its actual concentration is 0.0800 mol dm⁻³. Calculate the reported and actual acid concentrations, then the percentage error relative to the actual value.

Show hint

Perform the same calculation twice, changing only the alkali concentration. Use the actual acid concentration as the denominator of the percentage error.

Show the steps Optional working
Unit: dm³
Unit: mol
Unit: dm³
Unit: mol dm⁻³
Show the steps Optional working
Unit: dm³
Unit: mol
Unit: dm³
Unit: mol dm⁻³
Show the steps Optional working
Unit: mol dm⁻³
Unit: mol dm⁻³
Unit: %
Show answer

The reported concentration is:

c_reported = 0.100(0.0200)/0.0250 = 0.0800 mol dm⁻³

Using the actual alkali concentration gives:

c_actual = 0.0800(0.0200)/0.0250 = 0.0640 mol dm⁻³
percentage error = (0.0800-0.0640)/0.0640 × 100% = 25.0% high

Check your understanding 4 Would more repeats remove this error?

In the acid analysis above, the assumed sodium hydroxide concentration is 0.100 mol dm⁻³ but the actual value is 0.0800 mol dm⁻³. Explain whether repeating with that same alkali removes the error and how you would address it.

Show answer

The reported concentration is:

c_reported = 0.100(0.0200)/0.0250 = 0.0800 mol dm⁻³

Using the actual alkali concentration gives:

c_actual = 0.0800(0.0200)/0.0250 = 0.0640 mol dm⁻³
percentage error = (0.0800-0.0640)/0.0640 × 100% = 25.0% high

Repeating with the same incorrectly assumed concentration repeats the bias. Close agreement checks repeatability; it cannot verify the titrant concentration. Check that concentration with a suitable standard before reporting the result.

A useful evaluation connects cause → changed quantity → direction → matched improvement. “Repeat more often” can reduce random variation in a mean, but it cannot replace missing solute, correct dilution or undo a consistent overshoot.

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