Burette readings, titres and comparable repeats
Read a downward-increasing burette scale, subtract initial from final readings and evaluate repeated titres using a supplied comparison rule.
Core and Extended · Chemistry 0620, 2026–2028. These paper-based method exercises support supervised laboratory learning; they do not certify practical competence.
Before you begin
- Explain the titration endpoint and the burette's measuring role.
- Calculate differences and arithmetic means with units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which subtraction gives delivered volume?
What you will learn
- Find delivered volume from two correctly read burette values.
- Use comparable valid titres and explain why a rough or suspect trial is not automatically included in a mean.
Two readings define the delivered amount
A conventional burette scale increases downwards as liquid leaves. Read the bottom of the concave meniscus of the stated clear aqueous solution at eye level. The titre is the delivered volume: final reading minus initial reading. The burette need not start at zero; the initial reading must be recorded rather than assumed.
In the figure, the initial reading is 2.30 cm³ and the final reading is 27.10 cm³, so 24.80 cm³ was delivered. The drawings are scale close-ups, not two different physical burettes. Use the scale resolution and recording convention supplied. For a scale with 0.1 cm³ divisions, a typical school convention records readings to 0.05 cm³ and two decimal places; it does not prove uncertainty is zero.
Compare titres before taking a mean
A rough trial locates the endpoint and is generally not part of the final mean. Repeated accurate trials need the same aliquot, concentrations, indicator and endpoint rule. Concordant means sufficiently close under the stated criterion; a question can supply that criterion, so do not invent one universal tolerance for every investigation.
For accurate titres 24.80, 24.85 and 24.80 cm³, a supplied maximum spread of 0.10 cm³ is met: the spread is 0.05 cm³. Their mean is 24.8166… cm³, recorded as 24.82 cm³ to the stated two-decimal convention. A separate 26.00 cm³ result is not close to that set and should be investigated. Similar titres demonstrate consistency, not automatic freedom from shared calibration or concentration errors.
Pause and explain
Titres 24.80, 24.85 and 24.80 cm³ must have spread at most 0.10 cm³. Do they meet the supplied rule?
Worked example
Find the titre for an initial reading of 10.20 cm³ and final reading of 34.70 cm³. Explain why 34.70 cm³ alone is not the answer.
Show the worked solution
- The amount delivered is a difference between final and initial readings.
- Calculate 34.70−10.20=24.50 cm³.
- The final scale value includes the initial offset; it equals delivery only in the special case of a zero initial reading.
Answer 24.50 cm³ delivered; the final reading alone ignores the initial 10.20 cm³.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Use the two scale close-ups to calculate the titre from 2.30 to 27.10 cm³.
Give me a hint
Do not assume a zero starting reading.
Compare my reasoning
- Record initial 2.30 cm³ and final 27.10 cm³.
- Subtract 27.10−2.30.
- Keep the supplied two-decimal recording convention and volume unit.
24.80 cm³.
Look for these in your work
- The initial offset is included.
- The difference, unit and recording format are correct.
Accurate titres are 24.80, 24.85 and 24.80 cm³. Under a supplied maximum spread of 0.10 cm³, decide whether to average them and give the mean to two decimal places.
Give me a hint
Check the spread before averaging the comparable valid trials.
Compare my reasoning
- The spread is 24.85−24.80=0.05 cm³, within 0.10 cm³.
- The sum is 74.45 cm³ and the mean is 74.45/3=24.8166… cm³.
- To two decimal places the mean is 24.82 cm³.
The set meets the supplied rule; mean 24.82 cm³.
Look for these in your work
- The given comparison rule is applied.
- The mean is correctly calculated and rounded.
A rough trial is 25.50 cm³, accurate trials are 24.80, 24.85 and 24.80 cm³, and another trial is 26.00 cm³. Explain a defensible summary under the same 0.10 cm³ rule.
Give me a hint
Separate the rough trial, close accurate set and discrepant reading.
Compare my reasoning
- Use the rough trial to locate the endpoint, not automatically in the final mean.
- The three close accurate titres give 24.82 cm³ under the supplied rule.
- Investigate the 26.00 cm³ discrepancy, such as overshoot, before deciding on further valid trials; do not silently discard it or call the mean guaranteed accurate.
Report the close set's mean 24.82 cm³ with the stated selection rule; exclude the rough trial from that mean and investigate the discrepant 26.00 cm³ trial.
Look for these in your work
- Trial selection is explained transparently.
- Consistency is not confused with absence of systematic error.
Common mistakes
- Taking the final reading alone when the initial reading is non-zero.
- Subtracting in the wrong order for the downward-increasing scale.
- Including every rough or suspect result without checking comparability.
- Calling concordance proof of perfect accuracy.
What can you explain now?
A burette changes from 2.30 to 27.10 cm³. What is the titre?
Compare with the explanation
24.80 cm³.
Subtract initial from final; the starting value need not be zero.
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Make it stick
Read two burette positions, calculate a titre and explain which repeated readings are comparable under a supplied rule.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.