Chemistry 9701/31 — May/June 2011
Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme
Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis
FA 1 is sulfuric acid, , of approximate concentration .
FA 2 is sodium hydroxide.
You are also provided with phenolphthalein (indicator).
You will determine the exact concentration of FA 1 by titration.
Method
Dilution
- Pipette of FA 1 into the graduated (volumetric) flask labelled FA 3.
- Make the solution up to the mark using distilled water.
- Shake the flask to mix the solution of FA 3.
Titration
- Rinse out the pipette with distilled water and then with FA 3.
- Pipette of FA 3 into a conical flask.
- Add 5 drops of phenolphthalein indicator to the flask. The indicator should remain colourless.
- Fill the burette with FA 2.
- Titrate FA 3 with FA 2, until a permanent pale pink colour is obtained.
You should perform a rough titration.
In the space below record your burette readings for this rough titration.
The rough titre is ............ .
- Carry out as many accurate titrations as you think necessary to obtain consistent results.
- Record in a suitable form below all of your burette readings and the volume of FA 2 added in each accurate titration.
- Make sure that your recorded results show the precision of your practical work.
Answer
Rough titre: 23.50 cm³
Accurate titrations:
| Titration | 1 | 2 | 3 |
|---|---|---|---|
| Final burette reading / cm³ | 23.35 | 46.70 | 23.30 |
| Initial burette reading / cm³ | 0.00 | 23.35 | 0.00 |
| Volume of FA 2 added / cm³ | 23.35 | 23.35 | 23.30 |
All burette readings are recorded to the nearest 0.05 cm³. Titrations 1 and 2 are concordant (within 0.1 cm³ of each other). (Readings shown are representative examples; the candidate records their own.)
See working — representative readings: rough 23.50 cm³; accurate titres 23.35, 23.35, 23.30 cm³ (candidate-dependent)
Background Concept
Titration is a quantitative volumetric technique for determining the exact concentration of a solution by reacting a measured volume of it with a solution of known concentration (a standard solution). Here, sulfuric acid (FA 1, approximately 0.7 mol dm⁻³) is first diluted tenfold: 25.0 cm³ is pipetted into a 250 cm³ volumetric flask and made up to the mark with distilled water, giving FA 3 (approximately 0.07 mol dm⁻³). The diluted acid is then titrated against sodium hydroxide (FA 2, 0.150 mol dm⁻³) using phenolphthalein indicator. The endpoint is signalled by a permanent pale pink colour, which appears as soon as the sodium hydroxide is in very slight excess (the solution becomes alkaline).
The key practical skills being assessed are:
- Using a burette correctly and reading it to the nearest 0.05 cm³.
- Performing a rough titration to find the approximate endpoint quickly.
- Repeating accurate titrations until at least two concordant results (within 0.1 cm³ of each other) are obtained.
- Recording results clearly in a table with appropriate headings and units.
Understanding the Question
This part asks you to record the rough titre and then carry out accurate titrations, tabulating all burette readings and the volume of FA 2 added. The 7 marks reward the layout of the results (a clear table), the completeness of the recordings (initial and final readings plus titre for every accurate run), the precision of the readings (to 0.05 cm³), and the quality of the results (concordant titres within 0.1 cm³, and close agreement with the supervisor's value).
Approach
- Perform a rough titration first, adding FA 2 quickly until the colour changes, to find the approximate titre.
- Then carry out accurate titrations: record the initial burette reading, add FA 2 slowly with swirling until the pale pink colour persists, and record the final reading.
- Calculate each titre as final − initial.
- Repeat until at least two concordant titres (within 0.1 cm³) are obtained.
- Present all results in a clear table with headings and units.
Step-by-Step Reasoning
The rough titration is a quick first estimate. You run FA 2 into the acid in larger portions until the colour changes, giving an approximate titre (here 23.50 cm³). This tells you roughly where the endpoint is, so the accurate titrations can be done efficiently — you can add most of the alkali quickly and slow down near the endpoint.
For each accurate titration:
- Record the initial burette reading (e.g. 0.00 cm³).
- Add FA 2 from the burette, swirling the conical flask continuously, until the pale pink colour just persists.
- Record the final burette reading (e.g. 23.35 cm³).
- The titre is the difference: 23.35 − 0.00 = 23.35 cm³.
A burette graduated in 0.1 cm³ divisions can be read to the nearest 0.05 cm³ (half a division). All readings must be recorded to this precision.
Concordant titres are those within 0.1 cm³ of each other. In the example, titres of 23.35, 23.35 and 23.30 cm³ are concordant (the largest difference is 0.05 cm³). You should keep titrating until you have at least two concordant results.
Example table:
| Titration | 1 | 2 | 3 |
|---|---|---|---|
| Final burette reading / cm³ | 23.35 | 46.70 | 23.30 |
| Initial burette reading / cm³ | 0.00 | 23.35 | 0.00 |
| Volume of FA 2 added / cm³ | 23.35 | 23.35 | 23.30 |
Key Takeaways
- Always do a rough titration before accurate ones.
- Read the burette to the nearest 0.05 cm³.
- Aim for at least two concordant titres within 0.1 cm³.
- Record results in a clear, labelled table.
Common Mistakes
- Recording readings to only 1 decimal place (0.1 cm³) — the mark scheme requires 0.05 cm³ precision.
- Omitting initial readings or the titre volume.
- Using 50.00 cm³ as an initial burette reading — the mark scheme explicitly disallows this.
- Stopping after one accurate titration without checking concordance.
Things to Be Careful About
- Headings in the table must match the readings (e.g. "Volume of FA 2 added / cm³").
- No burette reading may exceed 50.00 cm³.
- The indicator should remain colourless before the titration begins; if it turns pink, the flask has been contaminated.
- The endpoint is a permanent pale pink — a transient pink that disappears on swirling is not the endpoint.
From your accurate titration results, obtain a suitable value to be used in your calculations.
Show clearly how you have obtained this value.
of FA 3 required ................ of FA 2.
Answer
Mean titre = (23.35 + 23.35 + 23.30) / 3 = 23.33 cm³
25.0 cm³ of FA 3 required 23.33 cm³ of FA 2.
23.33 cm³
Background Concept
A single titration reading is subject to random error, so the most reliable estimate of the true titre is the mean of several concordant readings. "Concordant" here means within 0.20 cm³ of each other (the mark scheme for part (b) uses 0.20 cm³, while part (a) uses 0.1 cm³ for the concordance quality mark).
Understanding the Question
From the accurate titrations in part (a), you must select a suitable value — the mean of the concordant titres — to use in the calculations. You must show how you obtained it, either as an expression or by ticking the titres used in the table.
Approach
- Identify the concordant titres (within 0.20 cm³ of each other).
- Calculate their mean.
- Give the mean to 2 decimal places (or to 1 decimal place if all readings were to 1 dp and the mean is exactly correct).
Step-by-Step Reasoning
With titres 23.35, 23.35 and 23.30 cm³, all three are within 0.20 cm³ of each other, so all are used:
This mean is the value used in all subsequent calculations. The mark scheme allows the mean to be rounded to the nearest 0.05 cm³ if desired, but 23.33 cm³ is acceptable.
Key Takeaways
- The mean of concordant titres is the best estimate of the true titre.
- Show which titres you averaged, clearly.
Common Mistakes
- Averaging titres that are not concordant (more than 0.20 cm³ apart).
- Using only one titre without justification.
- Not showing the mean calculation.
Things to Be Careful About
- The mark scheme requires the titres used to be clearly shown (in an expression or ticked in the table).
- If all readings are to 1 decimal place, the mean may be given to 1 decimal place only if it is numerically correct without rounding.
Calculations
Show your working and appropriate significant figures in the final answer to each step of your calculations.
Calculate how many moles of were present in the volume of FA 2 calculated in (b).
Working
Volume of FA 2 = 23.33 cm³ = 0.02333 dm³
Answer
mol
3.50 × 10⁻³ mol
Background Concept
The amount of substance (in moles) in a solution is given by:
Since burette volumes are measured in cm³, they must be converted to dm³ by dividing by 1000 before multiplying by the concentration.
Understanding the Question
You are asked to calculate the moles of NaOH in the volume of FA 2 (the mean titre from part b). FA 2 is 0.150 mol dm⁻³ sodium hydroxide.
Approach
Convert the mean titre from cm³ to dm³, then multiply by the concentration of FA 2.
Step-by-Step Reasoning
Mean titre = 23.33 cm³ = 23.33 / 1000 dm³ = 0.02333 dm³
The answer is given to 3 significant figures, consistent with the data (0.150 has 3 sf).
Key Takeaways
- Always convert cm³ to dm³ (divide by 1000) before using concentration in mol dm⁻³.
- Match the significant figures of the answer to those of the given data.
Common Mistakes
- Forgetting to divide the volume by 1000 (giving an answer 1000× too large).
- Using the concentration of FA 1 or FA 3 instead of FA 2.
Things to Be Careful About
- The volume must be in dm³ for the formula moles = c × V to work.
- Give the answer to 3 significant figures (3.50 × 10⁻³ mol), not 4 or more.
Calculate how many moles of were present in of FA 3.
Working
From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH.
Answer
mol
1.75 × 10⁻³ mol
Background Concept
The balanced equation gives the stoichiometric ratio between reactants:
1 mol of H₂SO₄ reacts with 2 mol of NaOH. Therefore, the moles of H₂SO₄ are half the moles of NaOH.
Understanding the Question
Using the moles of NaOH from part (i), calculate the moles of H₂SO₄ in 25.0 cm³ of FA 3 (the diluted acid).
Approach
Divide the moles of NaOH by 2, using the stoichiometric ratio from the equation.
Step-by-Step Reasoning
This is the amount of H₂SO₄ in the 25.0 cm³ sample of FA 3 that was pipetted into the conical flask.
Key Takeaways
- The stoichiometric ratio comes directly from the balanced equation.
- For a 1:2 acid:base ratio, moles of acid = moles of base ÷ 2.
Common Mistakes
- Using a 1:1 ratio (forgetting that H₂SO₄ is diprotic and the equation shows 1:2).
- Multiplying instead of dividing by 2.
Things to Be Careful About
- H₂SO₄ is a diprotic acid — each mole neutralises 2 moles of NaOH. The balanced equation confirms this.
Calculate how many moles of were present in of the undiluted solution FA 1.
Working
FA 3 is a 10-fold dilution of FA 1 (25.0 cm³ made up to 250 cm³).
Answer
mol
1.75 × 10⁻² mol
Background Concept
FA 3 was prepared by diluting FA 1: 25.0 cm³ of FA 1 was made up to 250 cm³ in a volumetric flask. This is a 10-fold dilution (250/25 = 10). The moles of H₂SO₄ in a given volume of the undiluted FA 1 are therefore 10 times the moles in the same volume of FA 3.
Understanding the Question
Find the moles of H₂SO₄ in 25.0 cm³ of the undiluted solution FA 1, given the moles in 25.0 cm³ of FA 3 from part (ii).
Approach
Multiply the moles in 25.0 cm³ of FA 3 by the dilution factor (10).
Step-by-Step Reasoning
Dilution factor = 250 cm³ / 25.0 cm³ = 10
Key Takeaways
- Dilution factor = final volume ÷ aliquot volume.
- The moles in the undiluted solution are the moles in the diluted solution times the dilution factor.
Common Mistakes
- Dividing by 10 instead of multiplying (getting 1.75 × 10⁻⁴ instead of 1.75 × 10⁻²).
- Confusing the dilution factor (10) with its reciprocal (0.1).
Things to Be Careful About
- The dilution is 25 → 250 cm³, a factor of 10, not 100.
Calculate the concentration, in , of in FA 1.
The concentration of in FA 1 was ............... .
Working
25.0 cm³ = 0.0250 dm³
Answer
mol dm⁻³
0.700 mol dm⁻³
Background Concept
Concentration is the amount of solute per unit volume of solution:
25.0 cm³ = 0.0250 dm³.
Understanding the Question
Calculate the concentration of H₂SO₄ in FA 1 (the undiluted acid), using the moles in 25.0 cm³ of FA 1 from part (iii).
Approach
Divide the moles in 25.0 cm³ of FA 1 by the volume in dm³ (0.0250 dm³).
Step-by-Step Reasoning
This matches the expected approximate concentration of 0.7 mol dm⁻³ given in the question, confirming the calculation is consistent.
Key Takeaways
- Concentration = moles ÷ volume in dm³.
- 25.0 cm³ = 0.0250 dm³.
Common Mistakes
- Using the volume in cm³ without converting to dm³ (giving 0.7 × 10⁻³ instead of 0.700).
- Using the diluted volume (FA 3) instead of the undiluted.
Things to Be Careful About
- Give the answer to 3 significant figures: 0.700 mol dm⁻³.
- The answer should be close to the stated approximate concentration (0.7 mol dm⁻³) — a wildly different value signals an error.
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