9701/31

Chemistry 9701/31May/June 2011

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

3
questions
40
marks
120
minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

FA 1 is sulfuric acid, H2SO4\text{H}_2\text{SO}_4, of approximate concentration 0.7 mol dm30.7\text{ mol dm}^{-3}.
FA 2 is 0.150 mol dm30.150\text{ mol dm}^{-3} sodium hydroxide.
You are also provided with phenolphthalein (indicator).

You will determine the exact concentration of FA 1 by titration.

H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})
(a)

Method

Dilution

  • Pipette 25.0 cm325.0\text{ cm}^3 of FA 1 into the 250 cm3250\text{ cm}^3 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 25.0 cm325.0\text{ cm}^3 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 ............ cm3\text{cm}^3.

  • 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.
7M
DifficultyMedium-Easy
Worked solution

Answer

Rough titre: 23.50 cm³

Accurate titrations:

Titration123
Final burette reading / cm³23.3546.7023.30
Initial burette reading / cm³0.0023.350.00
Volume of FA 2 added / cm³23.3523.3523.30

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.)

Final answer

See working — representative readings: rough 23.50 cm³; accurate titres 23.35, 23.35, 23.30 cm³ (candidate-dependent)

Detailed explanation

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

  1. Perform a rough titration first, adding FA 2 quickly until the colour changes, to find the approximate titre.
  2. 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.
  3. Calculate each titre as final − initial.
  4. Repeat until at least two concordant titres (within 0.1 cm³) are obtained.
  5. 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:

Titration123
Final burette reading / cm³23.3546.7023.30
Initial burette reading / cm³0.0023.350.00
Volume of FA 2 added / cm³23.3523.3523.30

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.
Techniques used
record burette readings to the nearest 0.05 cm³perform a rough titrationobtain concordant titres within 0.1 cm³prepare a tenfold dilution in a volumetric flask
(b)

From your accurate titration results, obtain a suitable value to be used in your calculations.
Show clearly how you have obtained this value.

25.0 cm325.0\text{ cm}^3 of FA 3 required ................ cm3\text{cm}^3 of FA 2.

1M
DifficultyEasy
Worked solution

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.

Final answer

23.33 cm³

Detailed explanation

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

  1. Identify the concordant titres (within 0.20 cm³ of each other).
  2. Calculate their mean.
  3. 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:

mean=23.35+23.35+23.303=23.33 cm3\text{mean} = \frac{23.35 + 23.35 + 23.30}{3} = 23.33 \text{ cm}^3

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.
Techniques used
calculate the mean of concordant titres
(c)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

4M
(i)

Calculate how many moles of NaOH\text{NaOH} were present in the volume of FA 2 calculated in (b).

DifficultyMedium-Easy
Worked solution

Working

Volume of FA 2 = 23.33 cm³ = 0.02333 dm³

moles NaOH=0.02333×0.150=3.50×103 mol\text{moles NaOH} = 0.02333 \times 0.150 = 3.50 \times 10^{-3} \text{ mol}

Answer

3.50×1033.50 \times 10^{-3} mol

Final answer

3.50 × 10⁻³ mol

Detailed explanation

Background Concept

The amount of substance (in moles) in a solution is given by:

moles=concentration (mol dm3)×volume (dm3)\text{moles} = \text{concentration (mol dm}^{-3}) \times \text{volume (dm}^3)

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

moles NaOH=0.02333×0.150=0.0034995 mol3.50×103 mol\text{moles NaOH} = 0.02333 \times 0.150 = 0.0034995 \text{ mol} \approx 3.50 \times 10^{-3} \text{ mol}

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.
Techniques used
convert volume from cm³ to dm³calculate moles from concentration and volume
(ii)

Calculate how many moles of H2SO4\text{H}_2\text{SO}_4 were present in 25.0 cm325.0\text{ cm}^3 of FA 3.

H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})
DifficultyMedium-Easy
Worked solution

Working

From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH.

moles H2SO4=3.50×1032=1.75×103 mol\text{moles H}_2\text{SO}_4 = \frac{3.50 \times 10^{-3}}{2} = 1.75 \times 10^{-3} \text{ mol}

Answer

1.75×1031.75 \times 10^{-3} mol

Final answer

1.75 × 10⁻³ mol

Detailed explanation

Background Concept

The balanced equation gives the stoichiometric ratio between reactants:

H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})

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

moles H2SO4=moles NaOH2=3.50×1032=1.75×103 mol\text{moles H}_2\text{SO}_4 = \frac{\text{moles NaOH}}{2} = \frac{3.50 \times 10^{-3}}{2} = 1.75 \times 10^{-3} \text{ mol}

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.
Techniques used
apply the stoichiometric ratio from the balanced equation
(iii)

Calculate how many moles of H2SO4\text{H}_2\text{SO}_4 were present in 25.0 cm325.0\text{ cm}^3 of the undiluted solution FA 1.

DifficultyMedium-Easy
Worked solution

Working

FA 3 is a 10-fold dilution of FA 1 (25.0 cm³ made up to 250 cm³).

moles H2SO4 in 25.0 cm3 FA 1=1.75×103×10=1.75×102 mol\text{moles H}_2\text{SO}_4 \text{ in 25.0 cm}^3 \text{ FA 1} = 1.75 \times 10^{-3} \times 10 = 1.75 \times 10^{-2} \text{ mol}

Answer

1.75×1021.75 \times 10^{-2} mol

Final answer

1.75 × 10⁻² mol

Detailed explanation

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

moles H2SO4 in 25.0 cm3 FA 1=1.75×103×10=1.75×102 mol\text{moles H}_2\text{SO}_4 \text{ in 25.0 cm}^3 \text{ FA 1} = 1.75 \times 10^{-3} \times 10 = 1.75 \times 10^{-2} \text{ mol}

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.
Techniques used
apply the dilution factor
(iv)

Calculate the concentration, in mol dm3\text{mol dm}^{-3}, of H2SO4\text{H}_2\text{SO}_4 in FA 1.

The concentration of H2SO4\text{H}_2\text{SO}_4 in FA 1 was ............... mol dm3\text{mol dm}^{-3}.

DifficultyMedium-Easy
Worked solution

Working

25.0 cm³ = 0.0250 dm³

concentration=1.75×1020.0250=0.700 mol dm3\text{concentration} = \frac{1.75 \times 10^{-2}}{0.0250} = 0.700 \text{ mol dm}^{-3}

Answer

0.7000.700 mol dm⁻³

Final answer

0.700 mol dm⁻³

Detailed explanation

Background Concept

Concentration is the amount of solute per unit volume of solution:

concentration (mol dm3)=molesvolume (dm3)\text{concentration (mol dm}^{-3}) = \frac{\text{moles}}{\text{volume (dm}^3)}

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

concentration=1.75×102 mol0.0250 dm3=0.700 mol dm3\text{concentration} = \frac{1.75 \times 10^{-2} \text{ mol}}{0.0250 \text{ dm}^3} = 0.700 \text{ mol dm}^{-3}

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.
Techniques used
calculate concentration from moles and volume

The rest of this paper

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  • Q3Qualitative Analysis · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation14M
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