Chemistry 9701/33 — May/June 2022

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

questions

marks

120

minutes

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

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

Acids donate protons, $\text{H}^+$ , in aqueous solution. The number of moles of $\text{H}^+$ donated per mole of acid is the proticity of the acid. In this experiment, you will carry out a titration to determine the proticity of phosphoric acid, $\text{H}_3\text{PO}_4$ , when it reacts with sodium hydroxide, $\text{NaOH}$ .

FA 1 is aqueous phosphoric acid, containing $6.86 \text{ g dm}^{-3}$ $\text{H}_3\text{PO}_4$ .
FA 2 is $0.150 \text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FA 3 is thymolphthalein indicator.

(a)

Method

Fill the burette with FA 2.
Pipette $25.0 \text{ cm}^3$ of FA 1 into a conical flask.
Add a few drops of FA 3.
Perform a rough titration and record your burette readings in the space below.

The rough titre is .............................. $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record in a suitable form below all your burette readings and the volume of FA 2 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Fill the burette with FA 2, run out any air below the tap, and record the initial burette reading.
Pipette 25.0 cm^3 of FA 1 into a conical flask and add a few drops of FA 3 (thymolphthalein).
Run FA 2 in from the burette, swirling the flask, until the indicator just changes colour; record the final burette reading and the titre.
Repeat accurate titrations until at least two titres agree within 0.10 cm^3 (concordant results).
Record all readings in a table with correct headings and units, e.g. initial burette reading / cm^3, final burette reading / cm^3, titre / cm^3.
Record every burette reading to the nearest 0.05 cm^3. Include the rough titration readings and titre as well.

Final answer

Candidate-dependent: two burette readings and a titre for the rough run, then at least two accurate titres within 0.10 cm^3, each reading recorded to 0.05 cm^3 with units in the table.

Detailed explanation

Background Concept

A titration measures the volume of a solution of known concentration needed to react exactly with a known portion of another solution. Here FA 2 (NaOH, 0.150 mol dm^{-3}) is run from a burette into a measured 25.0 cm^3 portion of FA 1 (H3PO4). The indicator thymolphthalein signals the end point — essentially the equivalence point — by a sharp colour change (colourless to blue in alkaline solution). Because the end point is detected visually, repeats are needed so the uncertainty of judgement is averaged out. The burette is graduated in 0.1 cm^3 divisions, so a careful reader records to the nearest 0.05 cm^3 (half a division). Consistency (concordance) between repeats is the evidence that the technique is reliable.

Understanding the Question

Part (a) is a hands-on practical task worth 7 marks. The candidate performs rough and accurate titrations and records all readings. The marks reward: recording two burette readings and the titre for the rough titration; recording initial and final readings for at least two accurate titrations; correct table headings with units; reading the burette to 0.05 cm^3; and achieving at least two accurate titres within 0.10 cm^3. Part of the accuracy is then judged against the supervisor's own mean titre, so the care shown actually earns extra marks.

Approach

The method has three stages. First, a rough titration, done quickly, finds the approximate end-point volume. Second, accurate titrations are performed carefully, slowing to drops near the end point so the colour change is judged precisely. Third, repeats continue until two titres agree within 0.10 cm^3. All data are then recorded in a neat table. This is the standard discipline for any acid–base titration in the practical examination.

Step-by-Step Reasoning

Fill the burette with FA 2, open the tap briefly to clear air from the jet, then record the initial reading — if the meniscus sits on a graduation, record that exact value; otherwise estimate to the nearest 0.05 cm^3.
Pipette 25.0 cm^3 of FA 1 into a clean conical flask and add a few drops of thymolphthalein. Swirl to mix.
For the rough run, add FA 2 fairly quickly until the indicator changes colour; note this titre as a guide.
For each accurate run, refill the burette, record the initial reading, and add FA 2 while swirling continuously. Near the end point, add one drop at a time until the solution just stays coloured.
Record the final reading and compute titre = final − initial.
Repeat until at least two accurate titres agree within 0.10 cm^3.
Present the results in a table with clearly labelled columns, each carrying a unit (cm^3): initial reading, final reading, titre.

Key Takeaways

The standard titration discipline — clear the jet, read at eye level to 0.05 cm^3, swirl constantly, add dropwise near the end, repeat until concordant, and record with units — is what earns the practical marks.

Common Mistakes

Recording only the titre and not the two burette readings for each run.
Missing units in the column headings.
Recording readings to 0.1 cm^3 instead of estimating to 0.05 cm^3.
Stopping after a single accurate titration instead of obtaining concordant results.
Reading the burette from an angle, giving parallax error.

Things to Be Careful About

The titre is always final minus initial. The rough titre is not part of the accurate set. If the burette tip is touched against the flask wall, that drop has been delivered and must not be wiped off. The target is concordance within 0.10 cm^3 for the accurate titres.

Techniques used

fill and clamp a buretteperform a rough titrationrepeat accurate titrations to concordancerecord burette readings to the nearest 0.05 cm^3construct a results table with headings and units

(b)

From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtained the mean value.

$25.0 \text{ cm}^3$ of FA 1 required .............................. $\text{cm}^3$ of FA 2.

DifficultyMedium-Easy

Worked solution

Working

Select the accurate titres that lie within 0.20 cm^3 of one another, e.g. 23.35, 23.30 and 23.35 cm^3.

\text{mean titre} = \frac{23.35 + 23.30 + 23.35}{3} = \frac{70.00}{3} = 23.33\text{ cm}^3

Answer

23.33 cm^3 (to 2 decimal places)

Final answer

Mean titre, e.g. 23.33 cm^3 (to 2 dp), from accurate titres within 0.20 cm^3 spread

Detailed explanation

Background Concept

The mean of several repeat titrations is more reliable than any single value because random errors — such as slight differences in judging the end point — tend to cancel out. Only concordant values should be averaged.

Understanding the Question

From the accurate titres recorded in part (a), choose the values that are consistent and calculate their mean to two decimal places, showing the working. The mark requires that the titres being averaged lie within a total spread of 0.20 cm^3.

Approach

Exclude the rough titre, identify the accurate titres that agree closely, add them together and divide by how many there are.

Step-by-Step Reasoning

With example accurate titres of 23.35, 23.30 and 23.35 cm^3, the spread is 23.35 − 23.30 = 0.05 cm^3, comfortably inside the 0.20 cm^3 limit.

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

The mean is quoted to 2 dp, a sensible precision for a mean of readings taken to 0.05 cm^3. If one reading is clearly inconsistent with the others (an outlier), it is discarded and the mean of the concordant set is used instead.

Key Takeaways

The mean titre smooths random error; concordance decides which values are averaged; the result is quoted to 2 dp.

Common Mistakes

Including the rough titre in the average; averaging values that differ by more than 0.20 cm^3; rounding the mean to one decimal place or quoting too many figures.

Things to Be Careful About

Show the addition and the division so the working is visible, and quote the final mean to 2 dp.

Techniques used

select concordant titres within 0.20 cm^3calculate the mean titre to 2 decimal places

(c)

Calculations

(i)

Calculate the amount, in mol, of sodium hydroxide present in the volume of FA 2 calculated in (b).

amount of $\text{NaOH}$ = .............................. mol

DifficultyEasy

Worked solution

Working

\text{amount of NaOH} = 0.150 \times \frac{23.33}{1000} = 3.4995 \times 10^{-3}\text{ mol}

Answer

$3.50 \times 10^{-3}$ mol (3 s.f.)

Final answer

3.50 x 10^-3 mol

Detailed explanation

Background Concept

Amount (mol) = concentration (mol dm^{-3}) × volume (dm^3). Because burette volumes are measured in cm^3, the volume must be divided by 1000 before multiplying by the concentration.

Understanding the Question

This is the first of four calculation parts. Using the mean titre V (in cm^3) from part (b), find the number of moles of NaOH delivered into the flask.

Approach

Substitute V into amount = 0.150 × (V/1000).

Step-by-Step Reasoning

With the example V = 23.33 cm^3:

\text{amount} = 0.150 \times \frac{23.33}{1000} = 3.4995 \times 10^{-3}\text{ mol}

The answer is quoted to 3 significant figures as $3.50 \times 10^{-3}$ mol; the mark scheme accepts 3 or 4 significant figures.

Key Takeaways

The cm^3 volume is divided by 1000 before multiplying by the concentration; the result is given to 3 or 4 significant figures.

Common Mistakes

Forgetting the ÷1000 conversion, which gives an answer 1000 times too large; using the rough titre instead of the mean titre.

Things to Be Careful About

Use the mean titre from part (b). If a candidate's own titre differs, the method is identical — error carried forward applies to this and the following parts.

Techniques used

convert volume from cm^3 to dm^3calculate amount in mol from concentration and volume

(ii)

Use the information on page 2 to calculate the amount, in mol, of phosphoric acid present in $25.0 \text{ cm}^3$ of FA 1.

amount of $\text{H}_3\text{PO}_4$ = .............................. mol

DifficultyMedium-Easy

Worked solution

Working

Concentration of FA 1:

\frac{6.86}{98.0} = 0.0700\text{ mol dm}^{-3}

Amount in 25.0 cm^3:

0.0700 \times \frac{25.0}{1000} = 1.75 \times 10^{-3}\text{ mol}

Answer

$1.75 \times 10^{-3}$ mol

Final answer

1.75 x 10^-3 mol

Detailed explanation

Background Concept

FA 1 is stated as a mass concentration in g dm^{-3}, meaning 6.86 g of H3PO4 in every dm^3. Converting to a molar concentration requires the molar mass of phosphoric acid, Mr(H3PO4) = 3(1.0) + 31.0 + 4(16.0) = 98.0 g mol^{-1}.

Understanding the Question

Find the amount of H3PO4 present in the 25.0 cm^3 portion that was pipetted into the flask. The solution in the flask has the same concentration as FA 1.

Approach

Convert g dm^{-3} to mol dm^{-3} by dividing by Mr, then multiply by the volume in dm^3.

Step-by-Step Reasoning

\text{concentration} = \frac{6.86}{98.0} = 0.0700\text{ mol dm}^{-3}

\text{amount in 25.0 cm}^3 = 0.0700 \times \frac{25.0}{1000} = 1.75 \times 10^{-3}\text{ mol}

Equivalently, 25.0 cm^3 is 1/40 of a dm^3, so amount = 6.86 × (1/40) / 98.0 = $1.75 \times 10^{-3}$ mol.

Key Takeaways

Molar concentration = mass concentration ÷ Mr, then amount = concentration × volume in dm^3.

Common Mistakes

Using the wrong Mr; forgetting to divide the 25.0 cm^3 volume by 1000; rounding too early and losing the $1.75 \times 10^{-3}$ value.

Things to Be Careful About

The Mr of H3PO4 is 98.0 g mol^{-1}. This value does not depend on the candidate's titre, so it must be the same for every candidate.

Techniques used

convert mass concentration to molar concentration using Mrcalculate amount in a fixed volume of solution

(iii)

Deduce whether phosphoric acid behaves as a monoprotic, diprotic or triprotic acid in this titration. Explain your reasoning.

$\text{H}_3\text{PO}_4$ is a .............protic acid.

explanation

DifficultyMedium-Easy

Worked solution

Answer

H3PO4 is a diprotic acid.

Mole ratio NaOH : H3PO4 = $3.50 \times 10^{-3} : 1.75 \times 10^{-3}$ = 2 : 1, so two H^+ ions are donated (neutralised) per molecule of acid.

Final answer

Diprotic

Detailed explanation

Background Concept

Proticity is the number of H^+ ions an acid molecule donates in a given reaction. NaOH supplies one OH^- per formula unit, so the number of moles of NaOH reacted equals the number of moles of H^+ neutralised. Comparing moles of NaOH with moles of H3PO4 therefore gives the number of H^+ donated per acid molecule — the proticity.

Understanding the Question

Deduce whether H3PO4 behaves as mono-, di- or triprotic in THIS titration, and justify the answer using the values obtained in (c)(i) and (c)(ii).

Approach

Compute the mole ratio NaOH : H3PO4 from the two amounts.

Step-by-Step Reasoning

Moles of NaOH = $3.50 \times 10^{-3}$ mol; moles of H3PO4 = $1.75 \times 10^{-3}$ mol.

\text{ratio} = \frac{3.50 \times 10^{-3}}{1.75 \times 10^{-3}} = 2

Two moles of NaOH react with one mole of H3PO4, so each acid molecule donates two H^+ ions and the acid is diprotic. (The third proton of H3PO4 is not removed in this titration, because the indicator changes colour at the point where two protons have been neutralised.)

Key Takeaways

The base-to-acid mole ratio directly reveals the number of H^+ donated, i.e. the proticity of the acid in the given reaction.

Common Mistakes

Saying triprotic simply because the formula H3PO4 contains three hydrogen atoms; failing to justify the deduction with the numerical ratio.

Things to Be Careful About

Use the candidate's own values from (c)(i) and (c)(ii) (error carried forward), and state the 2 : 1 ratio explicitly in the explanation.

Techniques used

compute the mole ratio of NaOH to H3PO4deduce proticity from stoichiometry

(iv)

Give the equation for this reaction of phosphoric acid, $\text{H}_3\text{PO}_4$ , with sodium hydroxide.

DifficultyEasy

Worked solution

Answer

\text{H}_3\text{PO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + 2\text{H}_2\text{O}

Final answer

H3PO4 + 2NaOH -> Na2HPO4 + 2H2O

Detailed explanation

Background Concept

A balanced neutralisation equation must conserve atoms and charge and must reflect the stoichiometry established in (c)(iii). For a diprotic acid, each molecule provides two H^+ which combine with two OH^- from the base to form two water molecules.

Understanding the Question

Write the equation for the reaction of H3PO4 with NaOH that follows from the diprotic behaviour deduced in part (c)(iii).

Approach

With H3PO4 donating two H^+, the anionic residue is HPO4^{2-}, which pairs with two Na^+ to give Na2HPO4, and two waters are formed.

Step-by-Step Reasoning

Each H3PO4 loses two protons, leaving HPO4^{2-}; this combines with two Na^+ to give Na2HPO4. The two NaOH supply two OH^-, which combine with the two H^+ to give 2H2O.

\text{H}_3\text{PO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + 2\text{H}_2\text{O}

Checking balance: H 5 = 5, O 6 = 6, Na 2 = 2, P 1 = 1, so the equation is balanced.

Key Takeaways

The equation must follow the proticity determined experimentally; every element must balance.

Common Mistakes

Writing Na3PO4 (which would imply a triprotic reaction) or NaH2PO4 (monoprotic); leaving an unbalanced equation.

Things to Be Careful About

The salt retains one proton as HPO4^{2-}, because only two of the three H^+ of phosphoric acid are neutralised in this titration.

Techniques used

write and balance the neutralisation equation

(d)

(i)

A student uses a pipette that is labelled $25.0 \pm 0.06 \text{ cm}^3$ to measure FA 1.

Calculate the maximum percentage error in the volume of FA 1. Show your working.

maximum percentage error = ..............................%

DifficultyEasy

Worked solution

Working

\text{maximum percentage error} = \frac{0.06}{25.0} \times 100 = 0.24\%

Answer

0.24%

Final answer

0.24%

Detailed explanation

Background Concept

The absolute uncertainty of a single measurement made with a calibrated piece of apparatus is the tolerance quoted by the manufacturer — here ±0.06 cm^3 for the pipette. Percentage error converts this into a relative measure: (uncertainty ÷ measured value) × 100.

Understanding the Question

Compute the maximum percentage error when a pipette of tolerance ±0.06 cm^3 is used to measure 25.0 cm^3 of FA 1.

Approach

Divide the tolerance by the volume and multiply by 100.

Step-by-Step Reasoning

\text{percentage error} = \frac{0.06}{25.0} \times 100 = 0.24\%

Because a pipette delivers a fixed volume in a single operation, the tolerance is not doubled.

Key Takeaways

For a fixed-volume pipette, percentage error = tolerance ÷ volume × 100.

Common Mistakes

Doubling the error (doubling applies to a burette, which is read twice, not to a pipette); omitting the × 100 so the answer comes out as 0.0024 rather than 0.24%.

Things to Be Careful About

0.06 ÷ 25.0 = 0.0024, and multiplying by 100 gives 0.24%.

Techniques used

calculate percentage error from tolerance and measured volume

(ii)

The student suggests it would be more accurate to measure the volume of FA 1 with a burette instead of the pipette.

State whether you agree with the student. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Disagree.

A burette is read twice (initial and final), so its total error is 2 × 0.05 = 0.10 cm^3, i.e. 0.40%. The pipette error is 0.06 cm^3, i.e. 0.24%. Since 0.40% > 0.24%, the pipette is the more accurate way to measure 25.0 cm^3 of FA 1.

Final answer

Disagree — burette error 0.10 cm^3 (0.40%) > pipette error 0.06 cm^3 (0.24%)

Detailed explanation

Background Concept

A burette is read twice to obtain a titre (initial and final), so its uncertainty is 2 × (±0.05) = ±0.10 cm^3. A pipette measures a fixed volume in one operation, so its tolerance of ±0.06 cm^3 is not doubled.

Understanding the Question

Decide whether measuring the 25.0 cm^3 of FA 1 with a burette instead of a pipette would be more accurate, and justify the answer.

Approach

Compare the percentage errors of the two ways of measuring 25.0 cm^3.

Step-by-Step Reasoning

Burette: uncertainty of 0.10 cm^3 gives 0.10/25.0 × 100 = 0.40%. Pipette: 0.06/25.0 × 100 = 0.24%. Because 0.40% > 0.24%, the burette is less accurate for delivering exactly 25.0 cm^3, so the student is incorrect. The fixed-volume pipette is the more precise tool for a specific volume.

Key Takeaways

Uncertainties add across multiple readings; a fixed-volume pipette is more precise than a burette for delivering one particular volume.

Common Mistakes

Agreeing with the student; quoting only one burette reading (±0.05 cm^3) instead of two; giving a conclusion without the numeric comparison.

Things to Be Careful About

State both the conclusion (disagree) and the quantitative reason (0.40% vs 0.24%) to secure the mark.

Techniques used

double the single burette reading uncertaintycompare percentage errors of pipette and burette

The rest of this paper

2 more questions

Q2Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis12M
Q3Qualitative Analysis · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation14M

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