Chemistry 9701/35 — May/June 2021

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

questions

marks

120

minutes

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationQualitative AnalysisFree sample

In this experiment you will carry out a titration to determine the relative formula mass, $M_r$ , of a monoprotic acid, HX.

FA 1 is HX, the monoprotic acid.
FA 2 is $0.0450\text{ mol dm}^{-3}$ sodium carbonate, $\text{Na}_2\text{CO}_3$ .
methyl orange indicator

(a)

Method

Preparing a solution of FA 1

Weigh the empty $250\text{ cm}^3$ beaker. Record the mass.
Transfer all the FA 1 into the beaker.
Weigh the beaker and FA 1. Record the mass.
Calculate and record the mass of FA 1 used.
Add approximately $100\text{ cm}^3$ of distilled water to the FA 1 in the beaker.
Stir the mixture with a glass rod until all the FA 1 has dissolved.
Transfer this solution into the $250\text{ cm}^3$ volumetric flask.
Wash the beaker with distilled water and transfer the washings to the volumetric flask.
Rinse the glass rod with distilled water and transfer the washings to the volumetric flask.
Make up the solution in the volumetric flask to the mark using distilled water.
Shake the flask thoroughly.
This solution of HX is FA 3. Label the flask FA 3.

Results

Titration

Fill the burette with FA 3.
Pipette $25.0\text{ cm}^3$ of FA 2 into a conical flask.
Add several drops of methyl orange indicator.
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 of your burette readings and the volume of FA 3 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

The exact values are the candidate's own readings. A model set of results is shown below.

Mass of FA 1

Reading	$\text{mass / g}$
mass of empty beaker	31.25
mass of beaker + FA 1	33.43
mass of FA 1 used	2.18

Titration results

	rough	accurate 1	accurate 2
final burette reading / $\text{cm}^3$	26.55	25.40	25.55
initial burette reading / $\text{cm}^3$	1.05	0.40	0.55
titre / $\text{cm}^3$	25.50	25.00	25.00

The rough titre is 25.50 $\text{cm}^3$ . All burette readings are recorded to 0.05 $\text{cm}^3$ , and the two accurate titres are concordant (within 0.10 $\text{cm}^3$ ).

Final answer

Candidate-dependent: mass FA1 = 2.18 g; rough titre = 25.50 cm3; accurate titres = 25.00 and 25.00 cm3 (example)

Detailed explanation

Background Concept

This is a quantitative volumetric titration. A solid acid FA 1 is weighed accurately, dissolved and made up to exactly 250 cm3 in a volumetric flask to give FA 3. A fixed volume of sodium carbonate FA 2 in a conical flask is titrated with FA 3 from a burette. Using methyl orange, the indicator changes colour at the end-point where the acid has neutralised the carbonate. An accurate relative formula mass depends on four things: a precise mass of FA 1, an accurate total volume, concordant burette readings and correct stoichiometry.

Understanding the Question

The command words are in the procedure, not in a question sentence. You are being asked to carry out the practical and to record your readings in the spaces provided. The marks test technique: headings with units, precise readings, more than one accurate titration and consistency of titres. There is no single correct answer; your own readings are used later.

Approach

Weigh the empty beaker, add all of FA 1 and weigh again. Stir the solid in about 100 cm3 distilled water until dissolved. Transfer the solution quantitatively to the 250 cm3 volumetric flask, rinsing the beaker and stirring rod with distilled water and adding the washings. Make up to the mark with distilled water and shake. Then pipette 25.0 cm3 of FA 2 into a conical flask, add methyl orange and titrate against FA 3 from the burette. Do a rough titration first, then accurate titrations until at least two are within 0.10 cm3. Record all readings with initial and final burette readings and the titre.

Step-by-Step Reasoning

Masses should be recorded to the full precision of the balance, for example 31.25 g and 33.43 g. The mass of FA 1 used is the difference, here 2.18 g. In the titration table, each column must have a heading and a unit; burette readings are recorded to 0.05 cm3, for example 25.40 cm3, and the titre is final minus initial. The rough titre is 25.50 cm3; the accurate titres are 25.00 and 25.00 cm3. Repeating until concordant values are obtained is essential; these two accurate titres are within 0.10 cm3 and can be averaged later.

Key Takeaways

Accurate titrations require headings and units, readings to the precision of the burette, and repeat titrations that agree closely. The rough titre is only for checking the volume range; the accurate titres are used in the mean.

Common Mistakes

Writing weight instead of mass; missing units from headings or values; recording only final burette readings without initial readings; performing only one accurate titration; using the rough titre as an accurate value; recording the titre heading as difference or total volume. All of these lose marks.

Things to Be Careful About

Read the burette at eye level at the bottom of the meniscus. Rinse the stirring rod and beaker into the flask so no acid is lost. Fill the volumetric flask to the bottom of the meniscus at the mark. Make sure the solid is fully dissolved before making up to the mark. Use a white tile to see the indicator colour change clearly.

Techniques used

measure mass by difference using a balanceprepare a solution in a volumetric flaskperform a titration with a burette and pipetterecord burette readings to 0.05 cm3repeat accurate titrations until concordant

(b)

From your accurate titration results, obtain a suitable value for the volume of FA 3 to be used in your calculations.
Show clearly how you obtained this value.

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

DifficultyEasy

Worked solution

Working

The two accurate titres are 25.00 $\text{cm}^3$ and 25.00 $\text{cm}^3$ ; they are concordant.

\text{mean titre} = \frac{25.00 + 25.00}{2} = 25.00\text{ cm}^3

Answer

25.00 $\text{cm}^3$ of FA 3 (candidate's own mean of concordant accurate titres).

Final answer

25.00 cm3 (example)

Detailed explanation

Background Concept

A mean titre is calculated only from accurate titres that agree with each other. Burette readings are taken to the nearest 0.05 cm3, and concordant values indicate a reliable end-point. The mean is used for all later calculations.

Understanding the Question

Part (b) asks for the volume of FA 3 to be used in the calculations, obtained from your accurate titration results. The marks are gained by showing how you selected the value, for example by ticking the two concordant readings or writing the calculation, and by quoting the mean to 0.01 cm3.

Approach

Choose two or more accurate titres within a total spread of no more than 0.20 cm3. Add them and divide by the number of readings. Write the mean with two decimal places.

Step-by-Step Reasoning

Here the accurate titres are 25.00 and 25.00 cm3. They are concordant. The mean is (25.00 + 25.00)/2 = 25.00 cm3. This value is then used in part (c)(iii). If the accurate titres were, for example, 24.90 and 25.10, the mean would still be 25.00, provided the spread is no more than 0.20 cm3.

Key Takeaways

Always average concordant accurate readings, not the rough titre. Quote the mean to 0.01 cm3 and show which readings were used.

Common Mistakes

Including the rough titre in the mean; averaging two titres that differ by more than 0.20 cm3; quoting the mean to only one decimal place; not showing which two readings were used.

Things to Be Careful About

Prefer the two accurate titres that are closest together. Use titres recorded to 0.05 cm3 and give the final mean to the nearest 0.01 cm3.

Techniques used

select concordant accurate titrescalculate the mean titre to two decimal places

(c)

Calculations

(i)

Calculate the number of moles of sodium carbonate present in the volume of FA 2 used in each titration.

\text{moles of } \text{Na}_2\text{CO}_3 = .............................. \text{ mol}

DifficultyEasy

Worked solution

Working

n(\text{Na}_2\text{CO}_3) = cV = 0.0450 \times \frac{25.0}{1000} = 0.001125\text{ mol}

Answer

$0.001125\text{ mol}$ ( $1.13 \times 10^{-3}\text{ mol}$ to 3 s.f.)

Final answer

0.001125 mol

Detailed explanation

Background Concept

Concentration is amount of solute per unit volume: n = cV, with volume in dm3. Since 1 cm3 = 1 × 10^-3 dm3, 25.0 cm3 = 0.0250 dm3.

Understanding the Question

FA 2 is 0.0450 mol dm^-3 sodium carbonate. A 25.0 cm3 pipette transfers this solution into the conical flask. The question asks for the number of moles of Na2CO3 present in each titration.

Approach

Convert the volume from cm3 to dm3 by dividing by 1000, then multiply by the concentration.

Step-by-Step Reasoning

n = 0.0450 × 25.0/1000 = 0.001125 mol. To three significant figures, consistent with the data, this is 0.00113 mol or 1.13 × 10^-3 mol. The mark scheme allows 0.001125 or 0.00113.

Key Takeaways

Always use dm3 in n = cV. Keep enough significant figures in intermediate values so later calculations are not distorted.

Common Mistakes

Forgetting to convert cm3 to dm3; using 25 instead of 25.0/1000; quoting too few significant figures.

Things to Be Careful About

Use the exact volume 25.0 cm3 as 0.0250 dm3, and keep the value 0.001125 for the later calculation even if you quote 0.00113 in the final line.

Techniques used

convert volume from cm3 to dm3calculate moles from concentration and volume

(ii)

Give the equation for the reaction of FA 2, $\text{Na}_2\text{CO}_3$ , with FA 3, HX.

\text{.....Na}_2\text{CO}_3 + \text{.....HX} \rightarrow \text{.....NaX} + ......................... + .........................

Use your answer to (c)(i) to deduce the number of moles of HX present in the volume you calculated in (b).

\text{moles of HX} = ...................... \text{ mol}

DifficultyMedium-Easy

Worked solution

Working

The balanced equation is:

\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HX}(\text{aq}) \rightarrow 2\text{NaX}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

Mole ratio $\text{Na}_2\text{CO}_3 : \text{HX} = 1 : 2$ .

n(\text{HX}) = 2 \times 0.001125 = 0.00225\text{ mol}

Answer

$0.00225\text{ mol}$ of HX

Final answer

0.00225 mol

Detailed explanation

Background Concept

A monoprotic acid releases one H+ ion per molecule of acid. Carbonate, CO3^2-, needs two H+ ions to form CO2 and H2O. Therefore 2 mol of a monoprotic acid react with 1 mol of sodium carbonate.

Understanding the Question

Part (c)(ii) first asks for the balanced equation for the reaction of Na2CO3 with HX. It then asks you to use that equation and the moles from part (c)(i) to calculate the moles of HX in the volume of FA 3 that you chose in part (b).

Approach

Write the neutralisation equation. The carbonate provides one CO3^2- ion, so two HX molecules are needed. Then multiply the moles of Na2CO3 by 2.

Step-by-Step Reasoning

The balanced equation is Na2CO3(aq) + 2HX(aq) → 2NaX(aq) + CO2(g) + H2O(l). From (c)(i), n(Na2CO3) = 0.001125 mol. Because the ratio is 1:2, n(HX) = 2 × 0.001125 = 0.00225 mol. This is the moles in the mean titre volume, not in the whole 250 cm3 flask.

Key Takeaways

For a monoprotic acid reacting with carbonate, the acid-to-carbonate mole ratio is 2:1. The equation must be balanced before using the ratio.

Common Mistakes

Writing a 1:1 ratio; writing an unbalanced equation; omitting state symbols when required; using the moles of Na2CO3 without doubling for HX.

Things to Be Careful About

If you quoted 0.00113 in part (c)(i), doubling gives 0.00226 mol; the mark scheme accepts 0.00226 as well. Use the value that is consistent with your own working.

Techniques used

write a balanced acid-carbonate equationuse the stoichiometric ratio to find moles of acid

(iii)

Use your answer to (c)(ii) and your data recorded on page 2 to calculate the relative formula mass, $M_r$ , of HX.
Show your working.

M_r \text{ of HX} = ..............................

DifficultyMedium-Easy

Worked solution

Working

Moles of HX in the whole $250\text{ cm}^3$ of FA 3:

n(\text{HX in }250\text{ cm}^3) = 0.00225 \times \frac{250}{25.00} = 0.0225\text{ mol}

Using mass of FA 1 = 2.18 g (candidate's value):

M_r = \frac{\text{mass of FA 1}}{\text{moles of FA 1 in }250\text{ cm}^3} = \frac{2.18}{0.0225} = 96.9\text{ (3 s.f.)}

Answer

$M_r(\text{HX}) = 96.9$ (candidate-dependent; use own mass and titre)

Final answer

96.9

Detailed explanation

Background Concept

Relative formula mass is defined as mass per mole: Mr = mass/amount, in g mol^-1. The mass of FA 1 that you weighed is the mass in the whole 250 cm3 solution. The moles calculated in (c)(ii) are only in the small titre volume, so you must scale them up to 250 cm3 before using them with the weighed mass.

Understanding the Question

This part asks you to use the moles of HX from (c)(ii) and the mass of FA 1 recorded on page 2 to calculate Mr. You must show the working and give the final value to 3 or 4 significant figures with the correct power of 10.

Approach

First scale the moles in the mean titre to the moles in 250 cm3 using the factor 250/mean titre. Then divide the mass of FA 1 by the total moles.

Step-by-Step Reasoning

The mean titre is 25.00 cm3 and n(HX) in that titre is 0.00225 mol. In 250 cm3, n = 0.00225 × 250/25.00 = 0.0225 mol. If the mass of FA 1 is 2.18 g, then Mr = 2.18/0.0225 = 96.9. If an earlier rounded value was used, the answer will be consistently slightly different; the method is what earns the marks.

Key Takeaways

Always scale the moles from the portion used in the titration to the total solution before calculating Mr. Keep 3 or 4 significant figures in the final answer.

Common Mistakes

Forgetting to scale up to 250 cm3; using the mass of the beaker instead of FA 1; dividing the mass by the moles in the titre; giving the answer to the wrong power of 10.

Things to Be Careful About

Use the mass and mean titre recorded by the candidate they are not known to the examiner. Show both steps clearly so that error-carried-forward marks can be awarded if an earlier value was slightly wrong.

Techniques used

scale moles from the titre volume to 250 cm3calculate relative formula mass from mass and total moles

(iv)

One molecule of HX contains one nitrogen atom, three oxygen atoms, three hydrogen atoms and one atom of another element, E.
The identity of E can be found by calculation.

Show this calculation and identify E.

\text{Element E is } ..............................

DifficultyMedium-Easy

Worked solution

Working

A_r(\text{E}) = M_r(\text{HX}) - [3(1) + 14 + 3(16)] = 96.9 - 65 = 31.9

The element with $A_r$ closest to 31.9 is sulfur, S ( $A_r = 32.1$ ).

Answer

Element E is S (sulfur).

Final answer

S (sulfur)

Detailed explanation

Background Concept

The formula HX contains 3 H atoms, 1 N atom, 3 O atoms and one atom of element E. The total atomic mass of the known atoms is 3(1) + 14 + 3(16) = 65. The remaining mass, Mr - 65, gives the atomic mass of E.

Understanding the Question

Given the Mr from part (c)(iii), identify the element E by calculation. This is a simple deduction: calculate the atomic mass left over and choose the element on the Periodic Table with that atomic mass.

Approach

Subtract 65 from Mr, then find the element whose Ar is closest to the result.

Step-by-Step Reasoning

Using Mr = 96.9, Ar(E) = 96.9 - 65 = 31.9. The Periodic Table gives sulfur Ar ≈ 32.1, so E is S. If the candidate's Mr is a little different, choose the element with the nearest Ar to the calculated value.

Key Takeaways

The difference between Mr and the sum of known atomic masses can identify an element. Always use the Periodic Table and look for the closest atomic mass.

Common Mistakes

Forgetting the three hydrogen atoms; subtracting only N + 3O = 62; choosing an element without checking its Ar; using an unrounded Mr and then failing to choose the nearest element.

Things to Be Careful About

Include 3 × 1 for H. Give the element symbol as well as the name, e.g. S and sulfur.

Techniques used

subtract known atomic masses from Mrmatch the remaining atomic mass to an element

(d)

What is the error in a single reading for the balance that you used?

\text{error} = .............................. \text{ g}

Calculate the maximum percentage error in the mass of FA 1 that you recorded on page 2.

\text{maximum percentage error} = .............................. \%

DifficultyEasy

Worked solution

Working

Single reading error of the balance (candidate's balance) is, for example, $\pm 0.005\text{ g}$ .

Mass of FA 1 used = 2.18 g.

\text{maximum percentage error} = \frac{2 \times 0.005}{2.18} \times 100 = 0.46\%

Answer

error = $\pm 0.005\text{ g}$ ; maximum percentage error = $0.46\%$ (candidate-dependent values)

Final answer

0.46% (example using ±0.005 g balance error and 2.18 g FA1)

Detailed explanation

Background Concept

A measuring instrument has an associated uncertainty. For a balance, each reading has an error, often half the smallest division or the value stated on the balance. The mass of FA 1 is found by taking two balance readings, so the total absolute error in the mass is 2 × balance error. Percentage error is absolute error divided by the measured quantity, multiplied by 100.

Understanding the Question

Part (d) asks you to state the error in a single reading for the balance you used, then calculate the maximum percentage error in the mass of FA 1 you recorded. The mark scheme rewards the expression (2 × balance error / mass FA1) × 100.

Approach

Identify the balance error, double it for two readings, divide by the mass of FA 1, and multiply by 100.

Step-by-Step Reasoning

For example, if the balance error is ±0.005 g and the mass of FA 1 is 2.18 g, then maximum percentage error = (2 × 0.005 / 2.18) × 100 = 0.46%. If the balance error on the candidate's balance is 0.01 g, the calculation gives 0.92%. The technique is the same.

Key Takeaways

A mass measured by difference carries the uncertainty of two separate readings, hence the factor of 2. Percentage error relates the absolute uncertainty to the size of the measured quantity.

Common Mistakes

Using only one balance error instead of two; forgetting to multiply by 100; using the mass of the beaker rather than the mass of FA 1.

Things to Be Careful About

Use the balance error appropriate to your balance. Keep the percentage to a sensible number of significant figures and include the % sign.

Techniques used

state the balance uncertaintycalculate maximum percentage error in a two-reading mass difference

(e)

Suggest and carry out an experiment using aqueous silver nitrate to determine whether the compound AgX is soluble or insoluble in water.

method .......................................................................................................................................

....................................................................................................................................................

observations ..............................................................................................................................

conclusion ..................................................................................................................................

DifficultyEasy

Worked solution

Answer

Method

Add a small volume of FA 3 (or an aqueous solution of FA 1) to a clean test tube.
Add a few drops of aqueous silver nitrate, $\text{AgNO}_3(\text{aq})$ .
Mix well and observe.

Observation

No precipitate forms; the solution remains colourless / no visible change.

Conclusion

Since no insoluble silver salt is formed, AgX is soluble in water.

Final answer

AgX is soluble in water (no precipitate with aqueous AgNO3)

Detailed explanation

Background Concept

Aqueous silver nitrate is used in qualitative analysis to test for halide ions: silver chloride is white, silver bromide is cream and silver iodide is yellow, all insoluble. If the silver salt of the anion in the acid is soluble, no precipitate forms when AgNO3 is added to a solution containing that anion.

Understanding the Question

The question asks you to suggest and carry out an experiment to determine whether AgX is soluble or insoluble. You must give a method, an observation and a conclusion. The mark scheme requires the reagent to be added to FA 3 or to a solution of FA 1, not to the solid directly.

Approach

Add aqueous silver nitrate to a solution containing the X- ion. If a precipitate appears, an insoluble silver salt is present; if no precipitate or no colour change occurs, the silver salt is soluble.

Step-by-Step Reasoning

Place a little FA 3 in a test tube and add a few drops of AgNO3(aq). If no precipitate appears and the solution stays colourless, no insoluble silver salt has formed. Therefore AgX is soluble in water. The conclusion must be clearly linked to the observation.

Key Takeaways

A solubility test is qualitative: the appearance or absence of a precipitate is the evidence. Always state the reagent, the observation and the conclusion in the correct order.

Common Mistakes

Adding AgNO3 to the solid FA 1 instead of a solution; saying no reaction without mentioning that no precipitate formed; concluding soluble without linking it to the observation.

Things to Be Careful About

Use aqueous silver nitrate and mix the contents of the test tube. Record the colour of the solution as well as the absence of a precipitate if you can.

Techniques used

add aqueous silver nitrate to a solutionobserve whether a precipitate formsdeduce solubility from the observation

(f)

Suggest why the use of methyl orange indicator might give an inaccurate titration result.

DifficultyEasy

Worked solution

Answer

Methyl orange changes colour gradually over a pH range (about pH 3.1–4.4), so the end-point colour change is indistinct / not sharp and the volume added at the end-point is uncertain.

Final answer

Methyl orange gives a gradual/indistinct end-point over a pH range, so the titre is uncertain.

Detailed explanation

Background Concept

An indicator is used to detect the equivalence point of a titration. For accuracy, the indicator should change colour sharply at the steepest part of the pH curve, where a very small addition of titrant causes a large pH change. If the colour change is gradual or occurs over a range of pH, the exact end-point is hard to judge.

Understanding the Question

This part asks why methyl orange might give an inaccurate titration result. The expected answer is that its colour change is not sharp, or that the pH range of the colour change may not coincide with the vertical part of the pH curve.

Approach

Describe the gradual colour change of methyl orange and link it to difficulty in judging the exact end-point, which increases the uncertainty in the titre.

Step-by-Step Reasoning

Methyl orange changes from red to yellow over approximately pH 3.1 to 4.4. Because this is a range rather than a single pH, the colour passes through intermediate shades over a significant volume of added acid. The exact point at which the titration should stop is therefore hard to identify, so the titre is less reliable.

Key Takeaways

A reliable indicator must have a sharp colour change at the equivalence point. If the colour change is gradual or the indicator range is not on the steep part of the pH curve, the end-point is indistinct and the titre is inaccurate.

Common Mistakes

Simply saying the indicator is wrong; saying there is no colour change; not explaining that the gradual change makes the volume uncertain.

Things to Be Careful About

Mention either the gradual colour change over a pH range or the idea that the colour change may not lie in the vertical part of the pH curve. Either wording is acceptable.

Techniques used

relate indicator colour-change range to end-point sharpnessexplain how an indistinct end-point causes titre uncertainty

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation7M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation15M

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