Cambridge International AS & A Level

CHEMISTRY 9701/33
Paper 3 Advanced Practical Skills 1 February/March 2025
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report f or
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes f or the February/March 2025 series f or most
Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level
components.

This document consists of 11 printed pages.

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Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alon gside the
specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these
marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond
the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assess ed by the
question as indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

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GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the fu ll mark range may
be limited according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

Science-Specific Marking Principles

1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be presen t, marks
should not be awarded if the keywords are used incorrectly.

2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for
any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the questi on should be ignored.

3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation fro m other
syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).

4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used i n a scientifically
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be include d in the mark scheme
where necessary and any exceptions to this general principle will be noted.

5 ‘List rule’ guidance

For questions that require n responses (e.g. State two reasons …):

• The response should be read as continuous prose, even when numbered answer spaces are provided.
• Any response marked ignore in the mark scheme should not count towards n.
• Incorrect responses should not be awarded credit but will still count towards n.
• Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this
should be treated as a single incorrect response.
• Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

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6 Calculation specific guidance

Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.

For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded
by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.

For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient ( a) to a value between 1
and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme.

Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded.
Exceptions to this general principle will be noted in the mark scheme.

7 Guidance for chemical equations

Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.

State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

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Question Answer Marks

1(a) I Six unambiguous headings for balance readings AND calculated values AND correct units 5

• (mass of) crucible (and lid)

• (mass of) crucible (and lid) and FA 1
• (mass of) crucible (and lid) and residue / contents after first heating
• (mass of) crucible (and lid) and residue / contents after second heating
• (mass of) FA 1
• (mass of) residue
• Units: / g, (g), in g (or g with every entry)

II Recording of all weighings

• All four weighings recorded to same number of decimal places (two or three).
• Fourth reading is within +0.02 and –0.05 g of third reading.

III Correct calculation of masses

Correct masses of FA 1 and residue
AND
both answers to min 2 sf.

Accuracy marks

Calculate candidate’s mass ratio (to 2 d.p.) = mass FA 1 / mass of residue.

IV award if ratio is within the range 1.70–2.00 (inclusive)

V award if ratio is within the range 1.80–1.90 (inclusive)

1(b)(i) Correct calculation of amount of water 1

amount of H2O = (mass of FA 1 – mass of residue) / 18

AND
answer to 2–4 s.f.

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Question Answer Marks

1(b)(ii) Correct use of amount of water 1

amount of potassium alum, FA 1 = (b)(i) / 12
AND
answer to 2–4 s.f.

1(b)(iii) Correct use of amount of anhydrous potassium alum 1

Mr of anhydrous alum = mass of residue / (b)(ii)

AND
answer to 2–4 s.f.

1(b)(iv) Formula is Al K(SO4)2 1

AND
(correct) working which shows use of (b)(iii)
AND
use of 27(.0) or 258.3.

1(c)(i) Correct expression 1

(Uncertainty U (for 2 d.p. balance) = 0.01)
% error in weighing FA 1 = 2  U / mass of residue  100
AND
answer given.

1(c)(ii) If anhydrous potassium alum decomposes: 1

mass loss will be higher / too high

OR no longer is only water lost
OR (calculated) moles of water / (b)(i) will be higher
AND
(calculated) amount / moles of residue / (anhydrous) potassium alum / (b)(ii) will be higher / too high.

(so Mr of residue / (b)(iii) will be too low).

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Question Answer Marks

2(a) I All the following data are recorded: 7

two burette readings AND
• titre for the rough titration
• initial and final burette readings for two (or more) accurate titrations.

II Titre values recorded for accurate titrations, and

correct headings and units in the accurate titration table

• initial / start AND (burette) reading / volume

• final / end AND (burette) reading / volume
• titre OR volume used / added OR FA 2 used / added
• unit: / cm3 OR (cm3) OR in cm3 (for each heading) OR cm3 unit given for each volume recorded.

III All accurate burette readings recorded to 0.05 cm3.

IV The final accurate titre recorded must be within 0.10 cm3 of any other accurate titre.

Accuracy marks
Round burette readings to the nearest 0.05 cm3 then check and correct titre subtractions where necessary.
Select the ‘best’ mean titre, using the following hierarchy:
• two (or more) accurate identical titres (ignoring any that are labelled “rough”) , then
• two (or more) accurate titres within 0.05 cm3, then
• two (or more) accurate titres within 0.10 cm3, etc.
The ‘best’ titres should be used to calculate the mean titre, expressed to the nearest 0.01 cm3.

Calculate the supervisor’s mean titre to 2 decimal places.

Calculate the candidate’s mean titre to 2 decimal places.
Calculate the difference  between the candidate’s titre and the supervisor’s titre.

V Award if  ⩽ 0.60 cm3

VI Award if  ⩽ 0.40 cm3
VII Award if  ⩽ 0.20 cm3

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Question Answer Marks

2(b) Correct calculation of the mean titre 1

• Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3.
• Working/explanation must be shown OR ticks must be put next to the two (or more) accurate readings selected.
• The mean should be quoted to 2 d.p. and be rounded to nearest 0.01 cm3.

2(c)(i) answers to parts (c)(ii), (c)(iii) and (c)(iv) are given to 3 or 4 significant figures. 1

2(c)(ii) Correct calculation of amount of sodium thiosulfate 1

amount of sodium thiosulfate used = 22.0 / 248.2  titre / 1000

2(c)(iii) Correct use of (c)(ii) 1

amount I2 = 0.5  (c)(ii)

2(c)(iv) Correct calculation of amount of FA 3 1

amount FA 3 = 0.0175  25 / 1000 = 0.0004375

2(c)(v) Correct use of (c)(iii) and (c)(iv) 1

amount I2 = (c)(iii) / (c)(iv)
AND
answer to 1 decimal place.

2(c)(vi) M1 Working to show increase in oxidation state for I (during reaction) 2

= (c)(v)  (+)2

M2 M changes from (2 + M1) to (+) 2

AND
both values are integers

2(d) Student is incorrect 1

AND
FA 5 / sulfuric acid is used in excess (so the exact volume does not matter).

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Question Answer Marks

FA 3 is KMnO4 FA 7 is (NH4)Fe(SO4)2 + H2SO4

3(a) Test 1 4
• (with NaOH) no change OR (solution) stays / remains purple
• (with Na2SO3) green solution (formed at first) / (solution) turns green
• (with Na2SO3) brown ppt / solid
• (with acid) turned / formed AND colourless (solution).

Test 2
• Purple turns colourless OR colourless solution formed
• Fizzing / effervescence

Test 3
• fizzing / effervescence
• (dark) brown solid / ppt forms
• (gas) re-lights glowing spill
• gas is O2

Two points needed for each mark, to a maximum of 4 marks.

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Question Answer Marks

3(b)(i) Test 1 6
• (with NaOH) brown / red-brown / rust precipitate
• (with NaOH) insoluble in excess (NaOH)
• (on warming) gas / NH3 turns (red) litmus blue
• (with Al) fizzing / effervescence

Test 2
• (with KI) (solution) turns darker yellow / yellow-brown / orange-brown / red-brown / brown
(with starch) turns dark blue / blue-black / black

Test 3
• fizzing / effervescence
(gas / H2) gives pop with lighted splint

Test 4
• (with Ba2+) white precipitate
• (with HCl) ppt remains / insoluble
• (with HCl) yellow solution formed

Test 5
• (with AgNO3) no change / no precipitate
• (with NH3) brown / red-brown / rust precipitate
• (with NH3) insoluble in excess

Two points needed for each mark, to a maximum of 6 marks.

3(b)(ii) Correct identification of ions 2

H+, NH4+, Fe3+, SO42–

4 ions correct = 2 marks

2 or 3 ions correct = 1 mark

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Question Answer Marks

3(b)(iii) One correct ionic equation: 1

Fe3+(aq) + 3OH–(aq) → Fe(OH) 3(s)

NH4+(aq) + OH–(aq) → NH3(g) + H2O(l or g)

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

Mg(s) + 2Fe3(aq) → 2Fe2+(aq) + Mg2+(aq)

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