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Edexcel a level
chemistry question
paper 3 june 2024 +
mark scheme

Please check the examination details below before entering your candidate information
Candidate surname

Centre Number

Other names

Candidate Number

Pearson Edexcel Level 3 GCE

Friday 21 June 2024
Morning (Time: 2 hours 30 minutes)

Chemistry

Paper
reference

9CH0/03
 

Advanced

PAPER 3: General and Practical Principles in Chemistry

You must have:
Scientific calculator, Data Booklet, ruler

Total Marks

Instructions

black ink or ball-point pen
...

• IfFillpencil
in
the
boxes at the top of this page with your name,
• centre number
and candidate number
...

• Answer
the questions in the spaces provided
• Answer
– there may be more space than you need
...

• The
marks for each question are shown in brackets
• The
– use this as a guide as to how much time to spend on each question
...

A Periodic Table is printed on the back cover of this paper
...

• Read
all your working in calculations and include units where appropriate
...


P74455A

©2024 Pearson Education Ltd
...

Write your answers in the spaces provided
...



(a) Define relative isotopic mass
...



...



...


(b) A sample of hydrogen gas is formed from atoms of three isotopes, 1H, 2H and 3H
...


(1)


...



The structures of four isomeric alkanes are shown
...




(a) Explain how London forces arise
...



...



...



...



...


(2)


...



...



...




*P74455A0332*

3

Turn over   





(c) The apparatus shown may be used in an experiment to compare how far, if at all,
a charged rod deflects different liquids
...

Cl

H

C

C

Cl

Cl
Cl

Cl

H
H

C

Cl
Cl

H

H
C
H

H
C
H

H
C

H
C

H

H
C
H
H

H

CCl4 CHCl3 C6H14


Explain whether or not each of these liquids will be deflected in this experiment
...



...



...



...




(Total for Question 2 = 8 marks)
4

*P74455A0432*




3 A student carried out an experiment to identify the Group 1 metal ion, M+, in a
sulfate, M2SO4
...
48 g sample of M2SO4 was added to about 50 cm3 of deionised water
in a beaker
...

Step 2  An excess of acidified barium chloride solution, BaCl2(aq), was added very
slowly to the solution from Step 1 to ensure the crystals of barium sulfate that
formed were as large as possible
...

Step 4  The crystals were rinsed using deionised water
...

Results


Mass of dry filter paper = 0
...
19 g



(a) Give a reason, with reference to the filtration in Step 3, why the formation
of extremely small crystals in Step 2 would lead to a lower mass of dry
barium sulfate in Step 5
...



...



...


(2)


...



...





...

M2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2MCl(aq)

[Mr of BaSO4 = 233
...






(i) Explain what effect this would have on the calculated value of the relative
formula mass of M2SO4
...



...



...




(ii) Describe how the student should amend the procedure
...



...





...




(i) Explain what causes the flame colour in such a test
...



...



...



...


(2)


...



...




(Total for Question 3 = 13 marks)



*P74455A0732*

7

Turn over   



4 This question is about the white crystalline solid, barium nitrate
...

2Ba(NO3)2(s) → 2BaO(s) + 4NO2(g) + O2(g)



Standard molar entropy data related to this reaction are shown
...
8

BaO(s)

 70
...
0

O2(g)

205
...


(5)





ΔH d = +1010 kJ mol–1

8

*P74455A0832*




(ii)
Calculate the minimum temperature, in °C, at which it is thermodynamically
feasible for barium nitrate to decompose
...


(3)

(b) Explain why calcium nitrate is less thermally stable than barium nitrate
...



...



...



...




(i) Complete the ionic equation for this redox reaction, using oxidation numbers
...

8Al +




...


OH– +


...


Al(OH)4– +

(ii) Describe a test to confirm that ammonia, NH3, has been produced
...


(1)

NH3

(1)


...




(Total for Question 4 = 13 marks)



*P74455A0932*

9

Turn over   



* 5 Compare and contrast the reactions of bromine with benzene and
of bromine with cyclohexene
...

Justify your answers
...


(6)


...



...



...



...



...



...



...



...



...



...





...



...



...



...



...



...



...



...



...



...



...



...



...




(Total for Question 5 = 6 marks)



*P74455A01132*

11

Turn over   



6 An ester Q has the molecular formula C8H16O2
...
07 g of Q formed a carboxylic acid with a 78 % yield
...




NaHCO3 + RCOOH → RCOO–Na+ + H2O + CO2
In this reaction, 269 cm3 of carbon dioxide gas was produced at room temperature
and pressure
...


(3)





12

*P74455A01232*






(b) The high resolution proton NMR spectrum of ester Q was obtained
...


H
proton environment A


O

H2C

C

C

O

H2C

CH3
H2
C

CH3

CH3

(i) Explain the expected splitting pattern for the peak due to proton
environment A circled in the structure
...



...



...

(iii) Predict the chemical shifts, relative peak areas and splitting patterns in the
high resolution proton NMR spectrum, due to the proton environments you
have labelled
...


(1)

(3)


...



...



...



...

Step 1
HBr

cyclohexene

Br

Step 2

Compound A

bromocyclohexane
Step 3

OH

Step 4
CH3CH2MgBr

1-ethylcyclohexan-1-ol
(a) Draw the mechanism for Step 1
...


cyclohexanone

(4)





O

14

*P74455A01432*




(b) Devise a reaction scheme for Step 2 and Step 3 of the synthesis
...


(5)







*P74455A01532*

15

Turn over   





(c) The reaction to produce the CH3CH2MgBr used in Step 4 requires bromoethane
and magnesium to be warmed under reflux with a suitable solvent
...


anhydrous calcium chloride

Tube X

glass wool

water out

water in

flask with solvent
and reactants
gentle heat


(i) Name the solvent used
...


16

*P74455A01632*






(ii) Explain why the anhydrous calcium chloride in Tube X is necessary
...



...



...



...


(1)


...





...




(i) Give the colour of the precipitate formed in this reaction
...




(ii) Complete the simplified mechanism for the first two steps of this reaction by
adding four curly arrows
...


R

_
O

R′

...

R


...


(4)


...



...



...



...



...




(Total for Question 7 = 22 marks)



*P74455A01932*

19

Turn over   



8 This question is about acids and bases
...






The student’s outline procedure is shown
...
0 cm3 of the ethanoic acid solution with a solution of ammonia
of known concentration, using phenolphthalein to find the end-point
...
0 cm3 of the same ethanoic acid solution to the mixture
from Step 1
...

Step 3  Record the pH of the solution from Step 2
...


(3)


...



...



...


(1)


...




(iii) Show that when the solution has been half-neutralised, the acid dissociation
constant is given by the expression
Ka = 10–pH



(3)

20

*P74455A02032*






(b) Calculate the mass of sodium ethanoate needed to be dissolved in 250 cm3 of
0
...
48
(5)



[Ka for ethanoic acid = 1
...
0 cm3 of 0
...
0 cm3 of
0
...






Calculate the pH of the resultant solution
...
00 × 10–14 mol2 dm–6]

(5)



(Total for Question 8 = 17 marks)

22

*P74455A02232*




9 Acidity in wine is mainly due to the dicarboxylic acid tartaric acid, C4H6O6
...
It also inhibits the growth of bacteria
...




(a) The procedure to find the total amount of tartaric acid in a white wine is shown
...
0 cm3 sample of the wine into a conical flask
...

Step 3  Boil the sample to remove any dissolved gases and allow it to cool
...
100 mol dm–3 sodium hydroxide solution, NaOH(aq), in a burette to
titrate the sample using a suitable indicator
...

Justify your answer
...



...



...


(2)


...



...




*P74455A02332*

23

Turn over   





(iii) A mean titre of 20
...




Calculate the concentration, in  g dm–3, of tartaric acid in the white wine
...
During the titration, the
air bubble escaped
...


(2)


...



...


24

*P74455A02432*






(b) A procedure to find the amount of sulfur dioxide, SO2 , in a wine is shown
...
0  cm3 of a white wine is added to a conical flask containing 5 cm3 of
dilute sulfuric acid and a few drops of starch indicator
...

Step 3  The contents of the flask are titrated rapidly using 0
...



The equation for the reaction that takes place is shown
...
80 cm3
...

Show by calculation that the concentration of sulfur dioxide in the white wine
is below the permitted maximum
...


(2)


...



...



...



...




(Total for Question 9 = 15 marks)

26

*P74455A02632*




10 The progress of the reaction between ethyl ethanoate and water with a
hydrochloric acid catalyst can be followed in an experiment using a titrimetric
method
...

Step 1  Add 100  cm3 of 0
...

Step 2  Using a pipette, add 5
...

Step 3  Immediately remove a 5
...

Step 4  Titrate the removed sample using 0
...

Step 5  Repeat Steps 3 and 4, removing 5
...

Step 6  After 60 minutes, heat the remaining reaction mixture under reflux for
15 minutes
...
200 mol dm–3 NaOH(aq)
...


(2)


...



...



...




(ii) Explain why Step 6 is carried out before the final sample is removed and
titrated in Step 7
...



...





...
Vfinal is the titre volume from Step 7,
Vt is the titre volume at time t
...
60

6
...
80

1
...
90

0
...
80

(i) Plot a graph of the Vfinal – Vt of sodium hydroxide against the time the reaction
sample is added to the crushed ice
...


(2)


...



...



...


(2)


...




(iv) Deduce the order of the reaction, using your answer from (b)(iii) to justify
your deduction
...



...




*P74455A02932*

29

Turn over   





(c) Ethyl ethanoate can also be hydrolysed under alkaline conditions
...

O–

O
+ OH–
O

Step 1
slow

O–

OH



OH
O

O

H2O +

O

–

O

Step 2
OH +

fast

Step 3
fast

HO

–

O

+ OH–

Deduce a rate equation that is consistent with this reaction mechanism
...


(2)


...



...



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Examiners must mark the
first candidate in exactly the same way as they mark the last
...
Candidates must be rewarded
for what they have shown they can do rather than penalised for omissions
...


•

There is no ceiling on achievement
...


•

All the marks on the mark scheme are designed to be awarded
...
e
...
Examiners should also be prepared to award zero marks if
the candidate’s response is not worthy of credit according to the mark
scheme
...


•

When examiners are in doubt regarding the application of the mark scheme
to a candidate’s response, the team leader must be consulted
...


•

Mark schemes will indicate within the table where, and which strands of
QWC, are being assessed
...
This does NOT mean giving credit for incorrect or
inadequate answers, but it does mean allowing candidates to be rewarded for answers showing correct application of
principles and knowledge
...

The mark scheme gives examiners:
• an idea of the types of response expected
• how individual marks are to be awarded
• the total mark for each question
• examples of responses that should NOT receive credit
...

( ) means that a phrase/word is not essential for the award of the mark, but helps the examiner to get the sense of the
expected answer
...

ecf/TE/cq (error carried forward) means that a wrong answer given in an earlier part of a question is used correctly in answer
to a later part of the same question
...
Make sure that the answer makes sense
...
Answers must be in the correct context
...

Full marks will be awarded if the candidate has demonstrated the above abilities
...


Question
Answer
Number
1(a)
An answer that makes reference to the following points:
•

(the relative isotopic mass is) the mass of an atom (of an
isotope)

•

relative to 1/12 of the mass of a carbon-12 atom
or
relative to a carbon-12 atom which has a mass of
(exactly) 12

Question
Answer
Number
1(b)
An answer that makes reference to the following point:
•

Additional Guidance

Mark
(2)

(1)

Allow weight for mass
Do not award reference to average/mean mass
Accept divided by 1/12 of the mass of a carbon-12
atom

(1)
Allow reference to moles such as
The mass of one mole of atoms of an isotope relative
to 1/12 of the mass of one mole of carbon-12 atoms

Additional Guidance

Mark
(1)

five / 5
(Total for Question 1 = 3 marks)

Question
Answer
Number
2(a)
An explanation that makes reference to the following points:

Additional Guidance
Allow reference to atoms throughout but for
M3 a second/different atoms should be clear

(Setting up of the dipole)
• (random) movement of electrons /
(temporary) uneven distribution of electrons /
Fluctuation of electrons

(1)

(Type of dipole)
• (this results in an) instantaneous/temporary dipole
(in the first molecule)

Allow oscillating dipole
(1) Do not award reference to permanent dipole

(Induction of a second dipole)
• (which) causes/induces a (second) dipole in a
neighbouring/adjacent molecule

Allow reference to electron density

Allow in/on another molecule for adjacent
(1) Do not award reference to permanent dipole
M3 is consequential on M2 or a near miss
Allow a labelled diagram for evidence of these
marks
Ignore any reference to attraction stated thereafter
Penalise reference to electronegativity once only

Mark
(3)

Question
Answer
Number
2(b)
An answer that makes reference to the following points:

Additional Guidance

Mark
(2)

•

compound A as London forces increase as branching decreases
or
compound A is unbranched and has greater/more London forces

Allow van der Waals’/dispersion forces/
induced dipole – dipole forces for London forces
(1) Allow ‘least branching’/ ‘longer chain’ for
unbranched

•

so surface area/points of contact (of molecules) increases

(1) Allow reference to packing of molecules more
closely/closer together
Allow compound for molecule
Ignore A molecules are more compact
Do not award ‘more electrons’
Do not award M2 if clear reference to covalent
bonds being broken/decomposition
Accept reverse argument e
...

• compound A, as the other compounds have
less London forces as branching increases
• so surface area/points of contact
(of molecules) decreases

Question
Answer
Number
2(c)
An explanation that makes reference to the following points:

Additional Guidance

Mark

Allow differences in electronegativities for references to
polar/non-polar bonds
CCl4,
X
✓

CHCl3
✓
✓

C6H14
X
*X

•

effect on the stream of liquid by CCl4, CHCl3 and C6H14 (1)

•

(because of) bond polarity in CCl4, CHCl3 and C6H14

(1)

•

(resulting in) molecular non-polar /polarity due to
shape

X
*(X)
✓
because
because
(1)
symmetrical non-symmetrical
Score either rows or columns, awarding the higher score

Deflection
Bond
polarity
Molecular
polarity

(3)

* For hexane only, allow reference to non-polar bonds to
imply non-polar molecule but not vice versa
Allow no dipole moment/vectors cancel for symmetry
Allow reference to dipoles cancel for symmetry
(Total for Question 2 = 8 marks)

Question
Number
3(a)

Answer
•

Additional Guidance

Mark
(1)

(crystals will be small enough) to get through (pores in) the filter paper

Question
Answer
Number
3(b)
An explanation that makes reference to the following points:

Ignore references to just not being filtered
Do not award references to crystals being
‘stuck’ on the filter paper or dissolving

Additional Guidance

Mark
(2)

•

to remove (excess) BaCl2(aq) / MCl(aq)

(1) Allow to remove remaining solution
Allow reference to removing filtrate
Ignore reference to residue

•

which would otherwise crystallise / add to mass (of sulfate)

(1) Allow reference to ‘change the mass’
Do not award a decrease in mass
If no other mark awarded then allow (1) for
to remove (soluble) impurities
Do not award removal of solid impurities

Question
Number
3(c)

Answer

Additional Guidance

•

calculation of moles of BaSO4

(1)

•

calculation of RFM of M2SO4

(1)

1
...
4 = 4
...
0043702 (mol)
0
...
3702 × 10–3 = 109
...
84 – (32
...
74 hence Ar of M = 6
...
9 ) so M = Li
Allow Li+
Ignore SF for M3
Allow TE from M1 to M2
Allow TE from M2 to M3 provided value is positive and linked to
an appropriate metal that can form a +1 ion
Correct answer without working scores (1)

Mark
(3)

Question
Answer
Number
3(d)(i)
An explanation that makes reference to the following points:
(some of) M2SO4 (solution) would be lost
(from the beaker)

•

so (calculated) moles/amount (of M2SO4) would be
less
and
(calculated) formula mass would be more/higher

•

Additional Guidance

Mark
(2)

Allow just some of the solution would be lost
(1)
Allow reference to moles of BaSO4 for M2SO4
Ignore less mass of BaSO4
(1)

Allow M2 for answers that show understanding of the
negligible change (of RFM)
M2 dependent upon M1 attempt
Allow M2 for a TE such as the moles of M2SO4 would be
more which results in a lower RFM provided the answer
states that the mass has increased

Question
Number
3(d)(ii)

Answer

Additional Guidance

Mark
(1)

•

rinse glass rod with deionised/distilled water
into the beaker

Allow wash for rinse
Allow ‘to the solution’ for “into the beaker”
Do not award rinsings go down the sink
References to the use of a magnetic stirrer can score if
rinsing is mentioned of the ‘flea’

Question
Answer
Number
3(e)(i)
An explanation that makes reference to the following points:

Additional Guidance
Electrons need to be mentioned once in M1 or M2

•

heat causes electrons to be promoted / excited / move
to a higher energy level

(1)

•

light is emitted when electrons fall back
(to lower level / ground state)

(1)

Mark
(2)

Allow just ‘energy (from flame) causes electrons to be
promoted/excited’
Accept reference to photons for light
Allow released/ given out for emitted
Ignore reference to just energy
Do not award reference to complementary
colours/absorption colour
Do not award reference to reflected/transmitted

Question
Answer
Number
3(e)(ii)
An answer that makes reference to the following points:
•

Advantage – simpler (process) / quicker (process)

•

Disadvantage – flame colours are subjective (by eye)
/ some metal (ions) produce very similar flame
colours (by eye) / often more than one colour may be
seen (due to contamination)

Additional Guidance

Mark
(2)

(1)

(1)

Allow just ‘easier’
Ignore just ‘cheaper’
Ignore references to more accurate
Allow several metal (ions) produce red flames / not all
metal (ions) produce a flame colour (in the visible
range)
If metals are named then their colours must be correct

(Total for Question 3 = 13 marks)

Question
Number
4(a)(i)

Answer

Additional Guidance
Example of calculation

•

calculation of ΔSosystem

(1)

((2 × 70
...
0) + 205
...
8 ) = 878
...
26 / 3389
...
38926 / 3
...
2+ (−3389
...
1 (J K–1 mol–1)
or
−2
...
8 scores (2) for M1 and M2 but the failure to convert to
consistent units means a positive ΔSototal is obtained that does not match the
question

Mark
(5)

Example of calculation
Alternative method using ΔG

((2 × 70
...
0) + 205
...
8 ) = (+) 878
...
2 = 261703
...
6
= (+) 748296
...
296 (kJ mol–1)
Penalise incorrect units for M4 only
Ignore SF except 1 SF
Allow TE from M1 to M4 provided M4 is positive

•

reason why thermally stable at 298 K

(1)

ΔG positive / > 0 so reaction is not feasible/ compound is stable (at 298 K)
Standalone mark but
Do not award on negative values for ΔGo
A negative value for ΔSosystem loses M3 to M4 so can only score M2 and M5
ΔG = − 260693
...
2)
= 1150
...

Marks are awarded for indicative content and for how the answer is
structured and shows lines of reasoning
...

Number of indicative marking Number of marks awarded for
points seen in answer
indicative marking points
6
4
5-4
3
3-2
2
1
1
0
0
The following table shows how the marks should be awarded for
structure and lines of reasoning
Number of marks awarded for
structure of answer and
sustained lines of reasoning
Answer shows a coherent logical
2
structure with linkages and fully
sustained lines of reasoning
demonstrated throughout
Answer is partially structured with
1
some linkages and lines of
reasoning
Answer has no linkages between
0
points and is unstructured

Additional Guidance

Mark

Guidance on how the mark scheme should be
applied:
The mark for indicative content should be added
to the mark for lines of reasoning
...

If there were no linkages between the points, then
the same indicative marking points would yield an
overall score of 3 marks (3 marks for indicative
content and zero marks for linkages)
...
g
...
011208 / 1
...
011208 × (100 ÷ 78) = 0
...
4370 × 10–2 (mol)

(1)
(1)

V (CO2) = 269 × (100 ÷ 78) = 344
...
87 ÷ 24000 = 0
...
437 × 10–2 (mol) = n (ester)

(1)

Mr of ester = 2
...
014370 = 144
...
07 ÷ 144 = 1
...
1
...
07
Ignore SF except 1 SF in M1 and M2
Ignore intermediate units even if incorrect
Note: Use of the formula pV=nRT with T =298 and P =1 × 105
gives a RMM=148 and scores (3)

Mark
(3)

Question
Answer
Number
6(b)(i)
An explanation that makes reference to the following points:

Additional Guidance
Ignore any comments about chemical shifts

•

(peak due to A) is a singlet

•

as there are no adjacent carbon atoms with hydrogen atoms/
as the carbon is (only) bonded to oxygen atoms

(1)

Question
Answer
Number
6(b)(ii)
An answer that makes reference to the following points:

C

B

Mark
(2)

Allow no splitting

Use of n+1 rule
(1)
Do not award if it is clear that the methanoate
carbon is being referred to as the adjacent carbon

Additional Guidance

Mark
(1)

Accept any clear means of labelling of the two different
hydrogen environments
Ignore labelling of just one CH3 and one CH2 unless it is clearly
stated that the other groups (of each sort) are equivalent

Question
Number
6(b)(iii)

Answer

Additional Guidance

Mark
(3)

•

chemical shifts

(1)

•

splitting patterns

(1)

•

relative peak areas

(1)

Chemical shift (∂)
/ ppm

Splitting
pattern of peak

Relative peak
area
6
B
0 − 1
...
9
triplet
Allow 3
Allow any single chemical shift value or range within the MS range
Allow four splits for quartet and three splits for triplet
Ignore reference to proton environment A
Additional proton environments max 1 for chemical shifts
If no other mark awarded then allow (1) for either B or C given correctly
for chemical shift and splitting and peak area (row)
(Total for Question 6 = 9 marks)

Question
Answer
Number
7(a)
An answer that makes reference to the following points:
•

curly arrow from C=C to ∂+ H in HBr

•

dipole on HBr

•

curly arrow from H-Br bond to, or just beyond, Br∂–

•

structure of carbocation

•

lone pair on Br−

•

curly arrow (from lone pair on) Br– to C+

Additional Guidance

Mark
(4)

Penalise use of half arrows once only in points 1, 3 and 6

Ignore drawing of hydrogen atoms on carbocation unless
the number of hydrogen atoms is too many

6 points scores (4)
5 points scores (3)
3 or 4 points scores (2)
2 points scores (1)
1 point scores (0)

Question
Number
7(b)

Answer

Additional Guidance

An answer that makes reference to the following points:

(5)
Each mark is standalone

• (reaction of) bromocyclohexane with KOH(aq) /NaOH(aq)

(1)

Accept aqueous ethanolic for (aq) or aqueous
Ignore reference to reflux / heating
Do not award just ethanolic KOH/ NaOH

• (nucleophilic) substitution

(1)

Accept hydrolysis

• structure of A (cyclohexanol)

(1)

Accept displayed formula
Ignore name of A, even if incorrect
• reflux / heat with sulfuric acid and sodium dichromate ((VI))

(1)

Allow acidified dichromate ((VI)) / H+ and Cr2O72–
Accept K2Cr2O7 for Na2Cr2O7
Ignore concentration
Allow distillation/distil for reflux/heat
Do not award HCl for sulfuric acid
Allow reference to oxidising agent/oxidised
Ignore redox

• oxidation

Mark

(1)

Questio
n
Number
7(c)(i)

Answer
•

Additional Guidance

(1)
Accept diethyl ether
Ignore any formulae, even if incorrect
Do not award if given with any additional substance

(dry) ether / ethoxyethane

Question
Answer
Number
7(c)(ii)
An explanation that makes reference to the following points:

Additional Guidance

Mark
(2)

•

to prevent moist air entering

(1)

Allow to keep the reactants dry
Allow reference to absorbing water
Ignore just drying agent
Do not award to remove water from the solution or
coming out from the solution

•

as Grignard reagent will not form in presence of water

(1)

Allow ‘to ensure Grignard Reagent doesn’t break
down (to an alkane)’ / Grignard reagents react with
water / avoid hydrolysis of Grignard reagents
Allow reference to ethane/alkane forming
Ignore reference to just reacting with the reactants

Question
Answer
Number
7(c)(iii)
An answer that makes reference to the following point:
•

Mark

to ensure efficient cooling / prevent air bubbles forming in condenser

Additional Guidance

Mark
(1)

Allow to fill the condenser with water

Question
Answer
Number
7(d)(i)
An answer that makes reference to the following point:
•

orange / yellow

Additional Guidance

Mark
(1)

Allow red
Allow any combination of these 3 colours
Ignore any shades

Question
Answer
Number
7(d)(ii)
An answer that makes reference to the following points:

Additional Guidance

(4)

•

arrow from C=O bond to O

(1)

•

arrow from lone pair on O− to H+

(1)

•

arrow from lone pair in O in H2O to H in NH2

(1)

•

arrow from N−H bond to N+

(1)

Exemplar diagram

Mark

Accept arrow to either of the two hydrogen atoms
Allow M2 to M4 to be drawn on the product of Step 1 if
not drawn in Step 2
Ignore any dipoles added, even if incorrect
Penalise any additional incorrect curly arrows over four

Question
Answer
Number
7(d)(iii)
An answer that makes reference to the following points:

Additional Guidance

Mark
(4)

•

solid dissolved in minimum amount of hot solvent

(1) Allow any named solvents

•

(hot filtration then) solution allowed to cool and
(re)crystallise/precipitate

M2 dependent on M1 or near miss
(1) Allow solid for crystal/ppt

•

solid filtered under reduced pressure

(1) Allow ‘Buchner filtration’ / suction filtration
Cooling and recrystalising follows M1 so M2 dependent
M3 and M4 must apply to crystals/solid being present
however derived
Filtering and drying of dissolved solid is nonsense so M3
and M4 would both be lost

•

rinsed with (cold) solvent
and
dried between (sheets of) filter paper

Allow desiccator / warm oven as alternative for (sheets of)
filter paper
(1) Allow drying with paper towels

(Total for Question 7 = 22 marks)

Question
Answer
Number
8(a)(i)
An explanation that makes reference to the following points:

Question
Number
8(a)(ii)

Additional Guidance

(3)

•

as the weak acid (is being titrated with) a weak base

(1)

•

as no rapid change in pH (to find volume at end-point)

(1)

Accept pH changes gradually around the end
point
Allow there is no vertical section in the titration
curve

•

so end-point cannot be determined

(1)

so a sharp colour change cannot be observed
Allow phenolphthalein would not change colour/
pKIn is too high

Answer
•

Mark

Additional Guidance

Mark
(1)

(replace ammonia with) a strong base such as NaOH / KOH

Allow name or formula of strong base
Allow use of a pH probe/pH meter
Do not award change of indicator

Question
Answer
Number
8(a)(iii)
An answer that makes reference to the following points:
(method 1)
• Ka expression for ethanoic acid
•

statement that at half-neutralisation concentration
of anion and acid are equal

•

Ka = [H+] (=10−pH)

Additional Guidance

(3)
(1) Ka = [CH3COO–][H+]
[CH3COOH]
(1)

or Ka = [A–][H+]
[HA]

[CH3COOH] = [CH3COO–] or [HA] = [A–]

(1) Allow pKa = pH (= − log [H+] = − log Ka)
Standalone mark
Do not award Ka = pKa
Penalise omission of square brackets once only
Penalise use of () throughout for [] once only

(method 2)
• Henderson-Hasselbalch expression

•

statement that at half-neutralisation concentration
of anion and acid are equal

•

pKa = pH

(so Ka = [H+] =10−pH)

Mark

(1) pH = pKa + log ( [A−] ) or pH = pKa + log ( [salt] )
[HA]
[acid]
or
pKa = pH − log ( [A−] )
[HA]
(1)
(1)

Question
Answer
Number
8(b)
A calculation that makes reference to the following points:

Additional Guidance
Example of calculation

•

(M1) calculation of [H+]

(1) 10–4
...
3113 × 10–5 (mol dm–3)

•

(M2) rearrangement of Ka expression

(1) Ka = ([H+][CH3COONa]) ÷ [CH3COOH]
[CH3COONa] = Ka × [CH3COOH] ÷ [H+]

•

(M3) calculation of [CH3COONa] in buffer

(1) [CH3COONa] = (1
...
52) ÷ 3
...
27325 (mol dm–3)

OR Use of Henderson Hasselbalch for M1 to M3
•

(M1) rearrangement of H−H expression

(1) log ([CH3COONa] ÷ [CH3COOH]) = 4
...
75945 = – 0
...
27945= 0
...
52547 × 0
...
27325 (mol dm–3)

--------------------------------------------------------------•

(M4) calculation of Mr of CH3COONa

(1) 82

•

(M5) calculate the mass of CH3COONa needed

(1) = (0
...
6015 (g) / 5
...
400 × 2 = 0
...
0 ÷ 1000) × 0
...
045 (mol)

•

calculate amount of excess OH–(aq) in mol

(1) = 0
...
016 = 0
...
029 ÷ (70 ÷ 1000) = 0
...
41429) = 13
...
6
or
pH = – log (1 × 10−14 ÷ 0
...
617 / 13
...
g
...
6 ÷ 1000) × 0
...
06 × 10–3 (mol)

calculation of amount of NaOH
and
amount of tartaric acid

(1)

= 2
...
03 × 10–3 (mol)

•

calculation of amount of tartaric acid in mol dm–3

(1)

1
...
0515 (mol dm–3)

•

calculation of amount of tartaric acid in g dm–3

(1)

= 0
...
725 (g dm–3)
Accept step 3 carried out before step 2
Ignore intermediate units even if incorrect
Ignore SF except 1SF
TE throughout

Question
Answer
Number
9(a)(iv)
An explanation that makes reference to the following points:

Additional Guidance

Mark
(2)

•

the titre value includes the volume of the air
bubble (as well as sodium hydroxide solution)

(1)

Allow some alkali/solution is used to fill the
air bubble/ burette tip/burette jet

•

so the titre value would be greater (than expected)

(1)

M2 is dependent upon M1 or a near miss

Question
Answer
Number
9(b)(i)
An explanation that makes reference to the following points:

Additional Guidance

Mark
(4)

•

calculation of amount of iodine

(1)

(11
...
01 = 1
...
000118 (mol)

•

calculation of concentration of SO2 in mol dm–3

(1)

1
...
36 × 10–3 / 0
...
36 × 10–3 × 64
...
15128 (g dm–3)

•

calculation of concentration of SO2 in mg dm–3
and
comment regarding allowable level

= 0
...
e

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