Chemistry 9701/52 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 5 Planning, Analysis and Evaluation, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Aqueous iron(II) ions, Fe2+(aq), are usually kept in acidic conditions to prevent them readily oxidising to aqueous iron(III) ions, Fe3+(aq).
Fe2+(aq) ions react with Ag+(aq) ions in a redox reaction. The following equilibrium is established.
Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s)
The concentration of Fe2+(aq) at equilibrium can be found by titration with a standard solution of aqueous potassium manganate(VII), KMnO4(aq). KMnO4(aq) is deep purple in colour. The equilibrium constant for the reaction can be found using the following equation.
[Fe3+(aq)]eqm
Kc =
[Fe2+(aq)]eqm × [Ag+(aq)]eqm
A student carries out the experiment using the following instructions.
step 1 Add 100.0 cm3 of 0.200 mol dm–3 AgNO3(aq) to 100.0 cm3 of 0.200 mol dm–3 Fe(NO3)2(aq) in a 500 cm3 conical flask and stopper the flask. Label the conical flask A.
step 2 Leave conical flask A for four hours, shaking intermittently. Then leave conical flask A untouched for one hour.
step 3 Use a pipette to transfer 25.00 cm3 of the solution from conical flask A into a clean
250 cm3 conical flask. Label this conical flask B.
step 4 Add 5 cm3 of 1.00 mol dm–3 NaCl (aq) to the solution in conical flask B. A white precipitate of silver chloride forms.
step 5 Use a measuring cylinder to add 20 cm3 of 1.00 mol dm–3 sulfuric acid to conical flask B.
step 6 Rinse a burette and fill it with a standard solution of KMnO4(aq).
step 7 Add KMnO4(aq) to the mixture in conical flask B until an end-point is reached.
step 8 Empty conical flask B and rinse it with distilled water ready for the next titration.
The student repeats the titration until concordant readings are achieved.
(a) The student records their results in Table 1.1.
Table 1.1
rough titration 1 titration 2 titration 3
final burette reading / cm3 10.60 20.35 30.25 9.85
initial burette reading / cm3 0.10 10.70 20.35 0.10
titre / cm3
(i) Complete Table 1.1. [1]
(ii) Calculate a suitable mean titre to be used in the student’s calculations.
Show clearly how you obtain the mean titre.
mean titre = …… cm3 [1]
(b) State what is meant by a standard solution in step 6.
…… [1]
(c) (i) Suggest why conical flask A is left for four hours in step 2.
…… [1]
(ii) Suggest why conical flask A is not shaken during the final hour in step 2.
…… [1]
(d) Suggest why a measuring cylinder is the most appropriate apparatus to use for measuring sulfuric acid in step 5.
…… [1]
(e) State what the burette should be rinsed with in step 6.
…… [1]
(f) State the change of colour seen in the mixture in conical flask B at the end-point in step 7.
from …… to …… [1]
(g) The student repeats the experiment using KMnO4(aq) at a lower concentration. The student obtains a larger mean titre.
Suggest one reason why a larger titre is better than a smaller titre.
…… [1]
(h) The equilibrium is shown.
Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s)
When another student carries out the titration with 0.0200 mol dm–3 KMnO4(aq), the mean titre is 21.10 cm3.
The ionic equation for the reaction between MnO4–(aq) and Fe2+(aq) is shown.
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) 5Fe3+(aq) + 4H2O(l) + Mn2+(aq)
(i) Calculate the concentration of Fe2+(aq) in the equilibrium mixture.
[Fe2+(aq)]eqm = …… mol dm–3 [2]
(ii) Suggest why it is not necessary to measure the concentration of Ag+(aq) ions in the equilibrium mixture experimentally.
…… [1]
(iii) Determine the decrease in concentration of Fe2+(aq) from the initial solution. Hence, determine the concentration of Fe3+(aq) in the equilibrium mixture.
If you were unable to obtain an answer to (h)(i), use [Fe2+(aq)]eqm = 0.0804 mol dm–3.
This is not the correct answer.
[Fe3+(aq)]eqm = …… mol dm–3 [1]
(iv) Determine the value of Kc . Include units in your answer.
Kc = ……
units = ……
[2]
[Total: 15]
Mark scheme
Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.
The reaction between iodide ions, I–(aq), and aqueous hydrogen peroxide, H2O2(aq), takes place in acidic conditions.
reaction 1 2I–(aq) + H2O2(aq) + 2H+(aq) I2(aq) + 2H2O(l)
A student carries out a series of experiments to investigate the order of reaction with respect to the concentration of I–(aq) ions. The student does this by measuring the time taken for a fixed amount of iodine to form.
A known amount of aqueous thiosulfate ions, S2O32–(aq), in the reaction mixture react with I2(aq) formed in reaction 1.
reaction 2 I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O62–(aq)
After all of the S2O32–(aq) ions have reacted in reaction 2, any further I2(aq) formed is detected using starch indicator.
The following materials are used:
● 50 cm3 beaker containing the correct mass of solid potassium iodide crystals needed to make
250.0 cm3 of 0.100 mol dm–3 KI(aq)
● 0.100 mol dm–3 Na2S2O3(aq)
● 0.100 mol dm–3 H2O2(aq)
● 0.200 mol dm–3 H2SO4(aq)
● starch indicator
● distilled water.
(a) A second student looked at the equation for reaction 1 and stated the order with respect to the concentration of I–(aq) ions must be second order because the balanced equation contains 2I–(aq).
Suggest why a balanced equation alone cannot be used to determine the order of a reaction.
…… [1]
(b) Describe how the student makes 250.0 cm3 of 0.100 mol dm–3 KI(aq) starting from the sample of solid potassium iodide in the 50 cm3 beaker.
Give the name and size of any key apparatus used. Describe how the student ensures the volume is exactly 250.0 cm3.
You may wish to write your answer using a series of numbered steps.
…… [3]
(c) The student carries out Experiment 1 using the following steps.
step 1 Add 25 cm3 of 0.200 mol dm–3 H2SO4(aq) to a conical flask.
step 2 Add 20.00 cm3 of distilled water to the conical flask from a burette.
step 3 Add 5.00 cm3 of 0.100 mol dm–3 KI(aq) to the conical flask from a burette.
step 4 Add 5.00 cm3 of 0.100 mol dm–3 Na2S2O3(aq) to the conical flask from a burette.
step 5 Add 2 cm3 of starch indicator to the conical flask.
step 6 Use a burette to add 10.00 cm3 of 0.100 mol dm–3 H2O2(aq) to a small beaker.
step 7 Add the contents of the beaker to the conical flask and start a timer immediately.
Stop the timer when the starch indicates the presence of I2(aq).
The student carries out a further six experiments by repeating steps 1 to 7, using the volumes shown in Table 2.1.
Table 2.1
experiment volume of volume of volume of volume of volume of time taken
H2SO4(aq) distilled KI(aq), V Na2S2O3(aq) indicator for colour
/ cm3 water / cm3 / cm3 / cm3 change, t
/ cm3 /s
1 25 20.00 5.00 5.00 2 257
2 25 17.50 7.50 5.00 2 120
3 25 15.00 10.00 5.00 2 112
Mark scheme
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 question should be ignored.
terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).
The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically
necessary and any exceptions to this general principle will be noted.
‘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.
• 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.
PUBLISHED
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
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
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.
25 0.00 25.00 5.00 2 44
(i) State how the student could improve the reliability of the experiment.
…… [1]
(ii) State the independent variable in Experiments 1 to 7.
…… [1]
(iii) In Experiment 5, the starch indicator changed colour immediately on adding H2O2(aq).
The student realised an error had been made.
Suggest which step was missed in Experiment 5.
…… [1]
(d) The rate equation is represented as rate = k [I–]n.
● [I–] is proportional to the volume of KI(aq)
● n is the order of reaction with respect to I–
● rate is proportional to 1 / t
● log (rate) = log k + nlog [I–]
(i) Use the results from the student’s experiments in (c) to complete Table 2.2.
V is the volume of KI(aq) and t is the time taken for the colour to change.
Give all values to three significant figures.
The results for Experiment 5 should not be used.
Table 2.2
experiment V / cm3 t/s log V log (1 / t )
1 5.00 257
2 10.00 120
3 12.50 112
4 15.00 76
5 17.50 1 X X
6 20.00 56
7 25.00 44
[2]
(ii) Plot a graph on the grid to show the relationship between log V and log (1 / t ). Use a cross
(x) to plot each data point. Draw a line of best fit.
–1.60
–1.80
–2.00
log (1 / t )
–2.20
–2.40
–2.60
0.60 0.80 1.00 1.20 1.40 log V
[2]
(iii) A timing error caused the most anomalous point on the graph.
Circle this point and explain the error in timing which led to this point.
…… [1]
(iv) Use your graph to determine the gradient of the line of best fit.
State the coordinates of both points you used in your calculation. These must be selected from your line of best fit.
Give the gradient to two decimal places.
coordinates 1 …… coordinates 2 ……
gradient = ……
[2]
(v) The total percentage error from measurements is determined to be 5.25%.
The true order of reaction is 1. Use this and your gradient from (d)(iv) to determine whether the error in the experiment could be accounted for by error from measurements or is caused by other factors.
Show any working.
…… [1]
[Total: 15]
Important values, constants and standards
molar gas constant R = 8.31 J K–1 mol–1
Faraday constant F = 9.65 × 104 C mol–1
Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K)
Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)
The Periodic Table of Elements
Group
1 2 13 14 15 16 17 18
1 2
H He hydrogen helium
Key 1.0 4.0
3 4 atomic number 5 6 7 8 9 10
Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
12
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – –
87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids
Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson
– – – – – – – – – – – – – – – – –
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
– 232.0 231.0 238.0 – – – – – – – – – – –
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Mark scheme
Guidance for chemical equations
PUBLISHED
Question Answer Marks
1(a)(i) All titres to nearest 0.05 cm3 (10.50, 9.65, 9.90, 9.75) 1
1(a)(ii) (9.65 + 9.75) ÷ 2 = 9.70 (cm3) 1
1(b) A solution of known concentration 1
1(c)(i) To allow to reach equilibrium 1
1(c)(ii) To allow solid to settle 1
1(d) Acid is in excess 1
1(e) aqueous potassium manganate(VII) 1
1(f) From colourless to pale pink 1
1(g) Lower percentage error 1
1(h)(i) M1 Mol MnO4–(aq) 2
= 0.0200 × 21.10 / 1000 = 4.22 × 10–4 mol
M2 Mol Fe2+(aq) = M1 × 5 = 2.11 × 10–3 mol
AND
Concentration Fe2+(aq)
= mol Fe2+ × 1000 / 25 = 0.0844 mol dm–3
1(h)(ii) Same as [Fe2+(aq)]eqm 1
1(h)(iii) 0.100 – 0.0844 = 0.0156 1
1(h)(iv) M1 Kc = 0.0156 / (0.0844 × 0.0844) = 2.19 2
M2 dm3 mol–1
PUBLISHED
Question Answer Marks
2(a) (not all) reactants may feature in the rate determining step 1
OR
order can only be determined experimentally
2(b) M1 Dissolve the solid / potassium iodide (in the beaker) (using a small volume of) (distilled) water 3
M2 Transfer / add to a (250 cm3 volumetric) flask
AND
Rinse (with distilled water)
M3 Top the 250 cm3 volumetric flask up to mark (with distilled water)
2(c)(i) Repeat the experiment to show the results are reproducible. 1
2(c)(ii) Volume of KI 1
2(c)(iii) Step 4 1
PUBLISHED
Question Answer Marks
2(d)(i) 2
Log V Log(1 / t)
0.699 –2.41
1.00 –2.08
1.10 –2.05
1.18 -1.88
X X
1.30 –1.75
1.40 –1.64
M1 Log V Correct values to 3 SF
M2 log(1 / t) Correct values to 3 SF
2(d)(ii) M1 6 points plotted correctly 2
M2 Line of best fit drawn
2(d)(iii) Point at log V = 1.10 1
and
(recorded time is too high because) watch started too early or stopped too late
2(d)(iv) M1 Co-ordinates read and recorded correctly 2
M2 Gradient calculated to 2 decimal places
PUBLISHED
Question Answer Marks
2(d)(v) percentage error of experimental value 1
= (2(d)(iv) – 1.00) / 1.00] × 100
(assuming 2(d)(iv) is 1.10)
= [(1.10 – 1.00) / 1.00] × 100 = 10%
(10% > maximum percentage error / 5.25% )
AND
other factors (beyond measurement caused the error)
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9701/52 February/March 2023 mark scheme?+
On this page, free: the mark scheme for AS & A Level Chemistry 9701/52, February/March 2023 (9701_m23_ms_52), and the question paper (9701_m23_qp_52), to read or download, with each question's marking points written out underneath.
How many marks is 9701/52 out of?+
Chemistry 9701/52 for February/March 2023 is out of 30 marks, with 1 hour 15 minutes to answer it, across 2 questions.
Can I get my 9701/52 answers marked?+
Yes. Upload your answers to 9701/52/F/M/23 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.