Cambridge AS & A Level Chemistry 9701/52 is the May/June 2026 Paper 5 (Planning, Analysis and Evaluation), variant 2. It is an A Level paper, lasts 1 hour 15 minutes and is worth 30 marks. It had two compulsory questions: a 16-mark preparation and titration of a polyiodide salt, and a 14-mark rate experiment with a log-log graph. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all five 9701 papers from this series, see A Level Chemistry 9701 May/June 2026 past papers: what came up and how it was marked.

9701/52 at a glance

Question	Context	Marks	What you do
1	The polyiodide salt (CH₃)₄NI₂ₙ₊₁: making methanolic iodine, preparing and drying crystals, titrating with thiosulfate	16	Calculate, describe, suggest, tabulate
2	Rate of sodium ethanedioate with sodium peroxydisulfate: collecting CO₂, a log-log graph for the order	14	Draw, identify, plot, calculate, evaluate
	Total	30

Question-by-question walkthrough

Question 1: the polyiodide salt (16 marks)

Tetramethylammonium iodide, (CH₃)₄NI, reacts with iodine dissolved in methanol to give (CH₃)₄NI₂ₙ₊₁. You planned parts of the preparation, then used titration data to find n.

• Mass of iodine (1 mark): 0.300 × 100.0 ÷ 1000 × 253.8 = 7.61 g. Using 127 or 254 for the molar mass, or giving 7.62, was rejected.
• Making 100.0 cm³ of 0.300 mol dm⁻³ I₂ in methanol (3 marks):
  • Dissolve the iodine in the beaker in a small volume (50 cm³ or less) of methanol (1).
  • Transfer the solution to a 100 cm³ volumetric flask, rinsing the beaker with methanol into the flask (1).
  • Make up to the mark with methanol, stopper and invert (1). “Mix” or “homogenise” without saying how was ignored, and stirring in the flask was rejected.
  • Using water as the solvent capped the answer at 2, and putting the solid straight into the volumetric flask also capped it at 2. Filling the flask to the mark first and then adding the solid scored at most 1.
• No Bunsen burner (1 mark): methanol is flammable. “It is flammable”, without saying what, was ignored, and so were explosions and the flask breaking.
• Washing the crystals (1 mark): to remove soluble impurities, such as unreacted iodine or (CH₃)₄NI. “To remove methanol” or “to remove solid impurities” was rejected.
• Making sure the crystals are dry (1 mark): leave them and weigh repeatedly until the mass is constant. Heating or an oven was rejected, because step 6 involves no heating, and “constant weight” was rejected for constant mass.
• Weighing the crystals (2 marks): a results table with rows for the mass of the evaporating basin, the basin plus crystals and the crystals, each in g (1); and a method that weighs the empty basin, transfers the dry crystals into it and reweighs (1). Weighing the basin with crystals, then rinsing, drying and reweighing the basin, was also accepted. Moving the crystals into a second container to weigh them did not score.
• Rinsing the burette (1 mark): with the 0.0200 mol dm⁻³ sodium thiosulfate solution itself. “Water or thiosulfate” was too vague, and rinsing with anything after the thiosulfate was rejected.
• Percentage error in titration 1 (1 mark): two readings, each ±0.05 cm³: (2 × 0.05) ÷ 13.35 × 100 = 0.749%. Working had to be shown, and 0.0749 was rejected.
• Mean titre (1 mark): titrations 1 and 3 (13.35 and 13.25 cm³) agree within 0.10 cm³, so the mean is 13.30 cm³. The rough titre and titration 2 (13.60 cm³) are left out. “13.3” did not score.
• Mass of iodine in solution A (3 marks):
  • Thiosulfate: 13.30 × 0.0200 ÷ 1000 = 2.66 × 10⁻⁴ mol (1).
  • I₂ in 10.00 cm³: ÷ 2 = 1.33 × 10⁻⁴ mol, so 1.33 × 10⁻³ mol in 100.0 cm³ (1).
  • Mass: 1.33 × 10⁻³ × 253.8 = 0.338 g (1). The correct answer on its own scored all three.
• The value of n (1 mark): 0.338 ÷ 253.8 = 0.00133 mol of I₂. The rest of the 0.472 g is (CH₃)₄NI: 0.134 ÷ 200.9 = 0.000667 mol. The ratio is 1.99, so n = 2. Working had to be shown. The fallback mass in the question, 0.374 g, gives n = 3.

Question 2: the rate of the ethanedioate-peroxydisulfate reaction (14 marks)

With sodium ethanedioate in excess, rate = k[Na₂S₂O₈]ᵐ. A graph showed the volume of CO₂ against time, levelling off at 10 cm³ by about 1400 s, with one point below the curve at 1000 s.

• Collecting the gas (2 marks): a labelled gas syringe connected directly to the side arm (1), and a sealed flask with a bung (1). Because CO₂ dissolves in water, a measuring cylinder over a named non-aqueous liquid could replace the syringe for the first mark. Adding a reactant through the bung in a way that pushes air into the syringe was rejected.
• Independent variable (1 mark): time. “Time intervals” was rejected.
• The anomalous point (1 mark): the volume was read too early, before 1000 s, so the recorded time is later than the true time. “The timer was started or stopped too early or late” was rejected.
• Higher temperature (1 mark): a smooth curve from the origin, to the left of the original without crossing it, levelling off at 10 cm³ before 1300 s and continuing to 1600 s.
• Completing the log table (2 marks): one mark per column, every value to 2 decimal places:

[Na₂S₂O₈] / mol dm⁻³	rate / mol dm⁻³ s⁻¹	log [Na₂S₂O₈]	log (rate)
1.00 × 10⁻²	0.02760	−2.00	−1.56
5.74 × 10⁻³	0.01160	−2.24	−1.94
3.67 × 10⁻³	0.00549	−2.44	−2.26
2.88 × 10⁻³	0.00398	−2.54	−2.40
2.08 × 10⁻³	0.00250	−2.68	−2.60
1.57 × 10⁻³	0.00169	−2.80	−2.77

• Plotting (2 marks): all six points within half a small square (1) and a straight line of best fit covering all the points (1). The scheme’s guide: the line should meet the right-hand edge of the grid above −1.6, and the point at (−2.24, −1.94) should not sit well above it.
• Gradient (2 marks): two points read from your line (table points only if they lie on it), at least 0.4 units apart on the x-axis (1), and the gradient correctly calculated and given to three significant figures (1). Your value depends on your own line, so there is no single answer.
• Finding k (1 mark): put your gradient and a point on the line into log(rate) = m log[Na₂S₂O₈] + log k, find log k, then k = 10 to the power of log k. If you had no gradient, the question let you use 1.55.
• Measurement error or other factors (1 mark): the true order is 3/2. Work out the percentage difference, (your gradient − 1.5) ÷ 1.5 × 100. If it is under 4.5%, the difference can be explained by measurement error; if it is over, other factors are involved. Your conclusion must match your number.
• Reliability (1 mark): reliable, because no points are anomalous and all lie on or very near the line. If your own plot showed an anomalous point, “not reliable” with that point identified also scored, as long as the answer matched your graph.

Where the marks are on 9701/52

• Analysis carried 14 marks: the iodine mass (1), the mean titre (1), the mass of iodine in solution A (3), n (1), the higher-temperature curve (1), the log table (2), plotting (2), the gradient (2) and k (1).
• Planning carried 12 marks: making the methanolic iodine (3), the three “suggest” parts on the preparation (3), the results table and weighing method (2), rinsing the burette (1), the gas collection apparatus (2) and the independent variable (1).
• Evaluation carried 4 marks: the percentage error (1), the timing error (1), measurement error against other factors (1) and reliability (1).
• Question 1 is mostly practical method and calculation; Question 2 is mostly graph work. The graph in 2(e) alone, from the log table to k, is worth 7 marks.

How to self-mark 9701/52 with the mark scheme

1. Learn the abbreviations first. A slash separates alternatives, A means accept, R means reject, I means ignore, ecf means error carried forward, ora means “or reverse argument” and owtte means “or words to that effect”.
2. Mark the method answer against the three steps. Dissolve in a little methanol, transfer with rinsings into a 100 cm³ volumetric flask, make up to the mark and invert. Then apply the caps: water as solvent, or solid straight into the flask, means 2 at most.
3. Watch the precise words. “Constant mass” not “constant weight”, “methanol is flammable” not “it is flammable”, “mass” in every heading of your table.
4. Check decimal places. The mean titre needs 13.30, the log values need 2 decimal places, and the gradient needs 3 significant figures.
5. Re-plot carefully before marking the graph. Points within half a small square, a single straight line through all six, and gradient points at least 0.4 apart on the x-axis.
6. Carry your own numbers forward. In 1(c)(iv), 1(c)(v), 2(e)(iv) and 2(e)(v), mark the method using your earlier answers, not the scheme’s.

Sit 9701/52 as a timed mock

The paper allows 75 minutes for 30 marks, two and a half minutes per mark. A workable plan:

Block	Marks	Time
Question 1(a) and (b): planning the preparation	9	20 minutes
Question 1(c): titration analysis and n	7	17 minutes
Question 2(a) to (d): apparatus and the volume graph	5	10 minutes
Question 2(e): log table, graph, gradient, k and evaluation	9	23 minutes
Check decimal places, units and the graph	–	5 minutes

Write the method in 1(a)(ii) as numbered steps, as the question asks, and name each piece of apparatus with its capacity.

Other May/June 2026 9701 papers

• 9701 Paper 12 (Multiple Choice)
• 9701 Paper 22 (AS Level Structured Questions)
• 9701 Paper 32 (Advanced Practical Skills)
• 9701 Paper 42 (A Level Structured Questions)

Practise and download

• Download 9701/52 and its mark scheme on the Learning Portal →
• All 9701 May/June 2026 papers: overview and marking patterns
• Try a free A Level Chemistry quiz

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Frequently asked questions

What topics were on 9701 Paper 52 May/June 2026? +

Two questions: preparing the polyiodide salt (CH₃)₄NI₂ₙ₊₁ and finding n by thiosulfate titration (16 marks), and the rate of reaction between sodium ethanedioate and sodium peroxydisulfate, including a log-log graph to find the order and k (14 marks).

How many marks is 9701/52 and how long is it? +

It is out of 30 marks and lasts 1 hour 15 minutes. It is the A Level planning, analysis and evaluation paper, and both questions are compulsory.

What was the value of n on 9701/52 2026? +

n = 2. The mean titre of 13.30 cm³ gives 0.338 g of iodine in 100 cm³ of solution A, leaving 0.134 g of (CH₃)₄NI. That is 0.00133 mol of I₂ to 0.000667 mol of (CH₃)₄NI, a ratio of 2.

What was the mean titre on 9701/52? +

13.30 cm³, the mean of titrations 1 and 3 (13.35 and 13.25 cm³), which are within 0.10 cm³ of each other. Writing 13.3 did not score, because burette values are given to two decimal places.

How do you find the order from the log-log graph on 9701/52? +

Plot log(rate) against log[Na₂S₂O₈]. Because log(rate) = m log[Na₂S₂O₈] + log k, the gradient is the order m, given to three significant figures, and log k comes from putting a point on the line into that equation. The question later gives the true order as 3/2.

Where can I download 9701/52 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Chemistry 9701 past papers page.

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