Papers Covered
9701/52 · Feb/March 2025
Q1: Copper in Brass (Iodometric Titration)
Q2: Unknown Ester Identification
Q2: Unknown Ester Identification
Brass dissolved in HNO₃; Cu²⁺ + KI → I₂ → Na₂S₂O₃. Q2: reflux, TLC, IR, ¹H NMR.
9701/51 & 53 · May/June 2024
Q1: Dissolved O₂ in River Water (Winkler)
Q2: Activation Energy (Clock Reaction)
Q2: Activation Energy (Clock Reaction)
Winkler method: MnSO₄ + KI + H₂SO₄ → I₂ → Na₂S₂O₃. Clock: HCl + Na₂S₂O₃; log(1/t) vs 1/T graph.
9701/52 · May/June 2024
Q1: Enthalpy of CaCO₃ Decomposition
Q2: Equilibrium K for Fe³⁺/SCN⁻
Q2: Equilibrium K for Fe³⁺/SCN⁻
Hess's law via calorimetry + temp-time extrapolation. Colorimetry + log K vs 1/T → ΔH.
9701/51 & 53 · May/June 2025
Q1: Grignard Synthesis (1-Phenylethanol)
Q2: Rate Constant for Ester Hydrolysis
Q2: Rate Constant for Ester Hydrolysis
PhBr + Mg + CH₃CHO → PhCH(OH)CH₃; separating funnel. log(V∞−Vt) vs time graph → k.
9701/52 · May/June 2025
Q1: Adsorption Isotherm (Activated Carbon)
Q2: Enthalpy of Combustion of Butane
Q2: Enthalpy of Combustion of Butane
Langmuir: mₑ/qₑ vs mₑ linear graph. Q2: gas burner + metal can calorimetry.
9701/54 · May/June 2025
Q1: Mᵣ of Volatile Liquid (Dumas Method)
Q2: Nernst Equation / Ion Charge
Q2: Nernst Equation / Ion Charge
pV = nRT applied. Q2: Electrode potential vs log[M^n+] graph; use gradient to find n.
9701/51 & 53 · Oct/Nov 2024
Q1: Aspirin Content in Tablets
Q2: Crystal Violet Colorimetry + Kinetics
Q2: Crystal Violet Colorimetry + Kinetics
Reflux + iodoform precipitation + gravimetry. Crystal violet: calibration curve + absorbance vs time.
9701/52 · Oct/Nov 2024
Q1: HCl Conc. by Na₂CO₃ Titration + Evaporation
Q2: Metal M + Sulfur Ratio (Graph)
Q2: Metal M + Sulfur Ratio (Graph)
Dual method: titration + mass of NaCl residue. Q2: mass S vs mass M → gradient → empirical formula.
9701/51 & 53 · Oct/Nov 2025
Q1: BaCl₂·xH₂O (AgNO₃ Mohr Titration)
Q2: Graham's Law of Effusion
Q2: Graham's Law of Effusion
Mohr method (K₂CrO₄ indicator); barium ions removed first. Rate of effusion vs √(1/M) graph.
9701/52 · Oct/Nov 2025
Q1: Ag⁺/Fe²⁺ Equilibrium — Kc
Q2: Conductimetric Titration (SO₄²⁻)
Q2: Conductimetric Titration (SO₄²⁻)
KSCN back-titration to find [Ag⁺]ₑq; calculate Kc. SO₄²⁻ in lake water using Ba(OH)₂ + conductivity meter.
9701/54 · Oct/Nov 2025
Q1: Winkler Method (Lake Water O₂ in mg dm⁻³)
Q2: Sn + I₂ Kinetics (Mass Loss Method)
Q2: Sn + I₂ Kinetics (Mass Loss Method)
Full Winkler with calculation of dissolved O₂ in mg dm⁻³. Tin block suspended in I₂/methylbenzene; rate = k[I₂]².
Complete Topic List — Everything They Could Ask
Analytical Techniques
TitrationIodometric titration (Na₂S₂O₃)appeared 4× across variants
TitrationAcid-base (HCl + Na₂CO₃, NaOH + HCl)appeared 3×
TitrationSilver nitrate / Mohr methodappeared 2×
TitrationKSCN back-titration (Volhard)appeared 1×
ColorimetryCalibration curve (absorbance vs [dye])appeared 2×
ColorimetryChoosing wavelength from spectrumappeared 2×
ChromatographyTLC — interpreting spots for purityappeared 1×
SpectroscopyIR — identifying functional groupsappeared 1×
Spectroscopy¹H NMR — chemical shift, splitting, areaappeared 1×
ConductimetryConductimetric titration (two-line intersection)appeared 1×
Kinetics
KineticsClock reaction (HCl + Na₂S₂O₃); 1/t as rate measureappeared 2×
KineticsActivation energy: log(1/t) vs 1/T graph → Eₐappeared 2×
KineticsRate constant k: log(V∞−Vt) vs time (1st order)appeared 1×
KineticsColorimetric monitoring (absorbance vs time)appeared 1×
KineticsMass loss method (Sn + I₂); rate = k[I₂]²appeared 1×
KineticsIced water to quench reactions (sampling)appeared 1×
Thermochemistry / Energetics
ThermochemistryTemperature-time graph + extrapolationappeared 2×
Thermochemistryq = mcΔT calculationappeared 3×
ThermochemistryHess's law energy cycle (ΔHr from two reactions)appeared 1×
ThermochemistryEnthalpy of combustion (gas burner + metal can)appeared 1×
ThermochemistryΔH from log K vs 1/T graph (van't Hoff)appeared 2×
Equilibrium
EquilibriumKc calculation from equilibrium concentrationsappeared 2×
Equilibriumlog K vs 1/T to determine ΔH and whether exo/endoappeared 2×
EquilibriumCalibration graph (colorimetry) to find [FeSCN²⁺]appeared 1×
Electrochemistry
ElectrochemistryNernst equation: E vs log[M^n+] graph → find nappeared 1×
ElectrochemistryElectrode potential; cell potential calculationappeared 1×
ElectrochemistrySalt bridge functionappeared 1×
Organic Chemistry in Paper 5
OrganicReflux apparatus — draw and labelappeared 2×
OrganicSeparating funnel — which layer contains productappeared 1×
OrganicGrignard reagent — anhydrous conditions, drying glasswareappeared 1×
OrganicEster hydrolysis — alkaline vs acid conditionsappeared 2×
OrganicTLC — spot pattern, evidence of incomplete reactionappeared 1×
OrganicIR spectrum — identify/rule out functional groupsappeared 1×
Organic¹H NMR — complete table, identify compoundappeared 1×
OrganicIndicator selection for ester hydrolysis monitoringappeared 1×
OrganicPercentage yield calculation (limiting reagent)appeared 1×
Experimental Technique & Planning
TechniqueMaking a standard solution (steps, apparatus)appeared 6×
TechniqueSerial dilution to make a new concentrationappeared 3×
TechniquePreparing a burette (rinsing, filling)appeared 2×
TechniqueWeighing by differenceappeared 3×
TechniqueReading burette from diagram (meniscus)appeared 1×
TechniqueApparatus selection (pipette vs burette vs measuring cylinder)appeared 5×
TechniqueDrawing a results table with units/headingsappeared 3×
TechniqueConcordant titres (choosing which to average)appeared 4×
TechniquePercentage error calculation (apparatus)appeared 8×
TechniqueHydrated salt: mass of hydrate needed for given conc.appeared 3×
SafetySafety precautions from hazard dataappeared 5×
PlanningDescribe alternative method (brief plan)appeared 1×
Graphs & Data Analysis
GraphsPlot graph + draw line of best fit (straight)appeared 10×
GraphsCalculate gradient (state coordinates used)appeared 7×
GraphsIdentify anomalous point + suggest reasonappeared 8×
GraphsUse graph to state whether data is reliableappeared 4×
GraphsTemperature-time extrapolation (calorimetry)appeared 2×
GraphsTwo intersecting lines (conductimetric)appeared 1×
GraphsSketch a new line (higher/lower temperature)appeared 1×
GraphsCurved line of best fitappeared 1×
Gases & Ideal Gas Law
GasespV = nRT to find Mᵣ (Dumas method)appeared 1×
GasesGraham's law of effusion: rate ∝ √(1/M)appeared 1×
GasesEffect of temperature on effusion rateappeared 1×
Inorganic / Environmental Applications
InorganicDissolved O₂ (Winkler method)appeared 3×
InorganicAspirin content via iodoform precipitationappeared 2×
InorganicMetal + sulfur empirical formula from mass dataappeared 1×
InorganicWater of crystallisation (x in MX·xH₂O)appeared 2×
EnvironmentalWhy Cl₂ in tap water interferes with Winkler methodappeared 1×
EnvironmentalSulfate concentration in lake water (conductimetric)appeared 1×
Topic Frequency Across 15 Papers
Tricky & Unique Questions — High Difficulty Highlights
Graph Interpretation — Often Misread
Distilled water should give 0 mol dm⁻³ dissolved O₂, but a student gets 2.26×10⁻⁵ mol dm⁻³. Suggest why, then use it to improve the river water answer.
Why tricky: Students think "error in procedure" but the question says the procedure was correct — the issue is O₂ absorbed from air during transfer. Part (ii) asks them to subtract this blank value from the river result.
Hint: The blank value is a systematic error (background O₂ contamination). Subtract it from the real result to correct for it.
Suggest why this Winkler method is unsuitable for tap water that contains Cl₂(aq).
Why tricky: Cl₂ is an oxidising agent that also oxidises I⁻ to I₂, giving a false high reading — it is not related to pH or indicator choice, which many students write.
Answer: Cl₂ oxidises I⁻ ions in the alkaline KI reagent, producing additional I₂ not related to dissolved O₂ — this gives an artificially high titre.
An anomalous point on the log K vs 1/T graph: suggest why the absorbance was lower than expected. No measurement errors were made.
Why tricky: Students must think about what causes lower absorbance specifically — the equilibrium has shifted left (less FeSCN²⁺) because the temperature was higher than recorded, or the mixture was not at equilibrium yet.
Answer: The mixture had not reached equilibrium (not left long enough at that temperature), so less FeSCN²⁺ formed, giving lower absorbance.
Ester X undergoes acid hydrolysis with H₂SO₄(aq). Suggest why NONE of the three indicators in Table 2.1 would change colour in this experiment.
Why tricky: Acid hydrolysis starts acidic AND ends acidic (ethanoic acid produced). Both reactant (ester in acid) and product mixtures are acidic — the pH never crosses any of the given indicator ranges (all start at pH 3–4 or above).
Answer: Both the initial solution and final mixture are acidic; no colour change occurs because pH remains below all the indicator transition ranges throughout.
Calculations — Unusual or Multi-Step
The actual vapour temperature is less than 100°C. Describe and explain the effect on the calculated Mᵣ.
Why tricky: If T is lower than assumed, then from pV=nRT, n = pV/RT is larger (more moles). So Mᵣ = mass/n is smaller — calculated Mᵣ is too low. Many students get the direction wrong.
Effect: calculated Mᵣ is too small (underestimate). Reason: lower T → n is overestimated (more moles) → Mᵣ = m/n is underestimated.
Use the Nernst gradient to calculate the charge n of M^n+. The equation is E = E° + (2.303RT/nF)·log[M^n+]. State what y, m, c correspond to.
Why tricky: You must recognise the slope = 2.303RT/nF, then rearrange to find n. Students often forget to convert T to Kelvin (25°C = 298 K). Many also misidentify which part is y and which is c.
Gradient = 2.303 × 8.31 × 298 / (n × 96500). So n = 2.303 × 8.31 × 298 / (gradient × 96500). y = E, m = 2.303RT/nF, c = E°.
Student incorrectly made Na₂CO₃ solution (wrong concentration = solution Y). They titrate with HCl (solution X), then use the same titre volume to find the NaCl residue mass. Explain how both concentrations are found from the same experiment.
Why tricky: Two unknowns determined from two independent measurements — the titre (moles of X via stoichiometry) and the residue mass (moles of NaCl = moles of Y). Students confuse which calculation gives which concentration.
From titre → mol HCl → mol Na₂CO₃ in 25 cm³ → [X]. From residue mass → mol NaCl → mol Na₂CO₃ in 25 cm³ → [Y].
Rate data for Sn + I₂: ratios show rate = k[I₂]². State whether the data supports this conclusion using values from the table.
Why tricky: You must calculate the ratio of rates and the ratio of [I₂]² for each pair. If rate doubles when [I₂] increases by √2 ≈ 1.41, that confirms second order — but students often just say "rate doubles when [I₂] doubles" (which would be first order).
Check: when [I₂] halves (0.400→0.200), rate should drop by factor 4. Calculate 4.76/2.35 ≈ 2.03 and compare with (0.400/0.200)² = 4. Show the working clearly.
q₀ is calculated as 78.1 mg g⁻¹ with 6.5% total experimental error. Data book value is 86.0 mg g⁻¹. Determine whether the discrepancy is due to experimental error or other factors.
Why tricky: You must calculate the acceptable range from the experimental error (78.1 ± 6.5%) and see if 86.0 falls within it, then state the conclusion explicitly.
78.1 × 1.065 = 83.2 mg g⁻¹. Since 86.0 > 83.2, it is outside the error range → the discrepancy cannot be explained by experimental error alone → other factors (e.g., model assumptions) contribute.
Planning & Procedure — Require Understanding, Not Memory
Describe a brief alternative method (not titration) to determine x in BaCl₂·xH₂O, using standard lab apparatus.
Why tricky: Open-ended planning question requiring students to design a dehydration experiment. Many forget to specify: weigh before AND after, heat until constant mass, calculate Δmass = mass of water, hence x.
Weigh sample → heat strongly (or use oven) → re-weigh → repeat until constant mass. Loss = mass of water. Moles of water = loss/18. Moles of BaCl₂ = original/208. x = mol H₂O / mol BaCl₂.
Why is the glass beaker reweighed in step 4 (after transferring brass to the conical flask)?
Why tricky: Many students write "to check the mass" or "to record the mass" — that's too vague for the mark. The specific reason is that not all of the brass powder may transfer completely to the flask; reweighing by difference gives the exact mass that actually entered the flask.
Answer: Not all brass may transfer from the beaker to the flask. Reweighing the beaker after transfer allows the exact mass added to be determined by difference: (mass beaker + brass) − (mass beaker after transfer) = exact mass in flask.
Give TWO reasons why iced water in step 4 decreases the rate of reaction.
Why tricky: Students give only one reason. Two marks require two distinct points: (1) lower temperature → fewer molecules with energy ≥ Eₐ → fewer successful collisions; (2) dilution lowers concentration of acid catalyst, reducing rate further.
Reason 1: Lower temperature decreases rate (Arrhenius). Reason 2: Dilution lowers [HCl] catalyst, decreasing rate. Both must be stated.
Deduce the theoretical gradient for metal M = strontium forming SrS, then use it to say if the experimental data supports this.
Why tricky: You must calculate the expected gradient (mass S / mass Sr for SrS) from molar masses, not measure it from the graph — then compare with the experimental gradient to draw a conclusion.
SrS: Mᵣ(Sr)=87.6, Mᵣ(S)=32.1. Gradient = 32.1/87.6 = 0.366. Compare with your experimental gradient; if close, data supports Sr.
Unique / Unusual Experimental Contexts
Conductivity drops then rises as Ba(OH)₂ is added to lake water containing SO₄²⁻ ions. Two lines of best fit intersect. Use the intersection volume to calculate [SO₄²⁻].
Why unique: Conductimetric titration is rarely tested. The correction formula (corrected conductivity = measured × total volume/initial volume) is given but interpreting the V-shaped graph is the core skill.
Intersection point = equivalence point. Before: SO₄²⁻ and other ions remain; conductivity falls as SO₄²⁻ and Ba²⁺ precipitate. After: excess Ba²⁺ and OH⁻ raise conductivity again.
Plot rate of effusion vs √(1/M) for noble gases. Use the graph to find M of natural gas. Then: does M > 16 (CH₄) suggest heavier gases are present?
Why unique: Graham's law applied experimentally is rare in Paper 5. The purging step (add 50 cm³, remove, then add 70 cm³) is to ensure the syringe contains pure gas — students must explain this specific procedural reason.
Purging removes residual air. The graph should be a straight line through the origin (Graham's law: rate ∝ √(1/M)). If M(natural gas) > 16, heavier gases like ethane/propane are present.
Why is it not necessary to know the mass of liquid Y injected into the flask? Why must the liquid be kept in boiling water for 3 minutes after all Y has evaporated?
Why unique: Both questions test deep understanding of the method. (1) Because all Y evaporates and fills the flask — the mass weighed after cooling is the mass of vapour that filled the flask, regardless of what was injected. (2) To ensure temperature of vapour equals 100°C (equilibrium) — not just time.
(1) Excess Y simply vaporises and escapes; only the flask volume of vapour remains. (2) To ensure all liquid has vaporised and vapour has reached 100°C (boiling water temperature) — equilibrium must be reached.
Why is hot distilled water NOT used to wash the precipitate C in step 7?
Why unique: Students know to wash precipitates with cold water but rarely think about why hot water is forbidden here. The iodoform precipitate (C₆H₂I₂O)₂ is slightly soluble when hot — using hot water would dissolve some, reducing the apparent mass.
Hot water increases the solubility of precipitate C, which would dissolve some of it — reducing its measured mass and giving a lower calculated % aspirin (underestimate).
Effect of Errors — "What Happens If…" Questions
Another student does not allow solid C to dry completely in step 8. State and explain the effect on the calculated percentage by mass of aspirin.
Why tricky: Students often say "the result will be inaccurate" — that gets zero. You must state the direction (higher or lower) AND give the reason as a chain of logic: wet solid is heavier → mol C appears higher → mol aspirin appears higher → % aspirin is higher than the true value.
Effect: % by mass is too high (overestimate). Reason: wet solid has greater mass → calculated mol of precipitate C is larger than true value → mass of aspirin calculated is larger → % by mass is larger.
State what happens to the calculated concentration of solution Y if not all the water is evaporated in step 7. Explain your answer.
Why tricky: Same chain-of-logic trap. Students say "it will be wrong" — you need: residue mass is higher (contains water) → calculated mol NaCl is higher → calculated mol Na₂CO₃ is higher → [Y] is calculated as higher than the true value.
Effect: [Y] is calculated as too high (overestimate). Reason: residue contains water → apparent mass of NaCl is larger → calculated moles of Na₂CO₃ in Y is larger → [Y] appears higher than it really is.
The burner in the butane combustion experiment is switched off after only 2 minutes instead of 3. Suggest why this (i) reduces accuracy AND (ii) increases accuracy of ΔHc.
Why tricky: Two opposite effects from the same change. Students usually only find one direction. (i) Less butane burns → smaller ΔT → less reliable q = mcΔT (larger relative error). (ii) Less time for heat to be lost to surroundings → the temperature rise is closer to the theoretical maximum → less systematic underestimate.
(i) Reduces accuracy: fewer moles of butane burned → smaller ΔT → larger relative error in ΔT measurement, so calculated ΔHc is less reliable. (ii) Increases accuracy: less time means less heat lost to surroundings during burning → ΔT is closer to the ideal value → ΔHc closer to true value.
Table: place a tick to show the effect of using a LARGER volume of alkaline KI on (a) the uncertainty of the measurement and (b) the percentage error of the measurement.
Why tricky: Students confuse uncertainty with percentage error. Uncertainty (the absolute ±0.5 cm³ for a syringe with 1 cm³ graduations) does NOT change when you use a larger volume — it is fixed by the instrument. But percentage error = uncertainty/volume × 100, so a larger volume means the percentage error is SMALLER.
Uncertainty: no effect (always ±0.5 cm³ for that syringe). Percentage error: smaller effect (same absolute uncertainty divided by a larger volume = smaller percentage).
More "Procedure Reasoning" — Questions that Need Understanding
Suggest why the enthalpy change for the thermal decomposition of CaCO₃ CANNOT be measured directly.
Why tricky: Students write "it is exothermic" or "it cannot be reversed" — both wrong. The real reason is that the decomposition requires extremely high temperatures (~900°C) to proceed at a measurable rate. Standard laboratory calorimetry cannot be used at these temperatures.
CaCO₃ decomposes only at very high temperatures (~900°C). Standard aqueous calorimetry (q = mcΔT) cannot be used at these temperatures — the apparatus would not withstand the heat and the water would have evaporated.
Identify the main weakness of the calorimetric procedure and suggest ONE improvement. (The weakness is NOT the type of thermometer used.)
Why tricky: Students immediately write "use a more accurate thermometer" — but the question specifically excludes this. The main weakness is heat loss to the surroundings, which causes the temperature rise to be lower than the theoretical value, giving a less exothermic (smaller magnitude) ΔH.
Main weakness: heat is lost to the surroundings from the glass beaker during the experiment. Improvement: use an insulated container (e.g., polystyrene cup with lid) to reduce heat loss to the surroundings.
(i) Why is flask A sealed with a bung in step 3? (ii) Why is flask A left to stand for twelve hours in step 4?
Why tricky: (i) Students say "to stop the reaction" — wrong. The bung prevents O₂ from the air entering and oxidising Fe²⁺ to Fe³⁺, which would change the equilibrium concentrations and make the Kc calculation incorrect. (ii) Students say "to reach equilibrium" in vague terms — you need to state the reaction is slow at room temperature and needs time.
(i) To prevent O₂ from the air oxidising Fe²⁺(aq) to Fe³⁺(aq) — this would change [Fe²⁺] and [Fe³⁺] at equilibrium, giving an incorrect Kc value. (ii) The reaction Ag⁺ + Fe²⁺ ⇌ Ag + Fe³⁺ is slow — 12 hours allows the equilibrium to be established.
Identify the precipitate in flask A in step 5 (when transferring the mixture carefully without disturbing it).
Why tricky: Students say "silver chloride" or "iron precipitate" without thinking. The equilibrium reaction is Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq) — the precipitate is metallic silver, Ag(s), produced by the reduction of Ag⁺ ions. This must be left undisturbed because it would react with the KSCN back-titration if included.
The precipitate is silver metal, Ag(s). It must not be disturbed or transferred because Ag(s) would react with the KSCN titrant, consuming it and giving a false (lower) titre for Ag⁺ in solution.
Suggest why a measuring cylinder is a suitable piece of apparatus to measure 40 cm³ of hydrochloric acid in step 5.
Why tricky: Students think any volume measurement always needs a pipette or burette — but here the HCl is added in excess to the Grignard mixture (reaction 3), so an exact volume is not required. A measuring cylinder is perfectly acceptable for "approximate" volumes where only excess is needed.
The HCl is added in excess — the exact volume does not need to be known precisely because more than enough is used to complete reaction 3. A measuring cylinder is accurate enough for this purpose; a pipette or burette would be unnecessary.
The IR spectrum of the 1-phenylethanol product shows a C=O absorption peak. Suggest why.
Why tricky: The product is an alcohol and should only show O–H and C–O peaks. A C=O peak means something has gone wrong. Students who don't think about the reaction mechanism write "the product contains a ketone" — the real reason is that incomplete reaction left unreacted ethanal (CH₃CHO), which contains a C=O group, present as an impurity in the extracted product.
The reaction was incomplete — unreacted ethanal (CH₃CHO) remains in the product. Ethanal contains a C=O (aldehyde) group which absorbs in the carbonyl region of the IR spectrum. The 1-phenylethanol itself should not give a C=O peak.
(i) A student suggests Kc is directly proportional to temperature T. State if the graph (Kc vs T) supports this. (iii) Use the graph to state if the forward reaction is exothermic or endothermic.
Why tricky (i): Direct proportionality requires the graph to pass through the origin AND be linear. The graph in the paper is linear but starts at a non-zero Kc value at T=280 K — so it does NOT support direct proportionality. Students often just say "it's linear, so yes" without checking the origin condition. Why tricky (iii): As T increases, Kc decreases (from the graph) — by Le Chatelier, this means increasing T shifts equilibrium to the LEFT, favouring the reverse reaction. So the forward reaction is EXOthermic.
(i) No — direct proportionality requires the line to pass through the origin. The graph is linear but does not pass through (0,0), so Kc is not directly proportional to T. (iii) Exothermic — as T increases, Kc decreases, so equilibrium shifts left (reverse direction). Increasing T favours endothermic direction (reverse), so forward must be exothermic.
More Tricky Procedure Questions
Suggest a reason why it is necessary to wait 10 minutes in step 5 (before mixing the two flasks in the clock reaction).
Why tricky: Students say "to let the reaction start" or "to wait for equilibrium" — neither is correct. The flasks are in a thermostatically controlled water bath; the 10-minute wait is specifically to allow both solutions to reach the same target temperature as the water bath. If mixed too soon, the measured temperature would not be the bath temperature, making the Arrhenius calculation wrong.
Answer: To allow both flasks A and B to reach the temperature of the water bath (thermalise). If mixed too soon, the reaction would not occur at the correct temperature, invalidating the rate measurement.
The procedure doesn't mention how to measure the reaction temperature. (i) State what temperature measurements should be taken and when. (ii) How do you get an accurate single temperature value for the mixture?
Why tricky: Students say "measure temperature after mixing" — but you need two measurements AND a method to combine them. (i) Measure the temperature of EACH flask separately just before mixing (after the 10-minute wait). (ii) Average the two temperatures to get the temperature of the mixture — because both were at the bath temperature but may differ slightly.
(i) Measure temperature of flask A and flask B just before step 7 (just before mixing). (ii) Calculate the mean of the two temperatures — this is the most accurate estimate of the mixture's temperature during the reaction.
Identify the substance used to rinse the burette before step 9 (titration of Cu²⁺ with Na₂S₂O₃).
Why tricky: Students write "distilled water" — this would dilute the Na₂S₂O₃ solution already in the burette, changing its concentration and making the titre inaccurate. The correct answer is to rinse with the Na₂S₂O₃ solution itself (the titrant), just like rinsing a burette with the solution it will contain.
Answer: Sodium thiosulfate solution, Na₂S₂O₃(aq) — the same solution that will fill the burette. Never rinse a burette with distilled water immediately before use; always rinse with the solution to be dispensed to avoid dilution.
Suggest why the activated carbon is placed in an oven at 120°C for three hours before use in the adsorption experiment.
Why tricky: Students say "to sterilise it" or "to activate it" — these are too vague. The specific reason is that activated carbon readily adsorbs water molecules and atmospheric gases onto its surface. Heating drives off any pre-adsorbed species, ensuring the surface is clean and the mass is consistent, so that only (COOH)₂ adsorption is measured.
Answer: To remove any water or gases previously adsorbed on the surface of the activated carbon. This ensures the carbon has a reproducible, known surface state before the experiment begins, so that only (COOH)₂ adsorption is measured.
Suggest why the student leaves the iodoform mixture to stand for 1 hour in step 6 (after adding alkaline aqueous iodine to the filtrate).
Why tricky: Students say "to let the reaction equilibrate" — you need to say the reaction is SLOW and incomplete if not given enough time. The iodoform precipitation reaction between the salicylate ion and iodine needs time to go to completion so that all the aspirin (via its salicylate intermediate) is converted to precipitate C.
Answer: The precipitation reaction is slow; leaving for 1 hour allows the reaction to go to completion so that the maximum amount of precipitate C is formed — giving a more accurate final mass and hence more accurate % aspirin.
In the Mohr titration for BaCl₂·xH₂O: suggest why barium ions are removed (using Na₂SO₄) before the titration with AgNO₃ is carried out.
Why tricky: Students say "Ba²⁺ would react with Ag⁺" — but they don't say how this causes a problem. Ba²⁺ would react with the chromate indicator (K₂CrO₄) to form a yellow BaCrO₄ precipitate, masking the red Ag₂CrO₄ endpoint colour, so the endpoint cannot be detected visually.
Answer: Ba²⁺ ions react with the K₂CrO₄ indicator to form a yellow precipitate of BaCrO₄. This would mask the red endpoint colour (Ag₂CrO₄) and make it impossible to detect when the endpoint has been reached.
(iii) Why is the evaporating basin NOT heated directly with a Bunsen burner in step 7? (iv) What should the student do to ensure all the water has evaporated before completing step 8?
Why tricky: (iii) Students say "it would crack the glass" — that's wrong for an evaporating basin (which can withstand direct heat). The real reason is that direct Bunsen heat would cause the NaCl solution to spit and lose material, reducing the measured residue mass. (iv) Students say "heat until dry" — you must specify the correct method: re-weigh, reheat, re-weigh again, and repeat until constant mass is achieved.
(iii) Direct Bunsen heat would cause the solution to spit, ejecting droplets of NaCl solution — the residue mass would be less than the true value. A water bath heats gently and evenly. (iv) Re-weigh the evaporating basin, heat again, re-weigh: repeat until two consecutive masses agree (constant mass = all water removed).
Suggest why solid Z (shown at the top of tube Y in the Grignard apparatus) is used. Also: give ONE reason why the apparatus does NOT have a bung at the end of tube Y.
Solid Z tricky: Students say "to dry the gas" which is vague. Solid Z is a drying agent (e.g., calcium chloride, CaCl₂) that absorbs moisture from the air before it enters the apparatus. Any water entering would destroy the Grignard reagent (Mg reacts with H₂O). Bung tricky: the reaction produces HBr and ethoxyethane vapour; a bung would build up pressure and could cause an explosion.
Solid Z: It is a drying agent (e.g., CaCl₂) to prevent moisture from the atmosphere entering the apparatus — water would react with Mg or the Grignard reagent and destroy it. No bung: gases (HBr, ethoxyethane vapour) are evolved; sealing the tube would build pressure dangerously.
k increases with temperature. A second experiment is carried out at a higher temperature. Sketch a suggested line of best fit on the log[(V∞−Vt)] vs time graph for the second experiment.
Why tricky: Students draw a line at the same y-intercept but steeper slope — this is wrong in two ways. (1) At higher T, k is larger so the slope (= −k/2.303) is MORE negative (steeper downward). (2) V∞ does not change (reaction goes to completion), but Vt increases faster, so (V∞−Vt) decreases faster. The y-intercept stays the same (same starting conditions), but the line is steeper (more steeply negative gradient).
The line should: start at the same y-intercept (same initial V∞−V₀); have a steeper negative gradient (larger |slope| because k is larger at higher T); and be otherwise straight. Do NOT shift the intercept.
In preparing alkaline KI solution (solution Y for Winkler method), NaOH is added in small portions and the mixture cooled to room temperature each time before adding more. Suggest why the solution is cooled.
Why tricky: Students say "to prevent evaporation of KI" — wrong. The real reason is that dissolving NaOH is highly exothermic and if the solution gets too hot, iodide ions (I⁻) can be oxidised by dissolved O₂ more rapidly at higher temperatures, generating unwanted I₂ before the experiment starts. Also, hot alkali would decompose KI.
Answer: Dissolving NaOH releases a lot of heat; at high temperatures, I⁻ ions are oxidised more readily by O₂ (or the heat causes KI decomposition). Cooling after each addition prevents unwanted oxidation of I⁻ and keeps the solution stable.
The student repeats the Ag⁺/Fe²⁺ titration using KSCN at a HIGHER concentration and obtains SMALLER titres. Suggest ONE reason why a larger titre is better than a smaller titre.
Why tricky: Students write "more accurate" without explaining why. A larger titre means the absolute uncertainty (±0.05 cm³ per burette reading) is a smaller fraction of the total volume — so the percentage error is smaller. Also, a larger titre makes reading the burette easier and reduces the relative impact of any systematic errors.
Answer: A larger titre has a smaller percentage error — the uncertainty in the burette reading (±0.05 cm³) is a smaller proportion of a larger titre. This gives a more precise and reliable result.
Key Exam Tips — What to Always Do
Percentage Error
Always use uncertainty / measured value × 100. For a burette: uncertainty = 0.05 cm³ per reading, but a titre uses two readings → total uncertainty = 0.10 cm³. For a measuring cylinder with 1 cm³ graduations: uncertainty = 0.5 cm³. Never forget the factor of 2 for burette titres.
Making a Standard Solution
The mark scheme always wants: (1) dissolve in distilled water in a beaker, (2) transfer to volumetric flask of the correct capacity (name it), (3) rinse beaker into flask 2–3 times, (4) add distilled water to below the graduation mark, (5) add drop by drop with an eye at graduation level, (6) stopper and invert.
Gradient Calculation
Always state both coordinates from the line (not data points unless they happen to lie on the line), use points far apart, show working as Δy/Δx, include units. Papers explicitly state: "coordinates must be from your line of best fit."
Anomalous Points
Say it is anomalous because it does not lie on/near the line of best fit. Give a specific procedural reason (e.g., sample not at equilibrium temperature, sample contaminated, timing error, not mixed properly). Never say "human error" without being specific.
Reliability from a Graph
Reliable = points are close to the line of best fit (small scatter). State this explicitly. If asked to justify: count scatter, mention R² or visual clustering. If one point is anomalous: state the results are still reliable because the majority of points lie close to the line.
Iodometric / Na₂S₂O₃ Titrations
The reaction chain is always tested in multi-step calculations: O₂ (or Cu²⁺) → I₂ → Na₂S₂O₃. The ratio is: 1 O₂ : 4 S₂O₃²⁻ (via the chain). For Cu: 2 Cu²⁺ : 1 I₂ : 2 Na₂S₂O₃. Starch is the indicator — turns blue-black with I₂, colourless at endpoint.
Concordant Titres
Use titres that agree within 0.10 cm³ of each other (Cambridge standard). Exclude the rough titration. If titrations 2 and 3 are used: average them. If asked "why only these two" — say they are concordant (within 0.10 cm³) and the rough titration is less precise.
Identify Independent / Dependent Variable
Independent = what is changed/controlled by the experimenter. Dependent = what is measured as a result. Controlled = what is kept constant. These 1-mark questions are free marks — be precise: "concentration of solution B" not just "concentration."
Apparatus Selection
For accurate volume: volumetric flask, volumetric pipette, or burette. For approximate volume: measuring cylinder is acceptable only when precision is not critical (e.g., 30 cm³ of excess iodine, 40 cm³ of HCl for Grignard). The mark scheme accepts measuring cylinder when the volume does not need to be exact.
Reflux Diagram
Always include: round-bottomed flask, condenser (vertical, water IN at bottom, water OUT at top), anti-bumping granules, heat source below. Water entry/exit arrows are always tested. Missing the direction of water flow = lost mark.
Quick Reference — Paper Summary Table
| Paper | Q1 Topic | Q2 Topic | Key Skills Tested |
|---|---|---|---|
| 9701/52 F/M/25 | Copper in brass (iodometric) | Ester identification (TLC, IR, NMR) | Multi-step iodometric calculation, NMR table completion, IR interpretation |
| 9701/51,53 M/J/24 | Dissolved O₂ (Winkler) | Activation energy (clock reaction) | Blank value correction, log(1/t) vs 1/T graph, Eₐ from gradient |
| 9701/52 M/J/24 | ΔH(CaCO₃ decomp.) via Hess | Equilibrium K (colorimetry) | Temp-time extrapolation, Hess cycle, log K vs 1/T → ΔH |
| 9701/51,53 M/J/25 | Grignard synthesis | Ester hydrolysis rate constant k | Dry apparatus, layer separation, log(V∞−Vt) vs t, k in s⁻¹ |
| 9701/52 M/J/25 | Adsorption isotherm (mₑ/qₑ vs mₑ) | ΔH combustion of butane | Langmuir plot, q₀ from gradient, q=mcΔT, percentage error |
| 9701/54 M/J/25 | Mᵣ by Dumas method (pV=nRT) | Nernst equation (find charge n) | Ideal gas law, density of water from graph, E vs log[M^n+] |
| 9701/51,53 O/N/24 | Aspirin in tablets (iodoform) | Crystal violet colorimetry + kinetics | Gravimetric analysis, calibration curve, absorbance-time kinetics |
| 9701/52 O/N/24 | HCl + Na₂CO₃ titration + evaporation | Metal + sulfur mass ratio | Dual concentration determination, empirical formula from gradient |
| 9701/51,53 O/N/25 | BaCl₂·xH₂O (AgNO₃ / Mohr) | Gas effusion (Graham's law) | Remove Ba²⁺ before titration, rate vs √(1/M), find M of gas mixture |
| 9701/52 O/N/25 | Ag⁺/Fe²⁺ equilibrium → Kc | Conductimetric titration (SO₄²⁻) | Back-titration chain, Kc with units, two-line conductimetry graph |
| 9701/54 O/N/25 | Winkler method (mg dm⁻³ O₂) | Sn + I₂ kinetics (mass loss) | Mass → mol → mg dm⁻³ conversion, rate = k[I₂]², verify from data |