Twelve required practicals run through AQA A-Level Chemistry, and every one of them is fair game in the written papers — there is no separate practical exam to revise for, because the practical marks are folded into Papers 1, 2 and 3. This page gathers all twelve in one place. For each, you get the method, the apparatus, the safety, the results you should see, the errors and uncertainties examiners keep coming back to, and real exam questions with mark schemes.
RP1–6 are the ones you meet across the AS year and they carry into the full A-level; RP7–12 are Year 13, A-level only. All twelve are set out below in the same shape: the aim, the method, the apparatus and safety, what you should see, how to process the data, the precision points that carry the marks, and real exam questions with their mark schemes.
You already carried out eight required practicals at GCSE — making salts, titration, electrolysis, rates, chromatography and the rest — and the exam habits are the same: know the method well enough to spot a fault in someone else’s, quote the observation precisely, and handle the numbers with their uncertainties. If you want the GCSE versions alongside these, they are all on the GCSE required practicals page.
How practicals are examined
At least 15% of the marks across the whole A-level assess knowledge and understanding of practical work, and Paper 3 devotes a large block — around 40 marks — to practical techniques and data analysis. A question can take any of the twelve practicals and ask you to describe a step, justify a choice of apparatus, spot the flaw in a method, or process a set of results. You are not expected to have memorised a script; you are expected to understand each practical well enough to reason about it.
Two things worth being clear about from the start:
- Your A-level grade (A*–E) comes entirely from the written papers. Your hands-on lab work is marked separately by your teacher as the practical endorsement — reported as a simple “pass” alongside your grade, but it doesn’t raise or lower the grade itself.
- Where a technique has options — heat with a water bath, electric heater or sand bath; measure pH with an indicator chart, meter or probe — treat every “or” as an “and” for the written papers. You need to understand all the alternatives, because a question can name any of them.
The same handful of practical skills turn up again and again — each practical below builds on some of these:
- taking accurate readings of mass, volume, temperature and time
- heating safely with a water bath, electric heater or sand bath
- measuring pH with an indicator chart, a pH meter or a probe
- titrating with a burette and pipette, and making a standard solution in a volumetric flask
- distillation and heating under reflux
- testing for ions and organic functional groups in test tubes
- filtering, including under reduced pressure
- purifying a solid by recrystallisation and a liquid with a separating funnel, then checking purity with a melting point
- thin-layer or paper chromatography
- setting up electrochemical cells and measuring voltage
- measuring reaction rates by more than one method
- handling corrosive, irritant, flammable and toxic chemicals safely
The twelve practicals
Jump to any practical — all twelve are built out in full below.
Uncertainties and data analysis
Paper 3’s data-analysis marks turn on a handful of rules that apply to every practical. Learn these once and they pay off across all six.
- Reading uncertainty (one judgement — e.g. a thermometer, or a single burette mark): at least ± half the smallest scale division.
- Measurement uncertainty (two judgements — e.g. a length off a ruler, or a titre from two burette readings): at least ± one whole scale division (each end adds ±½).
- Given values in a question: assume ±1 in the last significant figure (78 °C means ±1 °C).
- Percentage uncertainty = (uncertainty ÷ value) × 100. To combine several measured quantities, add their percentage uncertainties.
- Repeated readings: uncertainty = half the range; quote as mean ± half-range. Drop any anomaly before you average (just like non-concordant titres).
- Quote the uncertainty to the same number of decimal places as the measurement, and the final answer to the significant figures of the least accurate value.
The move examiners reward most: to reduce a percentage uncertainty, make the measured quantity bigger. A burette reads to ±0.05 cm3, but a titre is a difference of two readings, so it carries ±0.10 cm3; on a 22.40 cm3 titre that is (0.10 ÷ 22.40) × 100 = 0.45%, and a larger titre makes that fraction smaller.
Volumetric solution & acid–base titration
Full theory, worked titres & interactives: Amount of substance — Titrations
Aim: make up a solution of accurately known concentration — a standard solution — in a volumetric flask, then use it in an acid–base titration to find an unknown concentration or amount of substance. The two halves are examined together and separately, so learn both.
Part A — make the standard solution
- Weigh the solid accurately by difference — weigh the boat or bottle with the solid, tip it into a beaker, then re-weigh the empty boat. The difference is exactly how much you transferred.
- Dissolve it fully in the beaker in less than the final volume of distilled water, stirring with a glass rod.
- Transfer to the volumetric flask through a funnel, then rinse the beaker, rod and funnel into the flask so none of the solute is left behind.
- Make up to the mark: add distilled water until close to the graduation line, then add it dropwise with a teat pipette until the bottom of the meniscus sits on the line at eye level.
- Stopper and invert several times to mix into one uniform (homogeneous) solution.
Part B — the titration
- Rinse the burette with the solution it will hold, the pipette with its solution, and the conical flask with distilled water only.
- Pipette a measured volume (e.g. 25.0 cm3) into the conical flask using a safety filler; add 2–3 drops of a single indicator (phenolphthalein or methyl orange).
- Fill the burette, running the jet briefly to clear the air bubble below the tap, and take the initial reading (bottom of the meniscus, to 0.05 cm3).
- Run a rough titration first, then accurate runs — adding dropwise near the end point, swirling over a white tile.
- Repeat until you have concordant titres (agreeing within 0.10 cm3) and take the mean of those only.
Read the bottom of the meniscus with your eye level to the mark — parallax is the classic dropped mark. One reading is ± 0.05 cm³.
Apparatus & techniques
- accurate mass & volume readings
- titration technique
- volumetric flask — standard solution
- acid–base indicators
- safe handling of irritants
Balance, weighing boat, beaker, glass rod, wash bottle and filter funnel for the solution; a 250 cm3 volumetric flask, teat pipette, volumetric pipette with a safety filler, burette with a clamp and stand, conical flask and a white tile for the titration.
Safety. Wear eye protection throughout. The reagents are usually corrosive or irritant — sulfuric acid and sodium hydroxide especially — so handle them with care, fill the burette at a safe height, and rinse any splashes off the skin at once.
With the standard solution made and the flask holding acid plus a few drops of phenolphthalein (colourless), try the titration below — run the alkali in fast at first, then a drop at a time near the end point, and build up your own titre table.
Interactive — run the titration and build your own titre table
| Run | Initial / cm3 | Final / cm3 | Titre / cm3 |
|---|
What you should see: near the end point each drop gives a flash of colour that fades on swirling; the end point is the first permanent colour change from a single drop — phenolphthalein turns from colourless to pink (alkali run into acid) or pink to colourless (acid run into alkali); methyl orange goes yellow to red. Accurate titres end up within 0.10 cm3 of each other.
Both of the AQA titration indicators flip sharply at the end point — know the exact colour words, and which titration each one suits:
| Indicator | In acid | In alkali | End-point change |
|---|---|---|---|
| Phenolphthalein | colourless | pink | colourless → first permanent pale pink |
| Methyl orange | red | yellow | red → orange at the end point |
Use phenolphthalein for a weak acid–strong base titration and methyl orange for a strong acid–weak base; either works for a strong acid–strong base. (More on choosing indicators with pH curves in RP9.)
- A burette is read to ± 0.05 cm3, but a titre is the difference of two readings, so its uncertainty is 2 × 0.05 = ± 0.10 cm3.
- % uncertainty in the titre = (0.10 ÷ titre) × 100. A larger titre gives a smaller % uncertainty — so a more dilute solution in the burette, or a bigger sample in the flask, reduces it.
- Average concordant titres only (within 0.10 cm3); discard the rough run and any anomaly before you take the mean.
- Rinse the burette with the titrant — leaving water in it dilutes the titrant so the titre reads high. Rinse the conical flask with water only — titrant residue would add extra moles.
Only concordant titres go into the mean — the runs within 0.10 cm3 of each other. Pick the concordant ones from these results, then work out the mean:
Interactive — pick the concordant titres, then find the mean
Exam questions — RP1
Propanedioic acid contains two carboxylic acid groups and is a solid organic acid, soluble in water. Draw the skeletal formula of propanedioic acid.
Show answer
Skeletal HOOC–CH2–COOH: a central CH2 with a –COOH on each side, the O–H of both acid groups drawn in. 1 mark
Must be skeletal, and must show the H atoms of the OH groups.
Describe how to prepare 250 cm3 of an aqueous standard solution of propanedioic acid containing an accurately measured mass of the acid. Include essential practical details.
Show answer
Weigh the solid acid in a weighing boat or bottle. 1
Transfer to a beaker/flask, washing all the sample in (or re-weigh the empty container). 1
Dissolve in distilled/deionised water — a volume less than 250 cm3. 1
Transfer the solution, with washings, into a 250 cm3 volumetric flask. 1
Make up to the mark with distilled water. 1
Stopper and invert to mix (a homogeneous solution). 1
Mixing must be after making up to the mark. Warming to dissolve is not a scoring point; maximum 4 marks if any substance other than water is added.
Calculate the mass, in mg, of propanedioic acid (Mr = 104.0) needed to prepare 250 cm3 of a 0.00500 mol dm−3 solution.
Show answer
n = 0.00500 × 0.250 = 0.00125 mol
1 mark
mass = 0.00125 × 104.0 = 0.130 g = 130 mg
1 mark
Answer must be at least 2 s.f. Marks are carried forward if you slip earlier (“ECF”). Leaving it as 0.130 g (not mg) scores 1.
An unknown mass of sodium hydroxide is dissolved to make 200 cm3 of solution. A 25.0 cm3 sample is titrated with 0.150 mol dm−3 sulfuric acid: 2NaOH + H2SO4 → Na2SO4 + 2H2O. The concordant titres are 19.60 and 19.55 cm3. Calculate the mass of sodium hydroxide used to make the original solution.
Show answer
mean titre = (19.60 + 19.55) ÷ 2 = 19.575 cm3
1 mark
n(H2SO4) = 0.150 × 19.575/1000 = 2.936 × 10−3 mol
1 mark
n(NaOH) in 25 cm3 = 2 × 2.936 × 10−3 = 5.87 × 10−3 mol
1 mark
n(NaOH) in 200 cm3 = 5.87 × 10−3 × 8 = 0.0470 mol
1 mark
mass = 0.0470 × 40.0 = 1.88 g
1 mark
Correct alternative routes are fine, and a slip early on still earns the later marks (error carried forward).
The student fills the burette using a funnel and forgets to remove it before starting. Suggest why this might affect the titre recorded.
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Extra drops of solution can run from the funnel into the burette, lowering the reading, so the recorded titre is too low. 1 mark
Must imply solution from the funnel drips into the burette.
State one advantage of using a conical flask rather than a beaker for the titration.
Show answer
Less chance of splashing out / losing solution when swirling (easier to swirl without loss). 1 mark
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Measurement of an enthalpy change
Full theory, worked calorimetry & interactives: Energetics — Calorimetry
Aim: measure the heat energy released or taken in by a reaction and use it to find an enthalpy change (ΔH, in kJ mol−1). Two set-ups cover every case: reactions in solution (neutralisation, dissolving, displacement) run in an insulated polystyrene cup; combustion burns a fuel to heat a known mass of water.
In solution — polystyrene cup
- Measure a known volume of one solution into a polystyrene cup with a lid, and record its temperature every minute for a few minutes to fix a steady starting line.
- Add the second reactant — weighing any solid by difference — stir, and keep reading the temperature every minute through the peak and the cool-down.
- Find ΔT from the cooling-correction graph (below), then use q = mcΔT and divide by the moles reacting to get ΔH.
Combustion — spirit burner under a can
- Weigh the spirit burner; measure a known mass of water into a metal can and record the water’s start temperature.
- Burn the fuel to raise the water by about 10–20 °C, then reweigh the burner — the difference is the mass of fuel burned.
- Use q = mcΔT for the water, then divide by the moles of fuel burned to get ΔcH.
Apparatus & techniques
- accurate mass, volume & temperature readings
- weighing by difference
- insulated (polystyrene-cup) calorimetry
- plotting & extrapolating a cooling graph
- safe handling of flammable & corrosive substances
A polystyrene cup with a lid (or a thin metal can for combustion), a 0–50 °C thermometer, measuring cylinders (or a burette/pipette for better precision), a stopwatch, a balance reading to 0.01 g, a stirrer and graph paper.
Safety. Wear eye protection. Fuels such as ethanol are highly flammable — keep the spirit burner away from other flames and cap it to put it out. Acids and alkalis (hydrochloric acid, sodium hydroxide) are irritant or corrosive, so handle them with care and mop up any spills.
What you should get: neutralising a strong acid with a strong alkali gives ΔH ≈ −57 kJ mol−1 every time — it is always the same H+ + OH− → H2O reaction. Combustion values always come out less exothermic than the data book, because of heat loss and incomplete combustion.
Data analysis — the cooling correction
This is the RP2 skill the written papers really probe. A reaction is never instant and the cup is never perfect, so the mixture starts cooling the moment it starts reacting — the highest temperature you actually read is already an underestimate. The fix: record the temperature every minute before mixing, mix at a known time, keep recording through the peak and cool-down, then extrapolate the cooling line back to the moment of mixing and read ΔT there. Have a go:
Interactive — your turn: extrapolate the cooling line
- The thermometer is read to ± 0.1 °C, but ΔT is a difference of two readings, so its uncertainty is 2 × 0.1 = ± 0.2 °C. % uncertainty = (0.2 ÷ ΔT) × 100 — a bigger ΔT (higher concentrations) makes it smaller.
- The main systematic error is heat loss to the surroundings: a lid and insulation cut it down, and the extrapolation corrects for what is left.
- Two assumptions you must state: the solution’s density is taken as 1.00 g cm−3 and its specific heat capacity as water’s (4.18 J g−1 K−1) — only true for dilute solutions.
- Give the final answer to the significant figures of the least precise measurement.
Exam questions — RP2
50 cm3 of 0.400 mol dm−3 HCl is put in a beaker and its temperature recorded each minute for 3 minutes. At the 4th minute, 50 cm3 of 0.400 mol dm−3 NaOH is added and stirred; temperatures are recorded for a further 8 minutes. Describe how to use a temperature–time graph to find the temperature rise ΔT at the 4th minute.
Show answer
Plot temperature (y) against time, with a sensible scale that does not start at 0 °C. 1
Plot all points accurately. 1
Draw two straight best-fit lines — one through the 0–3 min readings, one through the 6–12 min cooling readings (ignore the 5th-minute point). 1
Extrapolate both lines back to the 4th minute (the time of mixing). 1
Read ΔT there: e.g. 21.9 − 19.8 = 2.1 °C. 1
A single "S-shaped" curve through all the points scores neither the best-fit nor the extrapolation mark.
Each temperature reading has an uncertainty of ± 0.1 °C. Calculate the percentage uncertainty in ΔT (= 2.1 °C).
Show answer
(2 × 0.1) ÷ 2.1 × 100 = 9.5%
1 mark
ΔT uses two readings, so the uncertainty doubles to ±0.2 before dividing.
Suggest one change that would minimise heat loss.
Show answer
Replace the glass beaker with a polystyrene cup / add a lid / insulate the beaker. 1 mark
Do not accept a copper or bomb calorimeter.
Suggest and explain another change that would decrease the percentage uncertainty using the same thermometer.
Show answer
Increase the size of the temperature change 1 by increasing the concentration of the acid and alkali. 1
A larger ΔT makes the fixed ±0.2 °C a smaller fraction.
25 cm3 of 0.80 mol dm−3 ethanedioic acid (HOOCCOOH) is added to 75 cm3 of 0.60 mol dm−3 KOH; the temperature rises 3.2 °C. Give the equation and calculate ΔH per mole of water formed. (c = 4.2 J K−1 g−1, density = 1.00 g cm−3.)
Show answer
HOOCCOOH + 2KOH → K2(COO)2 + 2H2O
1 mark (equation)
q = mcΔT = 100 × 4.2 × 3.2 = 1344 J
1 mark
n(acid) = 0.020 mol, n(KOH) = 0.045 mol
1 mark
acid limits (0.020 mol needs 0.040 mol of the 0.045 mol KOH) → moles of water = 2 × 0.020 = 0.040 mol
1 mark
ΔH = −1.344 ÷ 0.040 = −33.6 kJ mol−1
1 mark
Answer must be negative and to at least 2 s.f. Error carried forward allowed.
The enthalpy of neutralisation for sulfuric acid + KOH is −57.0 kJ mol−1. Explain why the ethanedioic-acid value differs.
Show answer
Ethanedioic acid is a weak acid — not fully dissociated — whereas sulfuric acid is strong (fully dissociated). 1
Energy is absorbed to complete the dissociation of the weak acid (it is endothermic), so less energy is released overall. 1
A student measures the enthalpy of combustion of heptane (spirit burner under a clamped copper calorimeter). Design a table to record all the readings needed.
Show answer
Temperature of water: initial, final (and ΔT) with units. 1
Mass of burner: before and after burning (and mass of heptane burned) with units. 1
Missing units on either data type caps this at 1 mark.
Suggest two disadvantages of using a glass beaker on a tripod and gauze instead of the clamped copper calorimeter.
Show answer
Glass is a poorer conductor of heat than copper. 1
The tripod and gauze increase heat loss (or fix the height above the flame). 1
Suggest two reasons why the experimental enthalpy of combustion is less exothermic than the data-book value.
Show answer
Heat loss to the surroundings / to the calorimeter. 1
Incomplete combustion of the fuel. 1
Suggest one addition to the apparatus that would improve the accuracy.
Show answer
A wind shield (or a lid, or insulation on the sides of the calorimeter) to reduce heat loss. 1 mark
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
How the rate of a reaction changes with temperature
Full theory, Maxwell–Boltzmann & interactives: Kinetics — Temperature
Aim: find out how the rate of a reaction changes as you raise the temperature, using a clock reaction — one you time to a single, fixed visible change.
- Stand a flask holding a fixed volume of sodium thiosulfate over a pencil cross on paper; add a measured volume of dilute hydrochloric acid and start the clock.
- Time how long the cloudiness (a fine precipitate of sulfur) takes to hide the cross. Shorter time means faster reaction, so rate ∝ 1/time.
- Repeat over about five temperatures set with a water bath — stand both solutions in the bath first, so the run starts at the temperature you record.
- Plot 1/time against the mean temperature.
Because the end point is judged by eye, everything except the temperature has to be pinned down:
- Change only the temperature — fix every volume and concentration.
- Judge the end point identically — same observer, same cross, and the same reaction vessel. A wider vessel spreads the mixture into a shallower layer, so you look through less solution and the cross takes longer to vanish; change the vessel and the times no longer compare.
- Allow for temperature drift — the mixture’s own temperature shifts as it reacts, so read it at the start and end of each run and plot the mean.
Apparatus & techniques
- accurate volume, temperature & time readings
- water bath for heating
- marked cross & stopclock
- 1/time vs temperature graphs
- safe handling of a toxic gas
Safety. The reaction gives off sulfur dioxide, which is toxic — work in a well-ventilated room (or a fume cupboard) and tip each cloudy mixture straight into a sodium carbonate “stop bath” to neutralise the acid and the gas. Dilute hydrochloric acid is an irritant.
What you should see: the colourless mixture slowly turns cloudy and pale yellow as solid sulfur forms, until the cross is completely hidden and you stop the clock. The higher the temperature, the shorter the time — so 1/time, your measure of rate, climbs steeply as the temperature rises.
Explaining the result. A small rise in temperature gives a surprisingly large rise in rate — a classic mark-scheme to get right. Build the full answer here:
Interactive — build the mark-scheme answer
“Explain why increasing the temperature increases the rate of a reaction.” Select every statement that earns a mark — and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
your answer, assembled
- The end point is judged by eye, so the real uncertainty is when the cross has “disappeared”. Record the time only to the nearest second — not the stopclock’s 0.01 s — because your judgement, not the clock, is the limit.
- Keep the depth of liquid and the vessel the same every run, or the cross is easier or harder to hide and the times are not comparable.
- Plot the mean of the start and end temperatures, and ignore any anomalous point when you draw the line.
Five runs of the thiosulfate clock, changing only the temperature, give these times for the cross to disappear. Work out 1/time for each — that is your measure of rate:
| Temperature / °C | Time / s | Rate = 1/time / s−1 |
|---|---|---|
| 20 | 200 | 1/200 = 0.0050 |
| 30 | 101 | 1/101 = 0.0099 |
| 40 | 52 | 1/52 = 0.0192 |
| 50 | 28 | 1/28 = 0.0357 |
| 60 | 15 | 1/15 = 0.0667 |
Plot 1/time (y) against mean temperature (x). The points rise ever more steeply — the rate roughly doubles for each 10 °C, so a small temperature rise gives a large rise in rate. To find the rate at any temperature, read its 1/time straight off the curve; to compare two runs, compare their 1/time values (never the raw times, which run the other way).
Processing the data. The exam questions below hand you real thiosulfate and manganate(VII) results to turn into 1/time, plot against temperature, and read off — the data-analysis this practical is built to test.
Exam questions
Increasing the temperature causes the rate of reaction to increase. Explain why a small increase in temperature causes a large increase in the rate of reaction.
Show answer
Many more particles now have energy equal to or greater than the activation energy. 1 mark
So there is a much greater frequency of successful collisions per unit time. 1 mark
The mark is for “many more particles have energy ≥ Ea” — the size of the effect matters. “Particles move faster” on its own scores nothing.
Draw the Maxwell–Boltzmann distribution curves for a fixed mass of a gas at two different temperatures. This gas decomposes when heated. By reference to these distribution curves, explain why the rate of decomposition of the gas increases at higher temperatures.
Show answer
The curves (as in the two-temperature figure above): label the vertical axis number of molecules and the horizontal axis (kinetic) energy. 1 mark Each curve starts at the origin and its tail approaches but never meets the energy axis. 1 mark
The higher-temperature curve has a lower peak shifted to the right, the same area beneath it, and crosses the first curve once. 1 mark
The explanation: at the higher temperature the molecules have more energy, so a greater proportion of molecules have energy ≥ the activation energy. 1 mark A greater proportion of collisions are therefore successful. 1 mark So the frequency of successful collisions increases and the rate increases. 1 mark
Do not accept a higher-temperature curve whose area is larger (the number of molecules is fixed), or one that touches the energy axis.
A student investigates the effect of temperature on the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid: Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + SO2(g) + S(s) + H2O(l). The student mixes the solutions in a flask placed on a paper marked with a cross, and records the time for the cross to disappear as the mixture becomes cloudy. The table shows the results.
| Temperature / °C | 22 | 31 | 36 | 42 | 49 | 54 |
|---|---|---|---|---|---|---|
| Time, t, for cross to disappear / s | 87 | 48 | 36 | 26 | 44 | 12 |
| 1/t / s−1 | 0.0115 | 0.0208 | 0.0278 | 0.0385 | 0.0227 |
The student uses a stopwatch to measure the time. The stopwatch shows each time to the nearest 0.01 s. Suggest why the student records the times to the nearest second and not to the nearest 0.01 s.
Show answer
It is hard to judge the exact moment the cross disappears (or: the mixture becomes too cloudy to judge that precisely / reaction time when stopping the clock). 1 mark
Vague appeals to “accuracy” are ignored unless qualified by the end-point judgement.
The rate of reaction is proportional to 1/t. Complete the table above.
Show answer
1/t = 1 ÷ 12 = 0.0833 s−1 1 mark
Plot the values of 1/t against temperature on the graph below. Draw a line of best fit.
Show answer
All six points plotted correctly (±½ small square). 1 mark
A smooth best-fit curve that misses the anomalous point at 49 °C and passes within one small square of the other five. 1 mark
The best-fit line is penalised if it is forced through (0, 0). Spot the anomaly, plot it, then ignore it.
Use your line of best fit to estimate the time for the cross to disappear at 40 °C. Show your working.
Show answer
from the line, 1/t at 40 °C ≈ 0.035 s−1, so t = 1 ÷ 0.035 ≈ 29 s 1 mark
Any answer consistent with your own line scores — but the working must show the 1/t read-off being inverted.
Suggest, by considering the products of this reaction, why small amounts of reactants are used in this experiment.
Show answer
SO2 is toxic — small amounts limit how much of the gas forms. 1 mark
“Harmful” or “causes acid rain” is not enough — the credited word is toxic/poisonous.
The student could do the experiment at lower temperatures using an ice bath. Suggest why the student chose not to carry out experiments at temperatures in the range 1–10 °C.
Show answer
The reaction would be very slow / take too long (and the gradual end-point becomes even harder to judge). 1 mark
Potassium manganate(VII) reacts with sodium ethanedioate in the presence of dilute sulfuric acid: 2MnO4−(aq) + 16H+(aq) + 5C2O42−(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g). The mixture is purple at the start and goes colourless when all the MnO4−(aq) ions have reacted, so the rate can be measured as 1000/t, where t is the time taken to go colourless. A student timed the reaction at different temperatures, with the same concentrations and volumes of each reagent every time.
| Temperature / °C | 32 | 38 | 44 | 54 | 67 |
|---|---|---|---|---|---|
| Time t / s | 155 | 85 | 50 | 22 | 9 |
| 1000/t / s−1 | 6.45 | 11.8 | 20.0 | 45.5 |
Complete the table above.
Show answer
1000 ÷ 9 = 111 s−1 (111.1) 1 mark
Any correctly rounded number of significant figures is allowed (110 is fine); a recurring-dot answer is not.
State the independent variable in this investigation.
Show answer
Temperature 1 mark
The student noticed that the temperature of each reaction mixture decreased during each experiment. Suggest how the student calculated the temperature values in the table above.
Show answer
Measure the temperature at the start and at the end (or at regular intervals) and take the mean. 1 mark
Use the data in the table to plot a graph of 1000/t against temperature.
Show answer
A vertical scale that uses more than half the axis for the five points. 1 mark
Points plotted correctly (±½ small square each). 1 mark
A smooth best-fit curve, rising throughout, within one small square of each point. 1 mark
Use your graph to find the time taken for the mixture to go colourless at 60 °C. Show your working.
Show answer
from the curve, 1000/t at 60 °C ≈ 68, so t = 1000 ÷ 68 ≈ 15 s 1 mark
Use the value from your line (construction lines shown on the graph); give the answer to at least 2 significant figures.
Apparatus is set up to measure the time for 20.0 cm3 of sodium thiosulfate solution to react with 5.0 cm3 of hydrochloric acid in a 100 cm3 conical flask at 20 °C. The timer is started when the thiosulfate is added and stopped when the cross can no longer be seen. What is likely to decrease the accuracy of the experiment? Tick (✓) one box.
The same experiment is repeated at 20 °C using a 250 cm3 conical flask instead of the 100 cm3 one. Which statement is correct about the time taken for the cross to disappear when using the larger flask? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Test-tube reactions to identify ions
Full theory: Group 2 — the sulfate test · Group 7 — the halide test
Aim: use simple test-tube reactions to identify common cations (Group 2 ions and the ammonium ion) and anions (halides, OH−, CO32− and SO42−) in unknown solutions. The marks are almost always in the observation — the precise colour, the precipitate, the gas — so learn those exactly.
Unlike GCSE, A-Level does not identify Group 2 metals by flame colour. Instead you use the opposite solubility trends of the Group 2 hydroxides and sulfates: hydroxides get more soluble down the group, sulfates get less soluble — so which one precipitates tells you where in the group the metal sits.
Here is every test, the reagent, and the observation that names the ion:
| Test for | Add… | Positive observation | Ionic equation |
|---|---|---|---|
| Group 2 cation (via hydroxide) | sodium hydroxide, dropwise | white precipitate — strongest for Mg2+, fading down the group (Ba2+ stays in solution, as Ba(OH)2 is soluble) | Mg2+ + 2OH− → Mg(OH)2 |
| Group 2 cation (via sulfate) | dilute sulfuric acid | white precipitate — strongest for Ba2+, fading up the group (Mg2+ stays in solution, as MgSO4 is soluble) | Ba2+ + SO42− → BaSO4 |
| Ammonium NH4+ | sodium hydroxide, then warm | a pungent gas turns damp red litmus blue (held at the mouth of the tube) | NH4+ + OH− → NH3 + H2O |
| Hydroxide OH− | red litmus paper | turns blue — the solution is alkaline | — |
| Carbonate CO32− | dilute acid; bubble the gas through limewater | effervescence; the gas (CO2) turns limewater milky | CO32− + 2H+ → CO2 + H2O |
| Sulfate SO42− | acidify (dilute HCl), then barium chloride | white precipitate | Ba2+ + SO42− → BaSO4 |
| Chloride Cl− | acidify (dilute HNO3), then silver nitrate; then ammonia | white precipitate — dissolves in dilute ammonia | Ag+ + Cl− → AgCl |
| Bromide Br− | cream precipitate — dissolves only in concentrated ammonia | Ag+ + Br− → AgBr | |
| Iodide I− | yellow precipitate — insoluble, even in concentrated ammonia | Ag+ + I− → AgI |
Apparatus & techniques
- test tubes & dropping pipettes
- qualitative tests for ions
- gas tests (litmus, limewater)
- recording observations clearly
- safe handling of toxic & corrosive reagents
Safety. Wear eye protection. Sodium hydroxide and silver nitrate are corrosive (silver nitrate also stains skin), barium chloride solution is toxic, and concentrated ammonia is an irritant — use it in a fume cupboard. Work with small volumes and rinse spills away.
Try the halide identifier — add the reagents in turn and read the precipitate and its behaviour in ammonia to name the ion:
Interactive — identify the unknown halide
A solution contains one of F−, Cl−, Br− or I−. Run the tests, then name it.
name the halide
To identify an unknown salt you run the anion tests in a sensible order, because some ions give a false positive in a later test:
- Carbonate first (add acid — fizz, limewater milky). It has to go first, because carbonate would also give a white precipitate with barium.
- Sulfate next (acidified barium chloride). Acidifying first removes any carbonate and sulfite.
- Halide last (acidified silver nitrate). Acidifying with nitric acid removes carbonate and hydroxide, which would also precipitate with silver.
And the acid you choose matters: acidify with nitric acid for the halide test — not hydrochloric (adds Cl−) and not sulfuric (adds SO42−) — or you introduce the very ion you are testing for.
Exam questions — RP4
Four unlabelled solutions are known to be ammonium nitrate, potassium sulfate, sodium carbonate and magnesium nitrate. Outline a series of test-tube reactions to identify each — include the expected observations and ionic equations for any reactions.
Show answer
Marked by levels of response — for full marks, all four are identified with a test, an observation linked to the right solution, and an equation.
Sodium carbonate — add dilute acid: it effervesces, and the gas turns limewater milky. CO32− + 2H+ → CO2 + H2O tests · obs · equations
Ammonium nitrate — add sodium hydroxide and warm: the gas turns damp red litmus blue. NH4+ + OH− → NH3 + H2O
Magnesium nitrate — add sodium hydroxide: a white precipitate. Mg2+ + 2OH− → Mg(OH)2
Potassium sulfate — add acidified barium chloride: a white precipitate. SO42− + Ba2+ → BaSO4
The fourth solution can be identified by elimination if only three tests are run. Do not put the red litmus in the solution — it is held at the mouth of the tube. Ignore state symbols; allow multiples.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Distillation of a product from a reaction
Full theory + animated rig: Alcohols — oxidation & distillation
Aim: separate a volatile product from a reaction mixture by distillation. The classic example is oxidising a primary alcohol and distilling off the aldehyde as it forms — taking it out of the flask before it can be oxidised any further to the carboxylic acid.
- Set up the distillation apparatus: a pear-shaped flask holding the reaction mixture with a few anti-bumping granules; a still head carrying a thermometer with its bulb level with the side-arm; a downward-sloping Liebig condenser with cooling water in at the bottom, out at the top; and an open receiver flask.
- Heat gently. The product has the lowest boiling point, so it vaporises first, passes the thermometer bulb, and condenses into the receiver.
- Collect the fraction that distils over at the product’s boiling range — the thermometer tells you when the right substance is coming across.
Distil or reflux? Distil to take a product out as it forms, so it cannot react further (an aldehyde, before it oxidises to the acid). Reflux to keep everything in and drive a reaction to completion (oxidising all the way to the carboxylic acid) — a vertical condenser, open at the top, returns the vapour to the flask. Both set-ups are drawn and animated in the figure.
Apparatus & techniques
- setting up glassware with clamps
- distillation & heating under reflux
- heating with a water bath / electric heater
- reading a thermometer to the boiling range
- safe handling of flammable & toxic reagents
Safety. Wear eye protection. Alcohols and their products (e.g. ethanol, ethanal) are flammable — heat with an electric heater or water bath, not a naked flame. Acidified potassium dichromate(VI) is toxic and a carcinogen, so avoid skin contact and work in a fume cupboard.
What you should see and collect: the thermometer holds steady at the product’s boiling point while it distils over — collect that fraction. Distilling ethanal off ethanol gives a colourless liquid in the receiver, while the flask itself turns from an orange solution to a green solution as the dichromate(VI) is reduced.
“What is wrong with this apparatus?” is the classic distillation question. Get these right:
- Thermometer bulb level with the side-arm — so it reads the temperature of the vapour actually entering the condenser, not the boiling flask.
- Condenser water in at the bottom, out at the top (counter-current) — so the jacket stays full and cold; the wrong way round and the product comes through as a gas.
- Anti-bumping granules — to give small, smooth bubbles and stop the mixture bumping up into the condenser.
- Never fully sealed, but closed at the flask with the thermometer/bung — open to the air at the receiver (so pressure cannot build), yet closed above the flask so only the product distils over.
Processing the data — percentage yield. The other half of RP5 marks is turning what you collect into a % yield, from the amount of limiting reactant you started with.
Exam questions — RP5
A student sets up apparatus to make propanone by distilling it out as propan-2-ol is oxidised. There are three problems with the set-up. For each, identify the problem, describe the issue it causes, and suggest how to solve it.
Show answer
Marked by levels of response — three faults, each with the problem, its effect and the fix:
1. No anti-bumping granules. The mixture can bump violently and jump up into the condenser — add anti-bumping granules. stage 1
2. Open system / no thermometer at the flask. Vapour (and product) escapes and you cannot control which substance distils — close it with a bung and thermometer so only propanone, in its boiling range, comes over. stage 2
3. Condenser water flowing the wrong way. The condenser is not full of cold water, so the product is not cooled and comes through as a gas — connect the water to flow in at the bottom, out at the top. stage 3
2.0 cm3 of propan-2-ol (density 0.786 g cm−3, Mr = 60.0) is oxidised with excess acidified dichromate and 0.954 g of propanone (Mr = 58.0) is collected. Calculate the percentage yield to the appropriate number of significant figures.
Show answer
mass of propan-2-ol = 2.0 × 0.786 = 1.572 g
1 mark
n(propan-2-ol) = 1.572 ÷ 60.0 = 0.0262 mol
1 mark
expected mass of propanone = 0.0262 × 58.0 = 1.52 g
1 mark
% yield = (0.954 ÷ 1.52) × 100 = 63%
1 mark
Answer to 2 s.f., following the 2.0 cm3. Allow ECF through the chain.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Tests for alcohol, aldehyde, alkene & carboxylic acid
Full theory + interactive identifier: Organic analysis — test-tube reactions
Aim: use four reliable wet tests to identify the common organic functional groups — alkene, alcohol, aldehyde and carboxylic acid. As with the ion tests, the marks are in the exact observation (the colour change or precipitate), not the name of the reagent.
Here are the four tests, colour-coded by what you see:
| Functional group | Add… | Positive observation | Negative (control) |
|---|---|---|---|
| Alkene (C=C) | bromine water, shake | decolourises: orange solution → colourless solution | an alkane stays orange |
| Alcohol (1° or 2°) | acidified potassium dichromate(VI), warm | orange → green | a 3° alcohol stays orange |
| Aldehyde –CHO | Tollens’ reagent, warm gently | a silver mirror forms | a ketone: no change |
| Aldehyde –CHO | Fehling’s solution, warm gently | blue solution → brick-red precipitate | a ketone stays blue |
| Carboxylic acid –COOH | sodium carbonate solution | effervescence — the gas (CO2) turns limewater milky | an alcohol: no bubbles |
bromine water · shake
The alkene adds Br2 across its C=C and uses it up — say decolourised, never “clear”
sodium carbonate solution
Only a carboxylic acid is acidic enough to free CO2 from a carbonate; the gas turns limewater cloudy
tollens’ reagent · warm gently
Fehling’s does the same job: blue solution → brick-red precipitate with the aldehyde, stays a blue solution with the ketone
acidified K2Cr2O7 · warm
1°/2° alcohols (and aldehydes) reduce orange Cr2O72− to green Cr3+; a 3° alcohol resists oxidation
Apparatus & techniques
- test tubes & a water bath
- qualitative tests for organic groups
- precise colour & precipitate observations
- distinguishing a pair by test & control
- safe handling of toxic & corrosive reagents
Safety. Wear eye protection. Bromine water is toxic and corrosive and acidified dichromate(VI) is toxic — use small volumes in a fume cupboard. Tollens’ reagent must be freshly prepared and not stored (it can form an explosive solid on standing).
Try the identifier — pick a group or distinguish a pair, and read the reagent and the observation that proves it:
Interactive — the test-tube identifier
- Acidify the dichromate (add dilute sulfuric acid) — plain dichromate will not oxidise the alcohol.
- Warm Tollens’ and Fehling’s gently in a water bath — left cold, no silver mirror or brick-red precipitate forms.
- Test the gas from the carbonate with limewater (it turns milky), not a lighted splint — CO2 puts a splint out.
- Always give the observation and the negative result (“stays orange”, “no change”) so the pair is actually distinguished — say decolourised, never “clear”.
Identifying an unknown. Run the tests in a sensible order — e.g. Tollens’ first to pick out an aldehyde, then acidified dichromate — so each result narrows down what the compound can be.
Exam questions
Give a reagent that could be added separately to each of CH3CH2CH2CHO and CH3CH2CH(OH)CH3 to distinguish between them. State what is observed in each case.
Show answer
Reagent: Tollens’ reagent (ammoniacal silver nitrate). 1 mark
With CH3CH2CH2CHO (aldehyde): silver mirror. 1 mark
With CH3CH2CH(OH)CH3 (secondary alcohol): no reaction / no visible change. 1 mark
Fehling’s (brick-red precipitate vs no change) is an equally good alternative. Do not accept acidified dichromate — it oxidises both.
Give a reagent that could be added separately to cyclohexane and cyclohexene to distinguish between them. State what is observed in each case.
Show answer
Reagent: bromine water. 1 mark
With cyclohexane: stays an orange solution / no visible change. 1 mark
With cyclohexene: decolourised (orange solution → colourless solution). 1 mark
Acidified KMnO4 (purple solution → colourless solution with the alkene) also scores. Ignore “clear” for colourless.
Three bottles contain either propan-1-ol, propanal or propanone. Each is warmed with Fehling’s solution. Identify the liquid that reacts with Fehling’s and give the observation, then suggest a further test to distinguish the remaining two liquids, with the observation for the one that reacts.
Show answer
Reacts with Fehling’s: propanal (aldehyde) — blue solution gives a brick-red precipitate. 1 mark
Further test: warm with acidified potassium dichromate(VI). 1 mark
Propan-1-ol (alcohol) turns the orange solution green; propanone gives no change. 1 mark
Source: AQA A-Level Chemistry past papers.
Measuring the rate of a reaction
Full theory, orders & rate–concentration graphs: Rate equations · Kinetics — measuring rate
Aim: measure the rate of a reaction by two different methods — an initial-rate method (the iodine clock) and a continuous-monitoring method (collecting a gas as it is produced) — and use each to find an order of reaction. AQA require both; you will have done one of each.
They answer the same question from opposite ends. An initial-rate method measures one number per run — how long a fixed, tiny amount of product takes to appear — then you change one concentration and run it again. A continuous-monitoring method follows one run all the way through, so a single experiment gives you a whole curve to take gradients from.
Method A — the iodine clock (an initial-rate method)
Hydrogen peroxide oxidises iodide ions to iodine. A small, fixed amount of thiosulfate is also present, and it mops up the iodine as fast as it forms:
H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)
2S2O32−(aq) + I2(aq) → 2I−(aq) + S4O62−(aq)
While any thiosulfate is left there is no free iodine, so the mixture stays colourless. The moment the thiosulfate runs out, iodine builds up and the starch turns the whole beaker blue-black — that is your clock stopping. Because the same fixed amount of iodine is made every time, 1/t is proportional to the initial rate.
- Rinse a burette with potassium iodide solution and fill it.
- Transfer 10.0 cm3 of hydrogen peroxide solution to a clean, dry 100 cm3 beaker — you add this last.
- Into a clean, dry 250 cm3 beaker put 25 cm3 of dilute sulfuric acid, then 20 cm3 of water, then about 1 cm3 of starch solution.
- Run 5.0 cm3 of potassium iodide solution in from the burette.
- Add 5.0 cm3 of sodium thiosulfate solution last, and stir.
- Tip in the hydrogen peroxide, start the timer immediately, and keep stirring.
- Stop the timer the instant the mixture turns blue-black. Record the time in a table of your own design.
- Rinse and dry the beaker, then repeat with 10.0, 15.0, 20.0 and 25.0 cm3 of potassium iodide — reducing the water to 15, 10, 5 and 0 cm3 so the total volume never changes.
Method B — collecting a gas (a continuous-monitoring method)
- Measure 50 cm3 of 0.8 mol dm−3 hydrochloric acid into a 100 cm3 conical flask.
- Clamp a 100 cm3 gas syringe and connect it to a bung and delivery tube (or collect the gas over water in an inverted measuring cylinder).
- Add one 6 cm strip of magnesium ribbon, push the bung firmly home and start the timer at the same moment.
- Record the volume of hydrogen every 15 s for 2.5 minutes.
- Repeat with 0.4 mol dm−3 acid, made by mixing 25 cm3 of the 0.8 mol dm−3 acid with 25 cm3 of water.
Apparatus & techniques
- measuring rate by two different methods
- burette & measuring cylinder to a stated precision
- collecting a gas in a syringe or over water
- stopwatch, and judging a colour end point
- tangents, gradients & rate–concentration graphs
Safety. Wear eye protection throughout. Dilute sulfuric and hydrochloric acids are irritants and the hydrogen peroxide solution is an irritant that bleaches skin and clothing. Hydrogen is flammable — keep the gas syringe away from flames, and never seal the apparatus so tightly that pressure can build up. Wash any splashes off at once.
What you should see. In the clock, nothing at all happens for most of the run — then the whole beaker goes blue-black in about a second. In the gas-collection run, the plunger moves fastest at the very start and slows steadily as the acid is used up, finally stopping when all the magnesium has gone.
Data analysis — getting an order out of each method
The two methods need different processing, and both are examined.
| Method | What you measure | How it becomes a rate | What you plot |
|---|---|---|---|
| Initial rate (clock) | time t for the colour to appear | rate ∝ 1/t, because the same amount of I2 is made every run | 1/t against the concentration you changed |
| Continuous (gas) | volume of gas at each time | gradient of a tangent to the curve | volume against time, then rate against concentration |
Reading a straight-line rate–concentration graph is the part that carries the marks: a horizontal line means zero order, a straight line through the origin means first order, and an upward curve means second order.
- Keep the total volume constant. The water is not padding — it is there so the only concentration that changes is the one you are investigating. Say this if asked why water is added.
- Add the thiosulfate last, and start the clock as the peroxide goes in. Any delay between mixing and starting the timer is a systematic error that makes every time too short.
- The clock measures the initial rate only because so little reactant is used before the colour appears — the concentrations are still effectively their starting values.
- Gas escapes before the bung is in. In the continuous method this loses volume from every reading, so the measured rate is too low. Add the magnesium and bung the flask in one movement.
- Draw the tangent at t = 0 to the curve itself, touching it at the origin, and take the gradient over a large triangle. A tangent drawn as a chord, or over two small squares, is the commonest lost mark on this practical.
- A loose cotton-wool plug, not a bung, if you are following mass loss on a balance — it stops acid spraying out but still lets the gas escape.
Exam questions — RP7
- Place a conical flask on a balance and add approximately 20 g of large marble chips.
- Add 50 cm3 of 0.4 mol dm−3 hydrochloric acid.
- Place a loose cotton wool plug in the neck of the flask.
- Zero the mass reading on the balance.
- Start a timer.
- Record the loss in mass (mt) every 30 seconds for 4 minutes.
- Wait for the reaction to finish and record the total mass loss (mtotal).
- Plot a graph of (mtotal − mt) against time.
Suggest why a loose cotton wool plug is placed in the neck of the flask, instead of leaving the flask open or inserting a bung.
Show answer
Instead of leaving the flask open: to avoid acid / solution / liquid escaping, or to avoid it splashing / spraying / spitting out. 1 mark
Instead of inserting a bung: to allow the gas / CO2 to escape. 1 mark
Ignore evaporation, spilling, impurities getting in, and “pressure would build up”. A wrongly named gas scores nothing.
20 g of large marble chips is a large excess of calcium carbonate. Suggest why using a large excess of calcium carbonate means that the rate is only affected by the changing concentration of the hydrochloric acid.
Show answer
So that the surface area / mass / amount of calcium carbonate stays approximately constant. 1 mark
Ignore “concentration stays constant”, “volume”, “so HCl is the limiting factor”, and simply restating that the rate is only affected by [HCl].
The mass of carbon dioxide produced in time t is equal to mt. The total mass of CO2 produced during the reaction is equal to mtotal. Explain why (mtotal − mt) is proportional to the concentration of hydrochloric acid remaining in the flask at time t.
Show answer
mt, the mass of CO2 produced in time t, is proportional to the amount / concentration of HCl that has reacted at time t. 1 mark
mtotal, the total mass of CO2 produced, is proportional to the total amount / concentration of HCl that reacted / was present initially — so (mtotal − mt) is proportional to the HCl present at time t. 1 mark
Equally accepted: (mtotal − mt) is the mass of CO2 still to be produced, and that is proportional to the HCl still to react. “Equal to” or “represents” are allowed for “proportional to”.
The rate of reaction, calculated from the gradient of the curve at five different times, is shown below, together with the corresponding value of (mtotal − mt).
| Rate of reaction / g s−1 | (mtotal − mt) / g |
|---|---|
| 23.0 × 10−4 | 0.340 |
| 19.0 × 10−4 | 0.280 |
| 15.7 × 10−4 | 0.225 |
| 11.5 × 10−4 | 0.170 |
| 6.67 × 10−4 | 0.100 |
Plot the rate of reaction (y-axis) against (mtotal − mt) (x-axis) and draw a line of best fit. Then state how your graph confirms that the rate equation for this reaction is rate = k[HCl].
Show answer
Scales chosen so the plotted points (and the origin, if shown) occupy more than half of each axis, with both axes labelled including units. 1 mark
Points plotted to within half a small square. 1 mark
A suitable straight line — within one square of every point. 1 mark
How it confirms the order: the graph is a straight line and it passes through the origin. 1 mark
“Constant gradient” is allowed for “straight line”, but the line must also be stated to go through the origin — “directly proportional” on its own is ignored. A reversed scale loses the first mark but the rest carry through by error.
In this experiment the variable measured is mass loss. The rate of this reaction at a constant temperature can be investigated in other ways. Suggest two other variables that can be measured instead of mass loss.
Show answer
Any two from: volume of gas / CO2 · pH · concentration of HCl / acid / H+ · conductivity. 2 marks
Temperature scores nothing. “Volume” or “concentration” unqualified scores nothing either — say of what.
Explain why the reaction is fastest at the start.
Show answer
The concentration of hydrogen peroxide / the number of H2O2 molecules is highest at the start. 1 mark
So there are more frequent successful collisions. 1 mark
The table below shows data from a similar experiment. Plot the data and draw a line of best fit, then use your graph to determine the order of reaction with respect to H2O2, stating how the graph shows this order.
| [H2O2] / mol dm−3 | Rate / mol dm−3 s−1 |
|---|---|
| 0.02 | 0.000 49 |
| 0.03 | 0.000 73 |
| 0.05 | 0.001 24 |
| 0.07 | 0.001 68 |
| 0.09 | 0.002 19 |
Show answer
Order = 1 (first order with respect to H2O2). 1 mark
How the graph shows it: the rate–concentration graph is a straight line through the origin, so the rate is directly proportional to [H2O2]. 1 mark
Check it yourself without plotting: 0.001 24 / 0.05 = 0.0248, and 0.002 19 / 0.09 = 0.0243 — rate / concentration is constant, which is exactly what first order means.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Measuring the EMF of an electrochemical cell
Full theory, cell notation & feasibility: Electrode potentials & cells · Redox
Aim: build a standard cell from two half-cells, measure its EMF with a high-resistance voltmeter, and then use one metal as a reference to rank several metals by their electrode potentials.
- Clean a piece of copper and a piece of zinc foil with emery paper or fine sandpaper.
- Degrease each metal with cotton wool and propanone.
- Stand the copper in a 100 cm3 beaker holding about 50 cm3 of 1.00 mol dm−3 CuSO4.
- Stand the zinc in a second beaker holding about 50 cm3 of 1.00 mol dm−3 ZnSO4.
- Plug one end of a U-tube with cotton wool, fill it with 2.0 mol dm−3 sodium chloride solution, then plug the free end with cotton wool soaked in the same solution.
- Invert the U-tube so a plugged end dips into each beaker — that is your salt bridge.
- Clip a lead to each metal and connect them to a digital (high-resistance) voltmeter. Read the voltage.
- Keep the copper half-cell from the first method — copper foil in 1.00 mol dm−3 CuSO4 — as your fixed reference, clipped to the positive terminal of the voltmeter.
- Clean and degrease the next metal exactly as before, and stand it in a beaker of 1.00 mol dm−3 solution of that metal’s own ions — zinc in ZnSO4, iron in FeSO4, magnesium in MgSO4, silver in AgNO3.
- Join the new half-cell to the copper reference with a fresh salt bridge — a U-tube, or a strip of filter paper, soaked in saturated potassium nitrate.
- Clip the second lead to the new metal, then note the voltage and its sign and record it in a table.
- Repeat with each metal provided, remaking the salt bridge each time, then write the conventional representation for every cell you have made.
- Use your readings to work out which pair would give the largest EMF.
Magnesium can be used as the reference instead of copper. Whichever you choose, every reading is measured against that one metal rather than against a standard hydrogen electrode, so the numbers rank the metals correctly without any of them being a true standard electrode potential.
Why each metal needs its own solution. An electrode potential is the position of the equilibrium Mn+(aq) + ne− ⇋ M(s). Stand the metal in anything else — sodium chloride solution, damp filter paper — and there are no Mn+ ions for that equilibrium to sit in, so there is nothing defined to measure and no cell you could write down. Use potassium nitrate, not sodium chloride, in the salt bridge if silver is one of your metals: chloride ions would precipitate AgCl on the electrode and kill the reading.
Apparatus & techniques
- preparing and using an electrochemical cell
- cleaning & degreasing electrodes
- making and placing a salt bridge
- reading a high-resistance voltmeter, with sign
- safe handling of solutions of transition-metal salts
Safety. Wear eye protection. Propanone is highly flammable — keep it away from flames and use it in a well-ventilated room. Copper(II) sulfate solution is harmful if swallowed and an irritant, and is dangerous to the environment, so it must not go down the sink; zinc sulfate solution is an irritant too. Emery paper leaves sharp metal edges — handle the cleaned foil by its flat faces.
What you should see. The standard zinc–copper cell reads about 1.10 V, with the copper positive and the zinc negative. If the leads are the other way round the meter simply shows −1.10 V, which tells you the same thing. A reading that drifts, or is far too low, almost always means the electrodes were not cleaned or the salt bridge is not making contact.
Data analysis — from two readings to a cell
The number the voltmeter gives you is the EMF. Turning it into chemistry means deciding which half-cell is the positive electrode, which is the negative, and writing the cell down in the conventional way:
E⦵cell = E⦵(more positive half-cell) − E⦵(more negative half-cell)
| Half-cell | Electrode reaction | E⦵ / V |
|---|---|---|
| Ca2+/Ca | Ca2+(aq) + 2e− → Ca(s) | −2.87 |
| Mg2+/Mg | Mg2+(aq) + 2e− → Mg(s) | −2.37 |
| Ti2+/Ti | Ti2+(aq) + 2e− → Ti(s) | −1.63 |
| Zn2+/Zn | Zn2+(aq) + 2e− → Zn(s) | −0.76 |
| Fe2+/Fe | Fe2+(aq) + 2e− → Fe(s) | −0.44 |
| Cu2+/Cu | Cu2+(aq) + 2e− → Cu(s) | +0.34 |
| Ag+/Ag | Ag+(aq) + e− → Ag(s) | +0.80 |
Zinc has the more negative electrode potential, so zinc is the negative electrode and copper the positive:
E⦵cell = (+0.34) − (−0.76) = +1.10 V
The zinc is oxidised (it is the better reducing agent, so it gives up electrons) and the copper(II) ions are reduced. Electrons flow through the wire from zinc to copper, and the conventional representation is written with the more negative half-cell on the left:
Zn(s) | Zn2+(aq) ‖ Cu2+(aq) | Cu(s)
Read it outwards from the double line: the oxidised species of each pair sits next to the salt bridge, the reduced species at the outside.
Build a cell from any two half-cells. The widget puts the more negative one on the left, marks the electrodes, works out the EMF and writes the conventional representation — check yourself against it before you look.
first half-cell
second half-cell
- Clean and degrease the electrodes. Emery paper takes off the oxide layer; propanone takes off grease. Either one left on gives a low, drifting reading.
- The voltmeter must be high-resistance (a digital one is). Almost no current then flows, so the concentrations stay at 1.00 mol dm−3 and the reading is the maximum potential difference — which is what EMF means.
- The salt bridge completes the circuit by letting ions move between the half-cells and keeping each solution electrically neutral. It must not react with either solution — potassium nitrate is the usual choice, and you must not use a chloride bridge with silver ions because silver chloride would precipitate.
- Standard conditions or it is not a standard EMF: all solutions 1.00 mol dm−3, 298 K, and 100 kPa for any gas. Adding water to one half-cell shifts its equilibrium and changes the reading.
- The size of the electrode makes no difference. A bigger piece of metal, or a bigger platinum electrode, changes how fast charge could flow, not the potential difference — a favourite one-mark question.
- Record the sign. A negative reading is a real result, not a mistake: it says the metal you clipped on second is the negative electrode.
Exam questions — RP8
A cell is set up to measure the standard electrode potential for the half-cell Mg2+(aq) + 2e− → Mg(s), using a magnesium electrode in magnesium chloride solution joined by a salt bridge to a standard hydrogen electrode. State the purpose of the salt bridge, and identify an ionic compound that could be used in it.
Show answer
Purpose: to complete the circuit, or to allow ions to move between the half-cells, or to maintain electrical neutrality. 1 mark
Identity: potassium nitrate or sodium nitrate — or any soluble ionic compound that does not react with H+, magnesium ions or chloride ions. 1 mark
“Allows electrons to flow” scores nothing — electrons travel through the wire, ions through the bridge.
State how, if at all, the EMF of this cell will change if the surface area of the platinum electrode is increased.
Show answer
No change. 1 mark
The standard electrode potential for the half-cell is E⦵ = −2.38 V. Water is added to the beaker containing the magnesium chloride solution. What is the effect on the magnitude of the EMF of the cell? Tick one box.
The voltmeter in this cell is replaced by a bulb. Give an equation for the overall reaction that occurs when the cell is operating.
Show answer
Mg + 2HCl → MgCl2 + H2
1 mark — Mg + 2H+ → Mg2+ + H2 is equally accepted, state symbols are ignored, and multiples are allowed.
Standard electrode potentials are measured by comparison with the standard hydrogen electrode. State the substances and conditions needed in a standard hydrogen electrode.
Show answer
H2(g) and 100 kPa. 1 mark
1 mol dm−3 and HCl / HNO3 / H+ — or 0.5 mol dm−3 H2SO4. 1 mark
Pt electrode and a temperature of 298 K. 1 mark
1 bar is allowed for 100 kPa; 1 atm and 101 kPa are not. Each mark needs both halves of the pair.
Describe an experiment a student does to show that the standard electrode potential for the TiO2+(aq) / Ti(s) electrode is −0.88 V. The student is provided with the Cu2+(aq) / Cu(s) electrode set up ready to use, solid titanium(IV) oxysulfate (Mr = 159.9), 0.50 mol dm−3 sulfuric acid, a strip of titanium, and laboratory apparatus and chemicals. Your answer should include how to prepare the solution, how to connect the electrodes, the measurements taken, and how they are used to calculate the electrode potential.
Show answer
Marked by levels of response across three stages. The indicative chemistry is:
Stage 1 — preparing the solution. Weigh 7.995 / 8.00 g of TiOSO4; dissolve it in the 0.50 mol dm−3 sulfuric acid; transfer to a volumetric flask and make up to the mark.
Stage 2 — setting up the cell. A piece of titanium immersed in the acidified TiO2+(aq); the two solutions connected by a salt bridge; the two metals connected through a high-resistance voltmeter. A labelled diagram earns these marks.
Stage 3 — measurement and calculation. Record the EMF of the cell; use Ecell = ERHS − ELHS, i.e. Ecell = Ecopper − Etitanium; the reading should be +1.22 V with copper on the right (or −1.22 V with copper on the left).
Level 3 — 5–6 marks all three stages covered, correct and virtually complete, with clear practical detail (weighing into a beaker or by difference plus washings, not straight into the volumetric flask; a suitable, saturated salt-bridge solution).
Level 2 — 3–4 marks all three stages covered but incomplete or containing inaccuracies, or two stages covered well.
Level 1 — 1–2 marks two stages covered but incomplete, or one stage covered well.
Where 7.995 g comes from: 50 cm3 of a 1.00 mol dm−3 solution needs 0.0500 mol, and 0.0500 × 159.9 = 7.995 g.
Give the half-equation for the electrode reaction in the TiO2+(aq) / Ti(s) electrode in acidic conditions.
Show answer
TiO2+ + 2H+ + 4e− → Ti + H2O
1 mark — the reverse reaction is allowed, state symbols are ignored, and multiples, fractions or an equilibrium arrow are all accepted.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
How pH changes in an acid–base titration
Full theory, Ka, buffers & indicators: Acids and bases — pH curves
Aim: follow the pH with a meter as alkali is run into acid, and plot the titration curve. AQA name two cases: a weak acid with a strong base, and a strong acid with a weak base. The curve you get tells you where the equivalence point is, which indicator would work, and — for a weak acid — its Ka.
- Rinse the probe thoroughly with deionised water and shake off the excess.
- Stand it in the pH 7.00 buffer with the bulb fully immersed, and record the reading.
- Rinse again, then repeat with the pH 4.00 and pH 9.20 buffers.
- Plot your reading (x) against the true pH of the buffer (y). The line may be straight or slightly curved — it is what you use to correct every reading in Part 2.
- Rinse a burette with 0.100 mol dm−3 ethanoic acid, fill it, and run exactly 20.0 cm3 into a clean 100 cm3 beaker. Label the burette so you do not mix the two up.
- Rinse and fill a second burette with 0.100 mol dm−3 sodium hydroxide solution.
- Rinse the probe, clamp it with the bulb fully immersed in the acid, stir gently and record the starting pH.
- Add the alkali in 2.0 cm3 portions up to 18 cm3, stirring and recording the pH after each addition.
- Switch to 0.20 cm3 portions from 18 cm3 to 22.0 cm3 — that is where the pH is changing fastest.
- Go back to 2.0 cm3 portions until 40 cm3 have been added, then rinse the probe.
Apparatus & techniques
- using a burette to a stated precision
- calibrating & using a pH meter
- sampling more often where the change is fastest
- plotting and interpreting a titration curve
- safe handling of acids & alkalis
Safety. Wear eye protection. Sodium hydroxide solution is corrosive — it attacks skin and, in particular, eyes; wash any splash off immediately with plenty of water. Ethanoic acid is an irritant with a strong smell, so keep the room ventilated. Fill burettes below eye level using a funnel, and clamp the pH probe rather than holding it.
What you should see. The pH climbs slowly at first, then barely moves at all through the middle of the titration — that flat stretch is a buffer. Around 20 cm3 it leaps by several pH units for a fraction of a cubic centimetre, which is why you switch to 0.20 cm3 portions there. After that it flattens out again as the excess alkali takes over.
Data analysis — reading the curve
Correct every reading with your calibration line first, then plot corrected pH (y) against volume of alkali (x) and join the points with a smooth curve. Four things come straight off it:
| Feature | Where to look | What it gives you |
|---|---|---|
| Starting pH | the y-intercept | High for a weak acid (about 2.8 here) — a strong acid of the same concentration would start near pH 1 |
| Buffer region | the flat stretch before the rise | A mixture of the weak acid and its salt resists pH change |
| Half-neutralisation | half the equivalence volume | pH = pKa, so Ka = 10−pH |
| Equivalence point | the middle of the vertical section | The volume needed, and a pH above 7 for weak acid + strong base |
| Vertical section | its top and bottom pH | An indicator only works if its range lies inside it |
- Why calibrate at all? A meter drifts, so its raw reading is not the true pH. Three buffers spanning the range you will measure let you correct every reading — and using three rather than one checks the response is linear.
- Rinse the probe with deionised water between every solution, and never wipe the bulb dry — carry-over from the last solution is a systematic error, and the glass bulb is fragile.
- Add dropwise near the equivalence point. The reason to give is that the pH changes very rapidly there for a very small addition of alkali — 2.0 cm3 portions would jump straight over the vertical section.
- Stir after every addition before reading, or the probe measures a local pocket rather than the bulk solution.
- Join the points with a smooth curve, not straight lines, and never force it through every point.
- The equivalence point of a weak acid with a strong base is above pH 7, because the salt formed (the conjugate base) is itself slightly basic. For a strong acid with a weak base it is below 7.
Exam questions — RP9
Propanoic acid is a weak acid. State the meaning of weak in this context.
Show answer
Only a small fraction of the molecules dissociate / ionise when added to water — it partially dissociates. 1 mark
“The reaction is reversible” scores nothing, and neither does “hydrogen ions do not fully dissociate”. Naming the wrong ions loses the mark.
Suggest why a student doing this experiment would add the sodium hydroxide solution dropwise around the equivalence point.
Show answer
Because there is a large pH change for a small addition of alkali there. 1 mark
The mark needs the idea of a rapid or large change; “the gradient is steep” is allowed. “So they do not miss the equivalence point” on its own is ignored.
Give an expression for Ka for propanoic acid (CH3CH2COOH). Use this expression to show that pH = pKa when half of the propanoic acid has reacted with sodium hydroxide.
Show answer
Ka = [H+][CH3CH2COO−] ÷ [CH3CH2COOH]
1 mark — [H3O+] is allowed for [H+]; round brackets for concentration are not.
At half-neutralisation, half the acid has been converted to its salt, so [CH3CH2COOH] = [CH3CH2COO−]. 1 mark
Those two terms cancel, leaving Ka = [H+], and taking −log of both sides gives pKa = pH. 1 mark
[HA] = [A−], or the ratio [HA]/[A−] = 1, earns the second mark just as well; log 1 = 0 earns the third.
Use the pH from the curve above, when half of the propanoic acid has reacted, to calculate Ka at 298 K.
Show answer
The equivalence volume is 20.0 cm3, so read the pH at 10.0 cm3: pH ≈ 4.55. 1 mark — anything from 4.4 to 4.7 is accepted.
Ka = 10−4.55 = 2.8 × 10−5 mol dm−3
1 mark — to at least 2 s.f. A pH in the accepted band gives Ka between 2.0 × 10−5 and 4.0 × 10−5.
Using the pH at the start (2.8) scores zero — that is the pH of the pure acid, not the half-neutralised mixture.
When sodium hydroxide solution is added to aqueous propanoic acid, the solution formed acts as a buffer when between 5 cm3 and 15 cm3 have been added. Explain why the pH stays approximately constant during this part of the experiment.
Show answer
The OH− added reacts with the propanoic acid (or with the H+ in solution). 1 mark
The ratio of [CH3CH2COOH] to [CH3CH2COO−] stays almost constant — equivalently, the dissociation equilibrium shifts right to replace the H+ removed. 1 mark
Methyl orange and universal indicator are not suitable indicators for the titration of propanoic acid with sodium hydroxide. State the reason why each is unsuitable.
Show answer
Methyl orange: it would not change colour at the equivalence point — its pH range does not lie inside the rapid pH change (it changes below the equivalence point, outside pH 6–11). 1 mark
Universal indicator: it gives a range of colours through the titration rather than one distinct colour change at the end point. 1 mark
Saying only that this is a weak acid–strong base titration is ignored — the mark is for what the indicator does.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Preparing a pure organic solid and a pure organic liquid
Full theory: Carboxylic acids & derivatives — esters and anhydrides · Alcohols — distillation · Organic synthesis
Aim: make an organic product, purify it, and show that it is pure. AQA name two preparations — a solid (aspirin, purified by recrystallisation and checked by melting point) and a liquid (an ester, purified by washing, drying and redistillation). The chemistry is only half the marks; the other half is the practical technique and the yield calculation.
Part A — a pure organic solid: aspirin
Aspirin is made by acylating salicylic acid (2-hydroxybenzenecarboxylic acid) with ethanoic anhydride. The –OH group on the ring becomes an ester:
HOOCC6H4OH + (CH3CO)2O → HOOCC6H4OCOCH3 + CH3COOH
Ethanoic anhydride is used rather than ethanoyl chloride because it is cheaper, less corrosive, less vigorously hydrolysed by water, and produces no HCl — a comparison AQA ask for directly.
- Weigh about 6.00 g of salicylic acid straight into a 100 cm3 conical flask, and record the exact mass.
- Add 10 cm3 of ethanoic anhydride from a measuring cylinder and swirl.
- Add 5 drops of concentrated sulfuric acid as the catalyst and swirl for a few minutes.
- Warm for about 20 minutes in a water bath at roughly 60 °C. Do not let the flask go above 65 °C.
- Cool the flask, then pour the contents into 75 cm3 of water, stirring well to precipitate the solid.
- Filter off the crude aspirin under reduced pressure on a double thickness of filter paper and leave it to dry.
- Recrystallise: dissolve the crude solid in the minimum volume of hot ethanol in a boiling tube standing in water at about 75 °C — never above ethanol’s boiling point of 78 °C.
- Pour the hot solution into about 40 cm3 of water and let it cool slowly; white needles separate. Scratching the flask with a glass rod, or cooling in ice, starts crystallisation if nothing appears.
- Filter under reduced pressure again, wash with a little cold ethanol, dry, and record the mass.
- Pack three melting-point tubes to about 0.5 cm and measure the melting point, heating slowly.
Crude aspirin carries no insoluble impurity, so this method goes straight from the hot solution to cooling. In the general recrystallisation technique — the one AQA can ask you to describe for any solid — you filter the hot solution through a warmed funnel first, which is what removes insoluble impurities; the soluble ones stay behind in the filtrate when the crystals form. The step-by-step version is on the carboxylic acids page.
Part B — a pure organic liquid: ethyl ethanoate
An alcohol and a carboxylic acid form an ester in an equilibrium catalysed by concentrated sulfuric acid:
CH3CH2OH + CH3COOH ⇌ CH3COOC2H5 + H2O
- Put a few anti-bumping granules in a pear-shaped flask, then add 10 cm3 of ethanol, 12 cm3 of glacial ethanoic acid and 15 drops of concentrated sulfuric acid in a fume cupboard.
- Clamp the flask in a beaker of water over a Bunsen, so the mixture is below the water line, and set up a condenser for heating under reflux. Do not put a stopper in the top.
- Boil gently for about 15 minutes, then turn off the heat and remove the water bath.
- Rearrange for distillation and collect the liquid that distils over.
- Shake the distillate with saturated sodium carbonate solution in a separating funnel, venting the carbon dioxide through the tap after each inversion.
- Let the layers settle. Ethyl ethanoate is less dense than water, so it is the upper layer. Run off and discard the lower aqueous layer.
- Add the ester to about 1 g of anhydrous sodium sulfate (or calcium chloride) in a dry boiling tube, and shake to take out the water.
- Decant into a dry flask, add fresh anti-bumping granules and redistil, collecting the fraction that comes over at 74–79 °C. Never distil to dryness.
- Weigh the receiver before and after to find the mass collected.
Apparatus & techniques
- heating under reflux & distillation
- filtration under reduced pressure
- recrystallisation
- using a separating funnel & a drying agent
- measuring a melting point and a boiling range
- percentage yield & atom economy
Safety. Wear eye protection throughout. Concentrated sulfuric acid is corrosive and so is ethanoic anhydride, whose vapour is a severe eye and respiratory irritant — dispense it in a fume cupboard. Ethanol and ethyl ethanoate are highly flammable, so every heating step uses a water bath or electric heater, never a naked flame near the solvent. Aspirin is harmful if swallowed; do not taste anything you make.
What you should see. The aspirin appears the moment the warm mixture hits cold water, as a fine white solid; after recrystallisation the same material comes back as larger, whiter needle-shaped crystals. In the ester preparation, the sweet, fruity smell is obvious as soon as the reflux starts, and the separating funnel gives two clear layers with a sharp boundary.
Data analysis — limiting reagent, yield and purity
Both halves finish with the same calculation: work out which reagent is limiting, use it to get the theoretical yield, and compare that with what you actually collected.
| Observation | What it means |
|---|---|
| Melts sharply, at the data-book value | Pure — the sample collapses to a liquid at essentially one temperature |
| Melts lower than the book value | Impure — impurities depress the melting point |
| Melts over a range of several degrees | Impure — and the wider the range, the more impurity |
| Does not melt, or chars | The compound decomposes on heating, so this method will not work for it |
For a liquid, the equivalent test is the boiling range: a pure liquid distils over a narrow range at its literature boiling point. Ethyl ethanoate boils at 77 °C, which is why the 74–79 °C fraction is the one to keep.
Which reagent runs out first? Pick a preparation and change what you start with. The limiting reagent is recalculated every time — “in excess” is something you work out from the amounts in moles, never something you can read off the volumes.
- Reweigh the empty container. Weighing the solid into a boat and then reweighing the empty boat is the only way to know the exact mass transferred — a favourite “what step is missing?” mark.
- Heat flammable solvents on a water bath, never with a Bunsen. Say the practical step, not just “ethanol is flammable” — the mark is for the precaution.
- Recrystallise from the minimum volume of hot solvent and keep the temperature below its boiling point, so the solvent does not boil away and the product does not stay in solution when it cools.
- Wash the crystals with a little cold solvent on the Büchner funnel. Cold, so the product does not redissolve; a little, so it only rinses off the soluble impurities left in the film of solution.
- Never seal a refluxing apparatus. A bung in the top of the condenser lets pressure build until it is forced out. The top is open by design.
- Condenser water in at the bottom, out at the top. The wrong way round and the jacket never fills, so the vapour is not condensed and product escapes.
- Vent the separating funnel. Sodium carbonate neutralises the acid and produces carbon dioxide, so pressure builds up — invert, then open the tap, every time.
- Say which layer and why. Two layers form because the ester is immiscible with water; the ester is on top because it is less dense. “Lighter” does not earn the mark.
Exam questions — RP10
- Add about 6 g of salicylic acid to a weighing boat.
- Place the weighing boat on a 2 decimal place balance and record the mass.
- Tip the salicylic acid into a 100 cm3 conical flask.
- __________________
- Add 10 cm3 of ethanoic anhydride to the conical flask and swirl.
- Add 5 drops of concentrated phosphoric acid.
- Warm the flask for 20 minutes.
- Add ice-cold water to the reaction mixture and place the flask in an ice bath.
- Filter off the crude aspirin from the mixture and leave it to dry.
- Weigh the crude aspirin and calculate the yield.
Describe the instruction that is missing from step 4 of the method. Justify why this step is necessary.
Show answer
Instruction: reweigh the empty weighing boat. 1 mark
Justification: so you can calculate the exact mass of salicylic acid actually added to the reaction mixture. 1 mark
Suggest a suitable piece of apparatus to measure out the ethanoic anhydride in step 5. Identify a hazard of using concentrated phosphoric acid in step 6.
Show answer
Apparatus: a 10 cm3 measuring cylinder — a pipette, burette, graduated pipette or 10 cm3 syringe are all accepted. 1 mark
Hazard: corrosive — skin burns or permanent eye damage. 1 mark
If you name a volume it must be between 10 and 50 cm3. “Irritant” and “toxic” are both ignored for the hazard.
A 6.01 g sample of salicylic acid (Mr = 138.0) is reacted with 10.5 cm3 of ethanoic anhydride (Mr = 102.0). In the reaction the yield of aspirin is 84.1%. The density of ethanoic anhydride is 1.08 g cm−3. Show by calculation which reagent is in excess, and calculate the mass, in g, of aspirin (Mr = 180.0) produced.
Show answer
n(salicylic acid) = 6.01 ÷ 138.0 = 4.36 × 10−2 mol
1 mark
mass of (CH3CO)2O = 10.5 × 1.08 = 11.34 g
1 mark
n((CH3CO)2O) = 11.34 ÷ 102.0 = 1.11 × 10−1 mol
1 mark
The reaction is 1 : 1, and there is far more anhydride, so ethanoic anhydride is in excess. 1 mark
mass of aspirin = 4.36 × 10−2 × 0.841 × 180.0 = 6.59 g
1 mark — 2 s.f. or more.
You must both show the two amounts and state that the anhydride is in excess. Later marks carry forward from an earlier slip.
Suggest two ways in which the melting point of the crude aspirin collected in step 9 would differ from the melting point of pure aspirin.
Show answer
The value would be lower. 1 mark
It would melt over a range of temperatures rather than sharply. 1 mark
The crude aspirin can be purified by recrystallisation using hot ethanol (boiling point 78 °C) as the solvent. Describe two important precautions when heating the mixture of ethanol and crude aspirin.
Show answer
Use a water bath rather than a Bunsen burner, because ethanol is flammable. 1 mark
Heat to a temperature below its boiling point, so the ethanol does not boil away — and use the minimum volume of solvent. 1 mark
The first mark needs a practical step. Simply writing “ethanol is flammable” states the hazard without giving the precaution, and scores nothing.
The pure aspirin is filtered under reduced pressure. A small amount of cold ethanol is then poured through the Büchner funnel. Explain the purpose of adding a small amount of cold ethanol.
Show answer
To remove soluble impurities — washing away the film of ethanolic solution left on the product, while the solvent being cold and in small amount stops the aspirin itself dissolving. 1 mark
A sample of the crude aspirin is kept to compare with the purified aspirin. Describe one difference in appearance you would expect to see between these two solid samples.
Show answer
The pure product has larger, needle-like crystals, or is lighter in colour. 1 mark
Whiter, less grey, more crystalline, less powdery, shinier or a single colour are all accepted — but the answer must be tied to the pure product (or the opposite point tied to the crude one).
- A mixture of ethanol, ethanoic acid, and concentrated sulfuric acid, with anti-bumping granules, is heated under reflux for 10 minutes.
- The apparatus is rearranged for distillation.
- The mixture is heated to collect the liquid that distils between 70 and 85 °C.
- The distillate is placed in a separating funnel. Aqueous sodium carbonate is added, and a stopper is placed in the funnel. The mixture is shaken, releasing pressure as necessary.
- The lower aqueous layer is removed and the upper organic layer is placed in a small conical flask.
- Anhydrous calcium chloride is added to the sample in the conical flask. The flask is shaken well and left for a few minutes.
- The liquid from the flask is redistilled and the distillate is collected between 74 and 79 °C.
State the role of the concentrated sulfuric acid in this reaction. The reaction mixture is flammable — suggest how it should be heated in step 1.
Show answer
Role: catalyst. 1 mark — “reduces the activation energy” is allowed; “speeds up the reaction”, “provides an alternative path”, “proton donor” and “dehydrating agent” are all ignored.
Heating: an electric heater, heat mantle or hot water bath. 1 mark — a heating plate, sand bath or oil bath are fine too. Any suggestion of direct heat from a Bunsen scores nothing.
The figure below shows how a student set up the apparatus for reflux in step 1. Assume the apparatus is clamped correctly. Identify two mistakes the student made, and state the problem caused by each.
Show answer
Mistake 1: there is a bung / stopper in the end of the condenser. 1 mark
Problem: pressure builds up — the stopper could be forced out. 1 mark
Mistake 2: the water goes the wrong way through the condenser (in at the top, out at the bottom). 1 mark
Problem: the condenser does not fill with water / is not cool enough, so the vapour is not condensed effectively and reactants or products escape. 1 mark
“The glass shatters” and “it explodes” are ignored, as is “the condenser is the wrong way round” without saying which way the water flows. An unsealed neck of the flask is an equally valid pair of marks.
State why sodium carbonate is added to the distillate in step 4. Explain why there is a build-up of pressure in the separating funnel.
Show answer
To neutralise / remove the acid — the ethanoic acid and / or the sulfuric acid. 1 mark
Carbon dioxide gas is produced — you see effervescence. 1 mark
Naming the wrong acid or the wrong gas loses the mark. “Neutralise the distillate” is ignored — say the acid.
Give a reason why two layers form in the separating funnel. Suggest why ethyl ethanoate forms the upper layer.
Show answer
Ethyl ethanoate is immiscible with / insoluble in water, so the aqueous and organic layers do not mix. 1 mark
It is less dense than water. 1 mark
“Lighter” is not accepted, and references to polarity or intermolecular forces are ignored.
State why anhydrous calcium chloride is added in step 6.
Show answer
To remove / absorb water — it is a drying agent. 1 mark
“Dehydrates” is ignored, and “to remove soluble impurities” or “to dry the reactants” score nothing.
A student adds 10.0 cm3 of ethanol (Mr = 46.0) to 5.25 g of ethanoic acid (Mr = 60.0) and obtains 5.47 g of ethyl ethanoate (Mr = 88.0). For ethanol, density = 0.790 g cm−3. Determine the limiting reagent and calculate the percentage yield. Then suggest a reason why the percentage yield is not 100%.
Show answer
mass of ethanol = 10.0 × 0.790 = 7.90 g
1 mark
n(ethanol) = 7.90 ÷ 46.0 = 0.172 mol n(ethanoic acid) = 5.25 ÷ 60.0 = 0.0875 mol
1 mark — both amounts must be shown.
The limiting reagent is ethanoic acid. 1 mark
maximum mass = 0.0875 × 88.0 = 7.70 g
1 mark
% yield = (5.47 ÷ 7.70) × 100 = 71.0%
1 mark — 70.6 to 71.1, to at least 2 s.f.
Why not 100%: the esterification is reversible and reaches equilibrium, so it never goes to completion — and product is lost at every transfer, in the washing and on redistillation. 1 mark
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Test-tube reactions to identify transition-metal ions
Full theory, every observation and equation: Reactions of aqueous ions · Transition metals
Aim: carry out simple test-tube reactions on unknown aqueous solutions of transition-metal (and other) ions, and record the observations. AQA are explicit about this one: in the task itself you are not required to identify any of the solutions or products — the practical is about seeing and describing accurately. The identifying comes in the exam.
You are given three solutions labelled Q, R and S, plus sodium hydroxide, sodium carbonate and silver nitrate solutions.
- Test 1(a). Put about 10 drops of solution Q in a test tube. Add sodium hydroxide solution dropwise, with gentle shaking, until in excess. Do not discard the mixture. Repeat with R and then S.
- Test 1(b). Half-fill a 250 cm3 beaker with freshly boiled water and stand the tubes from Test 1(a) in it for about 10 minutes. Start Test 2 while you wait.
- Test 2. Put about 10 drops of sodium carbonate solution in a fresh tube, add about 10 drops of solution Q and shake gently. Repeat with R and S.
- Test 3. Put about 10 drops of solution Q in a fresh tube, add about 10 drops of silver nitrate solution, shake gently, and let the tubes stand for about 10 minutes. Repeat with R and S.
Record what you see on dropwise addition, on addition to excess and on standing. Where nothing happens, write “no visible change” — a blank cell earns nothing.
Apparatus & techniques
- qualitative test-tube reactions
- dropwise addition, and addition to excess
- using a hot-water bath
- recording observations precisely
- safe handling of alkalis & silver nitrate
Safety. Wear eye protection. Sodium hydroxide solution is corrosive — wash any splash off immediately. Silver nitrate stains skin and clothing black and is harmful to the environment, so keep it off your hands and do not pour it down the sink. Transition-metal salts are harmful if swallowed. Use the hot water from a kettle or a beaker of previously boiled water — never boil solutions in a test tube over a flame.
What you should see. Every one of these ions gives a precipitate with a little alkali, because the hydroxide is insoluble. What separates them is what happens next: whether the precipitate redissolves in excess, whether it changes colour on standing, and whether the carbonate gives a gas.
Data analysis — the observations that separate the ions
| Ion in solution | Colour | Precipitate with a little OH− | Excess NaOH? | With Na2CO3 |
|---|---|---|---|---|
| [Fe(H2O)6]2+ | pale green | green, Fe(H2O)4(OH)2 — darkens to brown in air | Insoluble | Green FeCO3, no gas |
| [Fe(H2O)6]3+ | violet (looks yellow-brown) | red-brown, Fe(H2O)3(OH)3 | Insoluble | Red-brown precipitate and effervescence |
| [Cu(H2O)6]2+ | blue | blue, Cu(H2O)4(OH)2 | Insoluble | Blue-green CuCO3, no gas |
| [Al(H2O)6]3+ | colourless | white, Al(H2O)3(OH)3 | Dissolves → colourless [Al(OH)4]− | White precipitate and effervescence |
| [Mg(H2O)6]2+ | colourless | white, Mg(OH)2 | Insoluble | White MgCO3, no gas |
A 3+ ion is smaller and more highly charged than a 2+ ion, so it has a much higher charge density and polarises the O–H bonds in its water ligands far more strongly. That makes the 3+ aqua ion a much better proton donor — acidic enough to react with carbonate and release CO2:
2[Fe(H2O)6]3+ + 3CO32− → 2Fe(H2O)3(OH)3 + 3CO2 + 3H2O
The 2+ ions are not acidic enough. They simply swap partners and precipitate the carbonate itself, with no gas:
[Fe(H2O)6]2+ + CO32− → FeCO3 + 6H2O
So effervescence with carbonate is a 3+ test, and it is the fastest way to sort a 3+ ion from a 2+ one in an unknown.
- Every colour observation must say precipitate or solution. “Green” is not an observation; “a green precipitate” is. This is the single commonest lost mark in the whole practical.
- Add dropwise, then to excess — and record both. The aluminium precipitate only gives itself away when the excess redissolves it.
- Say “no visible change” rather than leaving a cell blank. It is a real result and it carries marks.
- Watch the iron(II) precipitate on standing. It darkens from green towards brown as it is oxidised by air — which is why AQA build a standing step into the method.
- Ammonia and sodium hydroxide give the same precipitates at first, because both provide OH−. They differ only in excess, where excess ammonia dissolves the copper precipitate to a deep blue solution but excess sodium hydroxide does not.
- Include the water ligands in A-level formulae. Fe(H2O)3(OH)3 is the expected answer, not Fe(OH)3 — though the simpler form is usually allowed.
Exam questions — RP11
Aqueous aluminium sulfate is added to aqueous sodium carbonate. What are the formulas of the precipitate and the gas formed?
Cobalt(II) chloride solution changes colour when an excess of concentrated hydrochloric acid is added. What type of reaction takes place?
When anhydrous aluminium chloride reacts with water, solution Y is formed that contains a complex aluminium ion, Z, and chloride ions. Give an equation for this reaction, and give an equation to show how the complex ion Z can act as a Brønsted–Lowry acid with water.
Show answer
AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl−
1 mark
[Al(H2O)6]3+ + H2O → [Al(H2O)5(OH)]2+ + H3O+
1 mark — an equation going on to [Al(H2O)4(OH)2]+ is equally acceptable.
Describe two observations you would make when an excess of sodium carbonate solution is added to solution Y. Give an equation for the reaction, including the formula of each complex aluminium species.
Show answer
Observation 1: a white precipitate forms. 1 mark
Observation 2: effervescence / bubbles / fizzing. 1 mark
2[Al(H2O)6]3+ + 3CO32− → 2[Al(H2O)3(OH)3] + 3CO2 + 3H2O
1 mark — multiples are accepted; spectator ions only if the equation still balances.
The two observations can be given in either order, but both are needed. “White” on its own does not score — it has to be a white precipitate.
Aqueous potassium hydroxide is added, until in excess, to solution Y. Describe two observations you would make, and for each give an equation for the reaction that occurs, including the formula of each complex aluminium species.
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Observation 1: a white precipitate forms. 1 mark
[Al(H2O)6]3+ + 3OH− → [Al(H2O)3(OH)3] + 3H2O
1 mark
Observation 2: the precipitate dissolves — a colourless solution forms. 1 mark
[Al(H2O)3(OH)3] + OH− → [Al(H2O)2(OH)4]− + H2O
1 mark — the 4-coordinate form [Al(OH)4]− + 3H2O is equally accepted; only 6- or 4-coordination is allowed.
| Test | Observations with L | Observations with M |
|---|---|---|
| Add ammonia solution slowly until in excess. | A red-brown precipitate forms that is insoluble in excess. | A green precipitate forms that is insoluble in excess. |
| Add sodium carbonate solution. | A red-brown precipitate forms. Effervescence is seen. | A green precipitate forms. |
| Add dilute nitric acid, then divide into two portions. | No change is seen. | No change is seen. |
| Add barium chloride solution to the first portion. | No change is seen. | A white precipitate forms. |
| Add silver nitrate solution to the second portion. | A white precipitate forms. | No change is seen. |
Identify L and M using the results in the table. In your answer: identify all precipitates; explain why effervescence is seen in the reaction of sodium carbonate with L but not with M; and give ionic equations for all reactions.
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Marked by levels of response across three stages.
Stage 1 — the identifications. Both red-brown precipitates from L are Fe(H2O)3(OH)3. The green precipitate from M is Fe(H2O)4(OH)2 with ammonia and FeCO3 with carbonate. The white precipitate with L is AgCl and with M is BaSO4. So L is FeCl3 and M is FeSO4.
Stage 2 — the effervescence. The gas from L is carbon dioxide. [Fe(H2O)6]3+ is more acidic than [Fe(H2O)6]2+, because Fe3+ is smaller and more highly charged — a greater charge density, so it polarises the O–H bonds more and is a better proton donor.
Stage 3 — the equations.
[Fe(H2O)6]3+ + 3NH3 → Fe(H2O)3(OH)3 + 3NH4+
2[Fe(H2O)6]3+ + 3CO32− → 2Fe(H2O)3(OH)3 + 3CO2 + 3H2O
[Fe(H2O)6]2+ + 2NH3 → Fe(H2O)4(OH)2 + 2NH4+
[Fe(H2O)6]2+ + CO32− → FeCO3 + 6H2O
Ag+ + Cl− → AgCl Ba2+ + SO42− → BaSO4
Level 3 — 5–6 marks all three stages covered, correct and virtually complete, communicated coherently with the equations.
Level 2 — 3–4 marks all three stages covered but incomplete or containing inaccuracies, or two stages covered well.
Level 1 — 1–2 marks two stages covered but incomplete, or one stage covered well.
Precipitates count as identified if they appear as products in your equations or in a table. Names are allowed provided the oxidation state of the iron is given.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Separation of species by thin-layer chromatography
Full theory, Rf and the interactive plate: Chromatography · Amino acids — two-way TLC
Aim: separate a mixture by thin-layer chromatography and identify its components from their Rf values. AQA’s exemplar analyses the active ingredients in common painkillers — aspirin, paracetamol, ibuprofen and caffeine — and asks you to work out what a combination tablet contains.
- Crush an aspirin tablet with a pestle and mortar and transfer it to a weighing boat.
- Dissolve about 0.1 g of the powder in 0.5 cm3 of ethanol.
- Repeat for the ibuprofen and paracetamol tablets.
- Crush the caffeine tablet and dissolve about 0.1 g in 7.0 cm3 of ethanol — a much more dilute solution, because a caffeine tablet is nearly all caffeine.
- Repeat for the combination tablet (one containing aspirin, paracetamol and caffeine together).
- Draw a faint pencil line 1 cm above the bottom of a TLC plate and mark five equally spaced spots along it.
- Use a capillary tube to put a tiny drop of each solution on a different spot, and let the plate air dry.
- Put about 10 cm3 of ethyl acetate in the developing chamber.
- Stand the plate in the chamber with the solvent level below the spotting line, and replace the lid so it seals.
- When the solvent reaches about 1 cm from the top, take the plate out and mark the solvent front in pencil straight away. Dry it in the fume cupboard.
- Visualise the spots under a UV lamp and draw round them lightly in pencil.
- Calculate the Rf value of every spot.
Apparatus & techniques
- thin-layer chromatography
- spotting with a capillary tube
- using a sealed developing chamber
- visualising with UV or a locating agent
- measuring and comparing Rf values
Safety. Wear eye protection. Ethanol and ethyl acetate are highly flammable and their vapours are irritating — keep them away from flames and work in a fume cupboard or a well-ventilated room. Never look directly at the UV lamp, and keep the plate flat under it rather than holding it. Handle plates by the edges only: fingerprints leave their own spots and grease ruins the run.
What you should see. Nothing at all, until the plate goes under the lamp — all five compounds are colourless. Under UV the silica fluoresces and the spots show up as dark patches. The combination tablet gives several spots, one lining up with each pure substance it contains, which is exactly how you read it.
Data analysis — Rf and what it can and cannot tell you
Rf = distance moved by the spot ÷ distance moved by the solvent front
Both distances are measured from the pencil origin line, and the spot is measured to its centre. Rf is a ratio, so it has no units and is always between 0 and 1.
| What you see | What it tells you |
|---|---|
| A spot in the mixture lane at the same height as a pure reference | That substance is probably present — same Rf, same conditions |
| A high Rf | More soluble in the moving phase, and less strongly adsorbed by the polar silica |
| A low Rf | Held tightly by the stationary phase — a more polar substance on silica |
| One spot that will not separate | Two substances may have the same Rf in this solvent — run it again in a different one |
| A streak rather than a spot | The spot was too concentrated, or the solvent was above the pencil line and washed it off |
Two different compounds can share an Rf value, so a match is evidence, not proof. It only means anything at all when the reference and the unknown are run on the same plate, in the same solvent, at the same temperature — which is why the pure references go on the plate beside the mixture rather than being looked up in a book. To be certain you either run a second solvent (two-way TLC) or take the separated substance away for NMR or mass spectrometry.
- Pencil, not pen. Ink is a mixture and would run up the plate with everything else.
- The solvent must start below the pencil line. If it covers the spots they dissolve straight into the solvent in the tank and never travel up the plate.
- Lid on the chamber. It keeps the atmosphere saturated with solvent vapour, so the solvent does not evaporate off the plate as it rises and the front stays level.
- Mark the solvent front the moment the plate comes out. Once the solvent evaporates you cannot tell where it got to, and every Rf is lost.
- Tiny, dry spots. A big wet spot streaks and the components overlap; let each application dry before adding the next.
- Measure to the centre of the spot, from the pencil origin line — not from the bottom edge of the plate.
- Colourless compounds need visualising — a UV lamp for anything that fluoresces or quenches, ninhydrin for amino acids, or iodine vapour.
Exam questions — RP12
Suggest a suitable reagent for the hydrolysis of a protein.
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Concentrated hydrochloric acid. 1 mark
Any concentration of 5 mol dm−3 or higher is allowed, and concentrated sulfuric acid or a concentrated strong alkali are both accepted.
Suggest how the positions of the amino acids on the TLC plate were located.
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Using ninhydrin or ultraviolet light. 1 mark
Iodine vapour is also allowed.
Deduce the minimum number of amino acids present in the original mixture.
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Seven. 1 mark
“Minimum” matters: two amino acids could still be sitting under one another in a single spot even after two solvents, so seven spots means at least seven amino acids.
Suggest why it was necessary to use two different solvents.
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Some of the amino acids did not separate in the first solvent — they had the same Rf value, or the same affinity for that solvent. 1 mark
Simply saying “amino acids have different Rf values in different solvents” is not accepted — the mark is for explaining that the first solvent left some of them unseparated.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
- RP1 · volumetric solution & titration — make a standard solution (weigh by difference, wash into a volumetric flask, make up to the mark), then titrate to concordant titres (within 0.10 cm3); the titre uncertainty is 2 × 0.05 = ± 0.10 cm3.
- RP2 · enthalpy change — q = mcΔT in an insulated cup or under a copper can; correct ΔT by extrapolating the cooling line back to mixing; the value comes out less exothermic than the data book (heat loss, incomplete combustion).
- RP3 · rate and temperature — thiosulfate “disappearing cross” clock; rate ∝ 1/time; plot 1/time against the mean temperature; SO2 is toxic, so ventilate.
- RP4 · ion tests — Group 2 by the hydroxide/sulfate solubility trends; NH4+ (NaOH + warm → damp red litmus blue); CO32− (limewater milky); SO42− (acidified BaCl2); halides (acidified AgNO3, then ammonia). Run carbonate → sulfate → halide, acidified with nitric acid.
- RP5 · distillation — distil a product out as it forms; thermometer bulb level with the side-arm, condenser water in at the bottom, anti-bumping granules, never fully sealed; then work out the % yield.
- RP6 · functional-group tests — bromine water (alkene: orange solution → colourless solution), acidified dichromate (alcohol: orange solution → green solution), Tollens’/Fehling’s (aldehyde: silver mirror / brick-red precipitate), sodium carbonate (carboxylic acid: fizz).
- RP7 · measuring rate — two methods. Initial rate: an iodine clock, where the total volume is kept constant so only one concentration changes and 1/t is the rate. Continuous monitoring: collect the gas or follow the mass loss on a balance, then take the gradient of a tangent at t = 0. A straight line through the origin on a rate–concentration graph means first order.
- RP8 · EMF of a cell — clean and degrease the electrodes, join the half-cells with a salt bridge that reacts with neither, and read a high-resistance voltmeter so almost no current flows. E⦵cell = more positive − more negative; electrode size makes no difference.
- RP9 · pH in a titration — calibrate the meter on three buffers (4.00, 7.00, 9.20), then add alkali in 2.0 cm3 portions but 0.20 cm3 near the equivalence point, where the pH moves fastest. Half-neutralisation gives pH = pKa; a weak acid with a strong base equalises above pH 7.
- RP10 · organic preparation — a solid (aspirin: warm at 60 °C, crash out in cold water, filter under reduced pressure, recrystallise, check the melting point) and a liquid (an ester: reflux, wash with Na2CO3 venting the CO2, dry with an anhydrous salt, redistil over a narrow boiling range). Both finish with limiting reagent → theoretical yield → % yield.
- RP11 · transition-metal ions — every colour observation must name a precipitate or a solution. Add dropwise and to excess: only Al3+ redissolves in excess NaOH, only Cu2+ gives the deep blue solution with excess NH3, and only the 3+ ions fizz with carbonate.
- RP12 · thin-layer chromatography — pencil line, solvent below it, lid on the chamber, and mark the solvent front the instant the plate comes out. Rf = spot distance ÷ solvent-front distance, measured from the origin to the centre of the spot. Colourless spots need UV or ninhydrin.