Whiteboard Chemistry with Joe White

Required Practicals

Every AQA A-Level required practical in one place — the method, the kit, the safety, the results you should get, and the errors and uncertainties examiners ask about, each with real exam questions and mark schemes.

AQA 7404/7405 All 12 required practicals

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.

The six practicals below (RP1–6) are the ones you meet across the AS year and they carry into the full A-level; RP7–12 are Year 13 and are on their way.

Building on GCSE

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 lab skills that keep coming up

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. RP1–6 are built out in full below; RP7–12 are Year 13 and coming soon.

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.

WHERE THE UNCERTAINTY COMES FROM A reading — one judgement ± ½ division A measurement — start to end, two judgements the measured value ±½ at each end = ± 1 whole division PERCENTAGE UNCERTAINTY uncertainty value × 100 bigger value → smaller %
A single reading is uncertain by ± half the smallest division; a measurement spans two readings, so it carries ± one whole division.
The uncertainty rules to know
  • 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 Required practical 1

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.

Method

Part A — make the standard solution

  1. 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.
  2. Dissolve it fully in the beaker in less than the final volume of distilled water, stirring with a glass rod.
  3. 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.
  4. 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.
  5. Stopper and invert several times to mix into one uniform (homogeneous) solution.

Part B — the titration

  1. Rinse the burette with the solution it will hold, the pipette with its solution, and the conical flask with distilled water only.
  2. 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).
  3. 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).
  4. Run a rough titration first, then accurate runs — adding dropwise near the end point, swirling over a white tile.
  5. Repeat until you have concordant titres (agreeing within 0.10 cm3) and take the mean of those only.
1 2 3 4 12.05 g weigh by difference dissolve in a little water transfer with washings make up to the line
Making a standard solution: the concentration is only exact if every bit of solid ends up in the flask — hence weighing by difference and transferring with washings.
READING A BURETTE 24 25 26 eye level reading = 25.45 cm³ bottom of the meniscus each reading ± 0.05 cm³

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.

Corrosive

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

0 50 Clamp Clamp stand Burette (alkali) read to 0.05 cm³ Tap Conical flask acid + indicator White tile

RunInitial / cm3Final / cm3Titre / 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:

IndicatorIn acidIn alkaliEnd-point change
Phenolphthaleincolourlesspinkcolourless → first permanent pale pink
Methyl orangeredyellowred → 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.)

Precision & uncertainty
  • 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-style questions — RP1
Q1a[1 mark]

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.

Q1b[6 marks]

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.

Q1c[2 marks]

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.

Q2a[5 marks]

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).

Q2b[1 mark]

The student fills the burette using a funnel and forgets to remove it before starting. Suggest why this might affect the titre recorded.

Show answer

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.

Q2c[1 mark]

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.


Measurement of an enthalpy change Required practical 2

Aim: measure the heat energy released or taken in by a reaction and use it to find an enthalpy changeH, 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.

Method

In solution — polystyrene cup

  1. 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.
  2. Add the second reactant — weighing any solid by difference — stir, and keep reading the temperature every minute through the peak and the cool-down.
  3. 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

  1. Weigh the spirit burner; measure a known mass of water into a metal can and record the water’s start temperature.
  2. Burn the fuel to raise the water by about 10–20 °C, then reweigh the burner — the difference is the mass of fuel burned.
  3. Use q = mcΔT for the water, then divide by the moles of fuel burned to get ΔcH.
thermometer reads the change: ΔT the solution the mass m being heated; c taken as water’s lid cuts evaporation & heat loss expanded-polystyrene cup insulator — keeps the heat in the solution being measured
Solution calorimetry: every term of q = mcΔT points at something physical.
water known mass thermometer copper can q heat that reached the water = q spirit burner — fuel weighed before and after heat that didn’t — lost around the can draught shield — the standard fix, but losses remain large
Combustion calorimetry leaks by design — only the heat that reaches the water is counted in q, so measured values come out less exothermic than the data book.

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.

Flammable
Corrosive

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:

temperature / °C time / minutes 2527293133 024681012 best-fit line — steady readings mix at t₁ observed maximum — already too low extrapolate the cooling line back to t₁ corrected ΔT bigger than any rise you observed
Cooling starts the moment the reaction does — so extrapolate the cooling line back to the time of mixing and read the corrected ΔT there.
Precision & uncertainty
  • 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-style questions — RP2
Q1a[5 marks]

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.

Q1b[1 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.

Q1c[1 mark]

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.

Q1d[2 marks]

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.

Q1e[5 marks]

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

KOH limits → moles of water = 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.

Q1f[2 marks]

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

Q2a[2 marks]

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.

Q2b[2 marks]

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

Q2c[2 marks]

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

Q2d[1 mark]

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.


How the rate of a reaction changes with temperature Required practical 3

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.

Method — the thiosulfate “disappearing cross” clock
  1. 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.
  2. Time how long the cloudiness (a fine precipitate of sulfur) takes to hide the cross. Shorter time means faster reaction, so rate ∝ 1/time.
  3. 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.
  4. 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.
Na₂S₂O₃(aq) + 2HCl(aq) → the mixture clouds with sulfur 0 s just mixed clear — cross visible 24 s sulfur forming cloudy — cross fading 47 s cross hidden stop the clock · rate ∝ 1/t
A clock reaction trades detail for a single clean timing. Describe the actual change — the mixture turns cloudy with sulfur — not just “it goes a colour”.

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
Toxic

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:

Precision & uncertainty
  • 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.
Worked example — turning clock times into a rate–temperature graph

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 / °CTime / sRate = 1/time / s−1
202001/200 = 0.0050
301011/101 = 0.0099
40521/52 = 0.0192
50281/28 = 0.0357
60151/15 = 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).

1/TIME RISES STEEPLY WITH TEMPERATURE 0 0.02 0.04 0.06 10 20 30 40 50 60 read the rate at 45 °C temperature / °C rate = 1/time / s⁻¹
Plot 1/time against the mean temperature: the curve climbs ever more steeply — the rate roughly doubles every 10 °C. Read the rate at any temperature straight off the curve.

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-style questions
Q1 [2 marks]

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.

Q2 [6 marks]

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.

Q3 [7 marks]

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 / °C223136424954
Time, t, for cross to disappear / s874836264412
1/t / s−10.01150.02080.02780.03850.0227 
Q3a [1 mark]

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.

Q3b [1 mark]

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

Q3c [2 marks]

Plot the values of 1/t against temperature on the graph below. Draw a line of best fit.

Show answer
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 25 30 35 40 45 50 55 Temperature / °C 1/t / s⁻¹ anomalous — ignore it at 40 °C: 1/t ≈ 0.035

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.

Q3d [1 mark]

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.

Q3e [1 mark]

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.

Q3f [1 mark]

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

Q4 [7 marks]

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 / °C3238445467
Time t / s1558550229
1000/t / s−16.4511.820.045.5 
Q4a [1 mark]

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.

Q4b [1 mark]

State the independent variable in this investigation.

Show answer

Temperature 1 mark

Q4c [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

Q4d [3 marks]

Use the data in the table to plot a graph of 1000/t against temperature.

Show answer
0 20 40 60 80 100 120 35 40 45 50 55 60 65 Temperature / °C 1000/t / s⁻¹ at 60 °C: 1000/t ≈ 68

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

Q4e [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.

Q5 [1 mark]

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.

Q6 [1 mark]

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 Required practical 4

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.

No flame tests here

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 forAdd…Positive observationIonic equation
Group 2 cation
(via hydroxide)
sodium hydroxide, dropwisewhite 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 acidwhite precipitate — strongest for Ba2+, fading up the group (Mg2+ stays in solution, as MgSO4 is soluble)Ba2+ + SO42− → BaSO4
Ammonium NH4+sodium hydroxide, then warma pungent gas turns damp red litmus blue (held at the mouth of the tube)NH4+ + OH → NH3 + H2O
Hydroxide OHred litmus paperturns blue — the solution is alkaline
Carbonate CO32−dilute acid; bubble the gas through limewatereffervescence; the gas (CO2) turns limewater milkyCO32− + 2H+ → CO2 + H2O
Sulfate SO42−acidify (dilute HCl), then barium chloridewhite precipitateBa2+ + SO42− → BaSO4
Chloride Clacidify (dilute HNO3), then silver nitrate; then ammoniawhite precipitate — dissolves in dilute ammoniaAg+ + Cl → AgCl
Bromide Brcream precipitate — dissolves only in concentrated ammoniaAg+ + Br → AgBr
Iodide Iyellow precipitate — insoluble, even in concentrated ammoniaAg+ + I → AgI
acidified BaCl₂ dense white precipitate Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)  white acidify first: dilute HCl or HNO₃ (never H₂SO₄) to remove CO₃²⁻ and SO₃²⁻ ions that would also give a white precipitate
A dense white precipitate with acidified BaCl2 means sulfate. The acid step is what makes the test reliable.
ADD DILUTE HNO₃, THEN AgNO₃(aq) AgCl · white dissolves indilute NH₃ AgBr · cream dissolves only inconcentrated NH₃ AgI · yellow insoluble, even inconcentrated NH₃ acidify with dilute HNO₃ first — never HCl (adds Cl⁻) or H₂SO₄ (adds SO₄²⁻), which would give false precipitates
The colour narrows it down; ammonia settles a close call between a white and a cream precipitate.

Apparatus & techniques

  • test tubes & dropping pipettes
  • qualitative tests for ions
  • gas tests (litmus, limewater)
  • recording observations clearly
  • safe handling of toxic & corrosive reagents
Corrosive
Toxic

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:

Identifying an unknown — the order matters

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:

  1. Carbonate first (add acid — fizz, limewater milky). It has to go first, because carbonate would also give a white precipitate with barium.
  2. Sulfate next (acidified barium chloride). Acidifying first removes any carbonate and sulfite.
  3. 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-style questions — RP4
Q1[6 marks]

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.


Distillation of a product from a reaction Required practical 5

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.

Method — distilling a product off
  1. 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.
  2. Heat gently. The product has the lowest boiling point, so it vaporises first, passes the thermometer bulb, and condenses into the receiver.
  3. 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.

DISTILLATION — collect the aldehyde as it forms REFLUX — nothing escapes water out water in condenses to liquid thermometer — bulb level with the side arm still head pear-shaped flask heat ethanol + limited acidified K₂Cr₂O₇ anti-bumping granules ethanal (colourless) collects — open flask → ethanal, out before round two water out water in open at the top — never seal it condenses & drips back heat ethanol + excess acidified K₂Cr₂O₇ anti-bumping granules → carries on to ethanoic acid
The drawing marks live in the details: thermometer bulb level with the side arm, condenser water in at the bottom, and a system that is never sealed — distillation lets the product escape; reflux sends it back.

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
Flammable
Toxic

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 orange to green as the dichromate(VI) is reduced.

Precision — the apparatus faults examiners look for

“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-style questions — RP5
Q1[6 marks]

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

Q2[4 marks]

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.


Tests for alcohol, aldehyde, alkene & carboxylic acid Required practical 6

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 groupAdd…Positive observationNegative (control)
Alkene (C=C)bromine water, shakedecolourises: orange → colourlessan alkane stays orange
Alcohol (1° or 2°)acidified potassium dichromate(VI), warmorange → greena 3° alcohol stays orange
Aldehyde –CHOTollens’ reagent, warm gentlya silver mirror formsa ketone: no change
Aldehyde –CHOFehling’s solution, warm gentlyblue solution → brick-red precipitatea ketone stays blue
Carboxylic acid –COOHsodium carbonate solutioneffervescence — the gas (CO2) turns limewater milkyan alcohol: no bubbles

bromine water · shake

+ alkene orange soln → colourless soln + alkane stays orange soln

The alkene adds Br2 across its C=C and uses it up — say decolourised, never “clear”

sodium carbonate solution

CO₂ ↑ + carboxylic acid effervescence + alcohol no bubbles

Only a carboxylic acid is acidic enough to free CO2 from a carbonate; the gas turns limewater cloudy

tollens’ reagent · warm gently

+ aldehyde silver mirror + ketone no change

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° or 2° alcohol orange soln → green soln + 3° alcohol stays orange soln

1°/2° alcohols (and aldehydes) reduce orange Cr2O72− to green Cr3+; a 3° alcohol resists oxidation

Always give the reagent and the observation for both substances — “no visible change” is a scoring answer; “nothing” is not.

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
Corrosive
Toxic

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:

Precision — the method mistakes examiners plant
  • 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-style questions
Q1 [3 marks]

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.

Q2 [3 marks]

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.

Q3 [3 marks]

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.

A2 A-level only from here. RP7–12 are taught in Year 13 — if you are sitting AS, you can stop here.

The A-level-only practicals — RP7 to 12 A-level only

The remaining six required practicals are taught in Year 13, and are on their way — each will get the same full treatment (method, apparatus, safety, data analysis and real exam questions with mark schemes). Here is what they cover and where the theory already lives:

RPPracticalStatus
RP7Measuring the rate of a reaction — by an initial-rate (clock) method and by continuous monitoringtheory in Kinetics · coming soon
RP8Measuring the EMF of an electrochemical cellcoming soon
RP9How pH changes when a weak/strong acid reacts with a weak/strong basecoming soon
RP10Preparing a pure organic solid (and testing its purity) and a pure organic liquidcoming soon
RP11Test-tube reactions to identify transition-metal ions in aqueous solutioncoming soon
RP12Separating a mixture by thin-layer chromatographycoming soon
The six AS required practicals — quick reference
  • 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 changeq = 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 → colourless), acidified dichromate (alcohol: orange → green), Tollens’/Fehling’s (aldehyde: silver mirror / brick-red), sodium carbonate (carboxylic acid: fizz).

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