There is no coursework in AQA GCSE Chemistry. Instead, the eight required practicals are assessed in the written papers — and questions drawing on practical work make up at least 15% of the marks. The examiner isn’t checking whether you once stood at a bench and did the experiment; they’re checking whether you can describe the method, justify each step, handle the variables and evaluate the results. That makes practicals one of the most learnable parts of the course — everything on this page can be revised like any other topic.
Every required practical can appear in your exams in four ways — learn each practical with all four in mind:
- Describe — write or order the method steps, name the apparatus, state the measurements taken.
- Explain — justify a step (why excess base? why a polystyrene cup? why pencil, not pen?). This is where most marks live.
- Apply — process results: plot graphs, calculate gradients, Rf values or means, spot anomalies.
- Evaluate — identify errors, suggest improvements, and comment on accuracy, repeatability and validity.
Paper 1 examines RP1–RP4 (topics C1–C5); Paper 2 examines RP5–RP8 (topics C6–C10). RP2 and RP7 are Triple Chemistry only.
Paper 1
Paper 2
1Making Salts
Aim: prepare a pure, dry sample of a soluble salt (e.g. copper sulfate) by reacting a dilute acid with an insoluble base (e.g. copper oxide).
📖 Full theory: C4 — Neutralisation & Making Salts
- Gently warm the dilute acid (e.g. sulfuric acid) with a Bunsen burner until it is almost boiling — this speeds up the reaction — then turn off the burner before adding anything.
- Add the insoluble base (copper oxide) a little at a time, stirring, until no more dissolves and some solid remains. This excess guarantees all the acid has been neutralised, giving the maximum yield of salt.
- Filter to remove the unreacted excess base. The filtrate is a solution of the pure salt.
- Pour the filtrate into an evaporating basin and evaporate gently over a water bath until about half the solution remains.
- Transfer the concentrated solution to a crystallising dish and leave it in a cool place (around 24 hours) so crystals form slowly, then pat them dry between filter paper.
CuO + H2SO4 → CuSO4 + H2O
What you should see: the colourless acid turns blue as copper sulfate forms; black copper oxide remains undissolved once the acid is used up; filtering leaves the black solid in the paper and a clear blue filtrate; slow cooling gives blue crystals.
Copper sulfate is only the textbook example. The exam can name any soluble salt — zinc chloride, magnesium sulfate, copper nitrate — and expect you to choose reactants that actually make it. Writing “add a base to an acid” is the classic dropped mark: the examiner wants the specific acid and the specific base, linked to the salt.
- The second word of the salt’s name tells you the acid: chloride → hydrochloric acid, sulfate → sulfuric acid, nitrate → nitric acid.
- The first word tells you the metal the insoluble base must contain — an oxide, hydroxide or carbonate of that metal.
So for zinc chloride: warm hydrochloric acid, then add zinc oxide (or zinc carbonate) to excess — same method as above, different reactants. Remember a carbonate also fizzes off carbon dioxide as it neutralises the acid.
It depends on whether it dissolves. An insoluble hydroxide behaves just like an oxide or carbonate — add it to excess, filter, crystallise. But a soluble hydroxide such as sodium hydroxide is an alkali: it’s already dissolved in solution. The whole method relies on being able to add an excess of solid and then filter the excess out — you can’t filter a dissolved excess, so any extra alkali stays mixed in with the salt and the sample isn’t pure.
Try it — here’s the skill the examiner is testing. Given a salt, choose the reactants that make it:
🧪 Which acid? Which base?
Read the salt’s name, then pick the acid and the insoluble base that would make it by the method above. The salt’s second name chooses the acid; its first name chooses the metal in the base.
Make this salt
- “An acid and a base.” If the question names the salt, name your reactants — dilute sulfuric acid and copper oxide, not “an acid and a base”. Generic reactants don’t show you can plan this preparation.
- “Evaporate to dryness.” Never write this — boiling off all the water drives out the water of crystallisation and spoils the crystals. Evaporate some water, then cool slowly.
- Skipping the excess. Adding a measured amount of base instead of an excess leaves unreacted acid mixed in with the salt — the sample isn’t pure.
- Steps out of order. The sequence is the mark: excess base → filter → crystallise → dry. In a 6-marker, a logically ordered method with these key details is what reaches Level 3.
AQA’s wording is precise: a Bunsen burner to heat the dilute acid, but a water bath or electric heater to evaporate the solution. A water bath can never exceed 100 °C, so the heating is gentle and even — the solution doesn’t spit, and the salt isn’t overheated and decomposed. “Why is a water bath used?” is a regular 1–2 mark question.
Required practical questions often end with: “Give one safety precaution the student should take. Explain why.” “Be careful” and “wear a lab coat” score nothing. Name the hazard, give the precaution that targets it, and say what harm it prevents: “wear safety goggles because the dilute acid is an irritant and could damage the eyes”. The dilute acids at GCSE are usually classed as irritant rather than corrosive — and the heating stages carry their own hazards: stop heating before the solution spits, and let the evaporating basin cool before moving it.
🧪 Exam-style questions
A student makes copper sulfate crystals by adding copper oxide to dilute sulfuric acid. Why is the copper oxide added in excess? Tick (✓) one box.
Why is a water bath used to evaporate the salt solution, rather than heating it directly with a Bunsen burner? Tick (✓) one box.
Describe how a student could make crystals of magnesium chloride from magnesium carbonate (an insoluble solid) and dilute hydrochloric acid.
Show the mark scheme
- Add magnesium carbonate to the (warm) dilute hydrochloric acid a little at a time until it is in excess — the fizzing stops and some solid remains. 1 mark
- Filter to remove the unreacted excess carbonate. 1 mark
- Evaporate some of the water from the filtrate (e.g. over a water bath). 1 mark
- Leave the solution to cool so crystals form slowly, then dry them (pat between filter paper). 1 mark
A carbonate also gives off carbon dioxide, so “no more fizzing” is the sign the acid is used up: MgCO3 + 2HCl → MgCl2 + H2O + CO2. Do not accept evaporating to dryness.
Describe a method to prepare a pure, dry sample of zinc sulfate crystals from zinc oxide and dilute sulfuric acid. In your method, name the apparatus used, give the purpose of each step, and include one safety precaution. This is a levels-of-response question — plan a logically ordered method, then compare with the model answer.
Show a model answer
How it is marked (levels of response):
- Level 3 (5–6): a logically ordered method with the key practical detail (excess base, filter, crystallise, dry), named apparatus and a sensible safety precaution.
- Level 2 (3–4): several correct steps, but with a gap or the order muddled.
- Level 1 (1–2): one or two relevant practical points.
Indicative content:
- Measure dilute sulfuric acid into a beaker and warm it gently with a Bunsen burner (tripod, gauze, heatproof mat) — to speed up the reaction — then turn off the burner.
- Add zinc oxide with a spatula, a little at a time, stirring with a glass rod, until some solid remains — the excess means all the acid has reacted.
- Filter through filter paper in a funnel into a conical flask — to remove the unreacted zinc oxide.
- Evaporate the filtrate gently in an evaporating basin over a water bath until about half remains — gentle heating avoids spitting.
- Leave in a crystallising dish in a cool place so crystals form slowly, then pat dry between filter paper.
- Safety (matched pair): wear safety goggles because the dilute acid is an irritant and could damage the eyes; allow hot apparatus to cool before handling.
Equation: ZnO + H2SO4 → ZnSO4 + H2O. “Be careful” earns nothing — name the hazard, the precaution and the harm prevented.
2Titration Triple Only
Aim: find the volumes of a strong acid and a strong alkali that exactly react with each other, using a titration.
📖 Full theory: C4 — Titrations · calculations in C3 — Quantitative Chemistry
- Use a pipette to measure a fixed volume of alkali (e.g. 25.0 cm3) into a conical flask, and add a few drops of a suitable indicator. Stand the flask on a white tile so the colour change shows up clearly.
- Fill a burette with the acid and record the starting reading — from the bottom of the meniscus.
- Do a rough (trial) titration first: run the acid in fairly quickly, swirling, until the indicator just changes colour. This gives an approximate titre.
- Repeat carefully: run the acid in quickly at first, then dropwise as you approach the end-point, until the indicator just changes colour. Record the final reading.
- Repeat until you have concordant results — within 0.10 cm3 of each other — and take a mean, ignoring the rough titre.
The required practical in two stages: first pipette a fixed volume of alkali into the flask. Add a few drops of indicator. Then run acid in from the burette drop by drop until the colour just changes.
The meniscus curves up at the glass walls and dips in the middle. Line your eye up with the lowest point and read from there — reading the higher edges gives a smaller, wrong value.
The volume of acid added from the burette.
The colour of the indicator — the change marks the end-point. The titre you record is the volume of acid at that moment.
The volume and concentration of the alkali (pipette), the indicator and the number of drops of it, and the same person judging the end-point.
- Universal indicator. Its gradual rainbow of colours gives no sharp end-point. Use a single-change indicator: phenolphthalein (pink → colourless) or methyl orange (yellow → red).
- Mixing up pipette and burette. The pipette measures one fixed volume every time (the alkali); the burette measures variable volumes (the acid — the unknown).
- Reading the meniscus wrong. Read from the bottom of the meniscus, at eye level. Reading the higher edges gives a smaller, wrong value.
- Averaging everything. Only concordant results (within 0.10 cm3) go into the mean — the rough titre and any outliers are ignored.
🧪 Exam-style questions
Which indicator is most suitable for a titration of a strong acid against a strong alkali? Tick (✓) one box.
A student’s titres are 24.05 cm3, 23.40 cm3, 23.45 cm3 and 23.90 cm3. Which titres should be used to calculate the mean? Tick (✓) one box.
How should the burette be read? Tick (✓) one box.
A student is given dilute sulfuric acid, potassium hydroxide solution, a pipette, a burette, a conical flask and a suitable indicator. Describe how the student should find the volume of dilute sulfuric acid that reacts exactly with 25.0 cm3 of the potassium hydroxide solution. Include the measurements the student should make. Do not describe any calculations. This is a levels-of-response question — plan, then compare with the model answer.
Show a model answer
How it is marked (levels of response):
- Level 3 (5–6): a logically ordered method covering the pipette, the indicator, the burette readings, dropwise addition at the end-point and concordant repeats.
- Level 2 (3–4): the right apparatus used, but a key measurement or the repeats missing.
- Level 1 (1–2): one or two relevant points.
Indicative content:
- Use the pipette to transfer 25.0 cm3 of potassium hydroxide solution into the conical flask, and add a few drops of indicator. Stand the flask on a white tile.
- Fill the burette with the dilute sulfuric acid and record the initial reading (from the bottom of the meniscus, at eye level).
- Do a rough titration first to find the approximate end-point.
- Repeat carefully: add the acid quickly at first, swirling, then drop by drop as the end-point approaches, until the indicator just changes colour.
- Record the final reading; the titre is final reading − initial reading.
- Repeat until there are concordant titres (within 0.10 cm3) — these are the values that would be averaged.
The question says do not describe calculations — describing the mean is fine as the measurements to take, but working through moles would waste time and earn nothing here.
3Electrolysis of Aqueous Solutions
Aim: investigate the products formed when different aqueous solutions are electrolysed using inert (graphite) electrodes. This is the specification’s named opportunity for developing a hypothesis — so predict the products at each electrode first, then test your prediction.
📖 Full theory: C4 — Electrolysis of Aqueous Solutions
- Pour the solution into a beaker, fit a lid, and insert two clean graphite (carbon) rod electrodes through the holes — the rods must not touch each other. Connect them to a low-voltage DC supply and switch on.
- Observe each electrode: bubbles at neither, one or both? A metal coating forming on the cathode? Record the observations.
- Test any gas at the anode by holding damp blue litmus paper next to it with tweezers (lifting the lid briefly): bleached white means chlorine.
- After no more than five minutes, switch off and examine the cathode for a metal coating.
- Clean the equipment and repeat with each solution (copper(II) chloride, copper(II) sulfate, sodium chloride, sodium sulfate), comparing the products with your predictions.
The two rules that predict every answer:
- At the cathode (−): hydrogen is produced unless the metal is less reactive than hydrogen — then the metal is produced (e.g. copper).
- At the anode (+): oxygen is produced unless the solution contains a halide ion (Cl−, Br−, I−) — then the halogen is produced.
| Solution | Cathode (−) | Anode (+) |
|---|---|---|
| Copper(II) chloride | Copper (brown coating on the electrode) | Chlorine (bleaches damp blue litmus) |
| Copper(II) sulfate | Copper (brown coating on the electrode) | Oxygen (does not bleach litmus) |
| Sodium chloride | Hydrogen (squeaky pop) | Chlorine (bleaches damp blue litmus) |
| Sodium sulfate | Hydrogen (squeaky pop) | Oxygen (does not bleach litmus) |
Two setups covering every electrode product: a metal or hydrogen at the cathode, oxygen or a halogen at the anode.
In AQA’s own method the gas volumes are usually too small to test with splints. Chlorine is identified directly (damp blue litmus bleaches white); a gas at the anode that does not bleach litmus is taken to be oxygen; any gas at the cathode is taken to be hydrogen. In an exam answer, still quote the full gas tests — pop, relight, bleach — when asked how to identify a gas.
The solution being electrolysed (which ionic compound is dissolved).
The product formed at each electrode, identified by observation and gas tests.
The same inert graphite electrodes, the same voltage and the same time — so any difference in products is due only to the solution.
- Vague gas tests. State the test and the precise result: oxygen relights a glowing splint (not “glows brighter”); chlorine bleaches damp blue litmus white (not “turns clear”). “Smells like a swimming pool” is a property, not a test — and chlorine is toxic.
- Reactive electrodes. The electrodes must be inert (graphite) so they don’t take part in the reaction and change the products.
- Safety. Chlorine is toxic — keep the room well ventilated and avoid breathing the gas at the anode.
🧪 Exam-style questions
Sodium chloride solution is electrolysed using inert electrodes. What is produced at the cathode? Tick (✓) one box.
Why are graphite electrodes used in this practical? Tick (✓) one box.
Copper(II) sulfate solution is electrolysed using inert electrodes. Predict the product at each electrode. Explain each prediction.
Show the mark scheme
- Cathode: copper (a brown coating forms on the electrode). 1 mark
- Because copper is less reactive than hydrogen, the metal is produced rather than hydrogen. 1 mark
- Anode: oxygen. 1 mark
- Because the solution contains no halide ions (sulfate is not a halide), so oxygen from the water is produced. 1 mark
The two rules answer every aqueous electrolysis question — check the metal against hydrogen for the cathode, and check for a halide for the anode.
A gas collects at the anode. Describe a test to show that the gas is chlorine, and give the result of the test if it is.
Show the mark scheme
- Hold a piece of damp blue litmus paper in the gas. 1 mark
- The litmus paper is bleached white. 1 mark
“Turns clear” is not accepted — the paper is bleached white. (It may flash red first, because chlorine forms an acidic solution with the water on the paper.)
4Temperature Changes
Aim: investigate the variables that affect the temperature change in reacting solutions — for example acid + metal, acid + carbonate, neutralisation, or the displacement of metals.
📖 Full theory: C5 — Energy Changes
- Stand a polystyrene cup inside a beaker (for support) and add a measured volume of the first solution.
- Measure and record the starting temperature.
- Add the second reactant, put a lid on, stir, and record the highest (or, for an endothermic reaction, the lowest) temperature reached.
- Calculate the temperature change = final temperature − starting temperature. Repeat while changing one variable at a time.
AQA’s own student sheet runs this as a neutralisation investigation: put 30 cm3 of dilute hydrochloric acid in the cup and record its temperature, then add sodium hydroxide solution 5 cm3 at a time (lid on, stir, record the highest steady temperature after each addition) up to 40 cm3. The first additions make the temperature rise; once all the acid is neutralised, further additions of cool alkali make it fall. Repeat the whole investigation and use mean values.
The graph is the exam point: plot total volume added (x) against mean maximum temperature (y), then draw two straight lines of best fit — one through the rising points, one through the falling points — and extend them until they cross. The intersection gives the volume of alkali that exactly neutralises the acid, and the maximum temperature reached. That’s the peak-and-dip shape explained below.
Try the practical here — all four reactions named in the specification:
🧪 Run the required practical — temperature change
Pick one of the four reactions named in the specification, choose your reactants, then add them to the polystyrene cup and watch the thermometer. Record the temperature change and decide: exothermic or endothermic?
Choose a reaction and its reactants above, then press Run experiment.
| Reaction | Final / °C | ΔT / °C | Result |
|---|
Every run starts from 21.0 °C (a control variable). Change one reactant at a time — the independent variable — and watch how ΔT responds.
What you should see: exothermic reactions (neutralisation, displacement, metals with acid) make the temperature rise; endothermic ones make it fall. In a displacement series, a more reactive metal added to copper sulfate solution gives a larger temperature rise — plot the results as a bar chart (the metal is a categoric variable) and use them to order the metals by reactivity.
Reading the chart: the more reactive the metal, the bigger the temperature rise — and copper, which cannot displace itself from the solution, gives none.
Whatever you are investigating — the type or concentration of a reactant, or the metal used in a displacement.
The maximum temperature change (final − start).
The volumes of solution, the insulation (same cup and lid), the starting temperature, and — for displacement — the mass and surface area of the metals.
The cup and lid are the key pieces of kit: polystyrene is a good insulator, so it reduces the energy transferred between the mixture and the surroundings — making the temperature change more accurate. For an exothermic reaction the insulation slows energy being lost; for an endothermic one it slows energy being gained — so for an endothermic reaction, write that it “prevents energy being gained”, not that it “prevents heat loss”.
And when a question asks why the cup makes the result more accurate, the answer must be comparative: “less energy is transferred to the surroundings than with a glass beaker”. Absolute claims — “no energy is lost” — are wrong (some always is) and score nothing.
- Random errors scatter readings either side of the true value — for example slightly misreading the thermometer or misjudging the highest temperature. Repeating the experiment and taking a mean reduces their effect.
- A systematic error shifts every reading the same way. Using a glass beaker instead of a polystyrene cup is systematic: glass is a poorer insulator, so energy is lost to the surroundings every time and the temperature change always comes out too small.
- The uncertainty in a mean is often estimated as half the range: uncertainty = (highest valid value − lowest valid value) ÷ 2, quoted as mean ± uncertainty.
A classic graph plots the temperature change against the mass or volume of one reactant you add to a fixed quantity of the other. Up to the point where the reactants are in their reacting proportions, the temperature change increases as you add more. What happens after that depends on what you added — and explaining it is where the marks are:
- add a solid and the line levels off (a plateau): the other reactant is now used up, so the extra solid just sits there unreacted and makes no further difference;
- add a solution and the line peaks then dips: the excess cold solution cools and dilutes the mixture, so the reading falls back.
Don’t be thrown if a question draws the plateau sloping gently downward rather than dead flat — once the reaction is over, the warm solution slowly cools toward room temperature, so the reading drifts down. The mark is still for spotting that the reaction stops adding heat once the limiting reactant is used up.
Add more of one reactant to a fixed quantity of the other and the temperature change rises at first, up to the reacting-proportions point. After that the two cases split: a solid is simply left over, so the line levels off; excess solution cools and dilutes the mix, so it dips.
Don’t mix that up with a temperature–time graph of a single run, where the across-the-bottom axis is time rather than the mass or volume you added. Here the temperature shoots up to a maximum as the reaction happens, then slowly falls back: once the reaction has finished no more energy is released, so the warm mixture loses heat to the cooler surroundings.
Temperature against time within a single experiment: a sharp rise to a maximum as the reaction gives out energy, then a slow fall as the warm mixture loses heat to the surroundings.
🧪 Exam-style questions
Why is the reaction carried out in a polystyrene cup rather than a glass beaker? Tick (✓) one box.
A student adds different metals to copper sulfate solution and records the temperature rise for each. What is the dependent variable? Tick (✓) one box.
Describe a method to investigate how the type of metal affects the temperature change when metals are added to copper sulfate solution. Include the variables that must be kept the same.
Show the mark scheme
- Measure a fixed volume of copper sulfate solution into a polystyrene cup (in a beaker, with a lid) and record the starting temperature. 1 mark
- Add the metal, stir, and record the highest temperature reached; calculate the temperature change. 1 mark
- Repeat with each metal, changing only the type of metal. 1 mark
- Control variables: the same volume and concentration of copper sulfate solution, the same mass (and surface area) of metal, and the same starting temperature. 1 mark
Say “the same mass” or “the same volume” — never “the same amount”, which is too vague to earn the mark.
Two students carry out the same neutralisation reaction. Student A uses a polystyrene cup with a lid. Student B uses a glass beaker with no lid. Student B’s temperature rises are lower every time. Explain why, and name this type of error.
Show the mark scheme
- Glass is a poorer insulator than polystyrene (and there is no lid), so energy is transferred to the surroundings. 1 mark
- So the maximum temperature recorded is too low every time. 1 mark
- This is a systematic error — it shifts every reading the same way. 1 mark
Repeating and taking a mean does not fix a systematic error — that works for random errors. The fix is to change the equipment or technique.
5Rates of Reaction
Aim: investigate how the concentration of a reactant affects the rate of reaction, following the reaction by two different methods. Like RP3, this practical is a named opportunity for developing a hypothesis: “increasing the concentration will increase the rate, because there are more particles in the same volume, so collisions are more frequent.”
📖 Full theory: C6 — Rates & Equilibrium
Sodium thiosulfate reacts with dilute hydrochloric acid to make a fine precipitate of sulfur, which slowly clouds the mixture:
Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO2(g) + H2O(l)
- Measure thiosulfate solution into a conical flask standing on a black cross drawn on paper.
- Add the acid, swirl, and start the stop clock.
- Look down through the solution; stop the clock the moment the cross is no longer visible.
- Repeat with different concentrations of thiosulfate — made by diluting the same stock solution with water so the total volume is always the same (e.g. 10 cm3 thiosulfate + 40 cm3 water, then 20 + 30, 30 + 20…). Record every time in seconds, repeat the whole investigation and take means, leaving out anomalies.
A shorter time means a faster rate — rate is proportional to 1/time.
React magnesium (or marble chips) with dilute hydrochloric acid and collect the gas in a gas syringe (or an upside-down measuring cylinder full of water). Record the volume every 10 seconds, repeat with a different acid concentration, and plot volume against time. The steeper the initial gradient, the faster the rate.
Two of the ways of following the rate in required practical 5. Method 1: time how long the black cross takes to vanish as the solution clouds. Method 2: read the gas volume off a syringe at regular times and plot the curve.
If a reaction gives off a gas, you can follow it on a balance. Stand the flask of reactants on the balance, plug the neck with cotton wool, and record the mass every few seconds. As gas escapes into the air the mass of the flask falls — a faster fall means a faster reaction.
CaCO3 + 2HCl → CaCl2 + H2O + CO2
- The usual reaction is a carbonate with acid — the carbon dioxide leaves the flask, so the reading drops.
- The cotton wool plug earns marks of its own: it lets the gas escape (so the mass can change) while stopping acid spray being thrown out as the mixture fizzes — spray leaving would make the mass fall too far.
- Mass only appears to be lost — no atoms are destroyed; the gas has simply left the flask. Seal the flask and the mass wouldn’t change at all.
- Mass loss suits a heavy gas like CO2 but is the wrong method for hydrogen — H2 is too light to move the balance reading reliably, so collect it in a gas syringe instead. Matching the method to the gas is itself a common exam point.
Following a reaction by loss of mass: a carbonate reacts with acid in a flask plugged with cotton wool. The CO₂ escapes into the air so the balance reading falls — while the cotton wool stops acid spray being lost with it.
The concentration of the reactant solution (the thiosulfate, or the acid).
The time for the cross to disappear, or the volume of gas at set times.
The total volume (and so depth) of solution, the temperature, the amount of the other reactant, and the same flask, cross and observer.
The mean rate over a period is volume ÷ time — but the rate at one moment is the gradient of the tangent to the curve at that time. Draw the tangent touching the curve at the point, make the gradient triangle large (small triangles magnify drawing errors), show your working on the graph, and quote the answer with its unit (e.g. cm3/s). Questions sometimes ask for the answer in standard form — e.g. 0.0025 = 2.5 × 10−3.
📐 Tangent explorer — the rate at any moment
Magnesium + dilute hydrochloric acid, with the hydrogen collected in a gas syringe. Slide along the curve and watch the tangent, the triangle and the gradient calculation update. Notice when the reaction is fastest — and what the gradient becomes once it finishes.
- Same depth of solution every time — you look down through the liquid, so a deeper mixture hides the cross sooner. Topping up with water when diluting keeps the total volume, and so the depth, the same.
- Measure volumes with a measuring cylinder, not a beaker — its finer scale (better resolution) makes the “same total volume” genuinely the same.
- State the observation, not an inference — timing stops when “the cross is no longer visible”. “The reaction finished” is an inference, and a false one: the reaction is still going; there’s just enough sulfur to block your view.
- The end-point judgement is the weakness — improve it with the same observer every time, or remove the judgement entirely with a light sensor and data logger.
- Safety — the reaction makes sulfur dioxide, a toxic gas: keep the room well ventilated.
🧪 Exam-style questions
A student reacts marble chips with dilute hydrochloric acid in a flask on a balance. The balance reading decreases during the reaction. Why? Tick (✓) one box.
A student investigates how concentration affects the rate of the sodium thiosulfate and acid reaction, using the disappearing cross. Which variables must be kept the same? Tick (✓) one box.
A reaction produces 48 cm3 of gas in the first 120 seconds. (a) Calculate the mean rate of reaction over this time. Give the unit. (b) The volume–time graph for this reaction is steepest at the start. Explain why, in terms of particles.
Show the mark scheme
- (a) Mean rate = 48 ÷ 120 = 0.4 1 mark
- Unit: cm3/s 1 mark
- (b) At the start the concentration of the reactants is greatest… 1 mark
- …so collisions are most frequent, giving the fastest rate (steepest gradient). 1 mark
Say more frequent collisions — not just “more collisions”. As the reactants are used up the collisions become less frequent, the curve flattens, and it goes horizontal when one reactant has run out.
An experiment with 20 g of large marble chips and excess acid is repeated with 20 g of smaller marble chips, with everything else unchanged. Describe how the volume–time curve for the smaller chips would compare, and explain why.
Show the mark scheme
- The curve is steeper at the start (faster rate). 1 mark
- It levels off at the same final volume of gas, sooner. 1 mark
- Smaller chips have a larger surface area (to volume ratio), so collisions at the surface are more frequent. 1 mark
The same mass of carbonate makes the same amount of CO2 — surface area changes how fast you get there, not how much you get. Sketch questions mark exactly these two features: steeper start, same plateau.
6Chromatography
Aim: use paper chromatography to separate and identify a mixture of coloured substances (e.g. food colourings), and calculate Rf values.
📖 Full theory: C8 — Chromatography & Rf Values
The one rule the practical is built on: the start line and spots must sit above the solvent, or the samples wash straight off into the beaker.
- Draw a pencil start line about 2 cm from the bottom of the chromatography paper (pencil is insoluble, so it won’t run).
- Use a fresh capillary tube to put a small spot (2–3 mm) of each known colouring on the line, plus a spot of the unknown mixture. Label each in pencil.
- Add solvent to the beaker to a depth of less than 1 cm, so the level is below the start line. Hang the paper (e.g. from a glass rod resting on the beaker’s rim) so its bottom edge dips in — the sides of the paper must not touch the beaker wall.
- Leave it undisturbed until the solvent has risen about three-quarters of the way up.
- Remove the paper, immediately mark the solvent front in pencil, and let it dry.
- Measure, in mm, the distance from the start line to the centre of each spot, and from the start line to the solvent front. Calculate Rf for each.
Rf = distance moved by substance ÷ distance moved by solvent
Analysis: compare each known colouring’s Rf (and spot position) with the unknown to see which dyes it contains. A substance that gives a single spot is pure; one that separates into several spots is a mixture. Rf values let you compare results between labs — but only if the same solvent was used, since Rf depends on the solvent.
The substance spotted on the start line (each different colouring or ink).
The distance each spot moves — used to work out its Rf value.
The same solvent, paper and start line for every sample; always measure to the centre of the spot.
- Pure / impure — categorical, not relative. A single spot means the substance is pure; more than one spot means impure. Writing “purer” or “less pure” misses the mark — tie the word to the number of spots.
- Why a dye travels further — the marks want both halves of the balance: it is more soluble in the solvent and has a weaker attraction to the paper. Solubility alone is usually not enough.
- What Rf really says — for a given substance, solvent and paper, the spot and the solvent travel in a constant proportion: that ratio is the Rf, which is why it identifies the substance regardless of how long the paper runs.
- Start line in ink. The ink dissolves and travels up with the samples, ruining the chromatogram. Always pencil.
- Solvent above the start line. The spots dissolve straight into the solvent and are lost — the one rule the practical is built on.
- Measuring to the top of a spot. Always measure to the centre.
- An Rf greater than 1, or with units. Rf is a ratio of two distances — no units, always less than 1. If yours is bigger than 1, you’ve divided the wrong way round.
🧪 Try the common mistakes
Run the chromatography three ways and watch what goes wrong with each mistake.
Correct setup: a pencil start line, solvent below the spots. Press play.
🧪 Exam-style questions
Why is the start line drawn in pencil? Tick (✓) one box.
A spot moves 3.6 cm and the solvent front moves 8.0 cm. What is the Rf value? Tick (✓) one box.
A chromatogram of an unknown mixture U is run alongside four known colourings A–D. U separates into three spots: two line up exactly with the spots from A and C, and one matches none of the known colourings. What two conclusions can be drawn about U?
Show the mark scheme
- U contains (the same dyes as) A and C. 1 mark
- U also contains at least one other substance that is not A, B, C or D. 1 mark
Matching spots means same height / same Rf in the same solvent. Note the careful wording — “at least one”: two different substances can sit at the same height, so one unmatched spot doesn’t prove only one extra dye.
A student wants to find out whether a green ink contains the same blue dye as a sample of pure blue dye. Describe a method the student could use. You should name the technique and say how the result would show the answer.
Show the mark scheme
- Use paper chromatography: draw a pencil line near the bottom of the paper and put a small spot of the green ink and a spot of the pure blue dye on the line. 1 mark
- Stand the paper in a small depth of solvent, with the solvent level below the start line. 1 mark
- Leave until the solvent has nearly reached the top; the green ink separates into its dyes. 1 mark
- If one of the green ink’s spots is at the same height (same Rf) as the pure blue dye’s spot, the ink contains that dye. 1 mark
The comparison only works because both samples ran on the same paper, in the same solvent, from the same start line — that’s what makes the Rf values comparable.
7Identifying Ions Triple Only
Aim: use chemical tests to identify the ions in a single unknown ionic compound — a cation test plus an anion test together name the compound.
📖 Full theory: C8 — Identifying Ions (sections 5–7)
Flame tests — metal cations
- Clean a nichrome (or platinum) wire loop by dipping it in dilute acid and holding it in a blue Bunsen flame until it adds no colour — this removes any ions left from before, so you test only your sample.
- Dip the clean loop in the sample and hold it in the edge of the blue flame.
- Observe the colour. Don’t let the wire glow red-hot — that orange glow can be mistaken for a flame colour.
The big limitation: in a mixture, a strong colour (sodium’s yellow especially) masks the others — which is exactly why flame emission spectroscopy is better for mixtures: it picks out every ion at once, quickly, sensitively and with a reading that can be matched to a database. If you name it, write flame emission spectroscopy in full — abbreviations like “F.E.S.” are not credited.
Sodium hydroxide precipitates — metal cations
Add a few drops of sodium hydroxide solution first (slowly), then add it in excess:
| Ion | Precipitate | In excess NaOH |
|---|---|---|
| Copper(II), Cu2+ | Blue | No change |
| Iron(II), Fe2+ | Green | No change |
| Iron(III), Fe3+ | Brown | No change |
| Aluminium, Al3+ | White | Dissolves (colourless solution) |
| Calcium, Ca2+ | White | Does not dissolve |
| Magnesium, Mg2+ | White | Does not dissolve |
Three coloured precipitates (copper blue, iron(II) green, iron(III) brown) and three that are all white (aluminium, calcium, magnesium) — the whites need a second test to tell apart.
Of the three whites, only aluminium re-dissolves in excess. To separate the remaining two, use a flame test: calcium is orange-red, magnesium gives no colour.
The three anion tests
| Anion | Add this… | Positive result |
|---|---|---|
| Carbonate (CO32−) | Dilute acid | Fizzes; the gas (CO2) turns limewater milky |
| Halide (Cl−, Br−, I−) | Dilute nitric acid, then silver nitrate | Silver halide precipitate — white (Cl), cream (Br) or yellow (I) |
| Sulfate (SO42−) | Dilute hydrochloric acid, then barium chloride | White precipitate (barium sulfate) |
The silver halides step through white → cream → yellow as you go down the group (Cl → Br → I). Barium sulfate is a separate white precipitate, made with barium chloride.
Carbonate ions would also form a precipitate with silver nitrate or barium chloride, giving a false positive. Adding acid first reacts away any carbonate (it fizzes off as CO2) so it can’t interfere. The catch is choosing an acid that doesn’t add the very ion you’re testing for: nitric acid for the halide test (hydrochloric would add chloride ions), and hydrochloric acid for the sulfate test (sulfuric would add sulfate ions).
- “Clear” when you mean “colourless”. “Clear” only means see-through — copper sulfate solution is clear and blue. Examiners penalise “clear” here.
- Adding NaOH too fast. Flood aluminium with excess straight away and the white precipitate forms and re-dissolves before you spot it — a few drops first.
- Chloride ions are not chlorine gas. The test for chloride ions is nitric acid + silver nitrate (white precipitate); the test for chlorine gas is damp blue litmus bleaching white (section 9). Examiners report students giving the gas test when the ion test is asked for — read which one the question wants.
- Stopping at the cation. A full identification needs both halves: e.g. green flame + white precipitate with barium chloride = copper sulfate.
Six-mark plan questions here are marked on whether the method would actually deliver an answer for the sample in the question. Students who list correct tests in the abstract — without saying they would carry them out on the sample, or in a workable order — can score nothing despite knowing every test. Structure the plan: take (a solution of) the sample → flame test for the cation → one anion test at a time, naming the reagent and the expected result for a positive.
Put the whole strategy together — run the bench tests and name the unknown compound:
🧪 Identify the unknown salt
You’re handed an unknown ionic compound. Run bench tests to find its cation (metal ion) and anion (non-metal ion), then name it — in as few tests as you can. For each result, try to recall what it tells you before you tap Reveal. This is exactly the Required practical 7 skill.
Pick a test to start.
| Test | Observation | What does it tell you? |
|---|---|---|
| No tests run yet — click a test above. | ||
🧪 Exam-style questions
Sodium hydroxide solution is added to a solution of a metal compound. A white precipitate forms, which dissolves when excess sodium hydroxide is added. Which metal ion is present? Tick (✓) one box.
What is the test for sulfate ions? Tick (✓) one box.
An unknown ionic compound gives a yellow flame in a flame test. A solution of the compound is acidified with dilute nitric acid, and silver nitrate solution is added: a cream precipitate forms. Name the compound. Explain how the results identify each ion, and why the solution is acidified before adding the silver nitrate.
Show the mark scheme
- Yellow flame = sodium ions (Na+). 1 mark
- Cream precipitate with silver nitrate = bromide ions (Br−) — silver bromide. 1 mark
- The compound is sodium bromide. 1 mark
- The nitric acid removes carbonate ions, which would also form a precipitate with silver nitrate and give a false positive. 1 mark
The silver halides step white → cream → yellow for Cl → Br → I. And note it must be nitric acid here — hydrochloric acid would add chloride ions to the solution being tested.
Flame emission spectroscopy is an instrumental method for identifying metal ions. Give two advantages of instrumental methods compared with flame tests.
Show the mark scheme
- Any two of: more sensitive (detects tiny amounts) / more accurate / faster / works on mixtures, where one flame colour would mask another / gives the concentration as well as the identity. 1 mark each
The mixtures point is the strongest pairing with this practical — a flame test can’t separate overlapping colours (sodium’s yellow especially), but each element’s line spectrum shows up independently.
8Water Analysis & Purification
Aim: analyse water samples for pH and dissolved solids, then purify a sample by distillation.
📖 Full theory: C10 — Potable Water
Part A — analysis.
- Measure the pH of each sample with universal indicator (or a pH probe).
- Find the dissolved solids: weigh an evaporating basin, add a known volume (e.g. 25 cm3) of the sample, evaporate the water, then re-weigh — the mass gained is the dissolved solid. Don’t overheat, or you decompose the solids.
Part B — purification by distillation.
- Heat the sample so the water boils off, then condense the vapour and collect it — the distillate is pure water; the dissolved solids stay behind in the flask.
- Test the distillate: pure water boils at exactly 100 °C and leaves no residue on evaporation.
Simple distillation: boil the salt water, condense the steam in a cooled tube, and collect pure water. The salt is left behind in the flask.
The water sample tested (fresh, sea or distilled) — or, in Part B, before vs after distillation.
The pH, the mass of dissolved solids per known volume, and whether the distillate is pure (boils at 100 °C, no residue).
The same volume of each sample, and the same apparatus and method each time (same balance, same basin, no overheating).
AQA’s student sheet tests the salt water chemically, before and after distilling — a neat link to the ion tests of RP7: a flame test (yellow = sodium ions) and dilute nitric acid + silver nitrate (white precipitate = chloride ions). The salt water gives both positives; the distillate — collected through a delivery tube into a test tube standing in ice water to condense the steam — gives neither. No yellow flame, no precipitate: the ions have been left behind, so the water is purified. A likely exam angle: “How could the student show the distilled water no longer contains chloride ions?”
- “Potable means pure.” No — potable water still contains dissolved substances; it’s just safe to drink. Pure water contains only H2O.
- Confusing the two treatment steps. Filtration removes insoluble solids; sterilisation (chlorine, ozone or UV) kills microbes. Filtering does not kill bacteria.
- Forgetting why desalination is a last resort. Distillation and reverse osmosis both need large amounts of energy, so they’re expensive — used only where fresh water is scarce.
🧪 Exam-style questions
Which result shows that a sample of water is pure? Tick (✓) one box.
In the treatment of fresh water for drinking, what is the purpose of sterilisation? Tick (✓) one box.
Sea water can be made into drinking water by distillation. Give one reason why distillation works for this, and one reason why it is only used where fresh water is scarce.
Show the mark scheme
- Only the water boils off and is condensed — the dissolved salt is left behind in the flask. 1 mark
- Boiling the water needs large amounts of energy, so it is expensive. 1 mark
The same applies to reverse osmosis — both desalination methods are energy-intensive, which is why they’re a last resort.
A student distils salt water in the lab. Describe how the student should collect the distilled water, and how they could show that the distilled water no longer contains chloride ions.
Show the mark scheme
- Heat the salt water (in a flask) so the water boils off as steam. 1 mark
- Condense the steam — e.g. lead it through a delivery tube into a test tube standing in ice water — and collect the distillate. 1 mark
- Add dilute nitric acid then silver nitrate solution to the distillate. 1 mark
- No white precipitate forms — showing the chloride ions have been left behind in the flask. 1 mark
The salt water itself gives a white precipitate with the same test — testing before and after is what proves the distillation worked. This is the RP7 halide test doing its job inside RP8.
9The Four Gas Tests
Not a required practical — but these four tests are needed inside RP3 and RP5, and they earn marks across both papers. Learn each as a method-and-result pair.
📖 Full theory: C8 — Tests for Common Gases
| Gas | Test | Positive result |
|---|---|---|
| Hydrogen, H2 | Hold a lighted splint at the open end of the tube | Burns rapidly with a squeaky “pop” |
| Oxygen, O2 | Place a glowing splint inside the tube | The splint relights |
| Carbon dioxide, CO2 | Bubble the gas through limewater | Limewater turns milky / cloudy |
| Chlorine, Cl2 | Hold damp blue litmus paper in the gas | Litmus is bleached white |
- “A splint” isn’t enough for oxygen. It must be a glowing splint — one that has just gone out, so only the ember still glows — and the oxygen relights the flame. “Splint” or “lit splint” earns nothing here: a lit splint is the hydrogen test.
- “CO2 puts out a lit splint” is not the test — other gases do this too. It’s a property, not an identification.
- Memory hooks: liGHted → Hydrogen; glOwing → Oxygen.
- State results precisely: “relights” not “glows brighter”; “milky/cloudy” not “white”; “bleached white” not “clear”.
10Separation Techniques
From C1 — these physical separation methods aren’t a required practical on their own, but they appear constantly: inside RP1 (filtering and crystallising), RP8 (distillation), and as standalone “describe how you would separate…” questions.
📖 Full theory: C1 — Atomic Structure & the Periodic Table
Filtration
Separates an insoluble solid from a liquid. The solid is trapped in the filter paper as the residue; the liquid passes through as the filtrate — using both terms correctly is often a mark in itself.
Crystallisation
Separates a dissolved solid from its solution. Evaporate some of the water, then cool slowly so crystals form — never evaporate to dryness, which spoils the crystals.
Simple distillation
Separates the solvent from a solution — e.g. pure water from salt water. The vapour is cooled in a condenser and collected; the dissolved solid stays behind in the flask.
Fractional distillation
Separates miscible liquids with different boiling points, using a fractionating column to give the clean split. The crude-oil version — with its own diagram — is in C7 — Organic Chemistry.
11Reactivity & Cells Experiments
Three experiments from C4 and C5 that exams treat almost like required practicals — each one is a way of putting metals in order of reactivity from experimental data.
📖 Full theory: C4 — The Reactivity Series · C5 — Energy Changes
Metals with dilute acid
Place equal-sized pieces of different metals in separate test tubes of the same dilute acid and compare the vigour of fizzing (hydrogen gas): magnesium fizzes rapidly, zinc steadily, iron slowly — and copper not at all, because it sits below hydrogen in the reactivity series. The rate of fizzing puts the metals in order: Mg > Zn > Fe >> Cu.
Displacement with temperature rise
Add the same mass and surface area of different metals to the same volume and concentration of copper sulfate solution, at the same starting temperature, and record the temperature rise (this doubles as a version of RP4). A more reactive metal gives a larger temperature rise — and the classic observations for zinc + copper sulfate are the blue solution fading and red-brown copper appearing.
Zn + CuSO4 → ZnSO4 + Cu
Simple cells T
Connect two different metals in an electrolyte (e.g. potassium nitrate solution) with a voltmeter between them: the bigger the difference in reactivity, the bigger the voltage. Two electrodes of the same metal give 0 V. Exams use cell voltages the same way as the experiments above — data to place metals in reactivity order.
A real exam question typically gives a table of voltages for pairs of metals (metal 1 down the side, metal 2 across the top) and defines a sign convention in the stem — for example: “if metal 2 is more reactive than metal 1, the voltage is positive; if metal 1 is more reactive, it is negative.” Read that convention from the question every time rather than memorising one — the examiner sets it, and it can differ between papers. Then three reads unlock the whole table:
- Same metal → 0 V — no reactivity difference, no voltage. These entries run down the table’s diagonal.
- The size of the voltage ranks the gap in reactivity: the pair furthest apart in the series gives the largest reading.
- The sign tells you which of the two metals is the more reactive one — apply the stem’s convention.
Here’s the format with example results — metal 1 down the side, metal 2 across the top, and only half the grid measured (swapping the metals just flips the sign, so the other half is redundant):
| Metal 1 ↓ Metal 2 → | Copper | Iron | Magnesium | Zinc |
|---|---|---|---|---|
| Copper | 0.0 V | |||
| Iron | −0.7 V | 0.0 V | ||
| Magnesium | −2.7 V | −2.0 V | 0.0 V | |
| Zinc | −1.1 V | not measured | +1.6 V | 0.0 V |
Reading it with the convention above: every copper-column voltage is negative, so copper is less reactive than everything — and magnesium’s row has the largest readings, so magnesium is furthest from the others. The full order falls out: magnesium > zinc > iron > copper.
And the classic follow-up — predict the voltage for the cell that wasn’t measured (zinc/iron): from copper’s column, zinc beats copper by 1.1 V and iron beats copper by 0.7 V, so the zinc–iron gap is about 0.4 V. Metal 2 (iron) is less reactive than metal 1 (zinc), so the convention makes it −0.4 V. Both the size and the sign earn marks. Questions often include unfamiliar metals (chromium, tin) precisely so you can’t rely on a memorised series — the data is the series.
Electrolysis of molten compounds
The molten version of RP3, with simpler rules (no water to provide competing ions): the metal forms at the cathode and the non-metal at the anode. For molten lead bromide: lead at the cathode, brown bromine vapour at the anode.
12Cracking & the Test for Alkenes
From C7 — cracking itself is usually examined as conditions and equations, but the bromine water test is a hands-on test exams expect you to describe precisely.
📖 Full theory: C7 — Cracking & the Test for Alkenes
- Catalytic cracking — heat the fraction to vaporise it, then pass the vapours over a hot catalyst.
- Steam cracking — mix the vaporised fraction with steam at a very high temperature.
- Asked for two conditions? High temperature + catalyst, or high temperature + steam, are the safe pairings. And why must air be kept out of the reactor? The hot hydrocarbon vapour would combust.
- The test for alkenes: shake the sample with bromine water. An alkene turns it from orange to colourless; an alkane leaves it orange. (Colourless — not “clear”.)
Bromine water: orange → colourless with an alkene; stays orange with an alkane.
13Practical Language That Earns Marks
Every practical question is marked in the same vocabulary. These definitions — and the safety “matched pair” — apply to all eight required practicals above.
- Repeatable — the same person, using the same method and equipment, gets similar results on repeating.
- Reproducible — different people, often with different equipment, get similar results. (The tougher test — it’s how science checks a claim.)
- Accurate — a measurement close to the true value.
- Precise — repeated measurements close together (a small spread). Not the same as accurate — tight readings can still all be off the true value.
- Valid — the experiment measures what it’s meant to, because the other variables were controlled (a fair test).
- Resolution — the smallest change an instrument can detect (a balance reading to 0.01 g has a higher resolution than one reading to 1 g).
- Anomalous result — a result that doesn’t fit the pattern. Look for a cause, and if it came from a poor measurement, leave it out of the mean.
- Interval — the gap between values of the independent variable (e.g. testing concentrations 10 g/dm3 apart). Choose a sensible range and intervals — too narrow a range and the pattern may not show.
- Categoric vs continuous — a categoric variable is a label (the type of metal); a continuous one is a measured number (temperature, volume). This choice decides the graph: categoric → bar chart, continuous → line graph.
- Calibration / zero error — checking an instrument against a known value. A balance reading non-zero with nothing on it is a zero error: tare it (or subtract the offset) before weighing.
Practical questions regularly award marks for how you record and present the data — easy marks if you know the checklist:
- Tables — the independent variable in the first column, headings with units in the heading (so the cells are just numbers), and a column for repeats and the mean.
- Choosing the graph — categoric independent variable (type of metal) → bar chart; continuous (concentration, temperature, time) → line graph.
- Drawing it — sensible scales that use more than half the grid, both axes labelled with quantity and unit (independent variable on the x-axis, dependent on the y), points plotted accurately (±half a small square), and a line of best fit — a smooth line or curve through the trend, not dot-to-dot, ignoring anomalous points, drawn once, cleanly (feathered, doubled or sketchy lines lose the mark).
- Trial runs — a quick rough go (like RP2’s rough titration) finds sensible values and the right range before you collect the real data.
- Significant figures — give calculated values (means, rates, Rf) to a sensible number of significant figures, matching the data: a mean of times measured to the nearest second shouldn’t be quoted to five figures.
- Straight-line graphs — a straight line means a linear relationship: this is the y = mx + c idea from maths, where the gradient (m) is how fast one variable changes with the other and the intercept (c) is the starting value. In chemistry you’re asked for the gradient and intercept in context, not the equation — and if the line passes through the origin, the two variables are directly proportional.
- Never write “amount”. Say exactly what you are changing, measuring or keeping the same — the mass, volume, concentration or number of moles. “The same volume of acid”, not “the same amount of acid”.
- Safety is a matched pair. Name the hazard, give the precaution that targets it, and say what harm it prevents.
- Observation, then inference. Write down what you would see (“the cross is no longer visible”, “a white precipitate forms”), and keep it separate from what it means.
- Right + wrong = wrong. A correct point contradicted elsewhere in the same answer is cancelled — mark schemes do not let a wrong extra ride along with a right one. Don’t scatter alternatives to “cover all eventualities”; commit to the correct one.
- Six-markers: task → sequence → judgement. Work out exactly what the question needs you to produce, put the steps in an order that would actually work, and — if the command implies a decision — state the judgement. Length adds nothing; structure is what moves an answer up a level.