Making 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.
Titration 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.
Electrolysis 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.)
Temperature 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 value2, 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.
Separation 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.
Reactivity & 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.