You might recognise a buffer but use the wrong amounts in the calculation. You might understand titrations but forget that the flask holds ten times the acid in the portion you tested. Or you might spot a functional group in a spectrum without checking whether your structure fits the rest of the evidence.
The hardest International A Level Chemistry questions rarely test one fact. They test whether you can connect several steps correctly. This guide shows you how: recognise the demanding question types, find a starting point, work through to the answer and check that it makes chemical sense.
About these examples
all seven questions are original practice questions written for this guide. They are not past-paper questions, and they are not a ranking of the hardest questions ever set. Every answer has been checked step by step.
The essentials
- Buffers: do the reaction first, then the equilibrium calculation.
- Electrode potentials: balance electrons by multiplying half-equations, never E° values.
- Back titrations: track "acid added", "acid left" and "acid used" separately, and scale portions up to the whole flask.
- Rates: find each order from a comparison where only one concentration changes.
- Spectroscopy: a structure must explain every clue, not just the first one you recognise.
- Practical errors: follow the error all the way through to the final answer.
- Thermodynamics and equilibrium: keep feasibility, equilibrium position and rate as three separate questions.
First, check which qualification you are taking
"International A Level Chemistry" can mean two different qualifications. Pearson calls its version International Advanced Level (IAL) Chemistry; Cambridge calls its version Cambridge International AS & A Level Chemistry.
| Qualification | How it is assessed | What this means for revision |
|---|---|---|
| Pearson Edexcel IAL Chemistry (2018 specification, units WCH11–WCH16) | Six written units, including practical-skills Units 3 and 6 | Practise explaining experimental methods and interpreting data in writing |
| Cambridge International Chemistry 9701 (2025–2027 and 2028–2030 syllabuses) | Papers 1–3 for AS, then Papers 4 and 5 for the full A Level. Paper 3 is a practical exam; Paper 5 tests planning, analysis and evaluation | Prepare for hands-on practical work as well as written practical reasoning |
Use the specification and past papers that match your qualification. Pearson has also redeveloped IAL Chemistry for first teaching in September 2027, so check which version you are sitting — a resource's publication year does not tell you.
You can practise with real papers on Starley: Edexcel International A Level Chemistry and Cambridge AS & A Level Chemistry.
What makes a Chemistry question hard?
A question becomes demanding when you have to decide what to do, instead of being told each step.
| Question type | Why students get stuck | Useful first move |
|---|---|---|
| Multi-step calculation | The answer is several conversions away from the data | Write a route from the measured quantity to the one asked for |
| Acid–base calculation | Mixing changes the amounts before equilibrium matters | Work out the reaction and the moles left first |
| Redox prediction | Half-equations, signs and electrons get mixed up | Decide what is reduced and what is oxidised |
| Rate data | More than one concentration changes between experiments | Find a comparison that changes only one variable |
| Structure determination | One convincing clue makes you ignore the others | Make an evidence table before choosing a structure |
| Practical evaluation | An error is named but its effect is not explained | Follow the error through the calculation |
| Extended explanation | Facts are listed without being connected | Link each piece of evidence to a conclusion |
Note
the boards label their assessment objectives differently. In Cambridge 9701, AO2 is "handling, applying and evaluating information" and AO3 is "experimental skills and investigations" — so not every evaluation question is AO3. Check your own specification's definitions rather than borrowing another board's labels.
1. Buffer calculations: react first, calculate pH second
The question
A buffer contains 0.0300 mol of ethanoic acid and 0.0200 mol of sodium ethanoate. A student adds 0.00300 mol of hydrochloric acid. Assume the added volume is negligible. For ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the new pH and explain why it changes only slightly.
Why students get stuck
There are two stages:
- The added strong acid reacts with ethanoate ions.
- The new mixture then sets up its acid–base equilibrium.
Using the original amounts of acid and ethanoate skips the first stage.
Step 1: work out what reacts
Ethanoate ions remove the added hydrogen ions in a 1 : 1 ratio:
Equation
CH₃COO⁻ + H⁺ → CH₃COOH
| Substance | Start / mol | Change / mol | After reaction / mol |
|---|---|---|---|
| CH₃COO⁻ | 0.0200 | −0.00300 | 0.0170 |
| CH₃COOH | 0.0300 | +0.00300 | 0.0330 |
Both buffer components are still present, so a buffer calculation is valid.
Step 2: use the equilibrium expression
Worked example
Ka = [H⁺][CH₃COO⁻] ÷ [CH₃COOH]
so [H⁺] = Ka × [CH₃COOH] ÷ [CH₃COO⁻]
Both substances are in the same volume, so the concentration ratio equals the mole ratio:
[H⁺] = 1.74 × 10⁻⁵ × (0.0330 ÷ 0.0170) = 3.38 × 10⁻⁵ mol dm⁻³
pH = −log₁₀(3.38 × 10⁻⁵) = 4.47
Step 3: explain the small change
Before the acid was added, pH = −log₁₀(1.74 × 10⁻⁵ × 0.0300 ÷ 0.0200) = 4.58. So the pH falls only from 4.58 to 4.47.
Model answer
Ethanoate ions react with most of the added H⁺ ions to form ethanoic acid. The hydrogen ion concentration therefore rises much less than it would without the buffer, so the pH falls only slightly.
Check your answer
- Adding a strong acid should lower the pH.
- Ethanoate should go down and ethanoic acid should go up.
- A buffer resists pH change; it does not prevent all change.
Remember
complete the reaction first, then use Ka. If one buffer component is used up completely, a buffer calculation no longer applies.
Your turn
Instead of the acid, the student adds 0.00200 mol of sodium hydroxide to the original buffer. Calculate the new pH.
AnswerShow answerHide answer
OH⁻ reacts with ethanoic acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Ethanoic acid: 0.0300 − 0.00200 = 0.0280 mol. Ethanoate: 0.0200 + 0.00200 = 0.0220 mol
[H⁺] = 1.74 × 10⁻⁵ × (0.0280 ÷ 0.0220) = 2.21 × 10⁻⁵ mol dm⁻³
pH = 4.65 — slightly higher than 4.58, as expected after adding a base.
2. Electrode potentials: predict a reaction without changing the rules
The question
Use these standard reduction potentials to predict whether Fe³⁺ ions can oxidise iodide ions and bromide ions under standard conditions. Write an overall equation for any feasible reaction, and explain why a positive E°cell does not mean a reaction will be fast.
| Reduction half-equation | E° / V |
|---|---|
| Fe³⁺ + e⁻ ⇌ Fe²⁺ | +0.77 |
| I₂ + 2e⁻ ⇌ 2I⁻ | +0.54 |
| Br₂ + 2e⁻ ⇌ 2Br⁻ | +1.07 |
Step 1: decide the roles
If Fe³⁺ oxidises a halide ion, Fe³⁺ itself is reduced (Fe³⁺ + e⁻ → Fe²⁺), and the halide ion is oxidised to the halogen.
Step 2: use one consistent calculation
Using the tabulated reduction potentials as written:
Rule
E°cell = E°(reduction half-cell) − E°(oxidation half-cell)
- Iodide: E°cell = 0.77 − 0.54 = +0.23 V, so the reaction is feasible under standard conditions.
- Bromide: E°cell = 0.77 − 1.07 = −0.30 V, so the reaction is not feasible under standard conditions.
Step 3: balance the electrons
Worked example
2Fe³⁺ + 2e⁻ → 2Fe²⁺
2I⁻ → I₂ + 2e⁻
Adding them: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Check the charge: +6 − 2 = +4 on the left; +4 on the right. Balanced.
Step 4: separate feasibility from rate
A positive E°cell shows the reaction is thermodynamically favourable under standard conditions. It tells you nothing about the activation energy, so it cannot tell you how fast the reaction is.
Mistakes to avoid
- Multiplying E° by two. Multiply the half-equations to balance electrons, never the electrode potentials.
- Changing signs twice. If you subtract, use both values exactly as tabulated reduction potentials.
- Ignoring conditions. Concentration and other non-standard conditions can change the prediction.
Remember
decide what is reduced and oxidised, calculate E°cell, balance electrons, then state your conclusion with its conditions.
3. Back titrations: acid added, acid left and acid used
The question
A 1.25 g sample contains calcium carbonate and impurities that do not react with hydrochloric acid. The sample reacts completely with 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid. The remaining solution is transferred to a volumetric flask and made up to 250.0 cm³. A 25.0 cm³ portion needs 18.00 cm³ of 0.0400 mol dm⁻³ sodium hydroxide for neutralisation. Calculate the percentage by mass of calcium carbonate in the sample. (M of CaCO₃ = 100.0 g mol⁻¹)
Why students get stuck
The titration measures the acid left over, not the calcium carbonate. And the portion you titrate is only one tenth of the solution in the flask.
Step 1: acid originally added
n(HCl) = c × V = 0.500 × (50.0 ÷ 1000) = 0.0250 mol
Step 2: acid in the titrated portion
HCl + NaOH → NaCl + H₂O is a 1 : 1 reaction.
n(NaOH) = 0.0400 × (18.00 ÷ 1000) = 0.000720 mol, so the 25.0 cm³ portion contains 0.000720 mol of leftover HCl.
Step 3: scale up to the whole flask
Leftover HCl in the flask = 0.000720 × (250.0 ÷ 25.0) = 0.00720 mol
Step 4: acid that reacted with the sample
HCl used = 0.0250 − 0.00720 = 0.0178 mol
Step 5: apply the reaction ratio
Worked example
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
n(CaCO₃) = 0.0178 ÷ 2 = 0.00890 mol
m(CaCO₃) = 0.00890 × 100.0 = 0.890 g
Percentage by mass = (0.890 ÷ 1.25) × 100 = 71.2%
Check your answer
The mass of calcium carbonate is less than the sample's mass, and the percentage is between 0% and 100%. An answer above 100% usually means a missing flask factor, a wrong reaction ratio or a subtraction error.
Remember
label every amount — "added", "left in portion", "left in flask" and "reacted". Labels prevent more mistakes than unlabelled arithmetic.
Your turn
Same experiment, but the 25.0 cm³ portion needs 15.00 cm³ of the sodium hydroxide. What is the percentage of calcium carbonate?
AnswerShow answerHide answer
HCl left in portion = 0.0400 × 0.01500 = 0.000600 mol
HCl left in flask = 0.000600 × 10 = 0.00600 mol
HCl used = 0.0250 − 0.00600 = 0.0190 mol
n(CaCO₃) = 0.0190 ÷ 2 = 0.00950 mol, so mass = 0.950 g
Percentage = (0.950 ÷ 1.25) × 100 = 76.0%. A smaller titre means less acid was left, so more reacted with the sample.
4. Rates: isolate each concentration before writing the rate equation
The question
A reaction between A and B gives these initial-rate results at constant temperature. Find the rate equation and calculate the rate constant, with units.
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 2.00 × 10⁻⁴ |
| 2 | 0.200 | 0.100 | 8.00 × 10⁻⁴ |
| 3 | 0.200 | 0.300 | 2.40 × 10⁻³ |
Step 1: order with respect to A
Compare experiments 1 and 2: [B] stays the same, [A] doubles and the rate increases four times. If the order in A is m, doubling [A] multiplies the rate by 2 to the power m. That factor is 4, so m = 2: the reaction is second order in A.
Step 2: order with respect to B
Compare experiments 2 and 3: [A] stays the same, [B] triples and the rate triples. The reaction is first order in B.
So the rate equation is rate = k[A]²[B].
Step 3: calculate k and its units
Worked example
Using experiment 1:
k = 2.00 × 10⁻⁴ ÷ ((0.100)² × 0.100) = 0.200
Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹
k = 0.200 dm⁶ mol⁻² s⁻¹
Experiments 2 and 3 give the same value — a useful check.
What if both concentrations change?
Do not put the whole rate change down to one reactant. Once you know one order, work out how much that reactant changes the rate, then divide the total rate factor by it to isolate the other reactant's effect.
Remember
reaction orders come from experimental evidence. Never take them from the coefficients in the overall equation unless the question gives you a valid reason.
Your turn
Predict the initial rate for experiment 4, where [A] = 0.300 mol dm⁻³ and [B] = 0.200 mol dm⁻³.
AnswerShow answerHide answer
rate = k[A]²[B] = 0.200 × (0.300)² × 0.200 = 3.60 × 10⁻³ mol dm⁻³ s⁻¹
5. Spectroscopy: choose a structure that explains every clue
The question
Compound X has the molecular formula C₄H₈O₂. Its infrared spectrum has a strong absorption near 1740 cm⁻¹ and no broad O–H absorption. Its proton NMR spectrum has three signals, shown below. Deduce the structure of X and explain your reasoning.
| Chemical shift / ppm | Relative integration | Splitting |
|---|---|---|
| 1.1 | 3 | Triplet |
| 2.3 | 2 | Quartet |
| 3.7 | 3 | Singlet |
Step 1: what the IR evidence tells you
The absorption near 1740 cm⁻¹ shows a C=O group. The lack of a broad O–H absorption rules out an alcohol or carboxylic acid here. An ester is a sensible candidate — but IR alone cannot give you the full structure.
Step 2: build fragments from the NMR data
- The 3H triplet and 2H quartet together suggest an ethyl group, CH₃CH₂–. The CH₃ protons have two neighbouring H atoms (a triplet); the CH₂ protons have three (a quartet).
- The CH₂ signal near 2.3 ppm fits a CH₂ next to C=O.
- The 3H singlet near 3.7 ppm suggests an O–CH₃ group, with no neighbouring H atoms on carbon to cause splitting.
Step 3: assemble the structure
The structure that fits is CH₃CH₂COOCH₃, methyl propanoate.
Step 4: check every clue
| Evidence | Explained by methyl propanoate |
|---|---|
| C₄H₈O₂ | Four C, eight H and two O atoms |
| Absorption near 1740 cm⁻¹ | Ester C=O group |
| No broad O–H absorption | No O–H bond |
| 3H triplet at 1.1 ppm | CH₃ next to CH₂ |
| 2H quartet at 2.3 ppm | CH₂ next to CH₃ and C=O |
| 3H singlet at 3.7 ppm | O–CH₃ group |
| Three signals | Three different proton environments |
Why guessing an isomer is not enough
Watch out
ethyl ethanoate, CH₃COOCH₂CH₃, has the same formula and also gives a triplet, a quartet and a singlet. But its O–CH₂ quartet would appear around 4 ppm, and its CH₃ next to C=O would give a singlet around 2 ppm. Those shifts do not match the data, so X is not ethyl ethanoate.
Remember
use integration, splitting and chemical shift together. A familiar splitting pattern is a starting point, not an identification.
6. Practical errors: follow the effect through the calculation
The question
A student uses a pipette to transfer 25.0 cm³ of hydrochloric acid of unknown concentration into a conical flask, then titrates it with standard sodium hydroxide. The pipette was rinsed with distilled water but not with the acid, so some water remained inside. Predict the effect on the calculated concentration of the acid. Would extra distilled water already in the conical flask have the same effect?
Step 1: find the first affected quantity
Water left in the pipette dilutes the acid drawn into it. The 25.0 cm³ delivered therefore contains fewer moles of HCl than 25.0 cm³ of the original solution.
Step 2: follow the effect
Fewer moles of HCl need fewer moles of NaOH, so the titre is smaller. The student then calculates c(HCl) = n(HCl) ÷ 0.0250, using a value of n(HCl) that is too low for the volume assumed. The calculated concentration is too low.
Step 3: compare with water in the conical flask
Extra water in the conical flask dilutes the acid after its amount has been measured out, so the number of moles of HCl does not change. The same amount of NaOH is needed, and the calculated concentration is unaffected.
The key distinction is when the dilution happens: before measuring the portion of acid, or after.
Extension: titre uncertainty
Worked example
Each burette reading has an uncertainty of ±0.05 cm³, and the titre is 20.00 cm³.
A titre is the difference between two readings, so both uncertainties count:
Titre uncertainty = 0.05 + 0.05 = ±0.10 cm³
Percentage uncertainty = (0.10 ÷ 20.00) × 100 = 0.50%
Repeating titrations helps you spot and reduce random variation. It does not correct a systematic error like a diluted sample, because every repeat is affected the same way.
Remember
name the measurement affected, say whether it is too high or too low, and trace the effect to the final result.
7. Synoptic explanations: keep feasibility, equilibrium and rate separate
The question
For N₂O₄(g) ⇌ 2NO₂(g), the forward reaction has ΔH° = +57.2 kJ mol⁻¹ and ΔS° = +175.8 J K⁻¹ mol⁻¹. Assume these values stay constant. Calculate the temperature above which the forward reaction has a negative ΔG°. Then explain the effects of increasing the temperature, increasing the pressure at constant temperature, and adding a catalyst.
Step 1: make the units consistent and calculate
Worked example
Convert entropy to kJ: ΔS° = 0.1758 kJ K⁻¹ mol⁻¹
ΔG° = ΔH° − TΔS°. At the boundary, ΔG° = 0, so:
T = ΔH° ÷ ΔS° = 57.2 ÷ 0.1758 = 325 K (about 52 °C)
Above 325 K, ΔG° is negative under these assumptions.
Step 2: increasing temperature
The forward reaction is endothermic, so increasing the temperature shifts the equilibrium towards NO₂ and increases the equilibrium constant. Temperature also speeds up both reactions, but that kinetic effect is separate from its effect on the equilibrium composition.
Step 3: increasing pressure
At constant temperature, increasing the pressure (by reducing the volume) favours the side with fewer moles of gas. There is 1 mole of gas on the left and 2 on the right, so the equilibrium shifts towards N₂O₄. The equilibrium constant does not change, because the temperature has not changed.
Step 4: adding a catalyst
A catalyst provides an alternative route with a lower activation energy. It speeds up how quickly equilibrium is reached, but does not change the equilibrium constant or the equilibrium composition at a fixed temperature.
Step 5: do not overstate the ΔG° result
A negative ΔG° does not mean all the N₂O₄ turns into NO₂, or that the reaction is fast. For the reaction as written, it corresponds to an equilibrium constant greater than 1. Which way a particular mixture actually changes also depends on its composition.
Remember
answer three separate questions.
- Is the direction thermodynamically favourable?
- What mixture exists at equilibrium?
- How quickly does the system change?
Your turn
Calculate ΔG° at 298 K. What does the sign tell you about the equilibrium constant at that temperature?
AnswerShow answerHide answer
ΔG° = 57.2 − (298 × 0.1758) = 57.2 − 52.4 = +4.8 kJ mol⁻¹
ΔG° is positive, so at 298 K the equilibrium constant is less than 1 — but some NO₂ is still present at equilibrium. 298 K is below the 325 K boundary, consistent with Step 1.
How to tackle a question when you cannot see the solution
1. Name the target
Write a short target next to the question: "percentage purity", "rate equation", "structure", "effect on calculated concentration". This stops you doing useful calculations and then stopping before the actual answer.
2. Find one secure starting point
You do not need to see the whole route at once. Good starting points include:
- a balanced equation;
- an equilibrium expression;
- moles from n = c × V;
- one clearly identified spectral feature;
- a comparison between two experiments.
A valid first step often shows you the next one.
3. Label your intermediate answers
"0.00720 mol HCl left in the whole flask" is much safer than "0.00720". Labels show you which numbers can be compared, subtracted or linked by a reaction ratio.
4. Explain links, not just facts
Rule
relevant feature → chemical effect → answer to the question
Model answer
Water left in the pipette dilutes the acid, so the portion delivered contains fewer moles of HCl. This gives a smaller titre, so the original concentration is underestimated.
Technical vocabulary helps when it makes your reasoning precise. Adding keywords without connecting them does not strengthen an answer.
5. Plan your time for the actual paper
Do not apply "one minute per mark" to every exam. Work out a starting budget instead:
Rule
minutes per mark = (paper length − planned checking time) ÷ total marks
Then adjust for the question types on that paper. If you are stuck, write down a valid first step, move on, and come back later.
6. Check the chemistry before the calculator digits
- Are atoms and charges balanced?
- Have you converted cm³ to dm³ where needed?
- Did you use kelvin and consistent energy units?
- Is this the amount in a portion or in the whole solution?
- Does the sign or direction make chemical sense?
- Have you answered the command word?
How to practise so unfamiliar questions become manageable
Use a cycle that tests whether you can reproduce the method on your own:
- Attempt the question without help. Write down your reasoning, even if it is incomplete.
- Find your first wrong decision. Was it the chemistry, the setup, a ratio, the arithmetic or the explanation?
- Study the correction and explain why that step is valid.
- Close the solution and redo the question.
- A few days later, try a different question that uses the same skill.
Keep a short error log:
| Error | Why it happened | Rule for next time |
|---|---|---|
| Used the original buffer amounts | Skipped the reaction with the added acid | React first, then calculate equilibrium |
| Forgot the flask factor | Treated a portion as the whole sample | Write both volumes next to the mole calculation |
| Chose the wrong ester | Used splitting but ignored chemical shifts | Check every signal against the structure |
| Said pressure changes K | Confused composition with the constant | Check whether the temperature changed |
Use topic practice to fix a weakness, then mixed questions to check you can spot the method without a topic heading. Starley's past papers are a good source of mixed practice.
When you mark official papers, use the matching mark scheme and examiner report. Read the guidance on alternative methods, consequential errors and significant figures — and do not assume every early mistake earns follow-through credit.
Frequently asked questions
What are the hardest International A Level Chemistry topics?
There is no single hardest topic for everyone. Good areas to test yourself on are acid–base equilibria, multi-step mole calculations, electrochemistry, rate equations, organic structure determination and experimental evaluation. Prioritise the skills that keep costing you marks in practice.
Can a short question be hard?
Yes. A one- or two-mark question can depend on an unfamiliar inference. The number of marks tells you how much credit is available, not how difficult the question is.
Can I get marks if my final answer is wrong?
Often, yes. Mark schemes may give credit for a correct equation, method or substitution, or for later work that correctly uses an earlier wrong answer. Show enough working for your method to be visible — but do not assume every error will be carried forward.
Do I have to copy the mark scheme's wording?
Aim for scientifically accurate meaning. Some definitions need precise wording, but copying isolated keywords will not rescue an incorrect explanation.
Should I only practise hard questions if I want an A*?
No. Secure the routine marks and full topic coverage alongside the demanding questions. Dropping easy marks through missing units, weak recall or poor timing can undo strong work elsewhere. Our guide to making an A Level revision timetable can help you balance both.
How do I know I have mastered a method?
Try a fresh question with different numbers, substances or presentation. If you can recognise the method and explain why each step works without looking at a solution, your understanding will transfer to the exam.