Hardest AQA A level Chemistry Questions of All Time
Every year exam boards intentionally put some questions that separate A scoring students from . Our team at brightmindtutors have combed through years of examiner reports to find the hardest A level chemistry questions which were lowest scoring and conceptually complex ever set. Our tutors have also expanded the solutions using simple steps for you.
Ques1) A molecule Q is ionised by electron impact in a TOF mass spectrometer. The Q+ ion has a kinetic energy of 2.09 10 -15 J .This ion takes 1.23 10–5 s to reach the detector. The length of the flight tube is 1.50 m .Calculate the relative molecular mass of Q.
K.E = 0.5mv2 where m = mass (kg) and v = speed (m/s)
The Avogadro constant, L = 6.022 x 1023 mol–1 [5 marks]
Solution –
Tutor Tip – You are given with time and distance and kinetic energy ,if we find velocity we can find mass and put in K.E equation to find mass and from there we will find Mr
Step1) Recall equation
= 1.50 m 1.23 10–5s
= 1.22 105 m/s (1mark)
Step2) Use K.E equation and rearrange to find mass
v2 = 1.49 x 1010 m/s
Putting these value in above formula
m = 2.8 x 10-25 kg (1mark)
Step3) Finding mass for 1 mole of Q ion
m x L = 2.8 x 10-25 kg x 6.022 x 1023 mol–1
= 0.169Kg/mol (1mark)
Step4)Convert it into grams and get required Mr
= 0.169 x 1000 = 169 g/mol (2mark)
⭐Exam Tip (AQA A Level Chemistry)
Always convert the ion's mass into relative molecular mass (Mr) as the final step
Remember that ions are accelerated by an electric field, so the electrical potential energy gained is equal to the kinetic energy gained. This relationship is the key to solving mass spectrometry calculation questions involving ion velocity, mass, and Mr
Ques2) A student prepared a buffer solution by adding 0.0136 mol of a salt KX to 100 cm3 of a 0.500 mol dm–3 solution of a weak acid HX and mixing thoroughly. The student then added 3.00 × 10–4mol of potassium hydroxide to the buffer solution. Calculate the pH of the buffer solution after adding the potassium hydroxide. For the weak acid HX at 25°C the value of the acid dissociation constant, Ka = 1.41 × 10–5mol dm–3.Give your answer to two decimal places. [6 marks]
Solution –
Tutor Tip – Try to think of reaction between an acid and alkali (neutralization). Then try to find moles of weak acid left after reaction with KOH and total new moles of salt formed. Lastly formulas of Ka and pH
Step1) Writing reaction between KOH & HX
KOH + HX → KX + H2O
Find the moles of KX formed and HX left after reaction
initial moles of KOH = 3.00 × 10–4 mol (limiting reactant)
initial moles of HX = 0.1 X 0.5 = 0.05 mol
moles of KX formed = 3.00 × 10–4 mol
moles of HX left(unreacted) = 0.05 – 3.00 × 10–4 = 0.0497 mol
Total moles of KX in solution now = initially present + formed after reaction
= 0.0136 + 3.00 × 10–4 = 0.0139mol (4marks)
Step2) Use dissociation constant equation –
Rearrange to find [H+]
Put the values
= 5.04 x 10-5
pH = -log10 5.04 x 10-5 = 4.30 (2marks)
⭐ Top Mark Tip:
For AQA buffer questions, do not use the initial acid concentration directly. First account for the neutralisation reaction, then use the equilibrium concentrations of the weak acid and conjugate base in the Ka expression.
This mirrors the AQA marking approach: reaction → mole changes → Ka expression → pH calculation, which helps students secure all available marks
Ques3)This question is about compounds containing ethanedioate ions.
A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4.2H2O) and an inert solid.
A volumetric flask contained 1.90 g of this solid mixture in 250 cm3 of aqueous solution. Two different titrations were carried out using this solution. In the first titration 25.0 cm3 of the solution were added to an excess of sulfuric acid in a conical flask. The flask and contents were heated to 60o C and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour. The equation for this reaction is
2MnO4 − + 5 C2O4 2− + 16H+ → 2Mn2+ + 8H2O + 10CO2
In the second titration 25.0cm3 of the solution was titrated with a 0.100 mol dm−3 solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed colour after the addition of 10.45cm3 of sodium hydroxide solution. The equation for this reaction is
H2C2O4 + 2OH− → C2O4 2 −+ 2H2O
Calculate the percentage by mass of sodium ethanedioate in the white solid. Give your answer to the appropriate number of significant figures. Show your working. [8 marks]
Solution-
Tutor Tip – This question involves multiple titrations for different components of white solid. Write to figure out what component reacts in titration
Step1) TITRATION1
Finding moles of ethanedioate from the redox titration (2)
(Remember there are 2 sources of ethanedioate in white solid : sodium ethanedioate & ethanedioic acid )
As evident from equation 2mol of MnO4− requires 5 mol of C2O4 2−
mol MnO4− = concentration x volume
= 0.02 x 0.0265 (convert volume to dm−3 )
= 5.30 ×10-4
mol of C2O4 2− = 1.325 x 10-3 (5/2 x 5.30 ×10-4 )
Step2) TITRATION 2 (2)
Finding the moles of sodium hydroxide = concentration x volume
= 0.1 x 0.01045 (convert volume to dm−3 )
= 0.001045
As evident from balanced equation 2 mol of NaOH needs 1 mol of H2C2O4
Hence 0.001045 needs = 0.001045 2 = 5.225 x 10–4
Remember here moles of acid are just coming from H2C2O4 only)
Step3) Now follow this simple trick (1)
White solid = (Na2C2O4) + (H2C2O4.2H2O) + inert solid(does not react in any titration).
KMnO₄ = measures everything
⬇️
NaOH = measures only the acid
⬇️
Subtract acid from total
⬇️
Leftover = sodium ethanedioate
Hence moles of salt(Na2C2O4) = 1.325 x 10-3 – 5.225 x 10–4
= 8.025x 10–4
Step4) These moles are in 25cm3 of solution we have to find in 250cm3 (1)
= 8.025 x 10–4 x 10 = 8.025 x 10–3
Step5) Finding the mass of (1)
= moles x Mr
= 8.025 x 10–3 x 134 = 1.075g
Step6) So % sodium ethanedioate in original sample (1)
=
⭐ Top AQA Exam Strategy
Before writing a single calculation, ask yourself:
✔ What species is this titration measuring?
✔ Does it measure the whole mixture or just one component?
Ques4)This question is about hydrogen peroxide, H2O2 The half-equation for the oxidation of hydrogen peroxide is
H2O2 + O2 → 2H+ + 2e–
Hair bleach solution contains hydrogen peroxide. A sample of hair bleach solution is diluted with water. The concentration of hydrogen peroxide in the diluted solution is 5.00% of that in the original solution. A 25.0 cm3 sample of the diluted hair bleach solution is acidified with dilute sulfuric acid. This acidified sample is titrated with 0.0200 mol dm−3 potassium manganate(VII) solution. The reaction is complete when 35.85 cm3 of the potassium manganate(VII) solution are added.
a)Give an ionic equation for the reaction between potassium manganate(VII) and acidified hydrogen peroxide. Calculate the concentration, in mol dm−3, of hydrogen peroxide in the original hair bleach solution. [5 marks]
Solution –
Tutor Tip – You just need to write the correct overall redox equation and apply basic maths at the end of question to find concentration of hydrogen peroxide in the end.
Step1) Writing overall ionic equation – (1)
2MnO4 – + 6H+ + 5H2O2 → + 2Mn2+ + 8H2O + 5O2
( Remember H2O2 dissociates differently in acidic and basic medium both type of redox reactions are important for A levels try learning them anyways)
Step2) moles of KMnO4 = concentration x volume(dm−3) (2)
= 0.02 x 0.03585 = 7.17 × 10–4 (mol)
Compare moles of KMnO4 and H2O2 from balanced equation
= 2 mol of KMnO4 needs 5 mol of H2O2
7.17 × 10–4mol KMnO4 needs = 5/2 x 7.17 × 10–4 = 1.793 × 10–3 (mol)
Step3) Finding concentration of =
= 0.0717 (mol dm–3 ) (1)
Step4) Finding concentration of in original concentration (1)
5% of original concentration(H2O2) = 0.0717 (mol dm–3 )
Original concentration = 0.07170 0.05 = 1.43 (mol dm–3)
⭐ AQA rewards method marks.
In AQA redox titrations, the balanced ionic equation is your roadmap—get it right first, and the rest of the calculation becomes much easier.
Ques5)Calcium hydroxide is almost insoluble in water, but it reacts with dilute hydrochloric acid. Ca(OH)2 (s) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
A student adds 100 cm3 of 0.100 mol dm–3 hydrochloric acid to 0.600 g of solid calcium hydroxide.The final mixture contains a saturated solution of Ca(OH)2 at 293 K.
At 293 K
• the solubility of Ca(OH)2 in this solution is 0.400 g dm–3
• Kw = 6.80 × 10–15 mol2 dm–6
Calculate the pH of this solution. Give your answer to two decimal places. [5 marks]
Solution –
Tutor Tip – You need to find the moles of OH– from concentration of Ca(OH)2 and then H+ from Kw and apply formula for pH
Step1) concentration of Ca(OH)2 in mol dm–3 = concentration (g dm–3 ) Mr (1)
= 0.4 74 = 0.00540 (mol dm–3 )
Step 2) concentration of OH – = 2 x 0.00540 = 0.0108 mol dm–3 (1)
Step3) Kw = H+ x OH–
6.80 × 10–15 = H+ x 0.0108
H+ = 6.80 × 10–15 0.0108 = 6.30 x 10–13 (mol dm–3 ) (1)
Step4) pH = -log( H+) ;
= -log( 6.30 x 10–13) = 12.2 (2)
⭐Common A level Examiner Mistakes
Using the initial mass of Ca(OH)₂ instead of the given solubility.
Forgetting to double the hydroxide ion concentration because Ca(OH)₂ dissociates to give 2OH⁻.
Rearranging the Kw expression incorrectly.
Omitting units or rounding the final pH too early.
