OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 22

11 marks · Medium difficulty · Structured Questions

Define a strong acid, write equations for reactions involving acids and copper oxide or ammonium carbonate, and perform stoichiometric calculations and titration analysis to identify an unknown Group 1 metal.

Practise this question

Question

Question text

22 This question is about reactions involving acids.

(a) Hydrochloric acid and nitric acid are classified as strong acids.

What is meant by a strong acid?

… [1]

(b) Write equations for the reactions below. State symbols are not required.

(i) The reaction of copper(II) oxide with dilute hydrochloric acid.

… [1]

(ii) The reaction of ammonium carbonate with dilute nitric acid.

… [2]

• The student reacts 2.62 g of the Group 1 metal, M, with water.

A solution of the alkali, MOH(aq), is formed.

• The student makes this solution of MOH(aq) up to 250.0 cm3 with water.

• The student pipettes 25.0 cm3 of this MOH(aq) solution into a conical flask.

• The student titrates this 25.0 cm3 volume of MOH(aq) with 0.165 mol dm–3 H SO (aq).

The equation is shown below.

2MOH(aq) + H2SO4(aq) M2SO4(aq) + 2H2O(l)

(i) Name the type of flask that the student should use to make up the 250.0 cm3 solution of

MOH(aq).

… flask [1]

(ii) The student takes burette readings to the nearest 0.05 cm3.

The student’s readings are shown in the table.

The rough titre has been omitted.

Complete the table below.

Final reading

3 20.25 40.85 25.85

/ cm

Initial reading

30.00 20.25 5.50

/ cm

Titre / cm3

[1]

(iii) Calculate the mean titre of H SO , to the nearest 0.05 cm3, that the student should use

to analyse the results.

mean titre = … cm3 [1]

(iv) Calculate the amount, in mol, of MOH in 25.0 cm3 of solution and determine the identity

of the Group 1 metal M.

metal M = … [4]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

22 (a) (Strong acid) completely/fully dissociates/ionises ✓ 1 AO1.1 DO NOT ALLOW easily dissociates

ALLOW ALL H+ ions are released

(b) (i) CuO + 2HCl → CuCl2 + H2O ✓ 1 AO2.6 ALLOW multiples

IGNORE state symbols

IGNORE charges, even if wrong

(ii) (NH4)2CO3 + 2HNO3 → 2NH4NO3 + CO2 + H2O 2 AO2.6 ALLOW multiples

2 IGNORE state symbols

Any 4 formulae correct ✓ IGNORE charges, even if wrong

All 5 formulae correct and balanced ✓

ALLOW H2CO3 for CO2 + H2O

Counts as 2 formulae for marking criteria

(ii) Final 1 AO1.2

3 20.25 40.85 25.85

reading/cm

Initial

30.00 20.25 5.50

reading/cm

Titre/cm3 20.25 20.60 20.35

DO NOT ALLOW 1 DP, e.g. 20.6 instead of 20.60

All 3 titres correct to 2 DP ✓

(iii) 20.25 + 20.35 3 1 AO2.8 ALLOW 20.3

mean titre = 2 = 20.30 (cm ) ✓ Missing ‘0’ already penalised in c(ii)

i.e. using concordant (consistent) titres

DO NOT ALLOW mean of all three titres,

20.25 + 20.60 + 20.35

i.e. 3 = 20.40

Question Answer 12 Marks Guidance

element

(iv) 4 ALLOW ECF throughout and from incorrect

concordant titres from 22c(iii)

20.30 –3 AO3.1

n(H2SO4) = 0.165 1000 = 3.35 10 (mol) ✓ –3

3 Calculator value = 3.3495 10

n(MOH) in 25.0 cm3 = 2 3.35 10–3

= 6.70 10–3 (mol) ✓ Calculator value = 6.699 10–3

n(MOH) in 250.0 cm3 = 10 6.70 10–3

= 6.70 10–2 (mol) ✓ Calculator value = 6.699 10–2

2.62

Ar of M = –2 = 39.1 AND M = potassium/K ✓ AO3.2 By ECF, ALLOW Group 1 metal nearest to

6.70 10

1 calculated value of Ar

COMMON ERRORS

Use of 20.4 from mean of all 3 titres ALL 4 MARKS Use of 25.0 (wrong volume) for n(H2SO4)

20.4 –3 25 –3

n(H2SO4) = 0.165 1000 = 3.366 10 (mol) ✓ from (c)(iii) n(H2SO4) = 0.165 1000 = 4.125 10 (mol)

n(MOH) in 25.0 cm3 = 2 3.366 10–3 n(MOH) in 25.0 cm3 = 2 4.125 10–3

= 6.732 10–3 (mol) ✓ = 8.25 10–3 (mol) ✓

n(MOH) in 250.0 cm3 = 10 6.732 10–3 n(MOH) in 250.0 cm3 = 10 8.25 10–3

= 6.732 10–2 (mol) ✓ = 8.25 10–2 (mol) ✓

2.62 2.62

Ar of M = –2 = 38.9…. OR 39 AND M = K ✓ Ar of M = –2 = 31.75….. AND M = K ✓

6.732 10 8.25 10

IF 10 is absent, Ar = 389 AND M = Cs OR Fr IF 10 is absent, Ar = 317.5 AND M = Cs OR Fr

How to answer it

Reactions Involving Acids & Titration Calculations

🔍 What this question tests

This multi-step question assesses your knowledge of acid-base definitions, writing balanced chemical equations with correct formulas, practical titration techniques (processing and recording raw data to correct decimal places), and performing complex multi-stage titration calculations combined with molar mass determination to identify an unknown Group 1 metal.

Part (a): Definition of a Strong Acid

What is meant by a strong acid? [1 mark]

✅ Correct Answer

Completely / fully dissociates / ionises.

❌ Common Errors & Examiner Penalties

DO NOT ALLOW: "Easily dissociates" (implies weakness or equilibrium shifts rather than completeness).
ALLOW: "All H⁺ ions are released."

Part (b): Writing Equations

Equations for reactions with acids (state symbols not required)

✅ Correct Answers

(i) Copper(II) oxide with dilute hydrochloric acid [1 mark]:
CuO + 2HCl → CuCl₂ + H₂O

(ii) Ammonium carbonate with dilute nitric acid [2 marks]:
(NH₄)₂CO₃ + 2HNO₃ → 2NH₄NO₃ + CO₂ + H₂O

💡 Key Knowledge & Examiner Tips

For part (ii), 1 mark is awarded for any 4 correct formulae, and the 2nd mark is for all 5 formulae correctly balanced.
H₂CO₃ is accepted in place of CO₂ + H₂O (counts as 2 formulae).
State symbols are explicitly not required by the prompt, but charges and formulae must be chemically sound.

Parts (c)(i) & (c)(ii): Practical Apparatus and Titration Data

Apparatus selection and completing titration tables [2 marks total]

✅ Correct Answers

(i) Flask type [1 mark]: Volumetric flask (allow graduated flask).

(ii) Titre completion [1 mark]:
Titre 1: 20.25
Titre 2: 20.60
Titre 3: 20.35

❌ Common Errors

For part (ii), all titrations must be recorded to 2 decimal places, ending in .0 or .5 because burettes are read to the nearest 0.05 cm³ . Writing 20.6 instead of 20.60 loses the mark!

Part (c)(iii): Mean Titre Calculation

Calculate the mean titre of H₂SO₄ using concordant results [1 mark]

✅ Correct Answers

mean titre = (20.25 + 20.35) / 2 = 20.30 cm³

🧠 Exam Technique

Always select only the concordant (consistent) titres—usually those within 0.10 cm³ of each other. Here, Titres 1 ( 20.25 ) and 3 ( 20.35 ) are averaged. Including the rough titre or Titre 2 ( 20.60 ) gives an incorrect mean ( 20.40 cm³ ) and loses you the mark.

Part (c)(iv): Multi-Step Titration Calculation

Determine the amount in mol of MOH and identify Group 1 metal M [4 marks]

📐 Step-by-Step Calculation Guide

Calculate moles of H₂SO₄ used in the titration:
n(H₂SO₄) = (concentration × volume) / 1000
n(H₂SO₄) = 0.165 × (20.30 / 1000) = 3.35 × 10⁻³ mol
Use the stoichiometric ratio to find moles of MOH in 25.0 cm³:
From equation 2MOH + H₂SO₄ → M₂SO₄ + 2H₂O , the ratio is 2 : 1.
n(MOH) in 25.0 cm³ = 3.35 × 10⁻³ × 2 = 6.70 × 10⁻³ mol
Scale up to find total moles in the 250 cm³ volumetric flask:
n(MOH) in 250 cm³ = 6.70 × 10⁻³ × (250 / 25.0) = 6.70 × 10⁻² mol (or 0.0670 mol )
Calculate the Molar Mass (A_r) and identify Metal M:
A_r = mass / moles = 2.62 g / (6.70 × 10⁻² mol) = 39.1 g mol⁻¹
Comparing with the periodic table, 39.1 corresponds to Potassium (K).

❌ Major Calculation Traps

Forgetting the 10x scale-up factor: Stopping after step 2 gives A_r = 391 , which incorrectly points to Caesium (Cs) or Francium (Fr).
Using the wrong volume: Using 25.0 cm³ instead of the mean titre 20.30 cm³ for the acid calculation propagates errors.

🌟 ECF (Error Carried Forward)

Examiners apply ECF throughout. If you used an incorrect mean titre from part (iii) (e.g., 20.40 cm³ ), you can still achieve full marks for the remaining steps provided your mathematical processing is consistent!

Topics

Module 2: Foundations in chemistry · Practical Activity Groups · PAG 2: Acid-base titration · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.