Edexcel A-Level Chemistry Paper 1, June 2024: Question 4

14 marks · Medium difficulty · Open Response

Explain acid-base concepts, identify a weak acid-strong base titration curve, compare acid pH values, and calculate blood buffer pH and equilibrium shifts.

Practise this question

Question

Question 4 on acids, bases, and buffers. Part (a) asks for the definition of a Brønsted–Lowry acid. Part (b) asks to select the correct titration curve (A, B, C, or D) for adding 0.100 mol dm⁻³ NaOH to 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH. Part (c) provides a table of 0.100 mol dm⁻³ solutions of HCl (pH 1.00), H₂SO₄ (pH 0.98), and propanoic acid (pH 2.94), asking to comment on these values without calculations. Part (d) provides the carbonic acid dissociation equilibrium, asking for the Ka expression and blood pH calculation given concentrations and Ka. Part (e) shows coupled equilibria of CO₂/H₂CO₃ and H₂CO₃/HCO₃⁻, asking to explain the effect of increased CO₂ during exercise and how the buffer restores blood pH.
Question text

4 This question is about acids, bases and buffers.

(a) State what is meant by a Brønsted−Lowry acid.

(1)

(b) Which could be the titration curve when 0.100 mol dm−3 NaOH(aq) is added to

25.0 cm3 of 0.100 mol dm−3 CH COOH(aq)?

(1)

A B

14 14

12 12

10 10

pH pH

0 10 20 30 40 50 0 10 20 30 40 50

Volume of NaOH / cm3 Volume of NaOH / cm3

C D

14 14

12 12

10 10

pH pH

0 10 20 30 40 50 0 10 20 30 40 50

Volume of NaOH / cm3 Volume of NaOH / cm3

(c) Some information about acids in aqueous solution is given.

Comment on these pH values. No calculations are required.

(4)

pH of a solution of

Name of acid Formula of acid −3

0.100moldm acid

hydrochloric acid HCl 1.00

sulfuric acid H2SO4 0.98

propanoic acid*P76895A0728*CH3CH2COOH2.94

(d) One of the systems controlling the pH of blood is the

8 carbonic acid–hydrogencarbonate buffer system.

*P76895A0828*HCO(aq) H+(aq)+HCO–(aq)

23 3

(i) Write the expression for the acid dissociation constant, Ka, for carbonic acid.

State symbols are not required.

(1)

(ii) A blood sample taken from an individual was analysed.

Calculate the pH of the blood sample.

Use your expression for Ka and the values shown.

K for carbonic acid = 4.50 × 10−7 mol dm−3

a

[HCO–] = 0.0240 mol dm−3 [H CO ] = 0.00200 mol dm−3

32 3

(3)

(e) The relevant equilibria that maintain the pH of blood are shown.

Equilibrium 1 CO2(aq) + H2O(l) H2CO3(aq)

Equilibrium 2 H CO (aq) H+(aq) + HCO−(aq)

23 3

(i) When a person exercises vigorously the concentration of carbon dioxide (aq)

in the blood increases.

Explain how this increase in the concentration of carbon dioxide affects the

pH of the blood.

Refer to the equilibria in your answer. No calculation is required.

(2)

… *P76895A0928*

(ii) Explain how the carbonic acid–hydrogencarbonate buffer system in

Equilibrium 2 acts to restore the pH of the blood after a person has exercised.

(2)

(Total for Question 4 = 14 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 4 detailing marking criteria: 4(a) proton donor (1 mark); 4(b) option B (1 mark); 4(c) points for HCl fully dissociated, H₂SO₄ diprotic, second dissociation of H₂SO₄ is partial/suppressed, propanoic acid is weak/partially dissociated (4 marks); 4(d)(i) Ka = [H⁺][HCO₃⁻]/[H₂CO₃] (1 mark); 4(d)(ii) rearrangement, [H⁺] = 3.75 × 10⁻⁸ mol dm⁻³, pH = 7.43 (3 marks); 4(e)(i) equilibria shift to the right, increasing [H⁺] and lowering pH (2 marks); 4(e)(ii) presence of large reservoir of HCO₃⁻, reacts with added H⁺ shifting equilibrium 2 left to restore pH (2 marks).

Question

Acceptable Answer Additional Guidance Mark

Number

4(a) An answer that makes reference to the following point: (1)

• ( a Brønsted-Lowry acid is a) proton donor / donator Allow donates protons / H+ (ions) / hydrogen ions

Question

Answer Mark

Number

4(b) The only correct answer is B (1)

( )

A is not correct because this is the titration curve of 0.100 mol dm−3 of a strong acid and 0.100 mol dm−3 of a strong

base

C is not correct because this is the titration curve of 0.100 mol dm-3 of a weak acid and 0.100 mol dm-3 of a weak

base

D is not correct because this is the titration curve of 0.100 mol dm-3 of a strong acid and 0.100 mol dm−3 of a weak

base

Question

Acceptable Answer Additional Guidance Mark

Number

4(c) An answer that makes reference to the following points: (4)

• hydrochloric acid is a strong acid (1) Allow HCl is (almost) fully dissociated /

ionised in solution

Allow HCl H+ + Cl−

Do not award reversible arrow

• sulfuric acid is diprotic / can donate two H+ ions / protons (1)

Ignore just H2SO4 is more strongly acidic

than HCl

Allow H SO 2H+ + SO 2−

24 4

Ignore reversible arrow

(1) Allow this shown in an equation

• the second ionisation of sulfuric acid is not complete / 2−

HSO − ⇌ H+ + SO

is suppressed by the first ionisation 4 4

Allow

• propanoic acid is a weak acid / partially dissociated (1) CH3CH2COOH ⇌ CH3CH2COO− + H+

Ignore just propanoic acid is weakly

acidic / weakest acid

Question

Acceptable Answer Additional Guidance Mark

Number

4(d)(i) An answer that makes reference to the following point: (1)

K = [H+] [HCO −]

a 3

• expression for Ka [H2CO3]

Ignore state symbols

Allow H O+ for H+

Do not award round brackets.

Question

Acceptable Answer Additional Guidance Mark

Number

4(d)(ii) Example of calculation (3)

(1) [H+] = K × [H CO ]

• rearrangement or substitution of values into a 2 3

[HCO −]

the equation 3

+ (1) [H+] = (4.5 × 10−7 × 0.0020)

• calculation of [H ]

0.024

= 3.75 × 10−8 (mol dm−3)

Allow TE from an incorrect expression in 4(c)(i) for 1 mark

pH = −log [H+]

• calculation of pH (1) 10

= − log (3.75 × 10−8) = 7.426 / 7.43 / 7.4

Standalone mark

Allow TE on incorrect value

Ignore SF except 1 SF

Question

Acceptable Answer Additional Guidance Mark

Number

4(e)(i) An answer that makes reference to the following points: (2)

• the position of the first and second equilibria are shifted (1) Allow reference to each specific equilibrium

to the right (to use up the CO2) and a comment about more of the relevant

products being produced if no mention of a shift

to the right.

• (so the) concentration of H+ / [H+] rises (1) M2 dependent on M1

and Ignore ‘becomes more acidic’ for lower pH

• (resulting in a ) lower pH

Question

Acceptable Answer Additional Guidance Mark

Number

4(e)(ii) An explanation that makes reference to the following points: (2)

• there is a (large) reservoir of HCO − (and H CO ) (1) Allow a large amount / high concentration

32 3

• (when the H+ ion concentration / [H+] increases )

the H+ ions / protons react with / the HCO − to form (1) Allow Equilibrium 2 shifts to the left hand side

more H2CO3 (so restoring the pH) (so restoring the pH)

Allow H+ + HCO − ⇌ H CO

32 3

Allow this equation with a forward arrow.

Do not award the H+ ions neutralise the HCO −

Ignore any mention of OH− ions shifting

equilibrium 2 to the left.

(Total for Question 4 = 14 marks)

How to answer it

Acids, Bases, and Blood Buffer Systems

WHAT THIS QUESTION TESTS

Core physical chemistry concepts assessed across 14 marks:

  • Acid-Base Fundamentals: Defining Brønsted-Lowry acids as proton donors.
  • Titration Curves: Differentiating between strong/weak acid-base combinations based on initial pH, buffer region, and equivalence point pH.
  • Comparing Acid Strengths: Interpreting pH differences between strong monoprotic (HCl), diprotic (H₂SO₄), and weak carboxylic acids without calculating.
  • Buffer Calculations: Writing the correct Ka expression and calculating buffer pH from component concentrations.
  • Equilibrium Shifts in Biological Buffers: Applying Le Chatelier's principle to the blood buffer system (CO₂ / H₂CO₃ / HCO₃⁻) under metabolic stress.
PART (a) • 1 MARK

Brønsted-Lowry Acid Definition

Defining acids in terms of proton transfer

✅ Correct Answer

A Brønsted-Lowry acid is a proton donor (or donates H⁺ ions / hydrogen ions).

1 mark: clear mention of "proton donor" or "donates H⁺"

❌ Common Errors

  • Writing "electron pair acceptor" (that is a Lewis acid, not Brønsted-Lowry).
  • Stating that it simply "contains H⁺ ions" rather than specifying that it donates them.
  • Confusing donor and acceptor roles.
PART (b) • 1 MARK

Identifying the Titration Curve

0.100 mol dm⁻³ NaOH added to 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH

✅ Correct Answer: B

Curve B is the only correct curve for a weak acid + strong base titration.

1 mark: Correct letter chosen (B)

💡 Curve Anatomy Breakdown

  • Initial pH: CH₃COOH is a weak acid, so its starting pH is roughly 2.8 to 3.0 (eliminates A and D, which start at pH 1.0).
  • Buffer Region: Gradual rise between 0 and 24 cm³ as the buffer system (CH₃COOH / CH₃COO⁻) forms.
  • Vertical Section: Jumps across pH 7 to ~11 (equivalence point pH > 7 due to salt hydrolysis).
  • Final pH: Levels out high at pH ~12.5–13.0 due to excess strong base NaOH (eliminates C, which levels off at ~10.5 for a weak base).

🧠 Exam Technique: Systematic Elimination

Always classify both components first:

CH₃COOH = weak acid (start ~pH 3) → rules out A & D.

NaOH = strong base (end ~pH 13) → rules out C (ends at pH 10, meaning weak base).

Therefore, only B can be correct.

PART (c) • 4 MARKS

Comparing Acid Strengths and pH Values

Qualitative analysis of HCl (1.00), H₂SO₄ (0.98), and propanoic acid (2.94)

✅ Marking Points Breakdown

  1. Hydrochloric acid is a strong acid: It completely (or almost fully) ionises/dissociates in aqueous solution ( HCl → H⁺ + Cl⁻ ), giving [H⁺] = 0.100 mol dm⁻³, so pH = -log(0.100) = 1.00.
  2. Sulfuric acid is diprotic: It can donate two protons / H⁺ ions per molecule ( H₂SO₄ → 2H⁺ + SO₄²⁻ ).
  3. Second ionisation of H₂SO₄ is incomplete: The second dissociation ( HSO₄⁻ ⇌ H⁺ + SO₄²⁻ ) is an equilibrium suppressed by H⁺ from the first step. Hence [H⁺] is only slightly greater than 0.100 mol dm⁻³, making the pH just slightly below 1.00 (0.98 instead of 0.70).
  4. Propanoic acid is a weak acid: It only partially dissociates ( CH₃CH₂COOH ⇌ CH₃CH₂COO⁻ + H⁺ ), so [H⁺] << 0.100 mol dm⁻³, giving a significantly higher pH of 2.94.
4 marks: 1 mark per bullet point

❌ Common Errors & Misconceptions

  • Vague descriptions: Stating simply "propanoic acid is weakly acidic" instead of stating it is a weak acid or partially dissociates.
  • Overlooking the second dissociation of H₂SO₄: Many candidates assume H₂SO₄ fully dissociates twice. If it did, [H⁺] would be 0.200 mol dm⁻³ (pH = 0.70). You must explain why pH is 0.98 (the second dissociation is partial/suppressed).
  • Using reversible arrows for HCl: Do not write an equilibrium arrow for HCl dissociation.
PART (d)(i) & (ii) • 4 MARKS TOTAL

Buffer Expression and pH Calculation

Blood buffer: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

✅ (d)(i) Ka Expression (1 mark)

Ka = [H⁺][HCO₃⁻] / [H₂CO₃]

Note: [H₃O⁺] is acceptable in place of [H⁺]. Square brackets are required; round brackets score 0.

📐 (d)(ii) Step-by-Step Calculation (3 marks)

  1. Rearrange for [H⁺]:
    [H⁺] = (Ka × [H₂CO₃]) / [HCO₃⁻]
    Mark 1: Rearrangement or correct substitution
  2. Substitute the data:
    [H⁺] = (4.50 × 10⁻⁷ × 0.00200) / 0.0240
    [H⁺] = 3.75 × 10⁻⁸ mol dm⁻³
    Mark 2: Correct evaluation of [H⁺]
  3. Calculate pH:
    pH = -log₁₀(3.75 × 10⁻⁸) = 7.426 (or 7.43)
    Mark 3: Standalone pH answer (allow TE from incorrect [H⁺])

❌ Common Errors in Part (d)

  • Using round brackets (H⁺) in equilibrium constant expressions. In A-Level Chemistry, square brackets denote concentration; round brackets denote partial pressures or simple grouping.
  • Inverting the ratio: calculating [HCO₃⁻] / [H₂CO₃] instead of [H₂CO₃] / [HCO₃⁻] when solving for [H⁺].
  • Rounding too early: keeping only 1 significant figure during intermediate steps causes rounding drift in the final pH.

🧠 Exam Technique: Sanity Check

Normal human blood pH is tightly regulated around 7.35 – 7.45. If your buffer calculation yields an answer like 3.2 or 10.4 for a blood sample, you immediately know an inversion error occurred in your algebra!

PART (e)(i) & (ii) • 4 MARKS TOTAL

Blood Buffer Equilibria in Action

Equilibrium 1: CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq)
Equilibrium 2: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

✅ (e)(i) Effect of Increased [CO₂(aq)] on Blood pH (2 marks)

  • Equilibrium Shifts: The positions of both Equilibrium 1 and Equilibrium 2 shift to the right (to consume the excess CO₂). 1 mark
  • Effect on pH: The concentration of H⁺ ions / [H⁺] increases, which causes the pH to fall / decrease / lower. 1 mark (dependent on M1)

✅ (e)(ii) How Buffer Restores pH After Exercise (2 marks)

  • Reservoir: There is a large reservoir / high concentration of HCO₃⁻ (and H₂CO₃) in the blood. 1 mark
  • Buffering Action: The excess H⁺ ions react with the reservoir of HCO₃⁻ to form H₂CO₃ ( H⁺ + HCO₃⁻ → H₂CO₃ ), shifting Equilibrium 2 to the left and restoring blood pH. 1 mark

❌ Critical Examiner Watch-outs for Part (e)

  • "Becomes more acidic": In (e)(i), writing "the blood becomes more acidic" does NOT award the mark for M2. You must specifically say the pH decreases / lowers.
  • Neutralisation vs Buffer action: In (e)(ii), the mark scheme explicitly rejects phrases like "H⁺ ions neutralise HCO₃⁻". Use precise chemical language: "H⁺ reacts with HCO₃⁻ to form H₂CO₃" or "shifts equilibrium to the left".
  • Irrelevant species: Mentioning OH⁻ ions shifting Equilibrium 2 to the left is chemically incorrect here and will lose marks.

Topics

Physical Chemistry · Topic 12: Acid-base Equilibria · Topic 10: Equilibrium I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.