Edexcel A-Level Chemistry Paper 3, June 2025: Question 8

20 marks · Medium difficulty · Practical Techniques and Data Analysis

Calculate Ka values, plot and interpret an acid-base titration curve, evaluate data logging, and calculate buffer concentrations and explain buffer action.

Practise this question

Question

Question 8 is a multi-part question worth 20 marks about weak acids and buffers. Part (a) asks to calculate Ka for a 0.0712 mol dm⁻³ weak acid HA with pH 3.25. Part (b) details the titration of 20.0 cm³ of 0.0500 mol dm⁻³ ethanoic acid (pH 3.03) with 0.0500 mol dm⁻³ NaOH up to 40.0 cm³: (i) calculate the pH of the NaOH solution; (ii) plot the titration curve on an axes grid with pH from 0 to 14 against Volume of NaOH(aq) from 0 to 40 cm³; (iii) determine the pH at the half-equivalence point on the curve; (iv) calculate Ka of ethanoic acid from this pH. Part (c) asks for three advantages of using a continuous data logger over a manual pH meter. Part (d) concerns a propanoic acid and sodium propanoate buffer: (i) calculate the concentration of propanoic acid in a 100 cm³ buffer containing 2.30 g sodium propanoate at pH 4.2; (ii) explain why pH remains virtually constant when a small amount of HCl(aq) is added.
Question text

8 This question is about weak acids.

(a) An aqueous solution of a weak acid, HA(aq), with a concentration of

0.0712 mol dm−3, has a pH = 3.25 at room temperature.

Calculate the value of Ka for the weak acid HA(aq).

(2)

(b) The value of Ka may be determined from experimental data, using a

titration curve.

Ethanoic acid, CH COOH(aq), with a concentration of 0.0500 mol dm−3, has a pH

of 3.03 at room temperature.

A 20.0 cm3 sample of the acid was titrated with sodium hydroxide solution,

NaOH(aq), with a concentration of 0.0500 mol dm−3.

During the experiment, 40.0 cm3 of sodium hydroxide solution was slowly added

to the ethanoic acid.

A pH probe connected to a data logger was used to monitor the pH throughout

the experiment.

(i) Calculate the pH of the sodium hydroxide solution.

Give your answer to one decimal place.

[K = 1.00 × 10−14 mol2 dm−6]

w

(1)

(ii) Draw a titration curve, on the axes shown, to show the change in pH

throughout the experiment.

(3)

pH

2 *P77834A02436*

0 10 20 30 40

Volume of NaOH(aq) / cm3

(iii) Give an estimate of the pH value at the half-equivalence point.

Show your working on the titration curve.

(1)

(iv) Calculate a value for Ka of ethanoic acid using your answer to (b)(iii).

You must show your working.

(3)

(c) Another method used to monitor the titration is to use a pH meter without a

data logger. The pH is measured and recorded after the addition of each 1.00 cm3

portionof NaOH(aq).

A student assessed the two methods and concluded that using the data logger

was a more effective way to carry out the experiment.

Give three reasons that support this conclusion.

(3)

… *P77834A02536*

(d) A solution containing propanoic acid and sodium propanoate forms a buffer.

(i) 2.30g of sodium propanoate is dissolved in propanoic acid solution.

This forms a buffer with pH = 4.2 and a volume of exactly 100 cm3.

Calculate the concentration of the propanoic acid, in mol dm–3, in this buffer.

[K (propanoic acid) = 1.34 × 10−5 mol dm−3 M (CH CH CO Na) = 96.0]

a r 3 2 2

(4)

(ii) Explain why there is no significant change in pH when a small amount ofhydrochloric acid, HC*P77834A02636*l(aq), is added to this buffer.

(3)

(Total for Question 8 = 20 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 8 detailing marks and accepted answers: 8(a) [H+] = 10^-3.25 = 5.6234 x 10^-4 mol dm^-3, Ka = 4.44 x 10^-6 mol dm^-3 (2 marks). 8(b)(i) [H+] = 2.00 x 10^-13, pH = 12.7 to 1 d.p. (1 mark). 8(b)(ii) Titration curve criteria: start pH 2.8-3.4, final pH at 40 cm³ between 12-12.8, vertical section at 20 cm³ between pH 6-11 (must include 7-10), characteristic weak acid curve shape with buffering region (3 marks). 8(b)(iii) Reading pH at 10 cm³ showing working (1 mark). 8(b)(iv) Working showing Ka = [H+] at half-neutralisation, giving Ka = 1.585 x 10^-5 mol dm^-3 (3 marks). 8(c) Three valid advantages for data logging including continuous data, reduced human/equipment reading error, and faster data collection (3 marks). 8(d)(i) Calculation of [salt] = 0.2396 mol dm^-3, [H+] = 6.31 x 10^-5, leading to [propanoic acid] = 1.13 mol dm^-3 (4 marks). 8(d)(ii) Explanation citing large reservoir of propanoate reacting with H+ from HCl, keeping acid to salt ratio essentially unchanged (3 marks).

Question

Answer Additional Guidance Mark

Number

8(a) Example of calculation (2)

• calculation of concentration of hydrogen ions (1) ([H+] =) 10–3.25 = 5.6234 × 10–4 (mol dm–3)

(1) (5.6234 × 10–4)2 ÷ 0.0712 = 4.4414 × 10–6 (mol dm–3)

• calculation of Ka

Allow TE from M1

Final answer with no working scores 2

Ignore SF except 1SF

If they round their answer to M1 to 5.6 × 10–4 they get an

answer of 4.4045 × 10–6 (mol dm–3)

Question

Answer Additional Guidance Mark

Number

8(b)(i) Example of calculation (1)

• calculation of pH [H+] = 1.00 × 10–14 ÷ [OH–]

= 1.00 × 10–14 ÷ 0.05

= 2.00 × 10–13 (mol dm–3)

–log (2.00 × 10–13) = 12.7

Do not award answers with more than 1 decimal

place (e.g. 12:70 scores 0)

Correct answer with no or incorrect working

scores 1

Question

Answer Additional Guidance Mark

Number

8(b)(ii) An answer that makes reference to the following points: Example of graph (3)

• starting pH between 2.8 and 3.4

and

pH at 40 cm3 between 12 and 12.8 (1)

no TE on answer to bi

• vertical section at 20 cm3 (1)

and

range of vertical section between 6-11, but

must include 7-10

(1)

• curved shape including steep rise at the start (as

example) between 0 and approx. 5 cm3

Question

Answer Additional Guidance Mark

Number

8(b)(iii) An answer that makes reference to the following points: Example of working on graph (1)

• working shown, and value of pH at 10 cm3

consistent with graph in 8b(ii)

Allow tolerance of ± 0.5 square

e.g. pH = 4.8

Question

Answer Additional Guidance Mark

Number

8(b)(iv) An answer that makes reference to the following points: (3)

• Ka expression for ethanoic acid (1) K = [CH COO−][H+] or K = [A−][H+]

a 3 a

or [CH3COOH] [HA]

at half−neutralisation [CH COOH] = [CH COO–]

/ [HA] = [A−]

• [H+] = K

a

(1)

or

pH = pKa

(1) e.g. Ka = 10–4.8 = 1.585 × 10–5 (mol dm–3)

• Value of Ka

Allow TE from 8biii

Question

Answer Additional Guidance Mark

Number

8(c) An answer that makes reference to three of the following points: (3)

Data logger likely to give most valid outcomes as

• time period / volume between readings is (much) smaller / (1)

greater resolution / more data points /

more frequent data-points

• (percentage) errors in using measuring equipment reduced / (1)

• data collection is faster (1)

• produces continuous data / graph / processed data can be seen (1)

in real time

• automatic collection of data means less chance of human error (1)

/ parallax error reduced

• data can be transferred (easily) to other software (e.g. (1)

spreadsheets) / stored Allow reverse arguments

Ignore a general comment about

accuracy e.g. more accurate

Question

Answer Additional Guidance Mark

Number

8(d)(i) Example of calculation (4)

• calculation of [H+] (1) 10–4.2 = 6.3096 × 10–5 (mol dm–3)

(1) (2.30 ÷ 96) × 10 = 0.23958 (mol dm–3)

• calculation of concentration of salt

(1) [CH CH COOH] = ([H+] × [CH CH COONa]) ÷ K

• rearrangement of Ka to find [CH3CH2COOH] 3 2 3 2 a

(1) (6.3096 × 10–5 × 0.23958) ÷ 1.34 × 10–5

• calculation of [CH3CH2COOH] –3

= 1.128 / 1.13 / 1.1 (mol dm )

Ignore SF except 1 SF

Allow TE throughout

Correct answer without working scores 4

Do not award 1.12 (incorrectly rounded)

Allow use of Henderson-Hasselbalch equation

pH = pKa + log10 [A-]

[HA]

M1: Rearrangement of HH equation

log10 [A−] = 4.2 – 4.873 = −0.673

[HA]

M2 : Calculation of [CH CH CO Na] = 10−0.673 = 0.2123

32 2

[CH3CH2COOH]

M3: [CH3CH2COONa] = (2.3 ÷ 96) ÷ 0.1 = 0.2396

(mol dm–3)

M4: [CH3CH2COOH] = 0.2396 ÷ 0.2123 =

1.128 / = 1.128 / 1.13 / 1.1 (mol dm–3)

TE at all stages

Question

Answer Additional Guidance Mark

Number

8(d)(ii) An explanation that makes reference to the following points: (3)

• (large) reservoir of sodium propanoate / propanoate ions (1) Allow a large amount / high concentration

(and propanoic acid)

• (large reservoir of) propanoate ions accept protons from / (1) Accept (acid) eqm moves to left (forming

react with HCl CH3CH2COOH)

CH CH CO – + H+ → CH CH COOH

32 2 3 2

NaCH3CH2CO2 + HCl →

CH3CH2COOH + NaCl

Do not award the H+ ions neutralise the

propanoate

• (small changes in [acid] and [salt] are negligible) so ratio (1)

of acid: salt remains (nearly) constant (and pH depends

on this ratio)

(Total for Question 8 = 20 marks)

How to answer it

Weak Acids, Titration Curves and Buffer Solutions

EXAM SPECIFICATION SUMMARY

What this question tests:

This 20-mark question covers the core physical chemistry topics of acid-base equilibria:

  • Calculating acid dissociation constants (Ka) from weak acid pH and concentration.
  • Determining the pH of a strong base using the ionic product of water (Kw).
  • Sketching and interpreting a weak acid–strong base titration curve, including the vertical equivalence section and buffer region.
  • Deducing Ka experimentally from the half-equivalence point where pH = pKa.
  • Evaluating digital instrumentation (data loggers) versus manual measurement methods.
  • Multi-step buffer calculations involving mass, molar mass, volume, and Henderson-Hasselbalch/Ka rearrangements.
  • Explaining buffer action using Le Chatelier's principle and reservoir equilibrium shifts.
PART (a) • 2 MARKS

Calculating Ka of a Monoprotic Weak Acid

Aqueous HA(aq) with concentration 0.0712 mol dm⁻³ has a pH of 3.25.

📐 Step-by-Step Calculation

Step 1: Calculate [H⁺]
[H⁺] = 10−pH = 10−3.25 = 5.6234 × 10⁻⁴ mol dm⁻³ [1 mark]
Step 2: Apply the weak acid approximation
For a weak acid HA ⇔ H⁺ + A⁻ where [H⁺] ≈ [A⁻] and [HA]eqm ≈ [HA]initial:
Ka = [H⁺]² / [HA]
Ka = (5.6234 × 10⁻⁴)² / 0.0712 = 4.44 × 10⁻⁶ mol dm⁻³ [1 mark]

❌ Common Errors & Examiner Tips

  • Rounding too early: Rounding [H⁺] to 5.6 × 10⁻⁴ gives Ka = 4.40 × 10⁻⁶ (allowed if working is shown, but best practice is keeping full calculator memory).
  • Significant figures: The mark scheme accepts 4.44 × 10⁻⁶ (3 s.f.) or unrounded values, but penalises 1 s.f. (e.g., 4 × 10⁻⁶).
  • Missing units: While units were not explicitly penalised here, standard units are mol dm⁻³ .
Mark Scheme Breakdown:
• M1: Correct calculation of [H⁺] = 5.6234 × 10⁻⁴ mol dm⁻³
• M2: Correct calculation of Ka = 4.44 × 10⁻⁶ mol dm⁻³ (allow TE from M1; final answer with no working scores 2 marks).
PART (b)(i) • 1 MARK

pH of a Strong Base (NaOH)

0.0500 mol dm⁻³ NaOH, with Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Answer to 1 decimal place.

📐 Calculation

NaOH is fully dissociated, so [OH⁻] = 0.0500 mol dm⁻³.
[H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³
pH = −log₁₀(2.00 × 10⁻¹³) = 12.7
Alternative method:
pOH = −log₁₀(0.0500) = 1.30
pH = 14 − 1.30 = 12.7

❌ Severe Decimal Trap

The question explicitly requested one decimal place.

  • Giving 12.70 scores 0 marks.
  • Always re-read the required precision stated in the command phrase!
PART (b)(ii) • 3 MARKS

Drawing the Weak Acid–Strong Base Titration Curve

20.0 cm³ of 0.0500 mol dm⁻³ CH₃COOH titrated with 0.0500 mol dm⁻³ NaOH up to 40.0 cm³.

Exact Examiner Features Required on the Graph:
  • Initial point (0 cm³ NaOH): Curve must start between pH 2.8 and 3.4 (pH 3.03 is stated in the question stem).
  • Buffer region (0 to ~5 cm³): An initial rapid rise followed by a flattening out (convex shape).
  • Equivalence volume: Since equimolar concentrations are used (0.0500 mol dm⁻³), 20.0 cm³ of acid requires exactly 20.0 cm³ of NaOH to neutralise.
  • Vertical section: Strictly vertical at V = 20.0 cm³, covering a range between pH 6 and 11, and must include pH 7 to 10.
  • Final point (40.0 cm³ NaOH): Flattens out and terminates between pH 12.0 and 12.8 (no error carried forward from (b)(i)).

🧠 Exam Technique for Titration Curves

  • Calculate the equivalence volume before putting pen to paper: Vbase = (cacid × Vacid) / cbase = 20.0 cm³ .
  • Plot the initial pH (3.03) and final pH (12.7) as target points first.
  • Draw the vertical line at 20.0 cm³ with a ruler (e.g. from pH 7 to 10.5).
  • Join smoothly with an S-shaped curve showing the characteristic weak acid buffer "hump".

❌ Common Drawing Mistakes

  • Making the vertical region centre at pH 7 (this is a weak acid–strong base titration, so equivalence pH is basic, >7).
  • Placing the vertical section at the wrong volume (e.g., 25 cm³ from habitual lab memory instead of calculated 20 cm³).
  • Curve ending above 13 or below 12.
PARTS (b)(iii) & (iv) • 4 MARKS

Determining Ka via the Half-Equivalence Point

✅ (b)(iii) Estimate pH at Half-Equivalence [1 mark]

Half-equivalence volume = 20.0 cm³ / 2 = 10.0 cm³.
Looking at the curve at V = 10.0 cm³, read the pH:

pH ≈ 4.8 (accept 4.6 – 5.0)

*Working must be clearly shown on the graph: draw dashed lines from 10.0 cm³ up to the curve and across to the pH axis.

💡 (b)(iv) Theory Behind the Calculation

At the half-equivalence point:

[CH₃COOH] = [CH₃COO⁻]

Substitute this into the Ka expression:

Ka = ([CH₃COO⁻][H⁺]) / [CH₃COOH]

The concentrations cancel out, giving:

Ka = [H⁺] ⇒ pKa = pH

📐 (b)(iv) Step-by-Step Working [3 marks]

Mark 1: State expression or relationship:
Ka = [CH₃COO⁻][H⁺] / [CH₃COOH] OR at half-neutralisation [CH₃COOH] = [CH₃COO⁻].
Mark 2: State equality:
[H⁺] = Ka OR pH = pKa.
Mark 3: Calculate Ka from reading:
If pH = 4.8 ⇒ Ka = 10−4.8 = 1.58 × 10⁻⁵ mol dm⁻³ (allow TE from b(iii)).
PART (c) • 3 MARKS

Evaluating Data Loggers vs. Manual pH Meters

Give three reasons supporting why a data logger is more effective than manual recording after each 1.00 cm³.

✅ Any 3 Valid Points from Mark Scheme

  • Continuous data / higher resolution: Time period/volume between readings is much smaller; records many more data points.
  • Lower error: Percentage/measurement errors in using measuring equipment are reduced.
  • Faster data collection: Saves time during experimental runs.
  • Real-time visualisation: Generates a continuous graphical display in real time.
  • Eliminates human error: Automated collection eliminates recording mistakes, parallax errors, and reaction-time delay.
  • Ease of data processing: Data can easily be exported to spreadsheet software for analysis and derivative plotting.

❌ Vague Statements That Score 0

  • "It is more accurate" — Too vague! The probe itself has the same sensor accuracy as a standard pH meter. The advantage is sampling frequency and automation.
  • "It is more reliable" — Needs qualifying (e.g., removes human transcription errors).
PART (d)(i) • 4 MARKS

Buffer Solution Calculation

2.30 g sodium propanoate (Mr = 96.0) dissolved in propanoic acid to make 100 cm³ of buffer.
Buffer pH = 4.2, Ka = 1.34 × 10⁻⁵ mol dm⁻³. Calculate [CH₃CH₂COOH].

📐 Structured 4-Step Solution

Step 1: Calculate [H⁺]
[H⁺] = 10−pH = 10−4.2 = 6.3096 × 10⁻⁵ mol dm⁻³ [1 mark]
Step 2: Calculate concentration of propanoate salt
Moles of CH₃CH₂COONa = mass / Mr = 2.30 / 96.0 = 0.023958 mol
Volume = 100 cm³ = 0.100 dm³
[CH₃CH₂COO⁻] = 0.023958 / 0.100 = 0.23958 mol dm⁻³ [1 mark]
Step 3: Rearrange the Ka expression for [acid]
Ka = ([H⁺][salt]) / [acid]  ⇒  [acid] = ([H⁺] × [salt]) / Ka [1 mark]
Step 4: Calculate final concentration
[CH₃CH₂COOH] = (6.3096 × 10⁻⁵ × 0.23958) / (1.34 × 10⁻⁵) = 1.13 mol dm⁻³ [1 mark]
(Accept 1.1, 1.128, 1.13)

🧠 Alternative Method: Henderson-Hasselbalch

pH = pKa + log₁₀([salt] / [acid])

pKa = −log₁₀(1.34 × 10⁻⁵) = 4.873

4.2 = 4.873 + log₁₀(0.23958 / [acid])

log₁₀(0.23958 / [acid]) = −0.673

[acid] = 0.23958 / 10−0.673 = 1.13 mol dm⁻³

❌ Critical Trap to Avoid

  • Volume omission: Forgetting to divide moles by 0.100 dm³ to get concentration gives salt conc = 0.024 mol dm⁻³, leading to an answer out by a factor of 10.
  • Premature rounding: Do not award 1.12 if it arises from incorrect rounding of intermediate figures.
PART (d)(ii) • 3 MARKS

Explaining Buffer Action Upon Addition of Acid

Explain why there is no significant change in pH when a small amount of HCl(aq) is added.

✅ 3-Point Model Answer

1. Identify the reservoir:
There is a large reservoir of propanoate ions, CH₃CH₂COO⁻ (and propanoic acid molecules). [1 mark]
2. State the reaction & equilibrium shift:
The added H⁺ ions react with the propanoate ions to form propanoic acid:
CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH
(Equilibrium shifts to the left, removing added H⁺ ions.) [1 mark]
3. Explain why pH stays constant:
Because the reservoir concentrations are very large, the changes in [acid] and [salt] are negligible, so the ratio of [acid] to [salt] remains almost constant (hence [H⁺] and pH remain constant). [1 mark]

❌ Examiner Pitfalls

  • Do NOT say: "H⁺ ions are neutralised by propanoate ions" — examiners specifically reject the word "neutralised" here; you must say they react with or are accepted by propanoate ions.
  • Missing the ratio: Simply saying "pH does not change" repeats the question. You must explain that the ratio of [HA] to [A⁻] remains virtually unchanged.

Topics

Physical Chemistry · Core Practicals · Topic 12: Acid-base Equilibria · Core Practical 9: Finding the Ka value for a weak acid

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.