OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 19
11 marks · Hard difficulty · Structured Questions
Calculate the pKa of chloroethanoic acid and determine the concentration of ethanoic acid using a titration curve and indicator selection.
Practise this questionQuestion
Question text
19 This question is about acids and bases.
(a) Chloroethanoic acid, ClCH2COOH, is a weak monobasic acid.
(i) Write the expression for the acid dissociation constant, Ka, of ClCH2COOH.
[1]
(ii) The expression for the acid dissociation constant, Ka, of ClCH2COOH can be simplified to:
[H+]2
Ka = Expression 19.1
[ClCH2COOH]
State one approximation that allows the expression from (a)(i) to be simplified to Expression
19.1.
… [1]
(iii) A student carries out an experiment to determine the pKa value of a solution of ClCH2COOH.
• The concentration of Cl CH COOH is 0.090 mol dm–3.
• The pH of ClCH2COOH is 1.95.
Use Expression 19.1 to calculate the pKa value of ClCH2COOH.
Give your answer to 2 decimal places.
pKa = … [3]
(b) A student titrates a 10.0 cm3 sample of ethanoic acid, CH COOH, against an aqueous solution of
0.0560 mol dm–3 Ba(OH) .
2CH3COOH + Ba(OH)2 Ba(CH3COO)2 + 2H2O
The student used a pH meter to measure the pH of the mixture after every addition of Ba(OH)2
throughout the titration.
The student’s results are shown below.
pH 6
01 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
volume of Ba(OH) / cm3
(i) Draw a best-fit curve on the graph and calculate the concentration of the CH3COOH solution.
CH COOH concentration = … mol dm–3 [5]
(ii) The end point of the titration can also be found by observing the colour change of an indicator.
The pH ranges of some indicators are shown in the table.
Indicator pH range
Malachite green 0.2 – 1.8
Bromophenol blue 2.8 – 4.6
Phenol red 6.8 – 8.4
Phenolphthalein 8.2 – 10.0
Identify the indicator in the table that would be suitable to observe the end point of the titration
between CH3COOH and Ba(OH)2.
… [1]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
19 (a) (i) [H⁺] [ClCH₂COO⁻] 1 DO NOT ALLOW without square brackets
(Ka) =
[ClCH₂COOH]
[H⁺]2
DO NOT ALLOW
[ClCH₂COOH]
[H⁺][A-]
DO NOT ALLOW [HA]
(ii) [H+] = [A–] 1 Answer must be in terms of concentration
OR [H+] from water is negligible
OR dissociation of water is negligible ALLOW [H+] ≈ [A–]
IGNORE HA ⇌ H+ + A- is a 1:1 mole ratio.
(iii) FIRST CHECK ANSWER ON ANSWER LINE 3 ALLOW ECF throughout
If answer = 2.85 OR 2.86 OR 2.87 award 3 marks
------------------------------------------------------------------------
([H+] =) 10–1.95
–2 ALLOW [H+] =1.1 10–2 up to calculator value
OR = 1.1(22…) 10
[H⁺]² ALLOW 2 sig figs up to calculator value.
(Ka) = ( )
[ClCH₂COOH]
ALLOW calculations based on finding the [HA]equ
(1.122… x10⁻²)² (1.12 x10⁻²)² (1.1 x10⁻²)²
(1.122… x10⁻²)² (1.12 x10⁻²)² (1.1 x10⁻²)² (0.079) OR (0.079) OR (0.079)
= (0.090) OR (0.090) OR (0.090)
=1.59 10–3 OR =1.59 10–3 OR = 1.53 10–3
=1.4(0) 10–3 OR =1.39 10–3 OR = 1.34 10–3
(pKa = –log10(Ka) =) 2.85, 2.86 OR 2.87 (2DP) (pKa = –log10(Ka) =) 2.80 OR 2.80 OR 2.81 (2DP)
Must be 2DP
Common error:
2 marks
0.90 (not using [H+]2 )
(b) (i) Smooth s-shaped curve using a best fit line that goes 5 DO NOT ALLOW point to point
through the majority of points.
DO NOT ALLOW tram/feather lines.
ALLOW Reading off x-axis from 12.4 – 12.6 cm3
Reading off x-axis at 12.5 cm3
12.5 ALLOW ECF throughout
n(Ba(OH)2) =) 0.0560 1000
= 7.00 10–4
ALLOW 3SF or more unless there is a trailing zero
n(CH3COOH =) 2 (moles Ba(OH)2)
= 1.40 10–3
1.4 x 10-3 Alternative answers:
(concentration =) (10/1000)
0.139 (mol dm–3) (from reading off x-axis at 12.4 cm3 )
= 0.14(0) (mol dm–3)
0.141 (mol dm–3) (from reading off x-axis at 12.6 cm3)
Common errors:
3 Marks
0.134 (Use of 12 cm3)
0.202 (use of 18 cm3)
Alternative method based on calculating pKa from the
half neutralisation point.
pH and [H+] reading will come from the candidates
graph and the data points provided.
e.g.
pH at half neutralisation
6.25 cm3 = pH 4.7 = pK
a
K = 10-4.7
a
= 1.995 x 10-5
ALLOW MP2 for K = 1.7 x 10-5 to 1.8 x 10-5 (knowledge of
a
actual Ka value)
[H+] at pH 3.3 (obtained from data on the graph
provided) ALLOW ECF from any quoted Ka
10-3.3 = 5.012 x 10-4 (mol dm-3)
[H⁺]²
[HA] = [K ]
a
(5.012 x 10-4)²
= (1.995 x 10-5)
= 0.0126 (mol dm-3)
(ii) Phenol red 1 Both indicators can change colour on the sharp vertical
OR section of the candidates curve.
Phenolphthalein
How to answer it
Acids, Bases, and pH Calculations Study Guide
What this question tests
This multi-part exam question assesses your mastery of weak acids, acid dissociation constant expressions (Ka and pKa), approximations used in weak acid calculations, processing complex titration curves, stoichiometry, and choosing suitable acid-base indicators based on vertical pH jumps.
Writing the Ka Expression
✅ Correct Answer
Ka = [H⁺][ClCH₂COO⁻] / [ClCH₂COOH]
❌ Common Errors
- Omitting square brackets (must use [ ] for concentration terms).
- Writing [H⁺]² / [ClCH₂COOH] prematurely (this is only true after applying the approximation in later parts).
Weak Acid Approximations
✅ Correct Answer
State that [H⁺] = [A⁻] (or that dissociation of water is negligible / [H⁺] from water is negligible).
💡 Key Knowledge
For a monobasic weak acid, we assume that every hydrogen ion originates from the dissociation of the acid molecule, ignoring the auto-ionization of water.
Calculating pKa from pH and Concentration
📐 Step-by-Step Calculation
- Find [H⁺]: [H⁺] = 10⁻¹·⁹⁵ = 1.122 × 10⁻² mol dm⁻³
- Rearrange Ka expression: Ka = [H⁺]² / [ClCH₂COOH]
- Substitute values: Ka = (1.122 × 10⁻²)² / 0.090 = 1.399 × 10⁻³
- Calculate pKa: pKa = -log₁₀(Ka) = -log₁₀(1.399 × 10⁻³) = 2.85
❌ Calculation Trap
Forgetting to square the [H⁺] term in the numerator is the most frequent place students drop marks in weak acid calculations.
Titration Curve Analysis & Concentration Calculation
🧠 Exam Technique & Curve Drawing
Draw a smooth, continuous S-shaped titration curve passing through the majority of plotted points. Do not use a point-to-point jagged line or feather/tramlines.
📐 Step-by-Step Calculation
- Determine equivalence volume from graph: Read inflection point on x-axis at 12.5 cm³ (acceptable range: 12.4 – 12.6 cm³).
- Calculate moles of Ba(OH)₂ added: n = 0.0560 × (12.5 / 1000) = 7.00 × 10⁻⁴ mol
- Use stoichiometry: From equation, 2 moles of CH₃COOH react with 1 mole of Ba(OH)₂.
n(CH₃COOH) = 2 × 7.00 × 10⁻⁴ = 1.40 × 10⁻³ mol - Calculate acid concentration: Volume of acid sampled was 10.0 cm³ .
Concentration = (1.40 × 10⁻³) / (10.0 / 1000) = 0.140 mol dm⁻³
Selecting a Suitable Indicator
✅ Correct Answer
Phenol red OR Phenolphthalein
💡 Key Knowledge
An indicator is suitable if its active pH range falls entirely within the sharp vertical pH jump (equivalence region) of the titration curve.
Topics
Module 5: Physical chemistry and transition elements · Practical Activity Groups · PAG 2: Acid-base titration · PAG 11: pH measurement · 5.1 Rates, equilibrium and pH
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.