OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 19
11 marks · Medium difficulty · Calculations
Calculate the pKa of ethanoic acid and percentage dissociation from pH, find the percentage purity of a sodium hydroxide drain cleaner from titration data, and draw the dot-and-cross diagram of a carbonate ion.
Practise this questionQuestion
Question text
19 This question is about acids and bases found in the home.
(a) Ethanoic acid, CH3COOH, is the acid present in vinegar.
A student carries out an experiment to determine the pKa value of CH3COOH.
• The concentration of CH COOH in the vinegar is 0.870 mol dm–3.
• The pH of the vinegar is 2.41.
(i) Write the expression for the acid dissociation constant, Ka, of CH3COOH.
[1]
(ii) Calculate the pKa value of CH3COOH.
Give your answer to two decimal places.
pKa = … [3]
(iii) Determine the percentage dissociation of ethanoic acid in the vinegar.
Give your answer to three significant figures.
percentage dissociation = … % [1]
(b) Many solid drain cleaners are based on sodium hydroxide, NaOH.
• A student dissolves 1.26 g of a drain cleaner in water and makes up the solution to
100.0 cm3.
• The student measures the pH of this solution as 13.48.
Determine the percentage, by mass, of NaOH in the drain cleaner.
Give your answer to three significant figures.
percentage = … % [4]
(c) Sodium carbonate, Na2CO3, is a base used in washing soda.
Na CO contains the carbonate ion, CO 2–, shown below.
23 3
O–
C
O O–
Draw the ‘dot-and-cross’ diagram for the carbonate ion.
Show outer electrons only and use different symbols for electrons from C and O, and any
‘extra’ electrons.
[2]
Mark scheme
Show the mark scheme
![The mark scheme provides the answers and guidance for question 19. Part (a)(i) gives Ka = [H+][CH3COO-] / [CH3COOH]. Part (a)(ii) shows calculation steps for [H+], Ka, and pKa giving 4.76. Part (a)(iii) shows percentage dissociation calculation giving 0.447%. Part (b) outlines steps using Kw and pH to find [OH-], calculating mass of NaOH as 1.21g, and the percentage as 95.9%. Part (c) displays the correct dot-and-cross diagram for the carbonate ion with bonding pairs around the central carbon and non-bonding electrons around the oxygen atoms, along with guidance notes for marking.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2FmarkSchemes%252FOCR.Q18.L01.19.SRC.ocr-a-alevel-1%25257C2018%25257CJUN%25257Cuntiered-1791652088773.png%3Falt%3Dmedia&w=1000)
Question Answer Marks Guidance
19 (a) (i) [H+] [CH COO–] 1 IGNORE state symbols
Ka = [CH COOH] Must be square brackets
IGNORE expressions with HA or with [H+]2
(ii) FIRST, CHECK ANSWER ON ANSWER LINE 3
IF answer = 4.76 award 3 marks ALLOW use of HA and A–
-------------------------------------------------------------------
[H+] = 10–pH ALLOW 3 SF up to calculator value of:
= 10–2.41 = 3.89 10–3 (mol dm–3) 3.89045145 10–3 correctly rounded
Ka
[H+]2 (3.89 10–3)2 K 1.739725573 10–3
a
= [CH COOH] = 0.870 –5 +
3 NOTE: 1.74 10 is same from unrounded [H ] calculator
–5 –3 value and 3 SF [H+] value
= 1.74 10 (mol dm )
pKa
= –log K = –log 1.74 10–5 = 4.76 2 DP required
a
(iii) [H+] 1 3 SF required
% dissociation = [CH COOH] 100
3.89 10–3
= 0.870 100 = 0.447(%)
Question Answer 16 Marks Guidance
(b) FIRST, CHECK ANSWER ON ANSWER LINE 4
IF answer = 95.9(%) award 4 marks ALLOW ECF throughout
-------------------------------------------------------------------
[H+] = 10–pH IGNORE rounding errors beyond 3rd SF throughout
= 10–13.48 = 3.31 10–14 (mol dm–3)
ALLOW 3.3 10–14 (mol dm–3)
[OH–] from K ALLOW 0.30
w
1.00 10–14 ALLOW 0.303 if 3.3 10–14 used in the first marking point
= = 0.302 (mol dm–3)
3.31 10–14
ALLOW pOH method:,
pOH = 14 – 13.48 = 0.52
[OH–] = 10–0.52 = 0.302 (mol dm–3)
ALLOW [OH–] 0.1 40
Mass of (NaOH)
= 0.302 1000 40.0 = 1.21 (g)
Rounding [OH–] to 0.3(0) gives 1.2/1.26 = 95.2%
% of NaOH to 3 SF Award 4 marks
Rounding [OH–] to 0.303 gives 1.212/1.26 = 96.2%
1.21
= 1.26 100 = 95.9 (%) Award 4 marks
(c) 2– 2 NOT REQUIRED
Charge (‘2–‘) IGNORE incorrect charges
O
Brackets
Circles
C
O O 17
IGNORE inner shells
Global rules ALLOW rotated diagram
C and O electrons must be shown differently,
e.g. • for C and × for O ALLOW diagram with missing C or O symbols.
Na electrons shown with different symbol
MARKING
Bonding around central C atom
4 electrons for C shown as • OR ×
4 electrons for O, different from C as • OR ×
C=O bond with 2 C electrons AND 2 O electrons In C=O bond, ALLOW sequence × × • •
Two C–O bonds with 1 C electron AND 1 O
electron In C–O bond, ALLOW ‘extra’ electron with different symbol
for O electron
Non-bonded (nb) electrons around 3 O atoms
C=O oxygen has 4 nb ‘O’ electrons ALLOW non-bonding electrons unpaired
Each C–O oxygen has 5 nb ‘O’ electrons
AND 1 ‘extra’ electron with different symbol ALLOW ‘extra’ electron as • OR × if it has been labelled
‘extra electron’ or similar
Total 11
How to answer it
Acids, Bases and Dot-and-Cross Diagrams
What this question tests
This multi-part exam question assesses your core knowledge of acid-base equilibria and atomic structure. Key competencies include writing acid dissociation constant expressions, performing multi-step pH, pKₐ, and percentage dissociation calculations for weak acids, manipulating ionic product of water expressions (K𝓌) for strong alkalis, determining mass percentages from titration-style purity data, and drawing accurate dot-and-cross diagrams for complex ions.
Acid Dissociation Expression
✅ Correct Answer
Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
💡 Key Knowledge
- Always use square brackets [] to represent equilibrium concentrations.
- Products go on the numerator (top), reactants on the denominator (bottom).
🧠 Exam Technique
State symbols are not required for Kₐ expressions. Alternative shorthand like HA and A⁻ are accepted by examiners.
Calculating the pKₐ Value
✅ Correct Answer
pKₐ = 4.76
📐 Step-by-Step Calculation
- Find [H⁺]: [H⁺] = 10⁻ᵖᴴ = 10⁻²·⁴¹ = 3.89 × 10⁻³ mol dm⁻³
- Assume [H⁺] = [CH₃COO⁻]: For a weak monoprotic acid, dissociation of H⁺ equals CH₃COO⁻. Therefore, Kₐ = [H⁺]² / [CH₃COOH]
- Substitute values: Kₐ = (3.89 × 10⁻³)² / 0.870 = 1.74 × 10⁻⁵ mol dm⁻³
- Calculate pKₐ: pKₐ = -log(Kₐ) = -log(1.74 × 10⁻⁵) = 4.76
❌ Common Errors
- Forgetting to round the final answer to two decimal places as requested.
- Rounding intermediate values too early, which drifts the final pKₐ value. Keep full calculator values until the end.
Percentage Dissociation
✅ Correct Answer
percentage dissociation = 0.447%
📐 Calculation & Technique
Use the formula:
% dissociation = ([H⁺] / [CH₃COOH]initial) × 100
= (3.89 × 10⁻³ / 0.870) × 100 = 0.447%
❌ Common Errors
- Failing to format the answer to three significant figures (e.g. writing 0.45% or 0.44712%).
Percentage by Mass of NaOH in Drain Cleaner
✅ Correct Answer
percentage = 95.9%
📐 Step-by-Step Calculation
- Find [H⁺] from pH: [H⁺] = 10⁻¹³·⁴⁸ = 3.31 × 10⁻¹⁴ mol dm⁻³
- Find [OH⁻] using K𝓌: [OH⁻] = (1.00 × 10⁻¹⁴) / (3.31 × 10⁻¹⁴) = 0.302 mol dm⁻³
- Calculate moles of NaOH in 100 cm³: Moles = 0.302 × (100 / 1000) = 0.0302 mol
- Calculate mass of NaOH: Mass = moles × Mᵣ = 0.0302 × 40.0 = 1.21 g
- Calculate percentage purity: (1.21 g / 1.26 g) × 100 = 95.9%
🧠 Exam Technique & ECF
Examiners apply Error Carried Forward (ECF) throughout. If your initial [OH⁻] calculation had a minor rounding variance (e.g. using unrounded pH values giving 95.2% or 96.2%), full marks are still awarded as long as the logical chain is correct and expressed to 3 SF.
Dot-and-Cross Diagram for Carbonate Ion
✅ Correct Answer Description
Central carbon atom bonded to three oxygen atoms: one double covalent bond to an oxygen atom (4 bonding electrons shared), and two single covalent bonds to oxygen atoms carrying extra electrons from the 2- charge (each with 2 shared bonding electrons and 3 lone pairs + 1 extra electron).
💡 Key Knowledge
- Use different symbols for electrons (e.g. dots ● for C, crosses × for O, and an alternate symbol like a square or dot inside a circle for extra electrons).
- Enclose the entire structure in square brackets with a superscript 2- charge outside.
❌ Common Errors
- Using identical symbols for electrons originating from different atoms.
- Omitting the square brackets or overall ionic charge.
- Incorrect electron counts around single-bonded versus double-bonded oxygens.
Topics
Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 5.1 Rates, equilibrium and pH · 2.2 Electrons, bonding and structure · 4.2 Alcohols, haloalkanes and analysis
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.