Edexcel A-Level Chemistry Paper 1, June 2023: Question 6
14 marks · Hard difficulty · Calculations
Assess various properties and reactions of benzoic acid including Brønsted-Lowry acid-base behavior, lattice properties of its salts, Ka calculations, indicators, and enthalpy changes of neutralisation.
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Question text
6 Benzoic acid is a weak acid found in cranberries.
OH
C
O
C6H5COOH – benzoic acid
(a) Which of these answers identifies the types of species present when benzoic acid
is mixed with nitric acid?
[K of benzoic acid = 6.3× 10–5 mol dm–3; K of nitric acid = 40mol dm–3]
a a
(1)
C H COOH + HNO C H COOH+ + NO–
65 3 6 5 2 3
A acid base acid base
B acid base base acid
C base acid base acid
D base acid acid base
(b) The ionic salts sodium benzoate and potassium benzoate are both used as
food preservatives.
Explain why the melting temperature of sodium benzoate is higher than the
melting temperature of potassium benzoate.
(2)
(c) The value of K for benzoic acid = 6.28× 10–5 mol dm–3.
a
(i) Write the expression for the acid dissociation constant, K , of benzoic acid. 11
*P71912A01128*a
(1)
(ii) Calculate the mass of benzoic acid needed to prepare 250cm3 of a solution
with a pH = 3.51
(4)
(d) Weak acids such as benzoic acid can be neutralised by sodium hydroxide solution.
C6H5COOH(aq) + NaOH(aq) → C6H5CO2Na(aq) + H2O(l)
(i) Which of these could be used to show the end‑point of a titration
of benzoic acid with sodium hydroxide solution?
(1)
A bromothymol blue
B litmus
C methyl orange
D phenolphthalein
(ii) Another weak acid found in cranberries is quinic acid, C6H7(OH)4COOH.
It is neutralised by sodium hydroxide solution in a similar way to benzoic acid.
A 25.0 cm3 sample of 0.500 mol dm–3 quinic acid solution was neutralised
under standard conditions in a polystyrene cup using 25.0 cm3 of
0.800 mol dm–3 of sodium hydroxide solution.
This resulted in a temperature rise of 2.9°C.
Calculate the standard enthalpy change of neutralisation, Δ H d, of
neut
quinic acid in kJ mol–1.
[Assume the density of both solutions is 1.0 g cm–3.
specific heat capacity of solution formed = 4.18 J g–1 °C–1]
(3)
(iii) The standard enthalpy change of neutralisation of the weak acid HCN
by sodium hydroxide is –11.7 kJ mol–1 while that of the strong acid HCl
is –57.9 kJ mol–1.
12 Explain the difference between these values.
*P71912A01228* (2)
(Total for Question 6 = 14 marks) 13
Mark scheme
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Question
Answer Mark
Number
6(a) The only correct answer is D (base acid acid base) (1)
A is not correct because HNO3 is the stronger acid
B is not correct because HNO3 is the stronger acid
C is not correct because C H COOH + is the conjugate acid, and NO – is the conjugate base
65 2 3
Question
Answer Additional Guidance Mark
Number
6(b) An explanation that makes reference to the following points Allow reverse argument (2)
• as the sodium ion / Na+ is smaller / has a smaller ionic radius (but has Allow (the sodium ion’s) charge density is greater
the same charge) (1) / the (oppositely charged) ions are closer together
Ignore sodium ion has 1 less shell
Ignore atomic radius
Ignore just ‘sodium is smaller’
Ignore Na+ has less shielding
Do not award sodium ions have a higher charge
• so sodium benzoate has stronger ionic bonds / electrostatic forces Allow stronger attraction between ions
between ions (so more energy required to break the bonds) (1) Allow just ‘sodium benzoate has stronger bonds’
if ion mentioned / formula given in M1
Ignore reference to polarising power of cation /
distortion of anion
Do not award attraction of electrons to nucleus /
electronegativity
Do not award M2 if reference to other types of
bonding
Question
Answer Additional Guidance Mark
Number
6(c)(i) Example of expression (1)
• expression for Ka
K = [C H COO−][H+]
a 6 5
[C6H5COOH]
Allow 6.28 x 10−5 instead of K
a
Allow [H O+] for [H+]
Ignore missing Ka
Ignore state symbols, even if incorrect
Do not award
K = [H+]2
a
[C6H5COOH]
Do not award expression including [H2O]
Ignore expression with HA / A− instead of full
formulae
Do not award round brackets instead of [ ]
Question
Answer Additional guidance Mark
Number
6(c)(ii) Example of calculation (4)
10−3.51 = 3.0903 × 10−4 (mol dm−3)
• calculation of concentration of H+ ions (1)
This can be subsumed in M2
• calculation of concentration of benzoic acid (1) 6.28 × 10−5 = (3.0903 × 10−4)2 ÷ [C H CO H]
65 2
[C H CO H] = (3.0903 × 10−4)2 ÷ 6.28 × 10−5
65 2
= 1.5207 × 10−3 (mol dm−3)
• calculation of moles of benzoic acid (1) = 1.5207 × 10−3 × (250÷1000)
= 3.8017 × 10−4 (mol)
• calculation of mass of benzoic acid (1) = 3.8017 × 10−4 × 122 = 0.046381 / 4.6381 x 10−2 (g)
Allow 46.381 mg
Alternative route for M2 to M4
• calculation of concentration of benzoic acid (1) [C H CO H] = 1.5207 x 10-3 = [C H CO H] - [H+]
65 2 eqm 6 5 2 int
[C6H5CO2H]int = [C6H5CO2H]eqm + [H+]
= 1.5207 x 10-3 + 3.0903 x 10-4
= 1.82973 x 10-3 (mol dm-3)
• calculation of moles of benzoic acid (1) 1.82973 x 10-3 / 4 = 4.5743 x 10-4 (mol)
• calculation of mass of benzoic acid (1) = 4.5743 x 10-4 × 122 = 0.055807 / 5.5807 x 10−2 (g)
Allow 55.807 mg
Allow TE throughout
Allow intermediate values shown as fractions
Ignore SF except 1SF
Ignore units
Final correct answer, with or without working scores 4 marks
Question
Answer Mark
Number
6(d)(i) The only correct answer is D (phenolphthalein) (1)
A is not correct because bromothymol blue does not change colour within the region of rapid change of pH of this weak acid-strong
base titration
B is not correct because litmus does not change colour within the region of rapid change of pH of this weak acid-strong base titration
C is not correct because methyl orange does not change colour within the region of rapid change of pH of this weak acid-strong base
titration
Question
Answer Additional guidance Mark
Number
6(d)(ii) Example of calculation (3)
• calculation of heat released (1) 50 × 4.18 × 2.9 = 606.1 (J)
Allow 0.6061 kJ
Ignore any signs
• calculation of moles of quinic acid (1) (25 ÷ 1000) × 0.500 = 0.0125 (mol)
Ignore calculation of moles of NaOH
−606.1 = −48.5 (kJ mol−1)
• calculation of standard enthalpy of neutralisation
and value in kJ mol−1 0.0125 x 1000
and correct sign (1) −1
Allow −30.305 (kJ mol ) as TE if M2 lost for only
calculating mol NaOH
Allow −48500 J mol−1 (units essential)
Allow TE throughout
Allow intermediate values shown as fractions
Ignore SF except 1 SF
Ignore missing / incorrect units in M1 and M2
Correct final answer with or without working scores (3)
Penalise incorrect rounding once only e.g. 0.012 mol
Question
Answer Additional Guidance Mark
Number
6(d)(iii) An explanation that makes reference to the following points Allow ionise for dissociate throughout (2)
• HCN / the weak acid is only partially dissociated (but HCl is fully Allow HCN is not fully dissociated
dissociated) (1) Allow HCN dissociates less than HCl
Allow HCN produces a lower concentration
of H+ ions (from the same concentration of
acid as HCl)
• (so releases less energy when neutralised as) energy is needed to break Allow (some) energy is needed to complete
H-C bond(s) (in order to completely dissociate) (1) the dissociation (of HCN)
Allow (some) energy is needed to remove /
separate all the H+ ions
Ignore references to enthalpy of hydration of
ions
Ignore just energy is needed to break bonds
Do not award fewer H+ ions so less NaOH
needed for neutralisation
Do not award incorrect bond broken
If no other mark is awarded allow 1 for HCl is
fully dissociated so no energy is required to
break the H-Cl bond
(Total for Question 6 = 14 marks)
Question
Answer Additional Guidance Mark
Number
How to answer it
Question 6: Properties and Reactions of Benzoic Acid
What this question tests
This comprehensive multi-part question assesses core physical and organic chemistry topics: Bronsted-Lowry acid-base equilibria, ionic lattice energies and ionic radii, expressions and calculations involving acid dissociation constants ( Ka ) and pH, selection of suitable indicators for titrations, calorimetry calculations for enthalpy of neutralisation ( ΔneutH ), and comparative thermochemistry between weak and strong acids.
Identifying Acid-Base Species in Equilibrium
✅ Correct Answer
D (base, acid, acid, base)
💡 Key Knowledge
Remember that Ka of nitric acid is 40 mol dm⁻³ , making it a much stronger acid than benzoic acid ( Ka = 6.3 × 10⁻⁵ mol dm⁻³ ). Therefore, nitric acid donates a proton to benzoic acid.
❌ Common Errors
Students often wrongly assume that organic carboxylic acids are always the strongest acid present in any mixture, ignoring numerical Ka data provided in the stem.
Explaining Melting Temperatures of Benzoate Salts
✅ Correct Answer
Sodium benzoate has a higher melting temperature because the Na⁺ ion has a smaller ionic radius than the K⁺ ion (with the same charge), resulting in stronger electrostatic forces of attraction between ions.
🧠 Exam Technique
Always structure ionic comparisons using the magic trio: ionic radius size → charge density → strength of electrostatic attraction / ionic bonds. Do not use incorrect terms like "atomic radius" or refer to "molecular forces".
❌ Common Errors
Failing to mention ionic radius or falsely stating that sodium ions have a higher charge. Both sodium and potassium ions form +1 ions.
Calculations Involving Ka and pH
(i) Expression for Ka
✅ Correct Answer
Ka = [C₆H₅COONa⁺ or C₆H₅COO⁻][H⁺] / [C₆H₅COOH] (Accept [H₃O⁺] instead of [H⁺] ).
❌ Common Errors
Leaving out square brackets or inverting numerator and denominator. Square brackets denote equilibrium concentrations.
(ii) Calculating Mass of Benzoic Acid
📐 Step-by-Step Calculation
- Find [H⁺] from pH:
[H⁺] = 10⁻³·⁵¹ = 3.0903 × 10⁻⁴ mol dm⁻³ - Set up the Ka expression to find [C₆H₅COOH] at equilibrium:
Assuming [H⁺] ≈ [C₆H₅COO⁻] :
[C₆H₅COOH] = ([H⁺])² / Ka = (3.0903 × 10⁻⁴)² / (6.28 × 10⁻⁵) = 1.5207 × 10⁻³ mol dm⁻³
(Note: using un-approximated initial concentration methods also fully accepted per mark scheme). - Calculate moles required in 250 cm³:
Moles = 1.5207 × 10⁻³ × (250 ÷ 1000) = 3.8017 × 10⁻⁴ mol - Convert moles to mass:
Molar mass of C₆H₅COOH = 122.0 g mol⁻¹
Mass = 3.8017 × 10⁻⁴ × 122 = 0.0464 g (or 46.4 mg )
Titration Indicators and Enthalpy of Neutralisation
(i) Indicator Selection
✅ Correct Answer
D (phenolphthalein)
💡 Key Knowledge
Titrating a weak acid with a strong base produces an alkaline equivalence point (pH roughly 8–10). Phenolphthalein changes colour sharply in this high pH range, unlike methyl orange or bromothymol blue.
(ii) Enthalpy of Neutralisation Calculation
📐 Step-by-Step Calculation
- Calculate heat energy released ( q = mcΔT ):
Total volume = 25.0 + 25.0 = 50.0 cm³ (mass m = 50.0 g )
q = 50.0 × 4.18 × 2.9 = 606.1 J = 0.6061 kJ - Calculate moles of quinic acid reacted:
Moles = (25 ÷ 1000) × 0.500 = 0.0125 mol - Calculate ΔneutH with correct sign ( − ):
ΔH = − q / moles = − 606.1 J / 0.0125 mol = − 48488 J mol⁻¹ = − 48.5 kJ mol⁻¹
(iii) Explaining Enthalpy Differences Between Weak and Strong Acids
✅ Correct Answer
HCN is a weak acid and is only partially dissociated (whereas HCl is fully dissociated). Therefore, energy is absorbed to break/complete dissociation of the H-C bond during neutralisation, reducing the net exothermic heat release.
🧠 Exam Technique
Examiners award 1 mark for stating the weak acid is only partially dissociated, and 1 mark for linking this to energy required/absorbed for further dissociation of bonds. Always explicitly mention bond breakage/dissociation.
Topics
Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Topic 2: Bonding and Structure · Topic 8: Energetics I · Topic 12: Acid-base Equilibria
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.