Edexcel A-Level Chemistry Paper 1, November 2021: Question 7
15 marks · Medium difficulty · Calculations
Calculate the pH of various acid and buffer solutions and explain related acid-base concepts.
Practise this questionQuestion
Question text
7 This question is about acids and buffer solutions.
(a) Ethanoic acid, CH3COOH, is a monobasic acid.
CH COOH + H O CH COO– + H O+
32 3 3
Give a reason why only the proton from the carboxylic acid group, and not from
the methyl group, is donated to a water molecule.
(1)
(b) The reaction of ammonia with water can be represented by
NH + H O NH+ + OH–
32 4
Which is the acid-conjugate base pair?
(1)
Acid Conjugate base
A NH OH–
B NH NH+
C H O OH–
D H O NH+
(c) A commercial nitric acid solution, HNO (aq), has a concentration of 15.9 mol dm–3.
A 15.0 cm3 sample was made up to 100 cm3 by adding deionised water.
Calculate the pH of this diluted solution.
(2)
(d) Propanoic acid is a weak acid.
(i) Calculate the pH of a 0.100 mol dm–3 solution of propanoic acid at 298 K.
Give your answer to an appropriate number of significant figures.
16 –5 –3
[Ka = 1.35×10 mol*P65463A01628*dmat 298K]
(3)
(ii) State two assumptions that you made in the calculation in d(i).
(2)
(e) A buffer solution was made using 20.0 cm3 of a butanoic acid solution, of
concentration 0.100 mol dm–3 and 30.0 cm3 of sodium butanoate solution,
of concentration 0.305 mol dm–3.
[K = 1.52 × 10–5mol dm–3 at 298K] 17
a *P65463A01728*
(i) Calculate the pH of this buffer solution at 298K.
(4)
*P65463A01828*
(ii) Explain why the pH of the buffer solution hardly changes when a few drops of
sodium hydroxide solution are added to it.
Include an equation or equations in your answer.
Use C3H7COOH as the formula for butanoic acid.
(2)
(Total for Question 7 = 15 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
7(a) An answer that makes reference to one of the following points (1)
• the loss of a hydrogen from the O−H group is made possible by the
delocalisation of charge of/stabilisation on the carboxylate ion
or
the loss of a hydrogen from a methyl group would produce a
carbanion with no stabilisation
or
similar electronegativies of carbon and hydrogen means Allow the C−H bond is not polar but the
that there is a lack of C−H bond polarity O−H bond is/ O−H bond is more polar
or
the enthalpy of hydration of the ions outweighs the energy needed
to break the O−H bond Do not award the O−H bond is weaker
than the C−H bond
Question
Answer Mark
Number
7(b) The only correct answer is C (H O and OH−) (1)
A is not correct because ammonia is acting as a base and not an acid
B is not correct because this is the base – conjugate acid pair
D is not correct because water and the ammonium ion are not an acid-conjugate base pair
Question
Answer Additional Guidance Mark
Number
(7c) Example of calculation (2)
• (M1) calculation of concentration of diluted acid (1) c=(15 x 15.9 / 100) = 2.385 (mol dm−3)
• (M2) calculation of pH (1) pH=−log(2.385) = −0.377/−0.38 / −0.4
TE on M1 provided answer is <7
Final answer without working scores (2)
Ignore SF
Question
Answer Additional Guidance Mark
Number
7(d)(i) Example of calculation (3)
• expression for K (1) K = [H+] x [A−]
a a
[HA]
+ [H+] = √( K x [HA]) = √(1.35 x 10−6 )
• calculation of [H ] (1) a
= 1.16... x 10−3 (mol)
• calculation of pH to 2/3 SF (1)
pH= -log(1.16... x 10−3) = 2.93/2.9
TE on M2 provided answer <7
Final answer without working scores (3)
Question
Answer Additional Guidance Mark
Number
7(d)(ii) An answer which makes reference to the following points ACCEPT assumptions in any order (2)
Allow HA for C2H5COOH
Allow A− for C H COO−
• (assumption 1) [C2H5COOH]initial=[C2H5COOH]eqm (1) Dissociation of propanoic acid is
negligible
Ignore propanoic acid is a weak acid
• (assumption 2) [H+]=[C H COO−] (1) ALLOW for M2
“Negligible [H+] from water”
Ignore reference to standard conditions
Question
Answer Additional Guidance Mark
Number
7(e)(i) Example of calculation (4)
• calculation of acid concentration (1) [Acid] = ((0.100 x (20 ÷ 50))
= 0.04 (mol dm−3)
[A−] = ((0.305 x (30 ÷ 50))
• calculation of salt concentration (1)
= 0.183 (mol dm−3)
[H+] = 1. 52 x 10−5 mol x (0.04 ÷ 0.183)
• calculation of hydrogen ion concentration (1)
= 3.322 x 10−6 (mol dm−3)
pH= -log(3.322 x 10−6)
• calculation of pH (1)
= 5.48/5.5
Correct answer without working scores (4)
Ignore SF except 1SF
Allow M3 and M4 if just moles and no volumes are used
Accept use of the Henderson-Hasselbalch equation
Question
Answer Additional Guidance Mark
Number
7(e)(ii) An answer which includes Example equation (2)
• suitable equation(s) (1) C3H7COOH + NaOH →C3H7COONa + H2O
OR
C H COOH + OH− →C H COO− + H O
37 3 7 2
Allow
OH− + H+ → H O followed by
C H COOH →C H COO− + H+
37 3 7
Allow use ⇌ of in all of above equations
• The pH stays approximately constant because there Allow (The pH stays approximately constant) as the
is a large reservoir of undissociated acid and so the hydroxide ions react to form water and butanoic acid
ratio of acid:salt does not change (1) dissociates to replace the hydrogen ions used up
(Total Question 7 = 15 marks)
How to answer it
Acids, Bases and Buffer Solutions Study Guide
What this question tests
This comprehensive question tests core physical chemistry topics including Brønsted-Lowry acid-base theory, conjugate pairs, strong and weak acid pH calculations, dilution factors, approximations for weak acid dissociation, and the composition and action of buffer solutions.
Part (a) — Acid Structure and Proton Donation
Give a reason why only the proton from the carboxylic acid group is donated.
✅ Correct Answer
- Loss of hydrogen from the O-H group is stabilised by the delocalisation of charge on the carboxylate ion ( CH₃COO⁻ ).
- Alternatively: The C-H bond in the methyl group is not polar, whereas the O-H bond is polar.
❌ Common Errors
- Stating incorrectly that "the O-H bond is weaker than the C-H bond" (this is not accepted by the mark scheme).
- Failing to mention charge delocalisation or ion stabilisation.
Part (b) — Acid-Conjugate Base Pairs
Identify the correct acid-conjugate base pair for the reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
✅ Correct Answer
C — Acid: H₂O | Conjugate base: OH⁻
🧠 Exam Technique
Remember that an acid donates a proton to become its conjugate base (differing by one H⁺ ion). Water donates a proton to become hydroxide ( OH⁻ ).
Part (c) — Strong Acid Dilution and pH Calculation
Calculate the pH of a 15.9 mol dm⁻³ nitric acid solution diluted from 15.0 cm³ to 100 cm³.
📐 Step-by-Step Calculation
- Step 1: Calculate new concentration ( c₁V₁ = c₂V₂ )
c = (15.0 × 15.9) / 100 = 2.385 mol dm⁻³ - Step 2: Calculate pH
pH = -log(2.385) = -0.38 (or -0.37 / -0.4)
❌ Common Errors & Traps
- Forgetting to adjust concentration when volume changes (using original 15.9 mol dm⁻³ directly).
- Panicking over negative pH values—strong concentrated acids can indeed have negative pH!
Part (d) — Weak Acid Calculations and Assumptions
(i) Calculate the pH of 0.100 mol dm⁻³ propanoic acid (Kₐ = 1.35 × 10⁻⁵ mol dm⁻³ at 298 K). Give answer to appropriate s.f.
📐 Step-by-Step Calculation
- Step 1: Kₐ expression
Kₐ = [H⁺][CH₃CH₂COO⁻] / [CH₃CH₂COOH] - Step 2: Calculate [H⁺]
Assuming [H⁺] = [A⁻] and [HA]eq ≈ [HA]initial :
[H⁺] = √(Kₐ × [HA]) = √(1.35 × 10⁻⁵ × 0.100) = 1.162 × 10⁻³ mol dm⁻³ - Step 3: Calculate pH
pH = -log(1.162 × 10⁻³) = 2.93 (2 or 3 significant figures accepted).
💡 Key Knowledge: (ii) Assumptions Made
- Assumption 1: Dissociation of propanoic acid is negligible (so [HA]initial ≈ [HA]equilibrium ).
- Assumption 2: Hydrogen ions come entirely from the acid dissociation, so [H⁺] = [A⁻] (ignoring auto-ionisation of water).
Part (e) — Buffer Solutions
(i) Calculate the pH of a buffer made from 20.0 cm³ of 0.100 mol dm⁻³ butanoic acid and 30.0 cm³ of 0.305 mol dm⁻³ sodium butanoate (Kₐ = 1.52 × 10⁻⁵ mol dm⁻³).
(ii) Explain why the pH hardly changes when a few drops of NaOH are added.
📐 Step-by-Step Calculation (Part i)
- Step 1: Acid moles / concentration
Moles = 0.100 × (20/1000) = 0.002 mol
New conc in 50 cm³ total volume: 0.002 / 0.050 = 0.040 mol dm⁻³ - Step 2: Salt moles / concentration
Moles = 0.305 × (30/1000) = 0.00915 mol
New conc in 50 cm³ total volume: 0.00915 / 0.050 = 0.183 mol dm⁻³ (or use moles directly in Kₐ expression since volumes cancel). - Step 3: Rearrange for [H⁺]
[H⁺] = Kₐ × ([HA] / [A⁻]) = (1.52 × 10⁻⁵ × 0.040) / 0.183 = 3.32 × 10⁻⁶ mol dm⁻³ - Step 4: Calculate pH
pH = -log(3.32 × 10⁻⁶) = 5.48 (Accept 5.5).
✅ Explanation (Part ii)
- Equation: C₃H₇COOH + OH⁻ → C₃H₇COO⁻ + H₂O
- Action: There is a large reservoir of undissociated acid and butanoate ions. Added hydroxide ions react with the acid, so the ratio of acid to salt remains virtually constant, keeping the pH stable.
Topics
Physical Chemistry · Topic 12: Acid-base Equilibria
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.