OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 21
12 marks · Hard difficulty · Structured Questions
Calculate the concentration of ethanoic acid from its pH and Ka, complete an acid-base equilibrium equation with conjugate pairs, calculate the pH of a buffer solution made from ethanoic acid and sodium ethanoate, and explain the effect of volume change on buffer pH.
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Question text
21 This question is about the properties and reactions of ethanoic acid, CH3COOH.
Ethanoic acid is a weak acid with an acid dissociation constant, K , of 1.75 × 10–5 mol dm–3 at
a
25 °C.
(a) A student uses a pH meter to measure the pH of a solution of CH3COOH at 25 °C.
The measured pH is 2.440.
Calculate the concentration of ethanoic acid in the solution.
Give your answer to three significant figures.
concentration = … mol dm–3 [3]
(b) Ethanoic acid is added to another weak acid, fluoroethanoic acid, FCH2COOH
(K = 2.19 × 10–3 mol dm–3). An equilibrium is set up containing two acid-base pairs.
a
Complete the equilibrium and label the conjugate acid-base pairs as A1, B1 and A2, B2.
CH3COOH + FCH2COOH … + …
[2]
(c) The student plans to prepare a buffer solution that has a pH of 4.50. The buffer solution will
contain ethanoic acid, CH3COOH, and sodium ethanoate, CH3COONa.
The student plans to add 9.08 g CH COONa to 250 cm3 of 0.800 mol dm–3 CH COOH. The
student assumes that the volume of the solution does not change.
(i) Show by calculation whether, or not, the student’s experimental method would produce
the required pH.
Show all your working.
[5]
(ii) When the student prepares the buffer solution, the volume of solution increases slightly.
Suggest whether the pH of the buffer solution would be the same, greater than, or less
than your calculated value in (c)(i).
Explain your reasoning.
… [2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
21 (a) FIRST, CHECK THE ANSWER ON ANSWER LINE 3
IF answer = 0.753, award 3 marks
--------------------------------------------------------------------------------- ALLOW use of HA and A–
[H+] = 10–pH = 10–2.440 = 3.63 10–3 (mol dm–3) ALLOW 3 SF up to calculator value of
3.630780548 10–3 correctly rounded
[H+]2 (3.63 10–3)2
[CH3COOH] = K OR –5
a 1.75 10
NOTE: Answer is same from unrounded [H+]
–3 calculator value and 3 SF [H+] value
= 0.753 (mol dm )
ALLOW 0.749 if [H+] has been subtracted
from [CH3COOH] for greater accuracy at end
(b) 2 Watch for opposite order on RHS, i.e.:
CH COOH + FCH COOH CH COOH + + FCH COO– FCH COO– + CH COOH +
32 3 2 2 2 3 2
B2 A1 A2 B1 Take great care matching labels
OR
B1 A2 A1 B2 ALLOW ECF for incorrect proton transfer as
i.e. labels other way round below. This is the ONLY ECF
CH COOH + FCH COOH CH COO– + FCH COOH +
32 3 2 2
A1 B2 B1 A2
OR
A2 B1 B2 A1 ECF
i.e. labels other way round
– 17
(c) (i) [CH3COO ] 5 ALLOW 2 sig fig
9.08 ALLOW use of HA and A–
n(CH3COONa) = 82.0 OR 0.111 (Calc: 0.1107317073)
– 9.08 1000 –3 Mark by ECF
[CH3COO ] = 82.0 250 = 0.443 (mol dm )
250 -----------------------------------------------------------
OR n(CH3COOH) = 0.800 1000 = 0.200 (mol)
[H+]
Alternative method
+ [CH3COOH] n(CH3COOH) (If both methods are attempted, mark the
[H ]= Ka – OR Ka –
[CH3COO ] n(CH3COO ) method which produces the higher mark)
[H+]
–5 0.800 –5 0.200 + –pH –4.50
= 1.75 10 0.443 OR 1.75 10 0.111 [H ] = 10 = 10
= 3.16 10–5 (mol dm–3)
= 3.16 10–5 (mol dm–3)
[CH COO–]
pH (must come from calculated [H+]) [CH COOH]
– 3
–5 [CH3COO ] = Ka +
pH = –log (3.16 10 ) = 4.50 [H ]
--------------------------------------------------------------- –5 0.800
OR 1.75 10 –5
LAST 3 marks are NOT available using 3.16 10
= 0.443 (mol dm–3)
Ka square root approach (weak acid pH)
–14 mass of CH3COONa
Kw /10 approach (strong base pH)
------------------------------------------------------------------------------------ mass CH3COONa = 0.443
1000
Henderson–Hasselbalch (HH) alternative
OR 0.111
pK = –log 1.75 10–5 = 4.757 (or 4.756961951..)
a
0.111 82.0 = 9.08 (g)
–
[CH3COO ] [CH3COOH]
pH = pKa + log OR = pKa – log –
[CH3COOH] [CH3COO ] -----------------------------------------------------------
0.443 0.800 Common errors
OR pKa + log 0.800 OR = pKa – log 0.443
4.64 Use of M(CH3COONa) = 60 4 marks
= pKa – 0.257
2.40 Use of Ka of FCH2COOH 4 marks
= 4.757 – 0.257 = 4.50
(ii) pH is the same/constant 18 2 M2 is dependent upon M1
ratio/proportion [HA]/[A–] is the same ALLOW Change in [HA] and [A–] is
proportional
Total 12
How to answer it
Properties and Reactions of Ethanoic Acid & Buffers
What this question tests
This question assesses your mastery of weak acid equilibria, pH calculations using acid dissociation constants (Kₐ), conjugate acid-base pairs, buffer solution preparation chemistry, and the understanding of how volume changes affect buffer ratios.
Calculating Acid Concentration from pH and Kₐ
✅ Correct Answer
concentration = 0.753 mol dm⁻³
📐 Step-by-Step Calculation
- Find [H⁺]: [H⁺] = 10⁻ᵖᴴ = 10⁻²·⁴⁴⁰ = 3.63 × 10⁻³ mol dm⁻³
- Rearrange Kₐ expression: Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]. Since [H⁺] ≈ [CH₃COO⁻], Kₐ = [H⁺]² / [CH₃COOH]
- Calculate concentration: [CH₃COOH] = (3.63 × 10⁻³)₂ / (1.75 × 10⁻⁵) = 0.753 mol dm⁻³
🧠 Exam Technique & Formatting
- Significant Figures: The question specifically requests 3 significant figures. Ensure trailing zeros are kept if applicable, and intermediate values are kept unrounded in your calculator.
❌ Common Errors
- Forgetting to square the [H⁺] term in the Kₐ expression.
- Using strong acid assumptions (assuming [H⁺] equals total acid concentration).
Completing Equilibria and Labelling Pairs
✅ Correct Answer
CH₃COOH + FCH₂COOH ⇌ CH₃COOH₂⁺ + FCH₂COO⁻
Pairs: CH₃COOH (B2) / CH₃COOH₂⁺ (A2) and FCH₂COOH (A1) / FCH₂COO⁻ (B1)
💡 Key Knowledge
- Acids are proton (H⁺) donors; bases are proton acceptors.
- In a competition between two weak acids, the stronger acid ( FCH₂COOH , higher Kₐ) acts as the acid, forcing the weaker acid ( CH₃COOH ) to accept a proton and act as a base.
❌ Common Errors
- Reversing the labels or failing to match the conjugate species correctly across the equilibrium arrow.
Evaluating Buffer pH from Mass and Concentration
✅ Correct Answer
Calculated pH = 4.50. Therefore, the student's method would produce the required pH.
📐 Step-by-Step Calculation
- Moles of salt: Mᵣ(CH₃COONa) = 82.0 g mol⁻¹. n(CH₃COONa) = 9.08 / 82.0 = 0.1107 mol
- Moles of acid: n(CH₃COOH) = 0.800 × (250 / 1000) = 0.200 mol
- Calculate [H⁺]: [H⁺] = Kₐ × (n(acid) / n(salt)) = 1.75 × 10⁻⁵ × (0.200 / 0.1107) = 3.16 × 10⁻⁵ mol dm⁻³
- Calculate pH: pH = -log(3.16 × 10⁻⁵) = 4.50
🧠 Exam Technique
- You can use either concentrations or moles directly in the Kₐ expression for buffers, because the total volume cancels out in the ratio.
- Clearly state your final conclusion ("would produce the required pH") to secure the final verification mark.
❌ Common Errors
- Using incorrect molar mass for sodium ethanoate (e.g., using 60 instead of 82.0).
- Using the Kₐ value of fluoroethanoic acid given in part (b) instead of ethanoic acid.
Effect of Total Volume Increase on Buffer pH
✅ Correct Answer
The pH is the same / constant because the ratio of [CH₃COOH] to [CH₃COO⁻] remains proportional/unchanged.
💡 Key Knowledge
- Buffer action depends on the ratio of weak acid to conjugate base concentrations, not their absolute individual volumes. If total volume increases slightly, both concentrations dilute by the exact same factor, leaving the ratio and therefore the pH unaffected.
❌ Common Errors
- ">
- Assuming that adding more solution volume automatically changes or ruins the buffer pH.
- Failing to link the constant pH to the constant acid-to-salt ratio.
Topics
Module 5: Physical chemistry and transition elements · 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 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.