Edexcel A-Level Chemistry Paper 1, June 2022: Question 6
17 marks · Hard difficulty · Open Response
Assess various aspects of acids, bases, pH calculations, ionic product of water variation with temperature, and titration curves including buffer action.
Practise this questionQuestion
Question text
6 This question is about acids and bases.
(a) State what is meant by a Brønsted-Lowry base.
(1)
(b) Write the ionic equation for the reaction between magnesium oxide and an acid.
State symbols are required.
(2)
–3
(c) Calculate the concentration of hydrogen ions, in moldm , in a solution with a
pH of 9.43
(1)
(d) The pH of two salt solutions, J and K, are
solution J pH = 5
solution K pH = 9
The solutions are equimolar.
Which acids and bases could form the salts in solutions J and K?
(1)
Acid and base forming the salt Acid and base forming the salt
in solution J in solution K
A HCl(aq) and NH (aq) CH COOH(aq) and NaOH(aq)
B HCl(aq) and NaOH(aq) CH COOH(aq) and NH (aq)
C CH COOH(aq) and NaOH(aq) HCl(aq) and NaOH(aq)
D CH COOH(aq) and NH (aq) HCl(aq) and NH (aq)
33 3
(e) The ionic product of water, Kw , varies with temperature as shown.
16 2 –6
*P67093RA01632*Temperature /°CKw/moldm
00.11 × 10–14
10 0.29 × 10–14
20 0.68 × 10–14
30 1.47 × 10–14
40 2.92 × 10–14
50 5.48 × 10–14
(i) Determine the value of Kw at 45°C by plotting a suitable graph.
You must show your working on the graph.
(3)
Kw at 45°C = …
*P67093RA01732*
(ii) The ionic product of water at 30 °C is 1.47 × 10–14 mol2 dm–6.
Calculate the pH of water at this temperature.
(3)
(f ) Hydrochloric acid, with a concentration of 0.100 mol dm–3, is added to 25.0 cm3 of
0.100 mol dm–3 aqueous sodium carbonate and the pH is measured.
The titration curve is shown.
12 X
first equivalence
10 point
pH
second equivalence
4 point
0 10 20 30 40 50
Volume HCl(aq) / cm3
The reaction takes place in two steps.
The equation for the reaction taking place in the first step is
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)
(i) Deduce a suitable indicator to identify the first equivalence point.
Justify your answer using values from the Data Booklet.
(2)
… 18
… *P67093RA01832*
(ii) Write the equation for the reaction taking place at the second
equivalence point.
State symbols are not required.
(1)
(iii) Explain how the solution at point X on the graph can act as a buffer solution.
(3)
(Total for Question 6 = 17 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
6(a) An answer that makes reference to the following (1)
point:
• (a Brønsted-Lowry base is a) proton acceptor Allow accepts protons / H+ (ions) / hydrogen
ions
Do not award additional references to
reacting with OH- / alkali
Question
Answer Additional Guidance Mark
Number
6(b) Example of equation (2)
• balanced equation (1) MgO(s) + 2H+(aq) → Mg2+(aq) + H O(l)
Allow multiples
• state symbols (1) Conditional on M1 or near miss e.g. Mg+
Allow a fully balanced equation with correct state symbols for 1 mark
e.g. MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)
e.g. MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l)
e.g. uncancelled spectator ions from the acid with (aq)
Do not award M1 for
Mg2+(s) + O2−(s) + 2H+(aq) → Mg2+(aq) + H O(l)
But M2 can be awarded for correct state symbols
Question
Answer Additional Guidance Mark
Number
6(c) Example of calculation (1)
• calculation of [H+(aq)] [H+(aq)] = 10−pH = 10−9.43
= 3.7154 x 10−10 / 3.715 x 10−10 / 3.72 x
10−10 /
3.7 x 10−10 (mol dm−3)
Do not award 3.71 X 10-10
Ignore units even if incorrect
Ignore SF except 1 SF
Correct answer with no working scores (1)
Question Answer Mark
number
6(d) (1)
The only correct answer is A (solution J: HCl(aq) and NH3(aq), solution K: CH3COOH(aq) and NaOH(aq))
B is incorrect because the salt formed from a strong acid (HCl) and a strong base (NaOH) will have pH 7
while that formed from a weak acid (CH3COOH) and a weak base (NH3) will have pH close to 7
C is incorrect because the salt formed from a weak acid and a strong base will have a pH of about 9 while
that formed from a strong acid and a strong base will have pH 7
D is incorrect because the salt formed from a weak acid and a weak base will have a pH of about 7 while
that formed from a strong acid and a weak base will have pH of about 5
Question
Answer Additional Guidance Mark
Number
6(e)(i) Example of graph (3)
• axes the correct way round, labelled, including
units
and
suitable scale with points covering at least half
the paper in both directions (1)
K x 1014
w
• points plotted correctly (±1/2 small square) / mol2 dm−6
and
smooth curve (1)
Temperature /°C
• value of K at 45°C (1) Allow K / 10−14 /mol2 dm-6 as units on y axis
w w
Allow K x 10−14/mol2 dm-6
w
4.0 x 10−14 (mol2 dm−6)
Allow 3.8 to 4.2 x 10−14 (mol2 dm−6) with no working
TE on their working from their graph
If they have converted Kw to pKw, drawn a graph with
correctly labelled axes and line of best fit then they
can access all three marks as long as their final
answer is Kw
Question
Answer Additional Guidance Mark
Number
6(e)(ii) Example of calculation (3)
• deduction of expression relating K and (K = [H+(aq)][OH−(aq)]
w w
[H+(aq)] (1) but [H+(aq)] = [OH−(aq)] so)
K = [H+(aq)]2
w
• calculation of [H+(aq)] (1) [H+(aq)]2 = 1.47 x 10−14
[H+(aq)] = 1.47 x 10−14
(so [H+(aq)] = 1.2124 x 10−7 (mol dm−3))
• calculation of pH (1)
pH = −log1.2124 x 10−7
= 6.9163 / 6.916 / 6.92 / 6.9
Do not award 1SF or final answer of 7 or answer
incorrectly rounded to 6.91
pH TE on [H+]
Correct answer with no working scores (3)
Allow alternative methods
Question
Answer Additional Guidance Mark
Number
6(f)(i) (2)
• phenolphthalein (1) Allow recognisable spellings
• pH at equivalence point / 9 is very close / ±1 to Allow indicator will change colour in the vertical
pKin / 9.3 section of the curve / at the end / equivalence
or point
pH range is (completely) within the (first) Accept correct reference to the pH range for
vertical jump in the titration curve / between phenolphthalein from the data book (8.2-10.0)
the range of (pH) 8.5 - pH9.5 (1) if there is a connection to the graph
Do not allow colourless to pink/red if the
colour change of phenolphthalein is mentioned
Question
Answer Additional Guidance Mark
Number
6(f)(ii) Example of equation (1)
• equation NaHCO3 + HCl → NaCl + H2O + CO2
or
HCO − + H+ → H O + CO
32 2
Allow
NaHCO3 + HCl → NaCl + H2CO3
Allow multiples
Ignore state symbols even if incorrect
Question
Answer Additional Guidance Mark
Number
6(f)(iii) • (solution at X) contains a large Allow there is a large amount of Na2CO3 and NaHCO3 (3)
amount of / reservoir of carbonate Allow solution at X contains a reservoir of an acid and its
ions / CO 2− conjugate base
and
hydrogencarbonate ions/ HCO − (1)
• carbonate ions / CO 2− react with Allow Na CO reacts with added hydrogen ions / H+ / acid
32 3
added hydrogen ions / H+ / acid to form NaHCO
or or
CO 2− + H+ → HCO − (1) CO 2− + HCl → HCO − + Cl-
33 3 3
or
A- + H+ → HA
• hydrogencarbonate ions / HCO − Allow NaHCO reacts with added hydroxide ions (to form
react with added hydroxide ions / Na2CO3 + H2O)
OH− / alkali Allow hydroxide ions react with hydrogen ions to form water
or and hydrogencarbonate ions dissociate to replace / form
HCO − + OH− → CO 2− + H O (1) hydrogen ions
33 2
or
OH− + H+ → H O and HCO − → CO 2− + H+
23 3
or
HA + OH- → A- + H O
Allow ⇌ in equations
Ignore state symbols
(Total for Question 6 = 17 marks)
How to answer it
Acids, Bases, and Ionic Equilibria Study Guide
What this question tests
This comprehensive Edexcel A-Level Chemistry question tests core understanding of Brønsted-Lowry acid-base theory, writing ionic equations with state symbols, calculating hydrogen ion concentrations and pH from Kw, interpreting salt hydrolysis and pH values, graphical analysis of temperature-dependent equilibria, and explaining buffer action mechanisms using titration curves.
Brønsted-Lowry Definition
✅ Correct Answer
A Brønsted-Lowry base is a proton acceptor (accepts H⁺ ions).
❌ Common Errors
Students often lose this mark by mentioning "reacting with OH⁻ ions" or stating it is an alkali that releases hydroxide. Keep the definition strictly focused on proton acceptance.
Ionic Equation for Magnesium Oxide and an Acid
✅ Correct Answer
MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l)
(Accepts alternative acid anions if balanced correctly, e.g., using HCl or H₂SO₄, provided spectator ions are omitted or correctly balanced).
💡 Key Knowledge
State symbols are strictly required. Magnesium oxide is a solid ( s ), hydrogen ions are aqueous ( aq ), magnesium ions are aqueous ( aq ), and water is a liquid ( l ).
❌ Common Errors
Failing to include state symbols or leaving spectator ions like Cl⁻ in the ionic equation. Do not write O²⁻ as a separate reactant unless showing the full ionic breakdown.
Calculating Hydrogen Ion Concentration from pH
📐 Step-by-Step Calculation
- Recall formula: pH = -log[H⁺], therefore [H⁺] = 10⁻ᵖᴴ
- Substitute values: [H⁺] = 10⁻⁹·⁴³
- Calculate result: 3.72 × 10⁻¹⁰ mol dm⁻³ (Accept values between 3.7 × 10⁻¹⁰ and 3.75 × 10⁻¹⁰).
🧠 Exam Technique
Make sure you know how to use the inverse log function on your calculator ( 10ˣ or shift log ). Examiner allows any number of significant figures here (even 1 SF), but standard practice is 2 or 3 SF.
Salt Solutions and pH Interpretation
✅ Correct Answer
Box A is correct.
Solution J (pH = 5, acidic): Formed from a strong acid ( HCl ) and a weak base ( NH₃ ).
Solution K (pH = 9, alkaline): Formed from a weak acid ( CH₃COOH ) and a strong base ( NaOH ).
💡 Key Knowledge
Strong acid + strong base = neutral (pH 7). Weak acid + weak base = approximately neutral. Strong acid + weak base = acidic salt solution (pH < 7). Weak acid + strong base = alkaline salt solution (pH > 7).
Ionic Product of Water (Kw) and Temperature
(i) Determine Kw at 45 °C from Graph
🧠 Exam Technique & Graph Skills
- Ensure axes are correctly oriented, labelled with physical quantities and units (e.g., Kw / 10⁻¹&sup4; mol² dm⁻⁶ ).
- Use a sensible scale occupying at least half the available grid space in both directions.
- Plot points accurately (± half a small square) and draw a smooth best-fit curve.
- Read off the value at 45 °C. Expected range: 4.0 × 10⁻¹&sup4; to 4.2 × 10⁻¹&sup4; mol² dm⁻⁶.
(ii) Calculate pH of Water at 30 °C
📐 Step-by-Step Calculation
- Identify expression: Kw = [H⁺][OH⁻]. Since water is neutral, [H⁺] = [OH⁻], so Kw = [H⁺]².
- Find [H⁺]: At 30 °C, Kw = 1.47 × 10⁻¹&sup4;. Therefore, [H⁺] = √(1.47 × 10⁻¹&sup4;) = 1.2124 × 10⁻⁷ mol dm⁻³.
- Calculate pH: pH = -log(1.2124 × 10⁻⁷) = 6.92 (Accept 6.9 to 6.92).
❌ Common Errors
Assuming water always has a neutral pH of 7 at all temperatures. As autoionization of water is endothermic, increasing temperature increases Kw, increasing [H⁺] and dropping neutral pH slightly below 7.
Titration Curves, Indicators, and Buffers
(i) Suitable Indicator for First Equivalence Point
✅ Correct Answer
Phenolphthalein.
Justification: The first equivalence point occurs at pH 9, which falls directly within the vertical jump range of phenolphthalein (pH 8.2 – 10.0) or matches its pKIn of 9.3.
(ii) Equation at Second Equivalence Point
✅ Correct Answer
NaHCO₃ + HCl → NaCl + H₂O + CO₂
(Or using ionic form: HCO₃⁻ + H⁺ → H₂O + CO₂). State symbols are not required.
(iii) Explanation of Buffer Action at Point X
💡 Key Knowledge & Explanation
- Composition: At point X, the solution contains a large reserve/mixture of unreacted carbonate ions ( CO₃²⁻ ) and formed hydrogen carbonate ions ( HCO₃⁻ ).
- Adding acid (H⁺): CO₃²⁻ + H⁺ → HCO₃⁻ (Carbonate ions react with added hydrogen ions to minimize pH change).
- Adding alkali (OH⁻): HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O (Hydrogen carbonate ions react with added hydroxide ions to neutralize them).
🧠 Exam Technique
To secure all 3 marks, you must explicitly state: 1) What species are present creating the buffer, 2) How added H⁺ is removed, and 3) How added OH⁻ is removed. Use clear chemical equations to earn chemical credit.
Topics
Physical Chemistry · Topic 12: Acid-base Equilibria
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.