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A Level H1 Chemistry Practice Paper 1
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TuitionGoWhere Practice Paper – Chemistry H1 A-Level
TuitionGoWhere Exam Practice (AI)
Subject: Chemistry H1
Level: A-Level
Paper: PRACTICE – Acids, Bases & Salts
Duration: 1 hour 30 minutes
Total Marks: 60
Version: 1 of 5
Name: _________________________
Class: _________________________
Date: _________________________
Instructions to Candidates
- This paper consists of three sections: Section A, Section B, and Section C.
- Answer all questions in the spaces provided.
- Show all working clearly for calculation questions. Marks are awarded for method.
- You may use a calculator.
- A Data Booklet is provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- You are advised to spend no more than 45 minutes on Section A, 30 minutes on Section B, and 15 minutes on Section C.
Section A: Structured Questions (30 marks)
Answer all questions in this section.
1. What is meant by the term weak acid? Illustrate your answer with a suitable equation, including state symbols. [2]
2. A student titrates 25.0 cm³ of ethanoic acid solution with 0.100 mol dm⁻³ sodium hydroxide solution. The volume of sodium hydroxide required to reach the end-point is 22.50 cm³.
(a) Write a balanced equation for the reaction between ethanoic acid and sodium hydroxide. [1]
(b) Calculate the concentration of the ethanoic acid solution in mol dm⁻³. [2]
3. Carbonic acid, H₂CO₃, is a weak diprotic acid found in carbonated drinks.
(a) Write an equation, including state symbols, for the first dissociation of carbonic acid in water. [1]
(b) Write the expression for the acid dissociation constant, Kₐ₁, for this dissociation. [1]
(c) The Kₐ₁ of carbonic acid is 4.3 × 10⁻⁷ mol dm⁻³ at 298 K. Calculate the pH of a 0.050 mol dm⁻³ solution of carbonic acid, assuming the second dissociation is negligible. [3]
4. Aluminium oxide, Al₂O₃, is described as an amphoteric oxide.
(a) Explain what is meant by the term amphoteric. [1]
(b) Write a balanced equation for the reaction of aluminium oxide with dilute hydrochloric acid. [1]
(c) Write a balanced equation for the reaction of aluminium oxide with aqueous sodium hydroxide. [1]
5. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid (CH₃COOH) with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate (CH₃COONa). The Kₐ of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³ at 298 K.
(a) Explain what is meant by a buffer solution. [1]
(b) Calculate the pH of this buffer solution. [3]
(c) Explain, using equations where appropriate, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [2]
6. The pH of a 0.010 mol dm⁻³ solution of hydrochloric acid is 2.0, while the pH of a 0.010 mol dm⁻³ solution of ethanoic acid is approximately 3.4.
(a) Calculate the hydrogen ion concentration, [H⁺], in the hydrochloric acid solution. [1]
(b) Explain why the pH values of these two solutions are different, despite having the same concentration. [2]
7. A student investigates the solubility of calcium hydroxide, Ca(OH)₂, in water at 298 K. A saturated solution of calcium hydroxide is filtered, and 25.0 cm³ of the filtrate requires 18.60 cm³ of 0.050 mol dm⁻³ hydrochloric acid for complete neutralisation.
(a) Write a balanced equation for the reaction between calcium hydroxide and hydrochloric acid. [1]
(b) Calculate the concentration of calcium hydroxide in the saturated solution in mol dm⁻³. [2]
(c) Hence, calculate the solubility product, Kₛₚ, of calcium hydroxide at 298 K, and state its units. [2]
Section B: Data-Based & Application Questions (20 marks)
Answer all questions in this section.
8. Methanoic acid, HCOOH, is the simplest carboxylic acid and is found in ant venom. It is a weak monoprotic acid with Kₐ = 1.78 × 10⁻⁴ mol dm⁻³ at 298 K.
(a) Calculate the pH of a 0.100 mol dm⁻³ solution of methanoic acid. State any assumption you make. [3]
(b) A 0.100 mol dm⁻³ solution of methanoic acid has a lower pH than a 0.100 mol dm⁻³ solution of ethanoic acid (Kₐ = 1.74 × 10⁻⁵ mol dm⁻³). Explain this observation. [1]
(c) Calculate the percentage dissociation of methanoic acid in the 0.100 mol dm⁻³ solution. [2]
9. The following table shows the pH ranges over which three acid-base indicators change colour.
| Indicator | pH range | Colour change (acid → alkali) |
|---|---|---|
| Methyl orange | 3.1 – 4.4 | Red → Yellow |
| Bromothymol blue | 6.0 – 7.6 | Yellow → Blue |
| Phenolphthalein | 8.3 – 10.0 | Colourless → Pink |
(a) A student titrates 0.100 mol dm⁻³ hydrochloric acid (a strong acid) against 0.100 mol dm⁻³ sodium hydroxide (a strong base). Which indicator from the table would be most suitable? Explain your choice. [2]
(b) Another student titrates 0.100 mol dm⁻³ ethanoic acid (a weak acid) against 0.100 mol dm⁻³ sodium hydroxide (a strong base). Which indicator from the table would be most suitable? Explain your choice. [2]
10. The concentration of ethanoic acid in a sample of vinegar can be determined by titration with standardised sodium hydroxide solution. A student dilutes 10.0 cm³ of vinegar to 100.0 cm³ with distilled water in a volumetric flask. 25.0 cm³ portions of this diluted solution are titrated against 0.100 mol dm⁻³ NaOH(aq). The average titre is 21.30 cm³.
(a) Calculate the concentration of ethanoic acid in the diluted vinegar solution. [2]
(b) Calculate the concentration of ethanoic acid in the original vinegar, in mol dm⁻³. [1]
(c) The density of the original vinegar is 1.01 g cm⁻³. Calculate the concentration of ethanoic acid in the original vinegar as a percentage by mass. [Mᵣ of CH₃COOH = 60.0] [3]
Section C: Free-Response Questions (10 marks)
Answer all questions in this section.
11. Discuss the role of acid-base chemistry in two of the following contexts:
- The action of antacid tablets in relieving indigestion
- The importance of buffer systems in maintaining blood pH
- The environmental impact of acid rain on limestone buildings
In your answer, you should include relevant chemical equations and explain the underlying chemical principles. [10]
END OF PAPER
Answers
TuitionGoWhere Practice Paper – Chemistry H1 A-Level
Answer Key and Marking Scheme
Paper: PRACTICE – Acids, Bases & Salts
Version: 1 of 5
Total Marks: 60
Section A: Structured Questions (30 marks)
1. What is meant by the term weak acid? Illustrate your answer with a suitable equation, including state symbols. [2]
Answer: A weak acid is an acid that only partially dissociates/ionises in aqueous solution. [1 mark]
Suitable equation, e.g.: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) Accept any valid weak acid equation with reversible arrow (⇌) and correct state symbols. [1 mark]
Marking notes:
- Award [1] for definition including "partially dissociates/ionises"
- Award [1] for correct equation with ⇌ and state symbols
- Do not accept "dilute acid" or equation with single arrow (→)
- Accept H₃O⁺ instead of H⁺
2. (a) Write a balanced equation for the reaction between ethanoic acid and sodium hydroxide. [1]
Answer: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l) [1 mark]
Marking notes:
- Accept CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
- State symbols not required but good practice
(b) Calculate the concentration of the ethanoic acid solution in mol dm⁻³. [2]
Answer: n(NaOH) = c × V = 0.100 × (22.50/1000) = 0.00225 mol [1 mark] Mole ratio CH₃COOH : NaOH = 1 : 1 n(CH₃COOH) = 0.00225 mol c(CH₃COOH) = n/V = 0.00225 / (25.0/1000) = 0.0900 mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct moles of NaOH
- Award [1] for correct final answer with units
- Accept 0.09 mol dm⁻³ (2 sf) or 0.0900 mol dm⁻³ (3 sf)
- Deduct [1] if volume not converted to dm³
3. (a) Write an equation, including state symbols, for the first dissociation of carbonic acid in water. [1]
Answer: H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq) [1 mark]
Marking notes:
- Must use reversible arrow (⇌)
- State symbols required: (aq) for all species except possibly H₂O(l) if included
- Accept H₃O⁺ instead of H⁺
(b) Write the expression for the acid dissociation constant, Kₐ₁, for this dissociation. [1]
Answer: Kₐ₁ = [HCO₃⁻][H⁺] / [H₂CO₃] [1 mark]
Marking notes:
- Square brackets essential
- Water not included in expression
- Accept [H₃O⁺] in place of [H⁺]
(c) Calculate the pH of a 0.050 mol dm⁻³ solution of carbonic acid. [3]
Answer: Kₐ = [H⁺]² / [H₂CO₃] (assuming [H⁺] = [HCO₃⁻]) [1 mark] 4.3 × 10⁻⁷ = [H⁺]² / 0.050 [H⁺]² = 4.3 × 10⁻⁷ × 0.050 = 2.15 × 10⁻⁸ [H⁺] = √(2.15 × 10⁻⁸) = 1.47 × 10⁻⁴ mol dm⁻³ [1 mark] pH = −log₁₀(1.47 × 10⁻⁴) = 3.83 [1 mark]
Marking notes:
- Award [1] for correct assumption and substitution
- Award [1] for correct [H⁺]
- Award [1] for correct pH (accept 3.83 or 3.8)
- Deduct [1] if assumption not stated but otherwise correct working shown
4. (a) Explain what is meant by the term amphoteric. [1]
Answer: An amphoteric substance is one that can react with both acids and bases. [1 mark]
Marking notes:
- Accept "a substance that can behave as both an acid and a base"
- Accept "reacts with both acids and alkalis"
(b) Write a balanced equation for the reaction of aluminium oxide with dilute hydrochloric acid. [1]
Answer: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) [1 mark]
Marking notes:
- Accept Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O
- Balanced equation required
(c) Write a balanced equation for the reaction of aluminium oxide with aqueous sodium hydroxide. [1]
Answer: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) OR Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq) [1 mark]
Marking notes:
- Accept Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄
- Accept formation of aluminate ion: Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻
- Balanced equation required
5. (a) Explain what is meant by a buffer solution. [1]
Answer: A buffer solution is one that resists changes in pH when small amounts of acid or base are added. [1 mark]
Marking notes:
- Must mention "resists changes in pH"
- Accept "maintains approximately constant pH"
(b) Calculate the pH of this buffer solution. [3]
Answer: After mixing: [CH₃COOH] = (50.0 × 0.200) / 100.0 = 0.100 mol dm⁻³ [1 mark] [CH₃COO⁻] = (50.0 × 0.200) / 100.0 = 0.100 mol dm⁻³ [1 mark] Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH] 1.74 × 10⁻⁵ = (0.100)[H⁺] / 0.100 [H⁺] = 1.74 × 10⁻⁵ mol dm⁻³ pH = −log₁₀(1.74 × 10⁻⁵) = 4.76 [1 mark]
Marking notes:
- Award [1] for correct diluted concentrations
- Award [1] for correct substitution into Kₐ expression
- Award [1] for correct pH (4.76)
- Accept use of Henderson-Hasselbalch equation: pH = pKₐ + log([salt]/[acid])
(c) Explain, using equations where appropriate, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [2]
Answer: When H⁺ ions are added (from HCl), they react with the conjugate base: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [1 mark] The added H⁺ ions are removed from solution, so the pH remains approximately constant. The ratio [CH₃COO⁻]/[CH₃COOH] changes only slightly. [1 mark]
Marking notes:
- Award [1] for correct equation showing H⁺ reacting with CH₃COO⁻
- Award [1] for explanation that H⁺ is removed/neutralised, maintaining pH
- Accept explanation in terms of equilibrium shift: CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts left
6. (a) Calculate the hydrogen ion concentration, [H⁺], in the hydrochloric acid solution. [1]
Answer: pH = 2.0 [H⁺] = 10⁻²·⁰ = 0.010 mol dm⁻³ [1 mark]
Marking notes:
- Accept 1.0 × 10⁻² mol dm⁻³
- Award mark for correct answer only
(b) Explain why the pH values of these two solutions are different, despite having the same concentration. [2]
Answer: HCl is a strong acid and dissociates completely in water, so [H⁺] = 0.010 mol dm⁻³. [1 mark] Ethanoic acid is a weak acid and only partially dissociates, so [H⁺] is much less than 0.010 mol dm⁻³, giving a higher pH. [1 mark]
Marking notes:
- Award [1] for stating HCl is strong/fully dissociates
- Award [1] for stating ethanoic acid is weak/partially dissociates
- Must link degree of dissociation to [H⁺] and pH
7. (a) Write a balanced equation for the reaction between calcium hydroxide and hydrochloric acid. [1]
Answer: Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) [1 mark]
Marking notes:
- Accept ionic equation: OH⁻ + H⁺ → H₂O (with correct stoichiometry)
- Balanced equation required
(b) Calculate the concentration of calcium hydroxide in the saturated solution in mol dm⁻³. [2]
Answer: n(HCl) = 0.050 × (18.60/1000) = 9.30 × 10⁻⁴ mol [1 mark] Mole ratio Ca(OH)₂ : HCl = 1 : 2 n(Ca(OH)₂) = 9.30 × 10⁻⁴ / 2 = 4.65 × 10⁻⁴ mol c(Ca(OH)₂) = 4.65 × 10⁻⁴ / (25.0/1000) = 0.0186 mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct moles of HCl
- Award [1] for correct concentration with units
- Accept 0.019 mol dm⁻³ (2 sf)
(c) Hence, calculate the solubility product, Kₛₚ, of calcium hydroxide at 298 K, and state its units. [2]
Answer: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) [Ca²⁺] = 0.0186 mol dm⁻³ [OH⁻] = 2 × 0.0186 = 0.0372 mol dm⁻³ [1 mark] Kₛₚ = [Ca²⁺][OH⁻]² = (0.0186)(0.0372)² = 2.57 × 10⁻⁵ Units: mol³ dm⁻⁹ [1 mark]
Marking notes:
- Award [1] for correct [OH⁻] and substitution
- Award [1] for correct Kₛₚ value and units
- Accept 2.6 × 10⁻⁵ mol³ dm⁻⁹
Section B: Data-Based & Application Questions (20 marks)
8. (a) Calculate the pH of a 0.100 mol dm⁻³ solution of methanoic acid. State any assumption you make. [3]
Answer: Assumption: [H⁺] << 0.100 mol dm⁻³, so [HCOOH]ₑq ≈ 0.100 mol dm⁻³ OR degree of dissociation is small. [1 mark] Kₐ = [HCOO⁻][H⁺] / [HCOOH] = [H⁺]² / 0.100 1.78 × 10⁻⁴ = [H⁺]² / 0.100 [H⁺]² = 1.78 × 10⁻⁵ [H⁺] = 4.22 × 10⁻³ mol dm⁻³ [1 mark] pH = −log₁₀(4.22 × 10⁻³) = 2.37 [1 mark]
Marking notes:
- Award [1] for stating assumption
- Award [1] for correct [H⁺]
- Award [1] for correct pH (2.37 or 2.4)
- Check assumption: 4.22 × 10⁻³ / 0.100 × 100% = 4.22% (< 5%, assumption valid)
(b) Explain this observation. [1]
Answer: Methanoic acid has a larger Kₐ value, so it dissociates to a greater extent, producing a higher [H⁺] and therefore a lower pH. [1 mark]
Marking notes:
- Must link Kₐ to degree of dissociation and [H⁺]
- Accept "methanoic acid is a stronger weak acid than ethanoic acid"
(c) Calculate the percentage dissociation of methanoic acid in the 0.100 mol dm⁻³ solution. [2]
Answer: [H⁺] = 4.22 × 10⁻³ mol dm⁻³ = amount dissociated [1 mark] Percentage dissociation = ([H⁺] / initial concentration) × 100 = (4.22 × 10⁻³ / 0.100) × 100 = 4.22% [1 mark]
Marking notes:
- Award [1] for correct formula/method
- Award [1] for correct answer (4.2% or 4.22%)
- Accept use of [HCOO⁻] instead of [H⁺]
9. (a) Which indicator from the table would be most suitable? Explain your choice. [2]
Answer: Any of the three indicators would be suitable because the pH change at the equivalence point for a strong acid-strong base titration is very large (from approximately pH 3 to pH 11). [1 mark] The vertical portion of the titration curve spans the pH ranges of all three indicators, so the colour change will occur sharply at the end-point regardless of which indicator is chosen. [1 mark]
Marking notes:
- Award [1] for recognising large pH change at equivalence point
- Award [1] for explaining that any indicator range falls within the vertical region
- Accept specific choice with valid reasoning (e.g., phenolphthalein is commonly used)
(b) Which indicator from the table would be most suitable? Explain your choice. [2]
Answer: Phenolphthalein is the most suitable indicator. [1 mark] For a weak acid-strong base titration, the equivalence point occurs at a pH > 7 (approximately pH 8-9 for ethanoic acid/NaOH). Phenolphthalein changes colour over the pH range 8.3-10.0, which matches the steep portion of the titration curve. Methyl orange and bromothymol blue change colour at pH ranges below the equivalence point. [1 mark]
Marking notes:
- Award [1] for identifying phenolphthalein
- Award [1] for explaining that equivalence point is alkaline and phenolphthalein range matches
- Accept reasoning based on titration curve shape
10. (a) Calculate the concentration of ethanoic acid in the diluted vinegar solution. [2]
Answer: n(NaOH) = 0.100 × (21.30/1000) = 2.13 × 10⁻³ mol [1 mark] Mole ratio CH₃COOH : NaOH = 1 : 1 n(CH₃COOH) in 25.0 cm³ = 2.13 × 10⁻³ mol c(CH₃COOH) diluted = 2.13 × 10⁻³ / (25.0/1000) = 0.0852 mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct moles of NaOH
- Award [1] for correct concentration
- Accept 0.085 mol dm⁻³
(b) Calculate the concentration of ethanoic acid in the original vinegar, in mol dm⁻³. [1]
Answer: Dilution factor = 100.0/10.0 = 10 c(original) = 0.0852 × 10 = 0.852 mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct answer with or without units
- Accept 0.85 mol dm⁻³
(c) Calculate the concentration of ethanoic acid in the original vinegar as a percentage by mass. [3]
Answer: Mass of CH₃COOH in 1 dm³ = 0.852 × 60.0 = 51.12 g [1 mark] Mass of 1 dm³ of vinegar = 1.01 × 1000 = 1010 g [1 mark] Percentage by mass = (51.12 / 1010) × 100 = 5.06% [1 mark]
Marking notes:
- Award [1] for mass of ethanoic acid in 1 dm³
- Award [1] for mass of 1 dm³ of vinegar using density
- Award [1] for correct percentage (5.1% or 5.06%)
- Accept alternative stepwise approach
Section C: Free-Response Questions (10 marks)
11. Discuss the role of acid-base chemistry in two of the following contexts. [10]
Marking scheme:
The answer should demonstrate comprehensive knowledge of acid-base chemistry applied to real-world contexts. Award marks according to the following criteria:
| Marks | Descriptor |
|---|---|
| 9–10 | Excellent discussion of two contexts with detailed chemical explanations, relevant balanced equations, and clear links between chemical principles and the real-world application. All chemistry is accurate and well-explained. |
| 7–8 | Good discussion of two contexts with mostly accurate chemical explanations and equations. Minor omissions or errors. Clear links to real-world application. |
| 5–6 | Adequate discussion of two contexts OR good discussion of one context. Some relevant equations and explanations. Some links to application may be superficial. |
| 3–4 | Limited discussion. One context discussed in some detail OR two contexts discussed superficially. Equations may be missing or incorrect. |
| 1–2 | Very limited discussion. Minimal relevant chemistry. Significant errors or omissions. |
| 0 | No relevant chemistry or no attempt. |
Expected content for each context:
Context 1: Antacid tablets
Chemical principles:
- Antacids contain bases such as magnesium hydroxide, aluminium hydroxide, calcium carbonate, or sodium hydrogencarbonate
- Neutralisation reaction: base + HCl(aq) → salt + water (+ CO₂ for carbonates)
- Excess stomach acid (HCl) causes indigestion/heartburn
Relevant equations (examples):
- Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)
- CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
- NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)
Key points for high marks:
- Explain that antacids neutralise excess HCl
- Discuss why Mg(OH)₂ and Al(OH)₃ are preferred (insoluble, slow-acting, long-lasting)
- Mention that CO₂ production can cause bloating/belching
- Link to pH change in stomach
Context 2: Buffer systems in blood
Chemical principles:
- Blood pH must be maintained at approximately 7.35–7.45
- Main buffer system: H₂CO₃/HCO₃⁻ (carbonic acid/hydrogencarbonate buffer)
- Also: H₂PO₄⁻/HPO₄²⁻ and protein buffers
Relevant equations:
- H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq)
- When acid added: HCO₃⁻(aq) + H⁺(aq) → H₂CO₃(aq)
- When base added: H₂CO₃(aq) + OH⁻(aq) → HCO₃⁻(aq) + H₂O(l)
- Role of lungs: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) (removal of CO₂ shifts equilibrium)
- Role of kidneys: regulation of HCO₃⁻ concentration
Key points for high marks:
- Explain why constant pH is vital (enzyme function, protein structure)
- Describe how the carbonic acid/hydrogencarbonate buffer works
- Explain the physiological responses (breathing rate, kidney function)
- Link equilibrium principles to buffer action
Context 3: Acid rain and limestone buildings
Chemical principles:
- Acid rain contains H₂SO₄ and HNO₃ (from SO₂ and NOₓ emissions)
- Also contains H₂CO₃ from dissolved CO₂ (natural rain pH ≈ 5.6)
- Limestone/marble is primarily CaCO₃
- Acid reacts with carbonate: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g)
Relevant equations:
- SO₂(g) + H₂O(l) → H₂SO₃(aq) then oxidation to H₂SO₄
- CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g)
- CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g)
Key points for high marks:
- Explain formation of acid rain (sulfur and nitrogen oxides from fossil fuels)
- Describe the chemical reaction between acid and limestone
- Discuss why damage occurs (soluble calcium salts wash away)
- Mention that CaSO₄ is slightly soluble, forming crust, but still causes damage
- Link to environmental and heritage concerns
General marking guidance:
- Award marks for quality of chemical explanation, not just quantity of text
- Equations must be balanced and relevant
- Look for clear links between chemical principles and the context
- Credit any additional relevant chemistry beyond the expected content
- Deduct marks for significant chemical errors
END OF ANSWER KEY