From Real Exams Exam Paper

A Level H1 Chemistry Practice Paper 1

Free A Level H1 Chemistry Practice Paper 1, DeepSeek Exam version, with questions, answers, and A Level-style practice for Singapore students.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H1 Chemistry From Real Exams Generated by DeepSeek V4 Pro Updated 2026-08-17

Back to Subject Browse Free Exam Papers

Questions

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

Free quiz and exam paper access

Enter your details to view this paper

Your access is remembered on this device.

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