A Level H1 Chemistry Practice Paper 4

TuitionGoWhere Practice Paper – Chemistry H1 A-Level (Version 4)

ANSWER KEY AND MARKING SCHEME

TuitionGoWhere Exam Practice (AI)

Field	Details
Subject:	Chemistry H1 (8873)
Level:	A-Level
Paper:	PRACTICE – Topic: Acids, Bases & Salts
Total Marks:	50

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Section A: Short Answer & Structured Response (15 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 one that only partially dissociates/ionises in aqueous solution. [1]

Equation (any valid weak acid accepted): CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1]

Marking notes:

• Award [1] for correct definition including "partially dissociates/ionises"
• Award [1] for correct equation with reversible arrow (⇌) and state symbols
• Accept other weak acids e.g. HCOOH, H₂CO₃, HF
• Do NOT accept single arrow (→) – this indicates complete dissociation
• Do NOT accept "dilute acid" as definition of weak acid

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2. Distinguish between a strong acid and a concentrated acid. [2]

Answer: A strong acid is one that completely dissociates/ionises in aqueous solution, whereas a concentrated acid contains a large amount of acid dissolved per unit volume of solution. [2]

Marking notes:

• Award [1] for correct definition of strong acid (complete dissociation)
• Award [1] for correct definition of concentrated acid (amount per volume)
• Key distinction: strength refers to degree of dissociation; concentration refers to amount of solute
• Accept: "Strength is about dissociation; concentration is about how much acid is present"

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3. Explain why the pH of 0.1 mol dm⁻³ ethanoic acid is higher than the pH of 0.1 mol dm⁻³ hydrochloric acid, even though both solutions have the same concentration. [2]

Answer: Ethanoic acid is a weak acid that only partially dissociates in water, producing a lower concentration of H⁺ ions. [1] Hydrochloric acid is a strong acid that completely dissociates, producing a higher concentration of H⁺ ions. Since pH = −log[H⁺], a lower [H⁺] gives a higher pH. [1]

Marking notes:

• Award [1] for stating ethanoic acid is weak (partial dissociation) and HCl is strong (complete dissociation)
• Award [1] for linking [H⁺] to pH
• Accept: "Fewer H⁺ ions in ethanoic acid solution → higher pH"

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4. Write an expression for the acid dissociation constant, Kₐ, of ethanoic acid, CH₃COOH. State the units of Kₐ. [2]

Answer: Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH] [1]

Units: mol dm⁻³ [1]

Marking notes:

• Award [1] for correct expression with products in numerator, reactant in denominator
• Award [1] for correct units
• Do NOT include [H₂O] in expression
• Accept: mol/dm³ or M

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5. A student measures the pH of a 0.050 mol dm⁻³ solution of a weak monoprotic acid, HA, and finds it to be 3.20. Calculate the Kₐ value of HA. [3]

Answer: [H⁺] = 10⁻³·²⁰ = 6.31 × 10⁻⁴ mol dm⁻³ [1]

HA ⇌ H⁺ + A⁻ At equilibrium: [H⁺] = [A⁻] = 6.31 × 10⁻⁴ mol dm⁻³ [HA] ≈ 0.050 − 6.31 × 10⁻⁴ ≈ 0.0494 mol dm⁻³ [1]

Kₐ = [H⁺][A⁻] / [HA] = (6.31 × 10⁻⁴)² / 0.0494 = 8.06 × 10⁻⁶ mol dm⁻³ [1]

Marking notes:

• Award [1] for correct [H⁺] from pH
• Award [1] for correct equilibrium concentrations (allow approximation [HA] ≈ 0.050)
• Award [1] for correct Kₐ value with units
• Accept: 7.9 × 10⁻⁶ to 8.1 × 10⁻⁶ mol dm⁻³ (depending on rounding)

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6. State and explain the effect on the pH of a buffer solution containing CH₃COOH and CH₃COONa when a small amount of dilute hydrochloric acid is added. [2]

Answer: The pH remains approximately constant / changes very little. [1]

The added H⁺ ions react with the conjugate base CH₃COO⁻ to form undissociated CH₃COOH: CH₃COO⁻ + H⁺ → CH₃COOH This removes the added H⁺ ions from solution, so [H⁺] and hence pH remain largely unchanged. [1]

Marking notes:

• Award [1] for stating pH remains approximately constant
• Award [1] for correct explanation with equation or description of H⁺ removal
• Accept: "The ethanoate ions 'mop up' the added H⁺ ions"

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7. Identify the Period 3 element that forms an amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [2]

Answer: Aluminium (Al) [1]

Equations: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) (reaction with acid) Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) (reaction with base) [1]

Marking notes:

• Award [1] for identifying aluminium
• Award [1] for both equations showing reaction with acid AND base
• Accept: Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O for acid reaction
• Accept: Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻ for base reaction
• Do NOT accept silicon (SiO₂ is acidic, not amphoteric)

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Section B: Calculation & Data Interpretation (20 marks)

8. A solution of sodium hydroxide was standardised by titrating it against 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. The volume of sodium hydroxide required was 23.40 cm³.

(a) Write a balanced equation for the reaction. [1]

Answer: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) [1]

Marking notes:

• Award [1] for correct balanced equation with state symbols
• Accept: OH⁻ + H⁺ → H₂O

(b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. [2]

Answer: n(HCl) = c × V = 0.100 × (25.0/1000) = 0.00250 mol [1]

From equation, 1:1 ratio: n(NaOH) = 0.00250 mol c(NaOH) = n/V = 0.00250 / (23.40/1000) = 0.107 mol dm⁻³ [1]

Marking notes:

• Award [1] for correct moles of HCl
• Award [1] for correct concentration with units
• Accept: 0.1068 mol dm⁻³ (3 s.f.)

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9. Benzoic acid, C₆H₅COOH, is a weak monoprotic acid with Kₐ = 6.3 × 10⁻⁵ mol dm⁻³ at 298 K.

(a) Write an equation for the dissociation of benzoic acid in water. [1]

Answer: C₆H₅COOH(aq) ⇌ C₆H₅COO⁻(aq) + H⁺(aq) [1]

Marking notes:

• Award [1] for correct equation with reversible arrow and state symbols
• Must use ⇌, not →

(b) Calculate the pH of a 0.020 mol dm⁻³ solution of benzoic acid at 298 K. [3]

Answer: Kₐ = [H⁺]² / [C₆H₅COOH] [H⁺]² = Kₐ × [C₆H₅COOH] = 6.3 × 10⁻⁵ × 0.020 = 1.26 × 10⁻⁶ [1] [H⁺] = √(1.26 × 10⁻⁶) = 1.12 × 10⁻³ mol dm⁻³ [1] pH = −log(1.12 × 10⁻³) = 2.95 [1]

Marking notes:

• Award [1] for correct substitution into Kₐ expression
• Award [1] for correct [H⁺]
• Award [1] for correct pH (2.95 or 2.9)
• Accept assumption that [C₆H₅COOH]eq ≈ 0.020 mol dm⁻³ (dissociation is small)

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10. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate. The Kₐ of ethanoic acid is 1.8 × 10⁻⁵ mol dm⁻³.

(a) Calculate the pH of this buffer solution. [2]

Answer: After mixing, total volume = 100 cm³ [CH₃COOH] = (0.200 × 50.0/1000) / (100/1000) = 0.100 mol dm⁻³ [CH₃COO⁻] = (0.200 × 50.0/1000) / (100/1000) = 0.100 mol dm⁻³ [1]

pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]) pKₐ = −log(1.8 × 10⁻⁵) = 4.74 pH = 4.74 + log(0.100/0.100) = 4.74 [1]

Marking notes:

• Award [1] for correct concentrations after mixing (or recognising ratio = 1)
• Award [1] for correct pH
• Accept: pH = pKₐ when [acid] = [conjugate base]

(b) Calculate the new pH after 5.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is added to 100 cm³ of the buffer solution. [4]

Answer: Initial moles in 100 cm³ buffer: n(CH₃COOH) = 0.100 × (100/1000) = 0.0100 mol n(CH₃COO⁻) = 0.100 × (100/1000) = 0.0100 mol [1]

Moles of H⁺ added = 0.100 × (5.0/1000) = 5.0 × 10⁻⁴ mol [1]

Added H⁺ reacts with CH₃COO⁻: New n(CH₃COO⁻) = 0.0100 − 5.0 × 10⁻⁴ = 0.0095 mol New n(CH₃COOH) = 0.0100 + 5.0 × 10⁻⁴ = 0.0105 mol [1]

New total volume = 105 cm³ (but ratio used, so volumes cancel) pH = 4.74 + log(0.0095/0.0105) = 4.74 + log(0.905) = 4.74 − 0.043 = 4.70 [1]

Marking notes:

• Award [1] for initial moles of acid and conjugate base
• Award [1] for moles of H⁺ added
• Award [1] for new moles after reaction
• Award [1] for correct new pH
• Accept: 4.69–4.70

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11. A student prepared a solution by dissolving 2.52 g of oxalic acid dihydrate, (COOH)₂·2H₂O, in distilled water and making the solution up to 250 cm³. Oxalic acid is a diprotic acid.

(a) Calculate the concentration of the oxalic acid solution in mol dm⁻³. [2] (Mᵣ of (COOH)₂·2H₂O = 126.0)

Answer: n = m/Mᵣ = 2.52 / 126.0 = 0.0200 mol [1] c = n/V = 0.0200 / (250/1000) = 0.0800 mol dm⁻³ [1]

Marking notes:

• Award [1] for correct moles
• Award [1] for correct concentration with units

(b) 25.0 cm³ of this oxalic acid solution required 20.50 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of the sodium hydroxide solution. [3]

Answer: Equation: (COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O n(oxalic acid) = 0.0800 × (25.0/1000) = 0.00200 mol [1]

From equation, 1:2 ratio: n(NaOH) = 2 × 0.00200 = 0.00400 mol [1] c(NaOH) = n/V = 0.00400 / (20.50/1000) = 0.195 mol dm⁻³ [1]

Marking notes:

• Award [1] for moles of oxalic acid
• Award [1] for correct mole ratio (1:2) and moles of NaOH
• Award [1] for correct concentration with units
• Accept: 0.1951 mol dm⁻³

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12. The pH of a 0.100 mol dm⁻³ solution of a weak base, B, is 11.20 at 298 K.

(a) Calculate the pOH of the solution. [1]

Answer: pOH = 14.00 − pH = 14.00 − 11.20 = 2.80 [1]

(b) Calculate the concentration of hydroxide ions in the solution. [1]

Answer: [OH⁻] = 10⁻²·⁸⁰ = 1.58 × 10⁻³ mol dm⁻³ [1]

(c) Calculate the base dissociation constant, Kբ, of B. [2]

Answer: B + H₂O ⇌ BH⁺ + OH⁻ [OH⁻] = [BH⁺] = 1.58 × 10⁻³ mol dm⁻³ [B] ≈ 0.100 − 1.58 × 10⁻³ ≈ 0.0984 mol dm⁻³ [1]

Kբ = [BH⁺][OH⁻] / [B] = (1.58 × 10⁻³)² / 0.0984 = 2.54 × 10⁻⁵ mol dm⁻³ [1]

Marking notes:

• Award [1] for correct equilibrium concentrations
• Award [1] for correct Kբ value with units
• Accept: 2.5 × 10⁻⁵ mol dm⁻³

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Section C: Context-Based & Integrated Questions (15 marks)

13. Carbonic acid, H₂CO₃, is formed when carbon dioxide dissolves in rainwater. It is a weak diprotic acid.

(a) Write an equation for the first dissociation of carbonic acid in water, including state symbols. [1]

Answer: H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq) [1]

Marking notes:

• Award [1] for correct equation with ⇌ and state symbols
• Must show first dissociation only

(b) Write the expression for the first acid dissociation constant, Kₐ₁, of carbonic acid. [1]

Answer: Kₐ₁ = [HCO₃⁻][H⁺] / [H₂CO₃] [1]

Marking notes:

• Award [1] for correct expression
• Do NOT include [H₂O]

(c) Explain why rainwater is naturally slightly acidic, even in unpolluted environments. [2]

Answer: Carbon dioxide in the atmosphere dissolves in rainwater to form carbonic acid: [1] CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) Carbonic acid is a weak acid that partially dissociates, producing H⁺ ions, making rainwater slightly acidic (pH ≈ 5.6). [1]

Marking notes:

• Award [1] for mentioning CO₂ dissolving to form H₂CO₃
• Award [1] for linking to H⁺ production and acidity
• Accept: equation for CO₂ + H₂O ⇌ H₂CO₃

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14. In the production of yoghurt, bacteria convert lactose into lactic acid, CH₃CH(OH)COOH. Lactic acid is a weak monoprotic acid with Kₐ = 1.4 × 10⁻⁴ mol dm⁻³.

(a) Calculate the pH of a 0.010 mol dm⁻³ solution of lactic acid. [3]

Answer: Kₐ = [H⁺]² / [HA] [H⁺]² = 1.4 × 10⁻⁴ × 0.010 = 1.4 × 10⁻⁶ [1] [H⁺] = √(1.4 × 10⁻⁶) = 1.18 × 10⁻³ mol dm⁻³ [1] pH = −log(1.18 × 10⁻³) = 2.93 [1]

Marking notes:

• Award [1] for correct substitution
• Award [1] for correct [H⁺]
• Award [1] for correct pH
• Accept: 2.92–2.93

(b) Explain why the pH of the yoghurt does not change significantly as more lactic acid is produced during fermentation, given that the yoghurt contains phosphate salts that act as a buffer. [2]

Answer: The phosphate salts act as a buffer, resisting changes in pH. [1] The added H⁺ ions from lactic acid react with the conjugate base (HPO₄²⁻ or PO₄³⁻) in the phosphate buffer, forming the weak acid component. This removes excess H⁺ from solution, so the pH remains relatively constant. [1]

Marking notes:

• Award [1] for stating buffer resists pH change
• Award [1] for explanation of how buffer works (H⁺ reacts with conjugate base)
• Accept: reference to equilibrium shifting to remove added H⁺

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15. A student investigates the reaction between magnesium ribbon and two different acids of the same concentration: 0.50 mol dm⁻³ hydrochloric acid (a strong acid) and 0.50 mol dm⁻³ ethanoic acid (a weak acid).

(a) State and explain one similarity and one difference the student would observe in the reactions. [2]

Answer: Similarity: Both reactions produce hydrogen gas / both reactions produce the same salt (magnesium chloride/ethanoate). [1]

Difference: The reaction with hydrochloric acid is faster / more vigorous than with ethanoic acid. This is because HCl is a strong acid that completely dissociates, giving a higher [H⁺], whereas ethanoic acid is weak and partially dissociates, giving a lower [H⁺]. [1]

Marking notes:

• Award [1] for valid similarity
• Award [1] for valid difference with explanation linking to [H⁺]
• Accept: "Same volume of H₂ produced eventually" as similarity
• Accept: "HCl reaction produces more effervescence initially" as difference

(b) The student repeats the experiment using 0.50 mol dm⁻³ sulfuric acid instead of hydrochloric acid. Explain why the rate of reaction with sulfuric acid is faster, even though both are strong acids. [2]

Answer: Sulfuric acid is diprotic (H₂SO₄), producing 2 moles of H⁺ per mole of acid, whereas HCl is monoprotic, producing only 1 mole of H⁺ per mole of acid. [1] Therefore, 0.50 mol dm⁻³ H₂SO₄ has [H⁺] = 1.0 mol dm⁻³, which is twice the [H⁺] of 0.50 mol dm⁻³ HCl (0.50 mol dm⁻³). The higher [H⁺] leads to a faster reaction rate. [1]

Marking notes:

• Award [1] for stating H₂SO₄ is diprotic
• Award [1] for linking higher [H⁺] to faster rate
• Accept: calculation showing [H⁺] in H₂SO₄ = 1.0 mol dm⁻³

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16. Aluminium hydroxide, Al(OH)₃, is used in some antacid tablets to neutralise excess stomach acid (hydrochloric acid).

(a) Write a balanced equation for the reaction between aluminium hydroxide and hydrochloric acid. [1]

Answer: Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(l) [1]

Marking notes:

• Award [1] for correct balanced equation with state symbols

(b) An antacid tablet contains 250 mg of Al(OH)₃. Calculate the volume of 0.100 mol dm⁻³ hydrochloric acid that can be neutralised by one tablet. [3] (Mᵣ of Al(OH)₃ = 78.0)

Answer: n(Al(OH)₃) = m/Mᵣ = 0.250 / 78.0 = 3.205 × 10⁻³ mol [1]

From equation, 1:3 ratio: n(HCl) = 3 × 3.205 × 10⁻³ = 9.615 × 10⁻³ mol [1]

V(HCl) = n/c = 9.615 × 10⁻³ / 0.100 = 0.0962 dm³ = 96.2 cm³ [1]

Marking notes:

• Award [1] for correct moles of Al(OH)₃ (must convert mg to g)
• Award [1] for correct mole ratio and moles of HCl
• Award [1] for correct volume with units
• Accept: 96.0–96.5 cm³

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17. The amino acid glycine, H₂NCH₂COOH, can act as both an acid and a base in aqueous solution.

(a) Write an equation to show glycine acting as an acid in water. [1]

Answer: H₂NCH₂COOH(aq) + H₂O(l) ⇌ H₂NCH₂COO⁻(aq) + H₃O⁺(aq) [1]

Marking notes:

• Award [1] for equation showing glycine donating H⁺ (from COOH group)
• Accept: H₂NCH₂COOH ⇌ H₂NCH₂COO⁻ + H⁺

(b) Write an equation to show glycine acting as a base in water. [1]

Answer: H₂NCH₂COOH(aq) + H₂O(l) ⇌ ⁺H₃NCH₂COOH(aq) + OH⁻(aq) [1]

Marking notes:

• Award [1] for equation showing glycine accepting H⁺ (on NH₂ group)
• Accept: H₂NCH₂COOH + H⁺ ⇌ ⁺H₃NCH₂COOH

(c) Explain why aqueous solutions of glycine can act as buffer solutions. [2]

Answer: Glycine is amphoteric – it can act as both a weak acid (donating H⁺ from COOH) and a weak base (accepting H⁺ on NH₂). [1] In solution, glycine exists as a zwitterion (⁺H₃NCH₂COO⁻) which can react with both added acid and added base, resisting changes in pH. [1]

Marking notes:

• Award [1] for identifying amphoteric nature
• Award [1] for explaining buffer action (reacts with both H⁺ and OH⁻)
• Accept: "The COO⁻ group neutralises added acid; the NH₃⁺ group neutralises added base"

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18. A student prepared a buffer solution by partially neutralising 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 10.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

(a) Calculate the number of moles of ethanoic acid remaining after the reaction. [2]

Answer: Initial n(CH₃COOH) = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol [1] n(NaOH) added = 0.100 × (10.0/1000) = 1.00 × 10⁻³ mol

CH₃COOH + NaOH → CH₃COONa + H₂O n(CH₃COOH) remaining = 2.50 × 10⁻³ − 1.00 × 10⁻³ = 1.50 × 10⁻³ mol [1]

Marking notes:

• Award [1] for initial moles of CH₃COOH
• Award [1] for correct remaining moles

(b) Calculate the number of moles of sodium ethanoate formed. [1]

Answer: n(CH₃COONa) = n(NaOH) = 1.00 × 10⁻³ mol [1]

Marking notes:

• Award [1] for correct moles (1:1 ratio with NaOH)

(c) Calculate the pH of the resulting buffer solution. [2] (Kₐ of ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³)

Answer: Total volume = 25.0 + 10.0 = 35.0 cm³ (ratio used, volumes cancel) pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]) pKₐ = −log(1.8 × 10⁻⁵) = 4.74 [1] pH = 4.74 + log(1.00 × 10⁻³ / 1.50 × 10⁻³) = 4.74 + log(0.667) = 4.74 − 0.176 = 4.56 [1]

Marking notes:

• Award [1] for pKₐ
• Award [1] for correct pH
• Accept: 4.56–4.57

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19. Explain why a mixture of aqueous ammonia and ammonium chloride can function as a buffer solution. Include relevant equations in your answer. [3]

Answer: Aqueous ammonia is a weak base: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) [1] Ammonium chloride provides the conjugate acid, NH₄⁺: NH₄Cl(aq) → NH₄⁺(aq) + Cl⁻(aq) [1]

When a small amount of acid (H⁺) is added, it reacts with NH₃: NH₃ + H⁺ → NH₄⁺ When a small amount of base (OH⁻) is added, it reacts with NH₄⁺: NH₄⁺ + OH⁻ → NH₃ + H₂O This removes the added H⁺ or OH⁻, so the pH remains approximately constant. [1]

Marking notes:

• Award [1] for identifying NH₃ as weak base with equilibrium equation
• Award [1] for identifying NH₄⁺ as conjugate acid
• Award [1] for explaining buffer action with both equations
• Accept: reference to equilibrium shifting to counteract changes

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20. A 0.0250 mol dm⁻³ solution of a weak monoprotic acid has a pH of 3.80. A student claims that this acid is ethanoic acid (Kₐ = 1.8 × 10⁻⁵ mol dm⁻³). By calculation, determine whether the student's claim is valid. [3]

Answer: [H⁺] = 10⁻³·⁸⁰ = 1.58 × 10⁻⁴ mol dm⁻³ [1]

Kₐ = [H⁺]² / [HA] = (1.58 × 10⁻⁴)² / 0.0250 = 2.50 × 10⁻⁸ / 0.0250 = 1.00 × 10⁻⁶ mol dm⁻³ [1]

The calculated Kₐ (1.00 × 10⁻⁶) is significantly different from the Kₐ of ethanoic acid (1.8 × 10⁻⁵). Therefore, the student's claim is NOT valid – the acid is much weaker than ethanoic acid. [1]

Marking notes:

• Award [1] for correct [H⁺] from pH
• Award [1] for correct calculation of Kₐ
• Award [1] for correct conclusion with justification (comparison of values)
• Accept: "No, because calculated Kₐ ≠ 1.8 × 10⁻⁵"

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END OF ANSWER KEY

Total marks: 50