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A Level H1 Chemistry Practice Paper 3
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Questions
A-Level Chemistry H1 Quiz - Acids Bases Salts
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- State units where appropriate.
- A Data Booklet is provided.
- Marks for each question are indicated in brackets.
Section A: Short Answer & Definitions (10 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. Explain why a solution of ethanoic acid has a higher pH than a solution of hydrochloric acid of the same concentration. [2]
3. Identify the Period 3 element that forms an amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [3]
4. Define the term buffer solution. [1]
5. State the colour change observed when methyl orange is added to a solution of sodium hydroxide, and then excess hydrochloric acid is added dropwise. [2]
Section B: Calculations & Data Interpretation (18 marks)
Answer all questions in this section. Show all working clearly.
6. A student titrated 25.0 cm³ of a solution of benzoic acid, C₆H₅COOH, against 0.100 mol dm⁻³ sodium hydroxide solution. The average titre was 22.50 cm³.
(a) Write a balanced equation for the reaction between benzoic acid and sodium hydroxide. [1]
(b) Calculate the concentration of the benzoic acid solution in mol dm⁻³. [2]
(c) Calculate the concentration of the benzoic acid solution in g dm⁻³. [2] (Mr of C₆H₅COOH = 122)
7. Carbonic acid, H₂CO₃, is a weak diprotic acid formed when carbon dioxide dissolves in rainwater.
(a) Write an equation, including state symbols, for the first dissociation of carbonic acid in water. [1]
(b) Write the expression for the first acid dissociation constant, Kₐ₁, of carbonic acid. [1]
(c) A sample of rainwater has a pH of 5.60. Calculate the concentration of hydrogen ions, [H⁺], in this sample. [1]
(d) Given that Kₐ₁ for carbonic acid is 4.3 × 10⁻⁷ mol dm⁻³, calculate the concentration of undissociated H₂CO₃ in the rainwater sample. State any assumption made. [3]
8. 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).
(a) Calculate the pH of this buffer solution. [3] (Kₐ for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³)
(b) Explain, using equations, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [2]
Section C: Structured & Applied Questions (12 marks)
Answer all questions in this section.
9. Calcium hydroxide, Ca(OH)₂, is sometimes added to fermentation tanks to prevent the build-up of lactic acid produced by bacteria.
(a) Explain why a build-up of lactic acid reduces the effectiveness of the enzymes involved in fermentation. [2]
(b) Write a balanced equation for the reaction between calcium hydroxide and lactic acid (CH₃CH(OH)COOH). [1]
(c) Suggest why calcium hydroxide, rather than sodium hydroxide, is preferred for this application. [1]
10. A student investigated the relative strengths of three acids: hydrochloric acid (HCl), ethanoic acid (CH₃COOH), and trichloroethanoic acid (CCl₃COOH). Equal concentrations of each acid were prepared, and their pH values were measured.
(a) The pH values recorded were 1.0, 2.9, and 1.3. Assign each pH value to the correct acid, explaining your reasoning. [3]
(b) Trichloroethanoic acid is a stronger acid than ethanoic acid. Explain this difference in terms of the structures of the two acids and the stability of their conjugate bases. [3]
(c) The student also measured the electrical conductivity of each acid solution. Rank the three acids in order of increasing conductivity, and explain your answer. [2]
END OF PAPER
Answers
A-Level Chemistry H1 Quiz - Acids Bases Salts: ANSWER KEY
Total Marks: 40
Section A: Short Answer & Definitions (10 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) [1 mark]
Marking notes:
- Award [1] for "partially dissociates/ionises" (not "dilute" or "does not dissociate").
- Award [1] for correct equation with reversible arrow (⇌) and correct state symbols.
- Accept any valid weak acid equation (e.g., H₂CO₃, HCOOH, HF).
- Do not award equation mark if single arrow (→) is used.
2. Explain why a solution of ethanoic acid has a higher pH than a solution of hydrochloric acid of the same concentration. [2]
Answer: Hydrochloric acid is a strong acid and fully dissociates in water, producing a higher concentration of H⁺ ions. [1 mark] Ethanoic acid is a weak acid and only partially dissociates, producing a lower concentration of H⁺ ions. Since pH = −log[H⁺], a lower [H⁺] results in a higher pH. [1 mark]
Marking notes:
- Award [1] for identifying HCl as strong/fully dissociated AND CH₃COOH as weak/partially dissociated.
- Award [1] for linking [H⁺] to pH (higher [H⁺] → lower pH, or lower [H⁺] → higher pH).
- Accept reference to degree of dissociation/ionisation.
3. Identify the Period 3 element that forms an amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [3]
Answer: Element: Aluminium / Al [1 mark]
Reaction with acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) [1 mark]
Reaction with base: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) [1 mark]
Marking notes:
- Award [1] for correctly identifying aluminium (accept Al).
- Award [1] for a correct equation showing reaction with acid (accept any strong acid; balanced equation required).
- Award [1] for a correct equation showing reaction with base (accept NaOH or other strong base; balanced equation required).
- Accept Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq) for the base reaction.
- Do not award marks for SiO₂ (acidic oxide) or other incorrect elements.
4. Define the term buffer solution. [1]
Answer: A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. [1 mark]
Marking notes:
- Award [1] for "resists changes in pH" (accept "maintains approximately constant pH").
- Must include reference to addition of small amounts of acid/base.
- Accept: "A solution containing a weak acid and its conjugate base (or weak base and its conjugate acid) that resists pH changes."
5. State the colour change observed when methyl orange is added to a solution of sodium hydroxide, and then excess hydrochloric acid is added dropwise. [2]
Answer: In sodium hydroxide (alkaline solution): methyl orange is yellow. [1 mark] After adding excess hydrochloric acid (acidic solution): methyl orange turns red/pink. [1 mark]
Marking notes:
- Award [1] for "yellow" in alkaline/NaOH solution.
- Award [1] for "red" or "pink" in acidic/excess HCl solution.
- Accept "orange" as intermediate colour if described as transition, but final colour must be red/pink.
- Do not accept "colourless" or other incorrect colours.
Section B: Calculations & Data Interpretation (18 marks)
6. Titration of benzoic acid with NaOH.
(a) Write a balanced equation for the reaction between benzoic acid and sodium hydroxide. [1]
Answer: C₆H₅COOH(aq) + NaOH(aq) → C₆H₅COONa(aq) + H₂O(l) [1 mark]
Marking notes:
- Award [1] for correct formulae and balanced equation.
- State symbols not essential but good practice.
- Accept C₆H₅COOH + NaOH → C₆H₅COO⁻Na⁺ + H₂O.
(b) Calculate the concentration of the benzoic acid solution in mol dm⁻³. [2]
Answer: n(NaOH) = c × V = 0.100 × (22.50/1000) = 0.00225 mol [1 mark] Mole ratio C₆H₅COOH : NaOH = 1 : 1 n(C₆H₅COOH) = 0.00225 mol c(C₆H₅COOH) = n/V = 0.00225 / (25.0/1000) = 0.0900 mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct calculation of moles of NaOH.
- Award [1] for correct final concentration with units.
- Accept 0.09 mol dm⁻³ (2 s.f.) or 0.0900 mol dm⁻³ (3 s.f.).
- Deduct [1] if volume not converted to dm³.
(c) Calculate the concentration of the benzoic acid solution in g dm⁻³. [2]
Answer: Mass concentration = molar concentration × Mr [1 mark] = 0.0900 × 122 = 11.0 g dm⁻³ (or 10.98 g dm⁻³) [1 mark]
Marking notes:
- Award [1] for correct method (c × Mr).
- Award [1] for correct answer with appropriate significant figures.
- Accept 11.0 g dm⁻³ or 10.98 g dm⁻³.
- Error carried forward (ECF) from part (b) accepted.
7. Carbonic acid in rainwater.
(a) Write an equation, including state symbols, for the first dissociation of carbonic acid in water. [1]
Answer: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) [1 mark]
Marking notes:
- Award [1] for correct equation with reversible arrow (⇌) and state symbols.
- Must show first dissociation only (not second dissociation).
- Do not award mark if single arrow (→) is used.
(b) Write the expression for the first acid dissociation constant, Kₐ₁, of carbonic acid. [1]
Answer: Kₐ₁ = [H⁺][HCO₃⁻] / [H₂CO₃] [1 mark]
Marking notes:
- Award [1] for correct expression.
- Square brackets required for concentrations.
- Do not include [H₂O] in expression.
(c) A sample of rainwater has a pH of 5.60. Calculate the concentration of hydrogen ions, [H⁺], in this sample. [1]
Answer: [H⁺] = 10⁻ᵖᴴ = 10⁻⁵·⁶⁰ = 2.51 × 10⁻⁶ mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for correct calculation.
- Accept 2.5 × 10⁻⁶ mol dm⁻³ (2 s.f.).
(d) Given that Kₐ₁ for carbonic acid is 4.3 × 10⁻⁷ mol dm⁻³, calculate the concentration of undissociated H₂CO₃ in the rainwater sample. State any assumption made. [3]
Answer: Kₐ₁ = [H⁺][HCO₃⁻] / [H₂CO₃] Assumption: [H⁺] = [HCO₃⁻] (from first dissociation only; contribution from second dissociation is negligible) [1 mark]
4.3 × 10⁻⁷ = (2.51 × 10⁻⁶)² / [H₂CO₃] [1 mark] [H₂CO₃] = (2.51 × 10⁻⁶)² / (4.3 × 10⁻⁷) = 6.30 × 10⁻¹² / (4.3 × 10⁻⁷) = 1.47 × 10⁻⁵ mol dm⁻³ [1 mark]
Marking notes:
- Award [1] for stating correct assumption ([H⁺] = [HCO₃⁻] or contribution from second dissociation/water autoprotolysis is negligible).
- Award [1] for correct substitution into Kₐ expression.
- Award [1] for correct final answer (accept 1.5 × 10⁻⁵ mol dm⁻³).
- ECF from part (c) accepted.
8. Buffer solution calculation.
(a) Calculate the pH of this buffer solution. [3]
Answer: After mixing, total volume = 100.0 cm³ [CH₃COOH] = (0.200 × 50.0/1000) / (100.0/1000) = 0.100 mol dm⁻³ [1 mark] [CH₃COO⁻] = (0.200 × 50.0/1000) / (100.0/1000) = 0.100 mol dm⁻³ [1 mark]
Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.8 × 10⁻⁵ = [H⁺] × 0.100 / 0.100 [H⁺] = 1.8 × 10⁻⁵ mol dm⁻³ pH = −log(1.8 × 10⁻⁵) = 4.74 [1 mark]
Marking notes:
- Award [1] for correct calculation of diluted concentrations (or recognition that [HA] = [A⁻] after mixing equal volumes and concentrations).
- Award [1] for correct substitution into Kₐ expression.
- Award [1] for correct pH (4.74 or 4.7).
- Alternative method using Henderson-Hasselbalch: pH = pKₐ + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74 is acceptable.
(b) Explain, using equations, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [2]
Answer: When H⁺ ions (from HCl) are added, they react with the conjugate base (ethanoate ions): CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [1 mark] The added H⁺ ions are removed from solution, so the pH remains approximately constant. [1 mark]
Marking notes:
- Award [1] for correct equation showing reaction of added H⁺ with CH₃COO⁻.
- Award [1] for explanation that H⁺ ions are consumed/removed, preventing pH change.
- Accept: "The equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts to the left, removing added H⁺."
Section C: Structured & Applied Questions (12 marks)
9. Calcium hydroxide in fermentation.
(a) Explain why a build-up of lactic acid reduces the effectiveness of the enzymes involved in fermentation. [2]
Answer: Enzymes have an optimal pH at which they function most effectively. [1 mark] A build-up of lactic acid lowers the pH, which can denature the enzymes. Denaturation changes the shape of the active site, so the substrate can no longer bind, and the enzyme loses its catalytic activity. [1 mark]
Marking notes:
- Award [1] for reference to optimal pH or pH sensitivity of enzymes.
- Award [1] for explanation involving denaturation and change in active site shape.
- Accept reference to disruption of ionic/hydrogen bonds in tertiary structure.
(b) Write a balanced equation for the reaction between calcium hydroxide and lactic acid (CH₃CH(OH)COOH). [1]
Answer: Ca(OH)₂(aq) + 2CH₃CH(OH)COOH(aq) → Ca(CH₃CH(OH)COO)₂(aq) + 2H₂O(l) [1 mark]
Marking notes:
- Award [1] for correct balanced equation.
- Accept ionic equation: OH⁻ + H⁺ → H₂O, with correct stoichiometry.
- Accept Ca(OH)₂ + 2CH₃CH(OH)COOH → Ca(CH₃CH(OH)COO)₂ + 2H₂O.
(c) Suggest why calcium hydroxide, rather than sodium hydroxide, is preferred for this application. [1]
Answer: Calcium hydroxide is less soluble/sparingly soluble, so it provides a more controlled/slow release of OH⁻ ions, preventing a sudden large increase in pH that could also denature the enzymes. [1 mark]
Marking notes:
- Award [1] for reference to lower solubility/controlled release/less likely to cause pH overshoot.
- Accept: "Calcium hydroxide is a weaker base/less corrosive/safer to handle."
- Accept: "Sodium hydroxide is too strong/caustic and may denature enzymes."
10. Relative strengths of acids.
(a) The pH values recorded were 1.0, 2.9, and 1.3. Assign each pH value to the correct acid, explaining your reasoning. [3]
Answer: HCl: pH = 1.0 (strongest acid; fully dissociates; highest [H⁺]) [1 mark] CCl₃COOH: pH = 1.3 (stronger weak acid due to electron-withdrawing Cl atoms) [1 mark] CH₃COOH: pH = 2.9 (weakest acid; least dissociation; lowest [H⁺]) [1 mark]
Marking notes:
- Award [1] for each correct assignment with valid reasoning.
- Reasoning must reference relative strength/dissociation.
- Accept: HCl is a strong acid (fully dissociated), so lowest pH. CCl₃COOH is stronger than CH₃COOH due to inductive effect of Cl atoms.
(b) Trichloroethanoic acid is a stronger acid than ethanoic acid. Explain this difference in terms of the structures of the two acids and the stability of their conjugate bases. [3]
Answer: In trichloroethanoic acid (CCl₃COOH), the three electronegative chlorine atoms exert an electron-withdrawing inductive effect (−I effect). [1 mark] This withdraws electron density from the O−H bond, making it easier for the H⁺ to be released. [1 mark] The resulting conjugate base (CCl₃COO⁻) is stabilised by delocalisation/dispersal of the negative charge over the electronegative chlorine atoms, making the dissociation more favourable. [1 mark]
Marking notes:
- Award [1] for identifying electron-withdrawing/inductive effect of Cl atoms.
- Award [1] for explaining effect on O−H bond polarity/weakening.
- Award [1] for explaining stabilisation of conjugate base (charge dispersal/delocalisation).
- Accept reference to pKₐ values or relative acid strengths.
(c) The student also measured the electrical conductivity of each acid solution. Rank the three acids in order of increasing conductivity, and explain your answer. [2]
Answer: Ranking (increasing conductivity): CH₃COOH < CCl₃COOH < HCl [1 mark] Explanation: Electrical conductivity depends on the concentration of ions in solution. HCl is fully dissociated (highest [H⁺]), CCl₃COOH is partially dissociated but stronger than CH₃COOH, and CH₃COOH is the weakest acid with the lowest degree of dissociation (lowest [H⁺]). [1 mark]
Marking notes:
- Award [1] for correct ranking.
- Award [1] for explanation linking conductivity to ion concentration/degree of dissociation.
- Accept reverse ranking if clearly stated as "decreasing conductivity" with correct order.
END OF ANSWER KEY