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A Level H1 Chemistry Acids Bases Salts Quiz

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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. Use appropriate significant figures. A Data Booklet is provided.

Section A: Short Answer (10 marks)

Answer all questions in this section.

1. Define the term Brønsted-Lowry acid. [1]

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

3. Write an equation, including state symbols, for the dissociation of ethanoic acid in water. [1]

4. Explain why the pH of 0.1 mol dm⁻³ hydrochloric acid is 1.0, while the pH of 0.1 mol dm⁻³ ethanoic acid is approximately 2.9. [2]

5. A student measures the pH of rainwater and finds it to be 5.6. Calculate the concentration of hydrogen ions, [H⁺], in the rainwater. [1]

Section B: Structured Questions (10 marks)

Answer all questions in this section.

6. State what is meant by the term buffer solution. [1]

7. Identify the conjugate base of H₂PO₄⁻. [1]

8. Explain why a solution of ammonium chloride, NH₄Cl, is acidic. Support your answer with an equation. [2]

9. A student titrates 25.0 cm³ of barium hydroxide, Ba(OH)₂, solution against 0.100 mol dm⁻³ hydrochloric acid. The average titre is 20.0 cm³.

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

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

10. The acid dissociation constant, Ka, of benzoic acid, C₆H₅COOH, is 6.3 × 10⁻⁵ mol dm⁻³.

(a) Write the expression for Ka of benzoic acid. [1]

(b) Calculate the pH of a 0.050 mol dm⁻³ solution of benzoic acid. State any assumption made. [3]

Section C: Data Analysis and Application (10 marks)

Answer all questions in this section.

11. A buffer solution is prepared by mixing 50.0 cm³ of 0.10 mol dm⁻³ benzoic acid with 50.0 cm³ of 0.10 mol dm⁻³ sodium benzoate. Calculate the pH of this buffer solution. [3]

12. The graph below shows the pH titration curve when 25.0 cm³ of an unknown monoprotic acid is titrated against 0.10 mol dm⁻³ sodium hydroxide solution.

Image pending generation for this question.

[The curve shows: initial pH ≈ 2.9, equivalence point at 25.0 cm³ with pH ≈ 8.5, final pH ≈ 12.5]

(a) Identify whether the acid is strong or weak. Explain your answer using information from the curve. [2]

(b) Determine the concentration of the acid. [1]

13. The table below shows the Ka values of three weak acids at 298 K.

Acid	Formula	Ka / mol dm⁻³
Ethanoic acid	CH₃COOH	1.8 × 10⁻⁵
Chloroethanoic acid	ClCH₂COOH	1.4 × 10⁻³
Dichloroethanoic acid	Cl₂CHCOOH	5.0 × 10⁻²

(a) State which acid is the strongest. Explain your answer. [1]

(b) Explain why chloroethanoic acid is a stronger acid than ethanoic acid. [2]

14. A student claims that dichloroethanoic acid is a strong acid because its Ka value is much larger than that of ethanoic acid. Evaluate this claim. [2]

15. A buffer solution is used to maintain the pH of blood at approximately 7.4. The buffer system involves carbonic acid, H₂CO₃, and hydrogencarbonate ions, HCO₃⁻.

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) Ka = 4.3 × 10⁻⁷ mol dm⁻³

(a) Using the Henderson-Hasselbalch equation or otherwise, calculate the ratio [HCO₃⁻] / [H₂CO₃] required to maintain a pH of 7.4. [2]

(b) Explain how this buffer system resists changes in pH when a small amount of acid is added to the blood. Support your answer with an equation. [2]

Section D: Further Applications (10 marks)

Answer all questions in this section.

16. Explain why the pH of pure water is 7.0 at 25 °C, but decreases when the temperature is raised. [2]

17. A student prepares a buffer solution by dissolving 2.10 g of sodium hydrogencarbonate, NaHCO₃ (Mr = 84.0), and 1.68 g of sodium carbonate, Na₂CO₃ (Mr = 106.0), in water to make 250 cm³ of solution. Calculate the pH of this buffer solution. [Ka of HCO₃⁻ = 4.8 × 10⁻¹¹ mol dm⁻³] [3]

18. Describe how you would prepare a sample of a sparingly soluble salt, such as barium sulfate, by precipitation. Include a brief explanation of why the salt is obtained in a pure form. [2]

19. A solution of a weak monoprotic acid, HA, has a pH of 3.0. The concentration of the acid is 0.10 mol dm⁻³. Calculate the Ka of the acid. [2]

20. Explain the difference between the terms end point and equivalence point in a titration. [1]

END OF QUIZ

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Answers

A-Level Chemistry H1 Quiz - Acids Bases Salts - ANSWER KEY

Total Marks: 40

Section A: Short Answer (10 marks)

1. Define the term Brønsted-Lowry acid. [1]

Answer: A Brønsted-Lowry acid is a proton (H⁺) donor. [1]

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

Answer: A strong acid is one that completely dissociates in water (e.g., HCl), while a concentrated acid contains a large amount of acid per unit volume of water. [1] Strength refers to the degree of dissociation; concentration refers to the amount of acid dissolved. [1]

3. Write an equation, including state symbols, for the dissociation of ethanoic acid in water. [1]

Answer: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1] (Must have equilibrium arrow and state symbols)

4. Explain why the pH of 0.1 mol dm⁻³ hydrochloric acid is 1.0, while the pH of 0.1 mol dm⁻³ ethanoic acid is approximately 2.9. [2]

Answer: HCl is a strong acid and completely dissociates, so [H⁺] = 0.1 mol dm⁻³, giving pH = –log(0.1) = 1.0. [1] Ethanoic acid is a weak acid and only partially dissociates, so [H⁺] is much less than 0.1 mol dm⁻³, resulting in a higher pH. [1]

5. A student measures the pH of rainwater and finds it to be 5.6. Calculate the concentration of hydrogen ions, [H⁺], in the rainwater. [1]

Answer: [H⁺] = 10⁻⁵·⁶ = 2.5 × 10⁻⁶ mol dm⁻³ [1] (Accept 2.51 × 10⁻⁶)

Section B: Structured Questions (10 marks)

6. State what is meant by the term buffer solution. [1]

Answer: A buffer solution is one that resists changes in pH when small amounts of acid or base are added. [1]

7. Identify the conjugate base of H₂PO₄⁻. [1]

Answer: HPO₄²⁻ [1]

8. Explain why a solution of ammonium chloride, NH₄Cl, is acidic. Support your answer with an equation. [2]

Answer: NH₄⁺ ions undergo hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) [1] This produces H₃O⁺ ions, making the solution acidic. [1]

9. Barium hydroxide titration.

(a) Balanced equation: [1]

Answer: Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l) [1]

(b) Concentration of Ba(OH)₂: [3]

n(HCl) = 0.100 × 20.0/1000 = 0.00200 mol [1]
From equation: 1 mol Ba(OH)₂ reacts with 2 mol HCl, so n(Ba(OH)₂) = 0.00200/2 = 0.00100 mol [1]
c(Ba(OH)₂) = 0.00100 / (25.0/1000) = 0.0400 mol dm⁻³ [1]

Ba(OH)₂ → Ba²⁺ + 2OH⁻, so [OH⁻] = 2 × 0.0400 = 0.0800 mol dm⁻³ [1]
pOH = –log(0.0800) = 1.10; pH = 14 – 1.10 = 12.9 [1]

10. Benzoic acid calculations.

(a) Ka expression: [1]

Answer: Ka = [C₆H₅COO⁻][H⁺] / [C₆H₅COOH] [1]

(b) pH of 0.050 mol dm⁻³ benzoic acid: [3]

Ka = [H⁺]² / [HA]; [H⁺] = √(Ka × [HA]) = √(6.3 × 10⁻⁵ × 0.050) [1]
[H⁺] = √(3.15 × 10⁻⁶) = 1.77 × 10⁻³ mol dm⁻³ [1]
pH = –log(1.77 × 10⁻³) = 2.75 [1]
Assumption: [H⁺] << [HA], so [HA]eq ≈ [HA]initial (dissociation is negligible) [1]

Section C: Data Analysis and Application (10 marks)

11. Buffer pH calculation: [3]

After mixing: [C₆H₅COOH] = 0.050 mol dm⁻³; [C₆H₅COO⁻] = 0.050 mol dm⁻³ [1]
Ka = [C₆H₅COO⁻][H⁺] / [C₆H₅COOH]; [H⁺] = Ka × [HA] / [A⁻] [1]
[H⁺] = 6.3 × 10⁻⁵ × 0.050 / 0.050 = 6.3 × 10⁻⁵ mol dm⁻³
pH = –log(6.3 × 10⁻⁵) = 4.20 [1]

12. Titration curve analysis.

(a) Acid identification: [2]

Answer: Weak acid. [1] The initial pH is approximately 2.9, which is higher than expected for a strong acid of similar concentration. The equivalence point pH is >7 (alkaline), characteristic of a weak acid-strong base titration. [1]

(b) Acid concentration: [1]

n(NaOH) = 0.10 × 25.0/1000 = 0.0025 mol; n(acid) = 0.0025 mol (1:1); c(acid) = 0.0025 / (25.0/1000) = 0.10 mol dm⁻³ [1]

Answer: Phenolphthalein (pH range 8.3–10.0). [1] The equivalence point pH (8.5) falls within this indicator's colour change range, so the colour change will be sharp at the endpoint. [1]

13. Weak acid comparison.

(a) Strongest acid: [1]

Answer: Dichloroethanoic acid. [1] It has the largest Ka value, indicating the greatest degree of dissociation.

(b) Chloroethanoic vs ethanoic acid: [2]

Answer: The electronegative chlorine atom withdraws electrons from the O–H bond, weakening it and making it easier for the acid to donate a proton. [1] This stabilises the conjugate base (ClCH₂COO⁻) through the inductive effect, shifting the equilibrium to the right. [1]

Ka = [H⁺]² / [HA]; [H⁺] = √(Ka × [HA]) = √(1.4 × 10⁻³ × 0.10) [1]
[H⁺] = √(1.4 × 10⁻⁴) = 0.0118 mol dm⁻³ [1]
pH = –log(0.0118) = 1.93 [1]

14. Evaluate claim: [2]

Answer: The claim is incorrect. [1] A strong acid is defined as one that completely dissociates in water. Dichloroethanoic acid, despite having a relatively large Ka, is still a weak acid because it only partially dissociates. Its Ka (5.0 × 10⁻²) is still much smaller than that of strong acids (Ka >> 1). [1]

15. Blood buffer system.

(a) Ratio calculation: [2]

pH = pKa + log([HCO₃⁻] / [H₂CO₃]); pKa = –log(4.3 × 10⁻⁷) = 6.37 [1]
7.4 = 6.37 + log([HCO₃⁻] / [H₂CO₃]); log([HCO₃⁻] / [H₂CO₃]) = 1.03
[HCO₃⁻] / [H₂CO₃] = 10¹·⁰³ = 10.7 ≈ 11 [1]

(b) Buffer action explanation: [2]

Answer: When acid is added, H⁺ ions react with the conjugate base: H⁺ + HCO₃⁻ → H₂CO₃ [1] The added H⁺ is consumed, so the pH remains approximately constant. [1]

Section D: Further Applications (10 marks)

16. pH of water and temperature: [2]

Answer: The dissociation of water is endothermic: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). [1] At higher temperatures, the equilibrium shifts to the right, increasing [H⁺] and [OH⁻], so pH decreases (although water remains neutral). [1]

17. Buffer pH calculation: [3]

n(NaHCO₃) = 2.10 / 84.0 = 0.0250 mol; n(Na₂CO₃) = 1.68 / 106.0 = 0.01585 mol [1]
[HCO₃⁻] = 0.0250 / 0.250 = 0.100 mol dm⁻³; [CO₃²⁻] = 0.01585 / 0.250 = 0.0634 mol dm⁻³ [1]
pKa = –log(4.8 × 10⁻¹¹) = 10.32; pH = 10.32 + log(0.0634 / 0.100) = 10.32 – 0.198 = 10.12 [1]

18. Preparation of sparingly soluble salt: [2]

Answer: Mix solutions of barium chloride and sodium sulfate. [1] The precipitate is filtered, washed with distilled water, and dried. The salt is pure because other ions remain in solution and are removed by washing. [1]

19. Ka calculation: [2]

[H⁺] = 10⁻³·⁰ = 1.0 × 10⁻³ mol dm⁻³ [1]
Ka = [H⁺]² / [HA] = (1.0 × 10⁻³)² / 0.10 = 1.0 × 10⁻⁵ mol dm⁻³ [1]

20. End point vs equivalence point: [1]

Answer: The equivalence point is when stoichiometrically equivalent amounts of acid and base have reacted. The end point is the point at which the indicator changes colour. [1]

END OF ANSWER KEY