A Level H2 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H2 Quiz - Acids Bases Salts

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

---

Section A: Short Answer & Structured Response (20 marks)

1. (a) Write a balanced equation for the reaction between ammonium chloride and sodium hydroxide.
[2 marks]

Answer:
NH₄Cl(s) + NaOH(aq) → NH₃(g) + NaCl(aq) + H₂O(l)

Marking notes:

• Correct formulae for all reactants and products [1 mark]
• Correct state symbols: (s) for NH₄Cl, (aq) for NaOH and NaCl, (g) for NH₃, (l) for H₂O [1 mark]
• Accept: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l) for ionic equation (award 2 marks if fully correct with state symbols)

---

1. (b) Describe a chemical test to confirm the identity of ammonia gas.
[2 marks]

Answer:
Hold a piece of damp red litmus paper near the mouth of the test tube / in the gas. [1 mark]
The damp red litmus paper turns blue. [1 mark]

Marking notes:

• Must specify "damp" red litmus paper (dry litmus will not work effectively)
• Accept: "moist" or "wet" red litmus paper
• Reject: "litmus paper" without specifying red
• Accept: use of universal indicator paper turning blue/purple

---

2. (a) Name a suitable indicator and state the colour change at the end-point.
[2 marks]

Answer:
Phenolphthalein [1 mark]
Colour change: pink to colourless (or colourless to pink, depending on which solution is in the flask) [1 mark]

Marking notes:

• Accept: methyl orange (yellow to orange/red)
• Accept: bromothymol blue (blue to yellow or yellow to blue)
• Indicator must be appropriate for strong acid-strong base titration (pH range 3-11)
• Colour change must be correctly stated for the indicator named

---

2. (b) Obtain a suitable volume of HCl(aq) from the titration results.
[2 marks]

Answer:
The rough titration (25.10 cm³) is excluded as it is a rough estimate. [1 mark]
Titrations 1, 2, and 3 are concordant (all 24.85 cm³).
Mean volume = (24.85 + 24.85 + 24.85) ÷ 3 = 24.85 cm³ [1 mark]

Marking notes:

• Must state that rough titration is excluded [1 mark]
• Must calculate mean of concordant results (24.85 cm³) [1 mark]
• Accept: identification that all three accurate titrations are identical, so mean = 24.85 cm³
• Award 1 mark if student correctly identifies concordant results but makes arithmetic error

---

3. (a) Explain what is meant by the term amphoteric.
[1 mark]

Answer:
An amphoteric substance is one that can react with both acids and bases / can behave as both an acid and a base.

Marking notes:

• Must mention reaction with both acids AND bases
• Accept: "reacts with both H⁺ and OH⁻"

---

3. (b) Write an ionic equation for the reaction of Al₂O₃ with hot aqueous sodium hydroxide.
[2 marks]

Answer:
Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)

Marking notes:

• Correct reactants and products [1 mark]
• Correct balancing and state symbols [1 mark]
• Accept: Al₂O₃(s) + 3H₂O(l) + 2OH⁻(aq) → 2[Al(OH)₄]⁻(aq)
• Reject: Al₂O₃(s) + 2OH⁻(aq) → 2AlO₂⁻(aq) + H₂O(l) (incorrect aluminate formula for A-Level)

---

4. (a) State the observation when a small amount of aqueous ammonia is added to aqueous copper(II) sulfate.
[1 mark]

Answer:
A blue precipitate is formed.

Marking notes:

• Must state "blue precipitate"
• Accept: "pale blue precipitate" or "light blue precipitate"

---

4. (b) State the observation when excess aqueous ammonia is added. Name the complex ion and write its formula.
[3 marks]

Answer:
Observation: The blue precipitate dissolves to form a deep blue solution. [1 mark]
Name: Tetraamminecopper(II) ion [1 mark]
Formula: [Cu(NH₃)₄]²⁺ [1 mark]

Marking notes:

• Observation must include "dissolves" and "deep blue solution"
• Complex ion name must be correct (tetraamminecopper(II))
• Formula must include square brackets and correct charge (2+)
• Accept: [Cu(NH₃)₄(H₂O)₂]²⁺

---

5. Describe how to use aqueous sodium hydroxide to distinguish between Al³⁺(aq), Fe²⁺(aq), and Fe³⁺(aq).
[5 marks]

Answer:
Add aqueous sodium hydroxide dropwise to each solution until in excess. [1 mark]

Observations:

• Al³⁺(aq): A white precipitate forms, which dissolves in excess NaOH(aq) to form a colourless solution. [1 mark]
• Fe²⁺(aq): A green precipitate forms, which is insoluble in excess NaOH(aq). On standing in air, the green precipitate turns brown at the surface. [1 mark]
• Fe³⁺(aq): A brown/reddish-brown precipitate forms, which is insoluble in excess NaOH(aq). [1 mark]

The three cations can be distinguished by:

• The white precipitate that dissolves in excess NaOH identifies Al³⁺.
• The green precipitate that turns brown identifies Fe²⁺.
• The brown precipitate identifies Fe³⁺. [1 mark for clear summary]

Marking notes:

• Must describe procedure (adding NaOH dropwise, then in excess) [1 mark]
• Must state correct colour and solubility for each cation [3 marks total, 1 per cation]
• Must explain how to distinguish (summary) [1 mark]
• Accept: Fe²⁺ precipitate described as "dirty green" or "dark green"
• For Fe²⁺: must mention oxidation to Fe³⁺ (green → brown) for full credit

---

Section B: Calculation & Data Interpretation (18 marks)

6. (a) Write a balanced equation for the reaction between ethanoic acid and sodium hydroxide.
[1 mark]

Answer:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

Marking notes:

• Correct formulae and balancing required
• State symbols not required for this mark but good practice
• Accept: CH₃COOH + NaOH → CH₃COONa + H₂O

---

6. (b) Calculate the number of moles of NaOH used.
[1 mark]

Answer:
n(NaOH) = c × V = 0.150 × (22.40 ÷ 1000) = 3.36 × 10⁻³ mol

Marking notes:

• Must convert cm³ to dm³
• Correct answer: 0.00336 mol or 3.36 × 10⁻³ mol
• Award mark for correct method even if minor arithmetic error

---

6. (c) Calculate the concentration of ethanoic acid in mol dm⁻³.
[2 marks]

Answer:
From equation: 1 mol CH₃COOH reacts with 1 mol NaOH
n(CH₃COOH) = n(NaOH) = 3.36 × 10⁻³ mol [1 mark]
c(CH₃COOH) = n ÷ V = 3.36 × 10⁻³ ÷ (25.0 ÷ 1000) = 0.1344 mol dm⁻³
= 0.134 mol dm⁻³ (3 s.f.) [1 mark]

Marking notes:

• Must use 1:1 mole ratio [1 mark]
• Correct calculation and significant figures [1 mark]
• Accept: 0.134 mol dm⁻³

---

6. (d) Calculate the concentration of ethanoic acid in g dm⁻³.
[2 marks]

Answer:
Mᵣ(CH₃COOH) = (2 × 12.0) + (4 × 1.0) + (2 × 16.0) = 60.0 [1 mark]
Concentration = 0.1344 × 60.0 = 8.064 g dm⁻³
= 8.06 g dm⁻³ (3 s.f.) [1 mark]

Marking notes:

• Correct Mᵣ calculation [1 mark]
• Correct multiplication and significant figures [1 mark]
• Accept: 8.06 g dm⁻³ or 8.07 g dm⁻³ depending on rounding

---

7. (a) Explain what is meant by a buffer solution.
[2 marks]

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

Marking notes:

• Must mention "resists changes in pH" [1 mark]
• Must mention "small amounts of acid/base added" or "on dilution" [1 mark]
• Accept: "maintains approximately constant pH"

---

7. (b) Calculate the pH of this buffer solution.
[3 marks]

Answer:
After mixing:
[CH₃COOH] = (0.200 × 50.0) ÷ 100.0 = 0.100 mol dm⁻³ [1 mark]
[CH₃COO⁻] = (0.200 × 50.0) ÷ 100.0 = 0.100 mol dm⁻³ [1 mark]

Using Henderson-Hasselbalch equation or Kₐ expression:
pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH])
pKₐ = -log(1.74 × 10⁻⁵) = 4.76
pH = 4.76 + log(0.100/0.100) = 4.76 + 0 = 4.76 [1 mark]

Marking notes:

• Correct calculation of diluted concentrations [1 mark]
• Correct use of Kₐ or Henderson-Hasselbalch equation [1 mark]
• Correct final pH value [1 mark]
• Accept: pH = 4.76 (or 4.759)

---

7. (c) Explain, with equations, how this buffer resists changes in pH when a small amount of hydrochloric acid is added.
[3 marks]

Answer:
The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base). [1 mark]
When H⁺ from HCl is added: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [1 mark]
The added H⁺ reacts with the conjugate base to form undissociated weak acid, so the [H⁺] remains approximately constant and pH changes very little. [1 mark]

Marking notes:

• Must identify buffer components [1 mark]
• Must write correct equation for reaction with added acid [1 mark]
• Must explain that H⁺ is removed, maintaining pH [1 mark]

---

8. Calculate the relative molecular mass, Mᵣ, of the acid HA.
[4 marks]

Answer:
Moles of NaOH used in titration = 0.100 × (20.00 ÷ 1000) = 2.00 × 10⁻³ mol [1 mark]
Moles of HA in 25.0 cm³ = 2.00 × 10⁻³ mol (1:1 reaction) [1 mark]
Moles of HA in 250.0 cm³ = 2.00 × 10⁻³ × 10 = 0.0200 mol [1 mark]
Mᵣ = mass ÷ moles = 1.26 ÷ 0.0200 = 63.0 g mol⁻¹ [1 mark]

Marking notes:

• Correct calculation of moles of NaOH [1 mark]
• Correct mole ratio [1 mark]
• Correct scaling to 250.0 cm³ [1 mark]
• Correct Mᵣ with units [1 mark]
• Accept: 63.0

---

Section C: Data-Based & Extended Response (12 marks)

9. (a) Define the term acid dissociation constant, Kₐ, for a weak acid HA. Write an expression for Kₐ.
[2 marks]

Answer:
Kₐ is the equilibrium constant for the dissociation of a weak acid in aqueous solution. [1 mark]
Kₐ = [H⁺][A⁻] / [HA] [1 mark]

Marking notes:

• Must mention equilibrium constant or equilibrium expression [1 mark]
• Correct expression with square brackets [1 mark]

---

9. (b) State and explain the trend in acid strength shown by the data in the table.
[3 marks]

Answer:
Acid strength decreases from methanoic acid to propanoic acid. [1 mark]
The Kₐ values decrease (methanoic > ethanoic > propanoic), indicating less dissociation. [1 mark]
The alkyl groups (CH₃, CH₃CH₂) are electron-donating, which destabilises the conjugate base (RCOO⁻) and shifts the equilibrium to the left, reducing acid strength. Longer alkyl chains have a greater electron-donating effect. [1 mark]

Marking notes:

• Correct trend stated [1 mark]
• Link to Kₐ values [1 mark]
• Explanation in terms of electron-donating effect of alkyl groups [1 mark]

---

9. (c) Calculate the pH of a 0.100 mol dm⁻³ solution of methanoic acid. State any assumption made.
[4 marks]

Answer:
Kₐ = [H⁺][HCOO⁻] / [HCOOH]
Assume [H⁺] = [HCOO⁻] and [HCOOH]eq ≈ 0.100 mol dm⁻³ [1 mark]
1.78 × 10⁻⁴ = [H⁺]² / 0.100 [1 mark]
[H⁺]² = 1.78 × 10⁻⁵
[H⁺] = √(1.78 × 10⁻⁵) = 4.22 × 10⁻³ mol dm⁻³ [1 mark]
pH = -log(4.22 × 10⁻³) = 2.37 [1 mark]

Marking notes:

• Correct assumption stated (degree of dissociation is small) [1 mark]
• Correct substitution into Kₐ expression [1 mark]
• Correct [H⁺] [1 mark]
• Correct pH (2.37 or 2.38) [1 mark]

---

10. (a) State and explain the trend in solubility of Group 2 hydroxides down the group.
[2 marks]

Answer:
Solubility increases down the group. [1 mark]
The lattice energy decreases more rapidly than the hydration energy due to increasing ionic size, making the dissolution process more exothermic/favourable. [1 mark]

Marking notes:

• Correct trend [1 mark]
• Explanation in terms of lattice energy and hydration energy [1 mark]

---

10. (b)(i) State the observation for each solution.
[2 marks]

Answer:
Mg²⁺(aq): White precipitate [1 mark]
Ca²⁺(aq): White precipitate (or slight white precipitate)
Ba²⁺(aq): No precipitate (or colourless solution remains) [1 mark]

Marking notes:

• Correct observation for Mg²⁺ and Ca²⁺ [1 mark]
• Correct observation for Ba²⁺ [1 mark]
• Accept: "faint white precipitate" for Ca²⁺

---

10. (b)(ii) Write an ionic equation, with state symbols, for any precipitation reaction that occurs.
[1 mark]

Answer:
Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s)
OR
Ca²⁺(aq) + 2OH⁻(aq) → Ca(OH)₂(s)

Marking notes:

• Correct ionic equation with state symbols [1 mark]
• Accept either Mg²⁺ or Ca²⁺ equation

---

10. (c) Explain why magnesium hydroxide is used as an antacid, whereas barium hydroxide is not suitable.
[2 marks]

Answer:
Mg(OH)₂ is sparingly soluble and provides a slow, controlled neutralisation of excess stomach acid without raising pH too high. [1 mark]
Ba(OH)₂ is soluble and strongly alkaline; it would cause a rapid and dangerous increase in pH, and Ba²⁺ ions are toxic. [1 mark]

Marking notes:

• Reference to solubility and controlled neutralisation for Mg(OH)₂ [1 mark]
• Reference to high solubility, strong alkalinity, and toxicity for Ba(OH)₂ [1 mark]

---

11. Define the term Brønsted-Lowry acid and give an example of a species that can act as a Brønsted-Lowry acid but not as an Arrhenius acid.
[2 marks]

Answer:
A Brønsted-Lowry acid is a proton (H⁺) donor. [1 mark]
Example: NH₄⁺ (ammonium ion) or HCl(g) [1 mark]

Marking notes:

• Correct definition (proton donor) [1 mark]
• Correct example that does not directly produce H⁺ in water (e.g., NH₄⁺, HCl gas) [1 mark]

---

12. Explain why the pH of pure water at 298 K is 7.00, but decreases when the temperature is increased to 323 K.
[2 marks]

Answer:
At 298 K, [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, so pH = 7.00. [1 mark]
The dissociation of water is endothermic. Increasing temperature shifts the equilibrium H₂O ⇌ H⁺ + OH⁻ to the right, increasing [H⁺], so pH decreases. [1 mark]

Marking notes:

• Reference to [H⁺] at 298 K [1 mark]
• Explanation using endothermic nature and equilibrium shift [1 mark]

---

13. Write an equation to show the amphoteric nature of the hydrogencarbonate ion, HCO₃⁻, by reacting with both H⁺ and OH⁻.
[2 marks]

Answer:
With acid: HCO₃⁻(aq) + H⁺(aq) → H₂O(l) + CO₂(g) [1 mark]
With base: HCO₃⁻(aq) + OH⁻(aq) → CO₃²⁻(aq) + H₂O(l) [1 mark]

Marking notes:

• Correct equation with H⁺ [1 mark]
• Correct equation with OH⁻ [1 mark]
• State symbols not essential but good practice

---

14. A solution of hydrochloric acid has a pH of 2.00. Calculate the concentration of the acid in mol dm⁻³.
[1 mark]

Answer:
[H⁺] = 10⁻²·⁰⁰ = 0.0100 mol dm⁻³
HCl is a strong acid, so [HCl] = [H⁺] = 0.0100 mol dm⁻³

Marking notes:

• Correct calculation and answer [1 mark]
• Accept: 0.01 mol dm⁻³

---

15. Explain why a solution of ammonium chloride, NH₄Cl, is acidic, whereas a solution of sodium chloride, NaCl, is neutral. Support your answer with relevant equations.
[3 marks]

Answer:
NH₄Cl dissociates: NH₄Cl(aq) → NH₄⁺(aq) + Cl⁻(aq).
NH₄⁺ is the conjugate acid of a weak base (NH₃) and undergoes hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq), producing H₃O⁺, so the solution is acidic. [1.5 marks]
NaCl dissociates: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Na⁺ is the cation of a strong base (NaOH) and Cl⁻ is the anion of a strong acid (HCl); neither hydrolyses, so [H⁺] = [OH⁻] and the solution is neutral. [1.5 marks]

Marking notes:

• Correct explanation and equation for NH₄Cl acidity [1.5 marks]
• Correct explanation for NaCl neutrality [1.5 marks]

---

Section D: Acids, Bases and Equilibria in Context (12 marks)

16. (a) Calculate the concentration of H⁺ ions in the solution.
[1 mark]

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

Marking notes:

• Correct calculation [1 mark]
• Accept: 6.31 × 10⁻⁴ mol dm⁻³

---

16. (b) Calculate the acid dissociation constant, Kₐ, of HX.
[3 marks]

Answer:
HX ⇌ H⁺ + X⁻
[H⁺] = [X⁻] = 6.31 × 10⁻⁴ mol dm⁻³ [1 mark]
[HX]eq ≈ 0.0500 mol dm⁻³ (assumption: dissociation is small) [1 mark]
Kₐ = (6.31 × 10⁻⁴)² / 0.0500 = 7.96 × 10⁻⁶ mol dm⁻³ [1 mark]

Marking notes:

• Correct [H⁺] and [X⁻] [1 mark]
• Correct assumption and [HX]eq [1 mark]
• Correct Kₐ value [1 mark]
• Accept: 7.96 × 10⁻⁶ to 8.00 × 10⁻⁶

---

16. (c) Calculate the pKₐ of HX.
[1 mark]

Answer:
pKₐ = -log(7.96 × 10⁻⁶) = 5.10

Marking notes:

• Correct pKₐ [1 mark]
• Accept: 5.10

---

17. Calculate the pH of this buffer solution.
[4 marks]

Answer:
Total volume = 30.0 + 20.0 = 50.0 cm³
[CH₃CH₂COOH] = (0.100 × 30.0) ÷ 50.0 = 0.0600 mol dm⁻³ [1 mark]
[CH₃CH₂COO⁻] = (0.150 × 20.0) ÷ 50.0 = 0.0600 mol dm⁻³ [1 mark]
pKₐ = -log(1.35 × 10⁻⁵) = 4.87 [1 mark]
pH = pKₐ + log([CH₃CH₂COO⁻]/[CH₃CH₂COOH]) = 4.87 + log(0.0600/0.0600) = 4.87 [1 mark]

Marking notes:

• Correct diluted concentrations [2 marks]
• Correct pKₐ [1 mark]
• Correct pH [1 mark]
• Accept: 4.87

---

18. Explain why the equivalence point of a titration between a weak acid and a strong base occurs at a pH greater than 7.
[3 marks]

Answer:
At the equivalence point, all the weak acid (e.g., CH₃COOH) has been converted to its conjugate base (CH₃COO⁻). [1 mark]
The conjugate base undergoes hydrolysis: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), producing OH⁻ ions. [1 mark]
The presence of excess OH⁻ makes the solution alkaline, so pH > 7. [1 mark]

Marking notes:

• Identification of conjugate base at equivalence point [1 mark]
• Correct hydrolysis equation [1 mark]
• Explanation linking OH⁻ to pH > 7 [1 mark]

---

19. (a) Calculate the concentration of OH⁻ ions in the saturated solution.
[2 marks]

Answer:
pH = 10.40, so pOH = 14.00 - 10.40 = 3.60 [1 mark]
[OH⁻] = 10⁻³·⁶⁰ = 2.51 × 10⁻⁴ mol dm⁻³ [1 mark]

Marking notes:

• Correct pOH [1 mark]
• Correct [OH⁻] [1 mark]

---

19. (b) Hence, calculate the solubility of Mg(OH)₂ in mol dm⁻³.
[2 marks]

Answer:
Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq)
[OH⁻] = 2.51 × 10⁻⁴ mol dm⁻³
Solubility = [Mg²⁺] = ½ × [OH⁻] = ½ × 2.51 × 10⁻⁴ = 1.26 × 10⁻⁴ mol dm⁻³ [2 marks]

Marking notes:

• Correct stoichiometric relationship [1 mark]
• Correct solubility value [1 mark]
• Accept: 1.26 × 10⁻⁴ mol dm⁻³

---

20. Describe, with the aid of equations, how the carbonic acid-hydrogencarbonate buffer system regulates the pH of blood. Explain why this buffer is essential for human health.
[3 marks]

Answer:
The buffer consists of H₂CO₃ (carbonic acid) and HCO₃⁻ (hydrogencarbonate ions).
When H⁺ is added: HCO₃⁻(aq) + H⁺(aq) → H₂CO₃(aq)
When OH⁻ is added: H₂CO₃(aq) + OH⁻(aq) → HCO₃⁻(aq) + H₂O(l) [1 mark for equations]
This maintains blood pH around 7.40. [1 mark]
It is essential because enzymes and biochemical processes are highly pH-sensitive; significant pH changes can lead to illness (acidosis/alkalosis) or death. [1 mark]

Marking notes:

• Correct equations for buffering action [1 mark]
• Reference to maintaining pH ~7.40 [1 mark]
• Explanation of importance (enzyme function, health consequences) [1 mark]

---

END OF ANSWER KEY