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A Level H1 Chemistry Practice Paper 1

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Questions

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TuitionGoWhere Practice Paper - Chemistry H1 A-Level

TuitionGoWhere Secondary School (AI)

Subject: Chemistry
Level: A-Level H1
Paper: Practice Paper — Acids, Bases & Salts
Version: 1 of 5
Duration: 60 minutes
Total Marks: 60

Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions

  • Answer ALL questions in the spaces provided.
  • Show all working for calculation questions. Answers without working may not receive full credit.
  • The number of marks for each question or part question is shown in brackets [ ].
  • You may use a calculator.
  • A copy of the Periodic Table and Data Booklet is provided.

Section A: Multiple Choice (10 marks)

Questions 1–10. Choose the ONE correct answer for each question. Write your answer in the space provided.

1. Which of the following is the correct expression for the ionic product of water, KwK_w, at 298 K?

A. Kw=[H2O]K_w = [\text{H}_2\text{O}]
B. Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]
C. Kw=[H+]+[OH−]K_w = [\text{H}^+] + [\text{OH}^-]
D. Kw=[H+][OH−]K_w = \dfrac{[\text{H}^+]}{[\text{OH}^-]}

Answer: _______________ [1]

2. A solution has a pH of 3.50 at 298 K. What is the concentration of OH−\text{OH}^-(aq) in this solution?

A. 3.16×10−11 mol dm−33.16 \times 10^{-11} \text{ mol dm}^{-3}
B. 3.16×10−4 mol dm−33.16 \times 10^{-4} \text{ mol dm}^{-3}
C. 3.16×10−3 mol dm−33.16 \times 10^{-3} \text{ mol dm}^{-3}
D. 3.16×10−1 mol dm−33.16 \times 10^{-1} \text{ mol dm}^{-3}

Answer: _______________ [1]

3. Which of the following salts will produce an acidic solution when dissolved in water?

A. Na2CO3\text{Na}_2\text{CO}_3
B. NH4Cl\text{NH}_4\text{Cl}
C. KNO3\text{KNO}_3
D. CH3COONa\text{CH}_3\text{COONa}

Answer: _______________ [1]

4. A buffer solution is prepared by mixing 50.0 cm350.0 \text{ cm}^3 of 0.100 mol dm−30.100 \text{ mol dm}^{-3} ethanoic acid with 50.0 cm350.0 \text{ cm}^3 of 0.100 mol dm−30.100 \text{ mol dm}^{-3} sodium ethanoate. What is the pH of this buffer? (KaK_a of ethanoic acid =1.74×10−5 mol dm−3= 1.74 \times 10^{-5} \text{ mol dm}^{-3})

A. 2.76
B. 4.76
C. 5.76
D. 7.00

Answer: _______________ [1]

5. In the titration of a strong acid with a strong base, the pH at the equivalence point is:

A. less than 7
B. equal to 7
C. greater than 7
D. dependent on the indicator used

Answer: _______________ [1]

6. Which of the following best describes a weak acid?

A. An acid that has a low concentration in solution.
B. An acid that partially dissociates in aqueous solution.
C. An acid that does not dissociate in aqueous solution.
D. An acid that reacts slowly with bases.

Answer: _______________ [1]

7. The KaK_a value for a weak acid HA is 2.5×10−4 mol dm−32.5 \times 10^{-4} \text{ mol dm}^{-3}. What is the pKa\text{p}K_a of this acid?

A. 3.60
B. 4.00
C. 10.40
D. 10.60

Answer: _______________ [1]

8. Which of the following combinations would NOT act as a buffer solution?

A. CH3COOH\text{CH}_3\text{COOH} and CH3COONa\text{CH}_3\text{COONa}
B. NH3\text{NH}_3 and NH4Cl\text{NH}_4\text{Cl}
C. HCl\text{HCl} and NaCl\text{NaCl}
D. H2CO3\text{H}_2\text{CO}_3 and NaHCO3\text{NaHCO}_3

Answer: _______________ [1]

9. During the titration of a weak acid with a strong base, the buffer region occurs:

A. before any base is added.
B. when half the volume of base required to reach the equivalence point has been added.
C. at the equivalence point.
D. after the equivalence point.

Answer: _______________ [1]

10. A 0.050 mol dm−30.050 \text{ mol dm}^{-3} solution of a weak acid has a pH of 3.00. What is the approximate KaK_a of this acid?

A. 2.0×10−5 mol dm−32.0 \times 10^{-5} \text{ mol dm}^{-3}
B. 1.0×10−5 mol dm−31.0 \times 10^{-5} \text{ mol dm}^{-3}
C. 2.0×10−4 mol dm−32.0 \times 10^{-4} \text{ mol dm}^{-3}
D. 1.0×10−3 mol dm−31.0 \times 10^{-3} \text{ mol dm}^{-3}

Answer: _______________ [1]

Section B: Structured Questions (30 marks)

11. Define the following terms:

(a) A Brønsted–Lowry acid.

_______________________________________________________________________________ [1]

(b) A Brønsted–Lowry base.

_______________________________________________________________________________ [1]

(c) A conjugate acid–base pair.

_______________________________________________________________________________ [1]

12. Ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) is a weak acid.

(a) Write an equation to show the dissociation of ethanoic acid in water. Include state symbols.
_______________________________________________________________________________ [1]

(b) Explain, with reference to your equation in (a), why ethanoic acid is described as a weak acid.

_______________________________________________________________________________ [2]

(c) The KaK_a of ethanoic acid at 298 K is 1.74×10−5 mol dm−31.74 \times 10^{-5} \text{ mol dm}^{-3}. Calculate the pH of a 0.100 mol dm−30.100 \text{ mol dm}^{-3} solution of ethanoic acid.

_______________________________________________________________________________ [3]

13. A student carries out a titration to determine the concentration of a solution of hydrochloric acid using 0.100 mol dm−30.100 \text{ mol dm}^{-3} sodium hydroxide.

(a) Write an equation for the reaction between hydrochloric acid and sodium hydroxide.
_______________________________________________________________________________ [1]

(b) The student titrates 25.0 cm325.0 \text{ cm}^3 of the hydrochloric acid and finds that 22.40 cm322.40 \text{ cm}^3 of sodium hydroxide is required to reach the end-point. Calculate the concentration of the hydrochloric acid.

_______________________________________________________________________________ [3]

(c) Suggest a suitable indicator for this titration. Justify your choice.

_______________________________________________________________________________ [2]

14. A buffer solution is made by mixing 25.0 cm325.0 \text{ cm}^3 of 0.200 mol dm−30.200 \text{ mol dm}^{-3} methanoic acid (HCOOH\text{HCOOH}) with 25.0 cm325.0 \text{ cm}^3 of 0.200 mol dm−30.200 \text{ mol dm}^{-3} sodium methanoate (HCOONa\text{HCOONa}). The KaK_a of methanoic acid is 1.78×10−4 mol dm−31.78 \times 10^{-4} \text{ mol dm}^{-3}.

(a) Calculate the pH of this buffer solution.

_______________________________________________________________________________ [3]

(b) Explain how this buffer solution resists changes in pH when a small amount of dilute hydrochloric acid is added.

_______________________________________________________________________________ [3]

15. The following pH curve was obtained during the titration of 25.0 cm325.0 \text{ cm}^3 of 0.100 mol dm−30.100 \text{ mol dm}^{-3} ethanoic acid with 0.100 mol dm−30.100 \text{ mol dm}^{-3} sodium hydroxide.

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: pH titration curve showing pH (y-axis, 0–14) against volume of NaOH added in cm³ (x-axis, 0–50). The curve starts at approximately pH 2.9, rises gradually, then steeply increases around 25.0 cm³ (equivalence point) to approximately pH 12. The buffer region is visible between roughly 5–20 cm³. The equivalence point is at pH ≈ 8.7. labels: y-axis: "pH", x-axis: "Volume of NaOH added / cm³", equivalence point marked at (25.0, ~8.7), initial pH ≈ 2.9, final pH ≈ 12 values: initial pH = 2.9, equivalence point volume = 25.0 cm³, equivalence point pH = 8.7, final pH = 12 must_show: y-axis labelled "pH" from 0 to 14, x-axis labelled "Volume of NaOH added / cm³" from 0 to 50, curve starting at pH ~2.9, steep rise near 25 cm³, equivalence point at pH ~8.7, buffer region visible, final pH ~12 </image_placeholder>

(a) Using the graph, state the pH at the equivalence point.
_______________________________________________________________________________ [1]

(b) Explain why the pH at the equivalence point is greater than 7.

_______________________________________________________________________________ [2]

(c) On the graph, mark with an arrow the region that corresponds to the buffer region.
[1]

(d) Calculate the KaK_a of ethanoic acid using information from the graph.

_______________________________________________________________________________ [3]

Section C: Free Response (20 marks)

16. Ammonia (NH3\text{NH}_3) is a weak base.

(a) Write an equation to show the behaviour of ammonia as a Brønsted–Lowry base in water.
_______________________________________________________________________________ [1]

(b) The KbK_b of ammonia at 298 K is 1.78×10−5 mol dm−31.78 \times 10^{-5} \text{ mol dm}^{-3}. Calculate the pH of a 0.100 mol dm−30.100 \text{ mol dm}^{-3} solution of ammonia.

_______________________________________________________________________________ [4]

(c) A solution of ammonium chloride (NH4Cl\text{NH}_4\text{Cl}) is acidic. Explain this observation with reference to the relevant equation.

_______________________________________________________________________________ [3]

17. A student wishes to prepare a buffer solution with a pH of 5.00 using ethanoic acid (Ka=1.74×10−5 mol dm−3K_a = 1.74 \times 10^{-5} \text{ mol dm}^{-3}) and sodium ethanoate.

(a) Calculate the required ratio of [CH3COO−][\text{CH}_3\text{COO}^-] to [CH3COOH][\text{CH}_3\text{COOH}] in the buffer.

_______________________________________________________________________________ [3]

(b) The student has 100 cm3100 \text{ cm}^3 of 0.500 mol dm−30.500 \text{ mol dm}^{-3} ethanoic acid. Calculate the mass of sodium ethanoate (CH3COONa\text{CH}_3\text{COONa}, Mr=82.0M_r = 82.0) that must be dissolved in this solution to achieve the buffer pH of 5.00.

_______________________________________________________________________________ [4]

18. Phosphoric acid (H3PO4\text{H}_3\text{PO}_4) is a triprotic acid with the following dissociation constants at 298 K:

Ka1=7.11×10−3 mol dm−3K_{a1} = 7.11 \times 10^{-3} \text{ mol dm}^{-3}
Ka2=6.32×10−8 mol dm−3K_{a2} = 6.32 \times 10^{-8} \text{ mol dm}^{-3}
Ka3=4.50×10−13 mol dm−3K_{a3} = 4.50 \times 10^{-13} \text{ mol dm}^{-3}

(a) Write an equation for the first dissociation of phosphoric acid.
_______________________________________________________________________________ [1]

(b) Explain why Ka1≫Ka2≫Ka3K_{a1} \gg K_{a2} \gg K_{a3}.

_______________________________________________________________________________ [3]

(c) Calculate the pH of a 0.100 mol dm−30.100 \text{ mol dm}^{-3} solution of phosphoric acid. (Assume only the first dissociation contributes significantly to [H+][\text{H}^+].)

_______________________________________________________________________________ [3]

19. A solution contains a mixture of 0.100 mol dm−30.100 \text{ mol dm}^{-3} hydrochloric acid and 0.100 mol dm−30.100 \text{ mol dm}^{-3} ethanoic acid.

(a) Explain why the pH of this mixture is approximately the same as that of 0.100 mol dm−30.100 \text{ mol dm}^{-3} hydrochloric acid alone.

_______________________________________________________________________________ [2]

(b) Describe how you could experimentally distinguish between the two acids in this mixture using a simple chemical test.

_______________________________________________________________________________ [2]

20. The solubility product (KspK_{sp}) of magnesium hydroxide, Mg(OH)2\text{Mg(OH)}_2, at 298 K is 5.50×10−12 mol3 dm−95.50 \times 10^{-12} \text{ mol}^3 \text{ dm}^{-9}.

(a) Write an expression for the solubility product of magnesium hydroxide.
_______________________________________________________________________________ [1]

(b) Calculate the solubility of magnesium hydroxide in water at 298 K, in mol dm−3\text{mol dm}^{-3}.

_______________________________________________________________________________ [3]

(c) Predict and explain the effect on the solubility of magnesium hydroxide when a small amount of sodium hydroxide is added to the solution.

_______________________________________________________________________________ [2]

END OF PAPER

Total Marks: 60

Answers

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TuitionGoWhere Practice Paper — Chemistry H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper — Acids, Bases & Salts | Version: 1 of 5 | Total Marks: 60

Section A: Multiple Choice (10 marks)

1. B [1]
Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-] is the definition of the ionic product of water. It is derived from the equilibrium H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}. At 298 K, Kw=1.00×10−14 mol2 dm−6K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}.
Common mistake: Choosing A — students confuse KwK_w with the concentration of water itself.

2. A [1]
Working:
pH=3.50⇒[H+]=10−3.50=3.16×10−4 mol dm−3\text{pH} = 3.50 \Rightarrow [\text{H}^+] = 10^{-3.50} = 3.16 \times 10^{-4} \text{ mol dm}^{-3}
Kw=[H+][OH−]=1.00×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14}
[OH−]=1.00×10−143.16×10−4=3.16×10−11 mol dm−3[\text{OH}^-] = \dfrac{1.00 \times 10^{-14}}{3.16 \times 10^{-4}} = 3.16 \times 10^{-11} \text{ mol dm}^{-3}
Teaching note: In any aqueous solution at 298 K, [H+][OH−]=Kw=1.00×10−14[\text{H}^+][\text{OH}^-] = K_w = 1.00 \times 10^{-14}. As [H+][\text{H}^+] increases, [OH−][\text{OH}^-] must decrease.

3. B [1]
NH4Cl\text{NH}_4\text{Cl} is a salt of a weak base (NH3\text{NH}_3) and a strong acid (HCl\text{HCl}). The NH4+\text{NH}_4^+ ion undergoes hydrolysis: NH4++H2O⇌NH3+H3O+\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+, producing H+\text{H}^+ ions and making the solution acidic.

  • Na2CO3\text{Na}_2\text{CO}_3: salt of strong base + weak acid → basic
  • KNO3\text{KNO}_3: salt of strong base + strong acid → neutral
  • CH3COONa\text{CH}_3\text{COONa}: salt of strong base + weak acid → basic

4. B [1]
Working:
When equal volumes of equal concentrations are mixed, the concentrations of acid and salt are halved, but their ratio remains 1:1.
Using the Henderson–Hasselbalch equation:
pH=pKa+log⁡10[salt][acid]\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]}
pKa=−log⁡10(1.74×10−5)=4.76\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76
pH=4.76+log⁡10(1)=4.76+0=4.76\text{pH} = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76
Key point: When [salt]=[acid][\text{salt}] = [\text{acid}], pH=pKa\text{pH} = \text{p}K_a.

5. B [1]
At the equivalence point of a strong acid–strong base titration, the salt formed (e.g., NaCl) does not hydrolyse. The solution is neutral with pH = 7.
Common mistake: Choosing D — the indicator choice does not affect the pH at the equivalence point; it only affects how accurately the end-point is detected.

6. B [1]
A weak acid is defined as an acid that partially dissociates in aqueous solution. This is a question of acid strength, not concentration.
Common mistake: Choosing A — "low concentration" describes a dilute acid, not a weak acid. A weak acid can be concentrated but still only partially dissociate.

7. A [1]
Working:
pKa=−log⁡10(Ka)=−log⁡10(2.5×10−4)=3.60\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(2.5 \times 10^{-4}) = 3.60
Teaching note: pKa\text{p}K_a is the negative logarithm of KaK_a. A smaller KaK_a (weaker acid) gives a larger pKa\text{p}K_a.

8. C [1]
A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid). HCl\text{HCl} is a strong acid and NaCl\text{NaCl} is its salt — there is no equilibrium to resist pH changes.

  • A: weak acid + its salt → buffer ✓
  • B: weak base + its salt → buffer ✓
  • D: weak acid + its salt → buffer ✓

9. B [1]
The buffer region occurs when significant amounts of both the weak acid and its conjugate base are present. This is most effective at the half-equivalence point, where exactly half the acid has been neutralised, giving [acid]=[salt][\text{acid}] = [\text{salt}].

10. A [1]
Working:
pH=3.00⇒[H+]=10−3.00=1.00×10−3 mol dm−3\text{pH} = 3.00 \Rightarrow [\text{H}^+] = 10^{-3.00} = 1.00 \times 10^{-3} \text{ mol dm}^{-3}
For a weak acid HA: Ka=[H+][A−][HA]≈[H+]2[HA]initialK_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \dfrac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}}
Ka=(1.00×10−3)20.050=1.00×10−60.050=2.0×10−5 mol dm−3K_a = \dfrac{(1.00 \times 10^{-3})^2}{0.050} = \dfrac{1.00 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}
Approximation: Since the acid is weak, [H+]≪[HA]initial[\text{H}^+] \ll [\text{HA}]_{\text{initial}}, so [HA]equilibrium≈[HA]initial[\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}}.

Section B: Structured Questions (30 marks)

11.

(a) A Brønsted–Lowry acid is a proton (H+\text{H}^+) donor. [1]
Marking note: Must mention "proton" and "donor" (or "donates"). Simply saying "produces H+\text{H}^+" is not sufficient for full credit at A-Level.

(b) A Brønsted–Lowry base is a proton (H+\text{H}^+) acceptor. [1]
Marking note: Must mention "proton" and "acceptor" (or "accepts").

(c) A conjugate acid–base pair is a pair of species that differ by one proton (H+\text{H}^+). [1]
Example: CH3COOH\text{CH}_3\text{COOH} and CH3COO−\text{CH}_3\text{COO}^- form a conjugate acid–base pair. The acid has one more proton than its conjugate base.

12.

(a) CH3COOH(aq)+H2O(l)⇌CH3COO−(aq)+H3O+(aq)\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}
or equivalently: CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} [1]
Marking note: The reversible arrow (⇌\rightleftharpoons) is essential. State symbols are required.

(b) Ethanoic acid is a weak acid because it only partially dissociates in water. [1] The equilibrium lies far to the left, meaning only a small fraction of ethanoic acid molecules donate protons to water. [1]
Teaching note: A strong acid (e.g., HCl) would have a single arrow (→\rightarrow) showing complete dissociation. The reversible arrow indicates incomplete dissociation.

(c) Working: [3]
Ka=[H+][CH3COO−][CH3COOH]=1.74×10−5K_a = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 1.74 \times 10^{-5}

Let [H+]=x[\text{H}^+] = x. Then [CH3COO−]=x[\text{CH}_3\text{COO}^-] = x and [CH3COOH]≈0.100−x≈0.100[\text{CH}_3\text{COOH}] \approx 0.100 - x \approx 0.100 (since xx is small).

1.74×10−5=x20.1001.74 \times 10^{-5} = \dfrac{x^2}{0.100}
x2=1.74×10−6x^2 = 1.74 \times 10^{-6}
x=1.74×10−6=1.32×10−3 mol dm−3x = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}

pH=−log⁡0(1.32×10−3)=2.88\text{pH} = -\log_{0}(1.32 \times 10^{-3}) = 2.88

Answer: pH = 2.88 [3]
Marking scheme:

  • [1] Correct KaK_a expression
  • [1] Correct substitution and calculation of [H+][\text{H}^+]
  • [1] Correct pH value (accept 2.88 ± 0.02)

13.

(a) HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}
or ionic: H+(aq)+OH−(aq)→H2O(l)\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)} [1]

(b) Working: [3]
Moles of NaOH used =22.401000×0.100=2.24×10−3 mol= \dfrac{22.40}{1000} \times 0.100 = 2.24 \times 10^{-3} \text{ mol}

From the equation, mole ratio HCl : NaOH = 1 : 1
Moles of HCl =2.24×10−3 mol= 2.24 \times 10^{-3} \text{ mol}

Concentration of HCl =2.24×10−325.0/1000=2.24×10−30.0250=0.0896 mol dm−3= \dfrac{2.24 \times 10^{-3}}{25.0/1000} = \dfrac{2.24 \times 10^{-3}}{0.0250} = 0.0896 \text{ mol dm}^{-3}

Answer: 0.0896 mol dm⁻³ [3]
Marking scheme:

  • [1] Correct moles of NaOH
  • [1] Correct use of 1:1 stoichiometry
  • [1] Correct final concentration (accept 0.0896 or 0.090 to 2 s.f.)

(c) Any of: methyl orange, methyl red, or bromophenol blue. [1]
For a strong acid–strong base titration, the equivalence point is at pH 7, and the pH change is very steep (approximately pH 3–11). Any indicator with a transition range within this steep portion is suitable. [1]
Note: Phenolphthalein is also commonly accepted, though its transition range (pH 8.2–10.0) is slightly above the equivalence point. The key justification is that the vertical portion of the curve spans the indicator's transition range.

14.

(a) Working: [3]
When equal volumes of equal concentrations are mixed, both concentrations are halved:
[HCOOH]=0.100 mol dm−3[\text{HCOOH}] = 0.100 \text{ mol dm}^{-3} and [HCOONa]=0.100 mol dm−3[\text{HCOONa}] = 0.100 \text{ mol dm}^{-3}

Using Henderson–Hasselbalch:
pKa=−log⁡10(1.78×10−4)=3.75\text{p}K_a = -\log_{10}(1.78 \times 10^{-4}) = 3.75
pH=pKa+log⁡10[HCOO−][HCOOH]=3.75+log⁡100.1000.100=3.75+0=3.75\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{HCOO}^-]}{[\text{HCOOH}]} = 3.75 + \log_{10}\dfrac{0.100}{0.100} = 3.75 + 0 = 3.75

Answer: pH = 3.75 [3]
Marking scheme:

  • [1] Correct pKa\text{p}K_a calculation
  • [1] Correct use of Henderson–Hasselbalch equation
  • [1] Correct final pH

(b) When HCl is added, the H+\text{H}^+ ions react with the methanoate ions (HCOO−\text{HCOO}^-) in the buffer: [1]
HCOO−(aq)+H+(aq)→HCOOH(aq)\text{HCOO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{HCOOH(aq)} [1]

The added H+\text{H}^+ is consumed by the conjugate base, converting it to the weak acid. Since both species are present in large amounts, the ratio [HCOO−]/[HCOOH][\text{HCOO}^-]/[\text{HCOOH}] changes only slightly, so the pH remains nearly constant. [1]
Teaching note: The buffer works because it contains significant amounts of both the weak acid (to react with added base) and its conjugate base (to react with added acid).

15.

(a) pH at equivalence point ≈ 8.7 [1]
Accept any value in the range 8.5–9.0 based on reading from the graph.

(b) At the equivalence point, all the ethanoic acid has been converted to sodium ethanoate (CH3COONa\text{CH}_3\text{COONa}). [1] The ethanoate ion (CH3COO−\text{CH}_3\text{COO}^-) is the conjugate base of a weak acid and undergoes hydrolysis: CH3COO−+H2O⇌CH3COOH+OH−\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-, producing OH−\text{OH}^- ions and making the solution slightly alkaline. [1]

(c) The buffer region should be marked on the graph between approximately 5 cm³ and 20 cm³ of NaOH added — the relatively flat portion of the curve before the steep rise. [1]
Specifically, the region where the curve rises gradually (before the equivalence point) should be indicated.

(d) Working: [3]
At the half-equivalence point (12.5 cm³ of NaOH added), half the ethanoic acid has been neutralised, so [CH3COOH]=[CH3COO−][\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-].
At this point, pH=pKa\text{pH} = \text{p}K_a (from Henderson–Hasselbalch: pH=pKa+log⁡10(1)=pKa\text{pH} = \text{p}K_a + \log_{10}(1) = \text{p}K_a).

From the graph, at 12.5 cm³, pH ≈ 4.76.
Therefore, pKa=4.76\text{p}K_a = 4.76 and Ka=10−4.76=1.74×10−5 mol dm−3K_a = 10^{-4.76} = 1.74 \times 10^{-5} \text{ mol dm}^{-3}.

Answer: Ka=1.74×10−5 mol dm−3K_a = 1.74 \times 10^{-5} \text{ mol dm}^{-3} [3]
Marking scheme:

  • [1] Correct identification that at half-equivalence point, pH=pKa\text{pH} = \text{p}K_a
  • [1] Correct reading of pH from graph at 12.5 cm³
  • [1] Correct calculation of KaK_a

Section C: Free Response (20 marks)

16.

(a) NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} [1]
Marking note: Reversible arrow and state symbols required. Ammonia accepts a proton from water, forming NH4+\text{NH}_4^+ and OH−\text{OH}^-.

(b) Working: [4]
Kb=[NH4+][OH−][NH3]=1.78×10−5K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.78 \times 10^{-5}

Let [OH−]=x[\text{OH}^-] = x. Then [NH4+]=x[\text{NH}_4^+] = x and [NH3]≈0.100−x≈0.100[\text{NH}_3] \approx 0.100 - x \approx 0.100.

1.78×10−5=x20.1001.78 \times 10^{-5} = \dfrac{x^2}{0.100}
x2=1.78×10−6x^2 = 1.78 \times 10^{-6}
x=1.78×10−6=1.33×10−3 mol dm−3x = \sqrt{1.78 \times 10^{-6}} = 1.33 \times 10^{-3} \text{ mol dm}^{-3}

pOH=−log⁡10(1.33×10−3)=2.88\text{pOH} = -\log_{10}(1.33 \times 10^{-3}) = 2.88
pH=14.00−2.88=11.12\text{pH} = 14.00 - 2.88 = 11.12

Answer: pH = 11.12 [4]
Marking scheme:

  • [1] Correct KbK_b expression
  • [1] Correct calculation of [OH−][\text{OH}^-]
  • [1] Correct calculation of pOH
  • [1] Correct conversion to pH (using pH+pOH=14\text{pH} + \text{pOH} = 14)

(c) Ammonium chloride dissociates completely in water to give NH4+\text{NH}_4^+ and Cl−\text{Cl}^- ions. [1] The NH4+\text{NH}_4^+ ion is the conjugate acid of the weak base NH3\text{NH}_3 and undergoes hydrolysis: [1]
NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)\text{NH}_4^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}

This produces H3O+\text{H}_3\text{O}^+ (or H+\text{H}^+) ions, making the solution acidic. [1]
Teaching note: Salts of weak bases and strong acids are always acidic because the conjugate acid of the weak base hydrolyses water to produce H+\text{H}^+ ions.

17.

(a) Working: [3]
Using the Henderson–Hasselbalch equation:
pH=pKa+log⁡10[CH3COO−][CH3COOH]\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}

pKa=−log⁡10(1.74×10−5)=4.76\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76

5.00=4.76+log⁡10[CH3COO−][CH3COOH]5.00 = 4.76 + \log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}
log⁡10[CH3COO−][CH3COOH]=0.24\log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 0.24
[CH3COO−][CH3COOH]=100.24=1.74\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 10^{0.24} = 1.74

Answer: [CH3COO−]:[CH3COOH]=1.74:1[\text{CH}_3\text{COO}^-] : [\text{CH}_3\text{COOH}] = 1.74 : 1 [3]
Marking scheme:

  • [1] Correct pKa\text{p}K_a value
  • [1] Correct substitution into Henderson–Hasselbalch
  • [1] Correct ratio (accept 1.74:1 or 1.7:1)

(b) Working: [4]
From (a), [CH3COO−][CH3COOH]=1.74\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 1.74

Moles of ethanoic acid in 100 cm³ of 0.500 mol dm⁻³ =1001000×0.500=0.0500 mol= \dfrac{100}{1000} \times 0.500 = 0.0500 \text{ mol}

Since the volume is the same for both species, the mole ratio equals the concentration ratio:
n(CH3COO−)n(CH3COOH)=1.74\dfrac{n(\text{CH}_3\text{COO}^-)}{n(\text{CH}_3\text{COOH})} = 1.74
n(CH3COO−)=1.74×0.0500=0.0870 moln(\text{CH}_3\text{COO}^-) = 1.74 \times 0.0500 = 0.0870 \text{ mol}

Mass of sodium ethanoate =0.0870×82.0=7.13 g= 0.0870 \times 82.0 = 7.13 \text{ g}

Answer: 7.13 g [4]
Marking scheme:

  • [1] Correct moles of ethanoic acid
  • [1] Correct use of ratio from (a) to find moles of sodium ethanoate
  • [1] Correct molar mass used
  • [1] Correct final mass (accept 7.1–7.14 g)

18.

(a) H3PO4(aq)+H2O(l)⇌H2PO4−(aq)+H3O+(aq)\text{H}_3\text{PO}_4\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{PO}_4^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}
or: H3PO4(aq)⇌H2PO4−(aq)+H+(aq)\text{H}_3\text{PO}_4\text{(aq)} \rightleftharpoons \text{H}_2\text{PO}_4^-\text{(aq)} + \text{H}^+\text{(aq)} [1]

(b) Ka1≫Ka2≫Ka3K_{a1} \gg K_{a2} \gg K_{a3} because: [3]

  • After the first dissociation, the remaining species (H2PO4−\text{H}_2\text{PO}_4^-) carries a negative charge. Removing a positively charged proton (H+\text{H}^+) from a negatively charged species is progressively more difficult due to increasing electrostatic attraction. [1]
  • Each successive dissociation removes H+\text{H}^+ from an increasingly negative ion, requiring more energy. [1]
  • The negative charge that builds up on the phosphate species after each dissociation makes it increasingly difficult to remove the next proton. [1]
    Teaching note: This is a general pattern for polyprotic acids. The first proton is easiest to remove; each subsequent proton is harder to remove because the remaining anion holds onto its protons more tightly.

(c) Working: [3]
Ka1=[H+][H2PO4−][H3PO4]=7.11×10−3K_{a1} = \dfrac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} = 7.11 \times 10^{-3}

Let [H+]=x[\text{H}^+] = x. Then:
7.11×10−3=x20.100−x7.11 \times 10^{-3} = \dfrac{x^2}{0.100 - x}

Since Ka1K_{a1} is relatively large, we cannot assume x≪0.100x \ll 0.100. Solving the quadratic:
x2+7.11×10−3x−7.11×10−4=0x^2 + 7.11 \times 10^{-3}x - 7.11 \times 10^{-4} = 0

Using the quadratic formula:
x=−7.11×10−3+(7.11×10−3)2+4(7.11×10−4)2x = \dfrac{-7.11 \times 10^{-3} + \sqrt{(7.11 \times 10^{-3})^2 + 4(7.11 \times 10^{-4})}}{2}
x=−7.11×10−3+5.06×10−5+2.844×10−32x = \dfrac{-7.11 \times 10^{-3} + \sqrt{5.06 \times 10^{-5} + 2.844 \times 10^{-3}}}{2}
x=−7.11×10−3+2.895×10−32x = \dfrac{-7.11 \times 10^{-3} + \sqrt{2.895 \times 10^{-3}}}{2}
x=−7.11×10−3+5.38×10−22=4.67×10−22=2.34×10−2x = \dfrac{-7.11 \times 10^{-3} + 5.38 \times 10^{-2}}{2} = \dfrac{4.67 \times 10^{-2}}{2} = 2.34 \times 10^{-2}

pH=−log⁡10(2.34×10−2)=1.63\text{pH} = -\log_{10}(2.34 \times 10^{-2}) = 1.63

Answer: pH = 1.63 [3]
Marking scheme:

  • [1] Correct Ka1K_{a1} expression
  • [1] Correct solution of quadratic (or valid approximation method)
  • [1] Correct pH value (accept 1.60–1.65)
    Note: If the student uses the approximation x≪0.100x \ll 0.100, they would get x=7.11×10−4=2.67×10−2x = \sqrt{7.11 \times 10^{-4}} = 2.67 \times 10^{-2}, giving pH = 1.57. This approximation is less accurate here because Ka1K_{a1} is relatively large. Award partial credit [2/3] for this approach.

19.

(a) Hydrochloric acid is a strong acid and dissociates completely, contributing 0.100 mol dm−30.100 \text{ mol dm}^{-3} of H+\text{H}^+ ions. [1] Ethanoic acid is a weak acid with Ka=1.74×10−5K_a = 1.74 \times 10^{-5}, so it contributes a negligible additional concentration of H+\text{H}^+ ions compared to the strong acid. The presence of the high [H+][\text{H}^+] from HCl also suppresses the dissociation of ethanoic acid (common ion effect). [1]
Teaching note: The common ion effect means that the H+\text{H}^+ from HCl pushes the ethanoic acid equilibrium even further to the left, making its contribution to [H+][\text{H}^+] even smaller.

(b) Add a small amount of solid calcium carbonate (or magnesium ribbon) to the solution. [1] The rate of effervescence (gas evolution) will be rapid initially due to the strong acid (HCl), then continue at a slower rate due to the weak acid (ethanoic acid). Alternatively, measure the pH over time — the pH will remain relatively constant for a period as the weak acid continues to react, whereas a solution of only strong acid would show a different reaction profile. [1]
Alternative acceptable answer: Measure the electrical conductivity — the mixture will have higher conductivity than ethanoic acid alone due to the fully dissociated HCl. Or: titrate with NaOH using a pH meter — the titration curve will show two distinct equivalence points (or a buffered region) indicating the presence of two acids.

20.

(a) Ksp=[Mg2+][OH−]2K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 [1]
Marking note: The expression must include the correct species with the correct powers (1 for Mg2+\text{Mg}^{2+}, 2 for OH−\text{OH}^-).

(b) Working: [3]
Let the solubility of Mg(OH)2=s mol dm−3\text{Mg(OH)}_2 = s \text{ mol dm}^{-3}.
Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}

[Mg2+]=s[\text{Mg}^{2+}] = s and [OH−]=2s[\text{OH}^-] = 2s

Ksp=(s)(2s)2=4s3=5.50×10−12K_{sp} = (s)(2s)^2 = 4s^3 = 5.50 \times 10^{-12}
s3=5.50×10−124=1.375×10−12s^3 = \dfrac{5.50 \times 10^{-12}}{4} = 1.375 \times 10^{-12}
s=1.375×10−123=1.11×10−4 mol dm−3s = \sqrt[3]{1.375 \times 10^{-12}} = 1.11 \times 10^{-4} \text{ mol dm}^{-3}

Answer: 1.11×10−4 mol dm−31.11 \times 10^{-4} \text{ mol dm}^{-3} [3]
Marking scheme:

  • [1] Correct relationship between solubility and ion concentrations ([OH−]=2s[\text{OH}^-] = 2s)
  • [1] Correct substitution into KspK_{sp} expression
  • [1] Correct final answer (accept 1.1×10−41.1 \times 10^{-4} to 1.11×10−41.11 \times 10^{-4})

(c) The solubility of magnesium hydroxide will decrease. [1] This is due to the common ion effect: adding NaOH\text{NaOH} increases [OH−][\text{OH}^-], shifting the equilibrium Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} to the left (Le Chatelier's principle), causing more Mg(OH)2\text{Mg(OH)}_2 to precipitate. [1]
Teaching note: The common ion effect is a direct application of Le Chatelier's principle. Adding a product ion (OH−\text{OH}^-) shifts the equilibrium towards the solid, reducing solubility.

END OF ANSWER KEY

Total Marks: 60