A Level H1 Chemistry Practice Paper 3

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TuitionGoWhere Practice Paper — Chemistry H1 A-Level

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TuitionGoWhere Secondary School (AI)

Subject:	Chemistry
Level:	A-Level H1
Paper:	Practice Paper 2 — Structured & Free Response
Version:	3 of 5
Duration:	2 hours
Total Marks:	80
Name:	______________________________
Class:	______________________________
Date:	______________________________

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Instructions

• Write your answers in the spaces provided.
• All questions are compulsory.
• Show all working for calculation questions. Answers without working may not receive full marks.
• Use appropriate units and significant figures where applicable.
• A Periodic Table with relative atomic masses is provided on the last page of this paper.
• The use of a scientific calculator is permitted.

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Section A: Multiple Choice [10 marks]

Answer ALL questions. Each question carries 1 mark. Write your answers in the spaces provided.

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1. Which of the following is a strong acid?

A. $CH_3COOH$
B. $H_2CO_3$
C. $HNO_3$
D. $H_3PO_4$

Answer: ______________

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2. A solution has a pH of 2.3. What is the concentration of $H^+(aq)$ in this solution?

A. $5.0 \times 10^{-3}$ mol dm$^{-3}$
B. $2.3 \times 10^{-1}$ mol dm$^{-3}$
C. $2.0 \times 10^{-2}$ mol dm$^{-3}$
D. $3.0 \times 10^{-3}$ mol dm$^{-3}$

Answer: ______________

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3. Which salt, when dissolved in water, produces an acidic solution?

A. $NaCl$
B. $KNO_3$
C. $NH_4Cl$
D. $Na_2CO_3$

Answer: ______________

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4. Which statement best describes a buffer solution?

A. A solution that always has a pH of exactly 7.
B. A solution that resists changes in pH when small amounts of acid or base are added.
C. A solution that contains only a strong acid and its conjugate base.
D. A solution that changes colour at the endpoint of a titration.

Answer: ______________

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5. In the reaction $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$, the $NH_3$ acts as a

A. Brønsted–Lowry acid.
B. Brønsted–Lowry base.
C. Lewis acid only.
D. neutral species.

Answer: ______________

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6. What volume of 0.100 mol dm$^{-3}$ $NaOH$ is required to neutralise 25.0 cm$^3$ of 0.200 mol dm$^{-3}$ $H_2SO_4$?

A. 25.0 cm$^3$
B. 50.0 cm$^3$
C. 100.0 cm$^3$
D. 12.5 cm$^3$

Answer: ______________

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7. Which of the following is the conjugate base of $H_2PO_4^-$?

A. $H_3PO_4$
B. $HPO_4^{2-}$
C. $PO_4^{3-}$
D. $H_2PO_4$

Answer: ______________

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8. The pH of a 0.050 mol dm$^{-3}$ solution of a weak acid, $HA$, is 3.00. What is the approximate $K_a$ of this acid?

A. $2.0 \times 10^{-5}$
B. $5.0 \times 10^{-6}$
C. $1.0 \times 10^{-3}$
D. $2.0 \times 10^{-3}$

Answer: ______________

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9. During a titration of a strong acid with a strong base, the pH at the equivalence point is

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

Answer: ______________

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10. Which of the following salts will produce a neutral solution when dissolved in water?

A. $NaF$
B. $KBr$
C. $Na_2SO_3$
D. $NH_4NO_3$

Answer: ______________

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Section B: Structured Questions [50 marks]

Answer ALL questions.

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11. [6 marks]

(a) Define the term strong acid. [2]

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(b) Write an equation to show the dissociation of hydrochloric acid in water. [1]

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(c) A student claims that a 0.001 mol dm$^{-3}$ solution of a strong acid has a pH of 3.0, but a 0.001 mol dm$^{-3}$ solution of a weak acid also has a pH of 3.0. Explain why this claim is incorrect. [3]

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12. [8 marks]

25.0 cm$^3$ of 0.150 mol dm$^{-3}$ $NaOH$ is titrated with 0.100 mol dm$^{-3}$ $HCl$.

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

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(b) Calculate the volume of $HCl$ required to reach the equivalence point. [3]

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(c) State a suitable indicator for this titration and explain your choice. [2]

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(d) Calculate the pH of the solution after 50.0 cm$^3$ of $HCl$ has been added. [2]

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13. [8 marks]

A buffer solution is prepared by mixing 50.0 cm$^3$ of 0.200 mol dm$^{-3}$ $CH_3COOH$ with 50.0 cm$^3$ of 0.100 mol dm$^{-3}$ $NaOH$.

(a) Explain why the resulting mixture acts as a buffer solution. [3]

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(b) Write an expression for the acid dissociation constant, $K_a$, of $CH_3COOH$. [1]

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(c) Given that $K_a$ of $CH_3COOH$ is $1.7 \times 10^{-5}$ mol dm$^{-3}$, calculate the pH of this buffer solution. [4]

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14. [10 marks]

The following data were collected during a titration of 25.0 cm$^3$ of a solution of a weak monoprotic acid, $HA$, with 0.100 mol dm$^{-3}$ $NaOH$:

Volume of NaOH added / cm³	pH
0.0	2.80
5.0	3.70
10.0	4.20
12.5	4.50
15.0	5.10
20.0	11.50
25.0	12.10
30.0	12.40

(a) From the data, estimate the $pK_a$ of the weak acid $HA$. Explain your reasoning. [3]

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(b) Calculate the concentration of the original $HA$ solution. [3]

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(c) Explain why the pH at the equivalence point is greater than 7. [2]

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(d) Sketch a titration curve for this experiment on the axes below. Label the equivalence point and the region where the solution acts as a buffer. [2]

<image_placeholder> id: Q14-fig1 type: graph linked_question: Q14 description: A set of axes for sketching a titration curve. The x-axis is labelled "Volume of NaOH added / cm³" ranging from 0 to 35 cm³. The y-axis is labelled "pH" ranging from 0 to 14. The graph should show a weak acid-strong base titration curve starting at approximately pH 2.8, rising gradually through a buffer region, with a steep rise near 25.0 cm³, and levelling off around pH 12.4. labels: x-axis: "Volume of NaOH added / cm³", y-axis: "pH", equivalence point marked at (25.0, ~8.5), buffer region shaded between approximately 5–20 cm³ values: Starting pH ~2.8, equivalence point volume = 25.0 cm³, pH at equivalence ~8.5, final pH ~12.4 must_show: Axes with labels and scales, a curve showing the characteristic weak acid-strong base shape, equivalence point labelled, buffer region indicated </image_placeholder>

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15. [8 marks]

(a) Define the term salt hydrolysis. [2]

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(b) Classify each of the following salts as producing an acidic, basic, or aqueous solution when dissolved in water. Explain each answer in terms of the acid and base from which the salt is derived. [6]

(i) $NH_4NO_3$ [2]

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(ii) $K_2CO_3$ [2]

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(iii) $NaCl$ [2]

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16. [10 marks]

A solution is prepared by dissolving 0.365 g of $HCl$ in water to make 500 cm$^3$ of solution.

(a) Calculate the concentration of the $HCl$ solution in mol dm$^{-3}$. [2]

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(b) Calculate the pH of this $HCl$ solution. [1]

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(c) 25.0 cm$^3$ of this $HCl$ solution is then mixed with 30.0 cm$^3$ of 0.050 mol dm$^{-3}$ $Ba(OH)_2$. Calculate the pH of the resulting mixture. [5]

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(d) Name the salt formed in the reaction in (c) and state whether its aqueous solution is acidic, basic, or neutral. [2]

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Section C: Free Response [20 marks]

Answer ALL questions.

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17. [10 marks]

Ethanoic acid, $CH_3COOH$, is a weak acid with $K_a = 1.7 \times 10^{-5}$ mol dm$^{-3}$.

(a) Write the expression for $K_a$ for ethanoic acid. [1]

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(b) Calculate the pH of a 0.100 mol dm$^{-3}$ solution of ethanoic acid. [4]

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(c) A solution is prepared by mixing 50.0 cm$^3$ of 0.100 mol dm$^{-3}$ $CH_3COOH$ with 50.0 cm$^3$ of 0.100 mol dm$^{-3}$ $NaOH$. Calculate the pH of the resulting solution. [5]

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18. [10 marks]

A student investigates the properties of three solutions: Solution X (0.100 mol dm$^{-3}$ $HCl$), Solution Y (0.100 mol dm$^{-3}$ $CH_3COOH$), and Solution Z (0.100 mol dm$^{-3}$ $NaOH$).

(a) Arrange the three solutions in order of increasing pH. Explain your reasoning. [3]

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(b) The student adds 1.0 g of magnesium ribbon to 50.0 cm$^3$ of each solution. Compare and explain the rate of hydrogen gas production in Solutions X and Y. [4]

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(c) The student performs a titration of Solution Y with Solution Z using phenolphthalein as indicator. Describe the colour change observed at the endpoint and explain why phenolphthalein is a suitable indicator for this titration. [3]

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End of Paper

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Periodic Table Data

Element	Symbol	Relative Atomic Mass
Hydrogen	H	1.0
Carbon	C	12.0
Nitrogen	N	14.0
Oxygen	O	16.0
Sodium	Na	23.0
Magnesium	Mg	24.3
Chlorine	Cl	35.5
Potassium	K	39.1
Calcium	Ca	40.1
Barium	Ba	137.3

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TuitionGoWhere Practice Paper — Chemistry H1 A-Level

Answer Key — Version 3 of 5

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Section A: Multiple Choice [10 marks]

1. Answer: C [1]

$HNO_3$ (nitric acid) is a strong acid because it completely dissociates in aqueous solution. $CH_3COOH$, $H_2CO_3$, and $H_3PO_4$ are all weak acids (partial dissociation).

Common mistake: Students confuse the number of ionisable hydrogens with acid strength. $H_3PO_4$ has three acidic protons but is still a weak acid because dissociation is incomplete.

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2. Answer: A [1]

$[H^+] = 10^{-pH} = 10^{-2.3} = 5.01 \times 10^{-3}$ mol dm$^{-3}$ ≈ $5.0 \times 10^{-3}$ mol dm$^{-3}$.

Common mistake: Students sometimes write the pH value itself as the concentration, or forget to use the inverse log function.

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3. Answer: C [1]

$NH_4Cl$ is formed from a weak base ($NH_3$) and a strong acid ($HCl$). The $NH_4^+$ ion undergoes hydrolysis to produce $H^+$ ions, making the solution acidic.

$NaCl$ and $KNO_3$ are salts of strong acids and strong bases → neutral. $Na_2CO_3$ is a salt of a strong base and weak acid → basic.

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4. Answer: B [1]

A buffer solution resists changes in pH upon addition of small amounts of acid or base. It typically contains a weak acid and its conjugate base (or weak base and its conjugate acid).

Common mistake: Students think buffers always have pH 7. Buffers can be prepared at various pH values depending on the weak acid/base pair used.

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5. Answer: B [1]

In this reaction, $NH_3$ accepts a proton ($H^+$) from $H_2O$ to form $NH_4^+$. A Brønsted–Lowry base is defined as a proton acceptor.

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6. Answer: C [1]

$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

Moles of $H_2SO_4$ = $0.200 \times \frac{25.0}{1000} = 0.00500$ mol

Moles of $NaOH$ needed = $2 \times 0.00500 = 0.0100$ mol

Volume of $NaOH$ = $\frac{0.0100}{0.100} = 0.100$ dm$^3$ = 100.0 cm$^3$

Common mistake: Students forget that $H_2SO_4$ is dibasic (1 mol reacts with 2 mol $NaOH$).

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7. Answer: B [1]

The conjugate base is formed when an acid loses a proton. $H_2PO_4^- \rightarrow H^+ + HPO_4^{2-}$. Therefore, $HPO_4^{2-}$ is the conjugate base of $H_2PO_4^-$.

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8. Answer: A [1]

$pH = 3.00$, so $[H^+] = 10^{-3.00} = 1.0 \times 10^{-3}$ mol dm$^{-3}$

For a weak acid: $K_a \approx \frac{[H^+]^2}{c} = \frac{(1.0 \times 10^{-3})^2}{0.050} = \frac{1.0 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5}$ mol dm$^{-3}$

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9. Answer: B [1]

For a strong acid–strong base titration, the salt formed (e.g., $NaCl$) does not hydrolyse. The equivalence point pH is exactly 7.

Common mistake: Students think the indicator determines the pH at the equivalence point. The indicator is chosen to change colour near the equivalence point pH, not to define it.

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10. Answer: B [1]

$KBr$ is derived from a strong acid ($HBr$) and a strong base ($KOH$). Neither ion hydrolyses, so the solution is neutral (pH = 7).

$NaF$ → strong base + weak acid → basic. $Na_2SO_3$ → strong base + weak acid → basic. $NH_4NO_3$ → weak base + strong acid → acidic.

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Section B: Structured Questions [50 marks]

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11. [6 marks]

(a) [2]

A strong acid is an acid that completely dissociates (or ionises) in aqueous solution.

• 1 mark: "completely dissociates" or "fully ionises"
• 1 mark: reference to "in aqueous solution" or "in water"

Common mistake: Saying "dissociates a lot" or "mostly dissociates" — this is not sufficient. The key word is "completely."

(b) [1]

$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

or equivalently: $HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$

Award 1 mark for correct species and a single (complete) arrow. State symbols are not required at H1 level but are good practice.

(c) [3]

A strong acid completely dissociates, so $[H^+] = 0.001$ mol dm$^{-3}$ and $pH = -\log(0.001) = 3.0$. ✓

However, a weak acid only partially dissociates. For a 0.001 mol dm$^{-3}$ weak acid, the $[H^+]$ will be less than 0.001 mol dm$^{-3}$ because not all acid molecules release $H^+$ ions. Therefore, the pH of the weak acid solution will be greater than 3.0 (i.e., less acidic).

• 1 mark: Strong acid fully dissociates → $[H^+] = 0.001$ mol dm$^{-3}$ → pH = 3.0
• 1 mark: Weak acid partially dissociates → $[H^+] < 0.001$ mol dm$^{-3}$
• 1 mark: Therefore pH of weak acid > 3.0, so the claim is incorrect

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12. [8 marks]

(a) [1]

$NaOH + HCl \rightarrow NaCl + H_2O$

(b) [3]

Moles of $NaOH$ = $0.150 \times \frac{25.0}{1000} = 3.75 \times 10^{-3}$ mol [1]

From the equation, mole ratio $NaOH : HCl = 1 : 1$

Moles of $HCl$ needed = $3.75 \times 10^{-3}$ mol

Volume of $HCl$ = $\frac{3.75 \times 10^{-3}}{0.100} = 3.75 \times 10^{-2}$ dm$^3$ = 37.5 cm$^3$ [1]

Correct answer with units [1]

(c) [2]

A suitable indicator is methyl orange (or phenolphthalein). [1]

This is a strong acid–strong base titration, so the equivalence point is at pH 7. Both methyl orange (pH range 3.1–4.4) and phenolphthalein (pH range 8.2–10.0) undergo a sharp colour change within the vertical region of the titration curve. [1]

Note: Any indicator with a transition range that falls within the steep portion of the curve is acceptable. Full credit for a valid indicator with a correct explanation.

(d) [2]

Total volume after adding 50.0 cm$^3$ $HCl$ = $25.0 + 50.0 = 75.0$ cm$^3$

Moles of $HCl$ added = $0.100 \times \frac{50.0}{1000} = 5.00 \times 10^{-3}$ mol

Moles of $NaOH$ initially = $3.75 \times 10^{-3}$ mol

Moles of excess $HCl$ = $5.00 \times 10^{-3} - 3.75 \times 10^{-3} = 1.25 \times 10^{-3}$ mol [1]

$[H^+] = \frac{1.25 \times 10^{-3}}{75.0 / 1000} = \frac{1.25 \times 10^{-3}}{0.075} = 0.01667$ mol dm$^{-3}$

$pH = -\log(0.01667) = 1.78$ [1]

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13. [8 marks]

(a) [3]

$NaOH$ reacts with $CH_3COOH$:

Moles of $CH_3COOH$ initially = $0.200 \times \frac{50.0}{1000} = 0.0100$ mol

Moles of $NaOH$ added = $0.100 \times \frac{50.0}{1000} = 0.00500$ mol

$NaOH$ neutralises half the $CH_3COOH$, producing $0.00500$ mol of $CH_3COONa$ (the conjugate base), with $0.00500$ mol of unreacted $CH_3COOH$ remaining. [1]

The resulting mixture contains a weak acid ($CH_3COOH$) and its conjugate base ($CH_3COO^-$) in comparable amounts. [1]

This combination constitutes a buffer solution because the weak acid can neutralise added base and the conjugate base can neutralise added acid, thereby resisting pH changes. [1]

(b) [1]

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

(c) [4]

Since the total volume is the same for both solutions (50.0 cm$^3$ + 50.0 cm$^3$ = 100.0 cm$^3$), and moles of $CH_3COOH$ remaining = moles of $CH_3COO^-$ formed = 0.00500 mol:

$[CH_3COOH] = [CH_3COO^-] = \frac{0.00500}{0.100} = 0.0500$ mol dm$^{-3}$ [1]

Using the Henderson–Hasselbalch equation:

$pK_a = -\log(1.7 \times 10^{-5}) = 4.77$ [1]

$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.77 + \log\frac{0.0500}{0.0500} = 4.77 + \log(1) = 4.77 + 0 = 4.77$ [1]

Correct answer: pH = 4.77 [1]

Alternative method using $K_a$ directly:

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

Since $[CH_3COO^-] = [CH_3COOH]$, $[H^+] = K_a = 1.7 \times 10^{-5}$ mol dm$^{-3}$

$pH = -\log(1.7 \times 10^{-5}) = 4.77$

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14. [10 marks]

(a) [3]

The $pK_a$ of a weak acid equals the pH at the half-equivalence point — the point at which exactly half the acid has been neutralised. [1]

From the data, the equivalence point occurs between 20.0 and 25.0 cm$^3$ (where the sharp pH rise occurs). The half-equivalence point is therefore at approximately 12.5 cm$^3$. [1]

At 12.5 cm$^3$, the pH = 4.50. Therefore, $pK_a$ ≈ 4.50 and $K_a$ ≈ $3.2 \times 10^{-5}$ mol dm$^{-3}$. [1]

Accept answers in the range pH 4.30–4.70 if justified with reference to the data.

(b) [3]

At the equivalence point, moles of $NaOH$ = moles of $HA$.

From the titration curve, the equivalence point occurs at approximately 25.0 cm$^3$ of 0.100 mol dm$^{-3}$ $NaOH$ (the steepest part of the curve is centred around 25 cm³ based on the data jump from pH 5.10 at 15.0 cm³ to pH 11.50 at 20.0 cm³ — the equivalence point is at approximately 20.0 cm³ where the sharp rise occurs). [1]

Re-reading the data: The pH jumps from 5.10 (at 15.0 cm³) to 11.50 (at 20.0 cm³). The equivalence point is therefore at approximately 20.0 cm³.

Moles of $NaOH$ at equivalence = $0.100 \times \frac{20.0}{1000} = 2.00 \times 10^{-3}$ mol [1]

Concentration of $HA$ = $\frac{2.00 \times 10^{-3}}{25.0 / 1000} = \frac{2.00 \times 10^{-3}}{0.0250} = 0.080$ mol dm$^{-3}$ [1]

(c) [2]

At the equivalence point, all the weak acid $HA$ has been converted to its conjugate base $A^-$ (as the salt $NaA$). [1]

The conjugate base $A^-$ undergoes hydrolysis:

$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)$

This produces $OH^-$ ions, making the solution basic (pH > 7). [1]

(d) [2]

The sketch should show:

• Starting pH ≈ 2.8 (at 0 cm³ $NaOH$) [½]
• A gradually rising curve with a buffer region (relatively flat section) between approximately 5–15 cm³ [½]
• A steep vertical rise centred at approximately 20.0 cm³ (the equivalence point) [½]
• The curve levelling off at high pH (~12.4) after the equivalence point [½]
• Equivalence point clearly labelled at approximately (20.0, ~8–9) [½ — included in above]
• Buffer region indicated/shaded on the curve [½ — included in above]

Award marks for correct shape, correct starting pH, correct equivalence point location, and labelled features.

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15. [8 marks]

(a) [2]

Salt hydrolysis is the reaction of the ions of a salt with water to produce $H^+$ or $OH^-$ ions, resulting in a solution that is not neutral (i.e., pH ≠ 7). [1]

More specifically, it occurs when the cation or anion (or both) of the salt reacts with water to form a weak acid or weak base, thereby altering the pH of the solution. [1]

(b) [6]

(i) $NH_4NO_3$ — Acidic solution [2]

$NH_4NO_3$ is derived from a weak base ($NH_3$) and a strong acid ($HNO_3$). [½]

The $NH_4^+$ ion (conjugate acid of the weak base) undergoes hydrolysis:

$NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$

This produces $H^+$ ions, making the solution acidic (pH < 7). [½]

The $NO_3^-$ ion does not hydrolyse because it is the conjugate base of a strong acid. [½]

Award [1] for correct classification with explanation, [½] for identifying parent acid/base, [½] for hydrolysis equation or reasoning.

(ii) $K_2CO_3$ — Basic solution [2]

$K_2CO_3$ is derived from a strong base ($KOH$) and a weak acid ($H_2CO_3$). [½]

The $CO_3^{2-}$ ion (conjugate base of the weak acid) undergoes hydrolysis:

$CO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HCO_3^-(aq) + OH^-(aq)$

This produces $OH^-$ ions, making the solution basic (pH > 7). [½]

The $K^+$ ion does not hydrolyse because it is the conjugate acid of a strong base. [½]

(iii) $NaCl$ — Neutral solution [2]

$NaCl$ is derived from a strong acid ($HCl$) and a strong base ($NaOH$). [½]

Neither $Na^+$ nor $Cl^-$ undergoes hydrolysis because they are the conjugate acid of a strong base and the conjugate base of a strong acid, respectively. Both ions are too weak as acids/bases to react with water. [½]

Therefore, the solution remains neutral (pH = 7). [½]

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16. [10 marks]

(a) [2]

Molar mass of $HCl$ = $1.0 + 35.5 = 36.5$ g mol$^{-1}$ [½]

Moles of $HCl$ = $\frac{0.365}{36.5} = 0.0100$ mol [½]

Concentration = $\frac{0.0100}{500 / 1000} = \frac{0.0100}{0.500} = 0.0200$ mol dm$^{-3}$ [1]

(b) [1]

$HCl$ is a strong acid, so it completely dissociates:

$[H^+] = 0.0200$ mol dm$^{-3}$

$pH = -\log(0.0200) = 1.70$ [1]

(c) [5]

Moles of $HCl$ = $0.0200 \times \frac{25.0}{1000} = 5.00 \times 10^{-4}$ mol [1]

Moles of $Ba(OH)_2$ = $0.050 \times \frac{30.0}{1000} = 1.50 \times 10^{-3}$ mol [½]

$Ba(OH)_2$ dissociates to give 2 $OH^-$ per formula unit:

Moles of $OH^-$ = $2 \times 1.50 \times 10^{-3} = 3.00 \times 10^{-3}$ mol [½]

The reaction is: $H^+ + OH^- \rightarrow H_2O$

Moles of excess $OH^-$ = $3.00 \times 10^{-3} - 5.00 \times 10^{-4} = 2.50 \times 10^{-3}$ mol [1]

Total volume = $25.0 + 30.0 = 55.0$ cm$^3$ = $0.0550$ dm$^3$

$[OH^-] = \frac{2.50 \times 10^{-3}}{0.0550} = 0.04545$ mol dm$^{-3}$ [½]

$pOH = -\log(0.04545) = 1.34$

$pH = 14.00 - 1.34 = 12.66$ [1]

(d) [2]

The salt formed is barium chloride, $BaCl_2$. [1]

$BaCl_2$ is derived from a strong acid ($HCl$) and a strong base ($Ba(OH)_2$). Neither ion undergoes hydrolysis, so the aqueous solution is neutral. [1]

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Section C: Free Response [20 marks]

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17. [10 marks]

(a) [1]

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

(b) [4]

For the dissociation: $CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)$

Let $[H^+] = x$ mol dm$^{-3}$ at equilibrium.

	$CH_3COOH$	$H^+$	$CH_3COO^-$
Initial	0.100	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.100 - x$	$x$	$x$

$K_a = \frac{x^2}{0.100 - x} = 1.7 \times 10^{-5}$ [1]

Since $K_a$ is small, $x \ll 0.100$, so $0.100 - x \approx 0.100$:

$x^2 = 1.7 \times 10^{-5} \times 0.100 = 1.7 \times 10^{-6}$ [1]

$x = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3}$ mol dm$^{-3}$ [1]

$pH = -\log(1.30 \times 10^{-3}) = 2.89$ [1]

Validation: $x / 0.100 = 1.3\% < 5\%$, so the approximation is valid.

(c) [5]

Moles of $CH_3COOH$ = $0.100 \times \frac{50.0}{1000} = 5.00 \times 10^{-3}$ mol

Moles of $NaOH$ = $0.100 \times \frac{50.0}{1000} = 5.00 \times 10^{-3}$ mol [1]

The acid and base react in a 1:1 ratio, so all the $CH_3COOH$ is completely neutralised:

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Moles of $CH_3COONa$ formed = $5.00 \times 10^{-3}$ mol [1]

Total volume = $50.0 + 50.0 = 100.0$ cm$^3$ = $0.100$ dm$^3$

Concentration of $CH_3COONa$ = $\frac{5.00 \times 10^{-3}}{0.100} = 0.0500$ mol dm$^{-1}$

The salt $CH_3COONa$ contains the conjugate base $CH_3COO^-$, which hydrolyses:

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$ [1]

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-5}} = 5.88 \times 10^{-10}$ mol dm$^{-3}$

Let $[OH^-] = y$:

$K_b = \frac{y^2}{0.0500} = 5.88 \times 10^{-10}$

$y^2 = 5.88 \times 10^{-10} \times 0.0500 = 2.94 \times 10^{-11}$

$y = \sqrt{2.94 \times 10^{-11}} = 5.42 \times 10^{-6}$ mol dm$^{-3}$ [1]

$pOH = -\log(5.42 \times 10^{-6}) = 5.27$

$pH = 14.00 - 5.27 = 8.73$ [1]

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18. [10 marks]

(a) [3]

Order of increasing pH: Solution X < Solution Y < Solution Z [1]

Solution X ($HCl$): $HCl$ is a strong acid and completely dissociates. $[H^+] = 0.100$ mol dm$^{-3}$, so $pH = 1.0$. [½]

Solution Y ($CH_3COOH$): $CH_3COOH$ is a weak acid and only partially dissociates. $[H^+] < 0.100$ mol dm$^{-3}$, so $pH > 1.0$ (approximately 2.89 for 0.100 mol dm$^{-3}$). [½]

Solution Z ($NaOH$): $NaOH$ is a strong base. $[OH^-] = 0.100$ mol dm$^{-3}$, $pOH = 1.0$, so $pH = 13.0$. [½]

(b) [4]

The initial rate of hydrogen gas production is faster in Solution X ($HCl$) than in Solution Y ($CH_3COOH$). [1]

This is because $HCl$ is a strong acid and completely dissociates, giving a much higher initial $[H^+]$ (0.100 mol dm$^{-3}$) compared to the weak acid $CH_3COOH$ (where $[H^+] \approx 1.3 \times 10^{-3}$ mol dm$^{-3}$ for 0.100 mol dm$^{-3}$). [1]

The rate of reaction between $Mg$ and acid depends on $[H^+]$:

$Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2(g)$ [1]

Since the $[H^+]$ in Solution X is much higher, the frequency of effective collisions between $Mg$ and $H^+$ ions is greater, resulting in a faster initial rate. [1]

Note: Both solutions contain the same total moles of acid (0.0050 mol in 50.0 cm$^3$ of 0.100 mol dm$^{-3}$), so the total volume of $H_2$ gas produced will be the same. However, the weak acid reacts more slowly because its $[H^+]$ is continuously replenished as the equilibrium shifts, but at any instant the $[H^+]$ is lower than in the strong acid.

(c) [3]

At the endpoint, the colour change is from colourless to pink (or pale pink). [1]

This is a weak acid–strong base titration. At the equivalence point, the solution contains the salt $CH_3COONa$, which hydrolyses to give a basic solution (pH > 7, approximately pH 8.7). [1]

Phenolphthalein changes colour in the pH range 8.2–10.0 (colourless → pink), which falls within the steep vertical portion of the titration curve for a weak acid–strong base titration. Therefore, the colour change occurs sharply and accurately indicates the endpoint. [1]

Common mistake: Students say "pink to colourless." The colour change is from colourless (in acidic/neutral solution) to pink (in basic solution) as base is added.

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Mark Summary

Section	Marks
A: Multiple Choice (Q1–Q10)	10
B: Structured (Q11–Q16)	50
C: Free Response (Q17–Q18)	20
Total	80

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