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A Level H2 Chemistry Practice Paper 5

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry Level: A-Level H2 Paper: Practice Paper — Acids, Bases & Salts Duration: 1 hour 45 minutes Total Marks: 60 Name: ___________________________ Class: ___________________________ Date: ___________________________

Instructions

Answer all questions in the spaces provided.
The number of marks for each question is shown in brackets [ ].
The total mark for this paper is 60.
You are advised to spend no more than 1 hour 45 minutes on this paper.
Essential working must be shown for calculation-based questions to earn full marks.
Use the Data Booklet where necessary.
A Periodic Table and relevant data are provided on the last page of this paper.

Section A: Multiple Choice [15 marks]

Questions 1–15: Choose the most appropriate answer (A, B, C, or D).

1. Which of the following is the conjugate base of $HSO_4^-$ ?

A. $H_2SO_4$ B. $SO_4^{2-}$ C. $H_3O^+$ D. $H_2SO_3$

[1 mark]

2. A solution has a pH of 3.40. What is the concentration of $OH^-$ ions in this solution at 25 °C?

A. $2.51 \times 10^{-4}$ mol dm $^{-3}$ B. $3.98 \times 10^{-4}$ mol dm $^{-3}$ C. $2.51 \times 10^{-11}$ mol dm $^{-3}$ D. $3.98 \times 10^{-11}$ mol dm $^{-3}$

[1 mark]

3. Which of the following salts will produce an aqueous solution with pH > 7?

A. Ammonium chloride, $NH_4Cl$ B. Sodium chloride, $NaCl$ C. Potassium nitrate, $KNO_3$ D. Sodium ethanoate, $CH_3COONa$

[1 mark]

4. The $K_a$ of a weak acid $HA$ is $4.7 \times 10^{-6}$ mol dm $^{-3}$ . What is the pH of a 0.050 mol dm $^{-3}$ solution of $HA$ ?

A. 2.32 B. 3.33 C. 3.83 D. 4.33

[1 mark]

5. In the titration of 25.0 cm $^3$ of 0.100 mol dm $^{-3}$ $NaOH$ with 0.100 mol dm $^{-3}$ $HCl$ , what is the pH at the equivalence point?

A. 1.0 B. 5.0 C. 7.0 D. 13.0

[1 mark]

6. Which statement best describes the action of a buffer solution?

A. It resists changes in pH by neutralising all added acid or base completely. B. It contains a strong acid and its conjugate base in equal concentrations. C. It maintains a relatively constant pH by shifting the equilibrium position when small amounts of acid or base are added. D. It always has a pH of exactly 7.0.

[1 mark]

7. The solubility product, $K_{sp}$ , of $Mg(OH)_2$ is $5.6 \times 10^{-12}$ mol $^3$ dm $^{-9}$ at 25 °C. What is the solubility of $Mg(OH)_2$ in mol dm $^{-3}$ ?

A. $1.1 \times 10^{-4}$ B. $1.8 \times 10^{-4}$ C. $2.2 \times 10^{-4}$ D. $1.1 \times 10^{-6}$

[1 mark]

8. Which of the following is a correct expression for $K_w$ at 25 °C?

A. $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-7}$ B. $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$ C. $K_w = \frac{[H_3O^+]}{[OH^-]} = 1.0 \times 10^{-14}$ D. $K_w = [H_3O^+] + [OH^-] = 1.0 \times 10^{-14}$

[1 mark]

9. A 0.200 mol dm $^{-3}$ solution of a weak monoprotic acid has a pH of 2.72. What is the approximate $K_a$ of this acid?

A. $1.8 \times 10^{-5}$ B. $3.6 \times 10^{-5}$ C. $1.8 \times 10^{-3}$ D. $3.6 \times 10^{-3}$

[1 mark]

10. During a titration of ethanoic acid with sodium hydroxide, the pH at half-equivalence point is equal to:

A. 7.00 B. $pK_a$ of ethanoic acid C. $pK_b$ of ethanoate ion D. $pK_w$

[1 mark]

11. Which salt undergoes hydrolysis to produce an acidic solution?

A. $Na_2CO_3$ B. $KNO_3$ C. $FeCl_3$ D. $CH_3COOK$

[1 mark]

12. The $K_{sp}$ of $AgCl$ is $1.8 \times 10^{-10}$ mol $^2$ dm $^{-6}$ at 25 °C. What is the concentration of $Ag^+$ in a saturated solution of $AgCl$ ?

A. $1.3 \times 10^{-5}$ mol dm $^{-3}$ B. $1.8 \times 10^{-10}$ mol dm $^{-3}$ C. $3.6 \times 10^{-10}$ mol dm $^{-3}$ D. $9.0 \times 10^{-11}$ mol dm $^{-3}$

[1 mark]

13. Which indicator is most suitable for a titration of a weak acid with a strong base where the equivalence point pH is approximately 8.7?

Indicator	pH range
Methyl orange	3.1 – 4.4
Bromothymol blue	6.0 – 7.6
Phenolphthalein	8.2 – 10.0
Thymol blue	1.2 – 2.8

A. Methyl orange B. Bromothymol blue C. Phenolphthalein D. Thymol blue

[1 mark]

14. A buffer solution is prepared by mixing 50.0 cm $^3$ of 0.400 mol dm $^{-3}$ $CH_3COOH$ with 50.0 cm $^3$ of 0.200 mol dm $^{-3}$ $NaOH$ . What is the pH of the resulting buffer? ( $K_a$ for $CH_3COOH = 1.8 \times 10^{-5}$ mol dm $^{-3}$ )

A. 3.74 B. 4.44 C. 4.74 D. 5.04

[1 mark]

15. Which of the following statements about the common ion effect is incorrect?

A. Adding a common ion decreases the solubility of a sparingly soluble salt. B. The common ion effect is an application of Le Chatelier's principle. C. Adding $NaCl$ to a saturated $AgCl$ solution increases the concentration of $Ag^+$ . D. The common ion effect shifts the solubility equilibrium to the left.

[1 mark]

Section B: Structured Questions [30 marks]

16. (a) Define the term Brønsted–Lowry acid and Brønsted–Lowry base.

[2 marks]

.......................................................................................................................

(b) For the following reaction, identify the conjugate acid–base pairs:

$H_2PO_4^-(aq) + NH_3(aq) \rightleftharpoons HPO_4^{2-}(aq) + NH_4^+(aq)$

[2 marks]

.......................................................................................................................

(c) Explain why $H_2PO_4^-$ can act as both an acid and a base. Give the formula of the species formed when $H_2PO_4^-$ acts as an acid.

[2 marks]

.......................................................................................................................

(d) Calculate the pH of a 0.150 mol dm $^{-3}$ solution of $HNO_3$ .

[1 mark]

.......................................................................................................................

(e) Calculate the pH of a 0.150 mol dm $^{-3}$ solution of $Ba(OH)_2$ .

[2 marks]

.......................................................................................................................

(f) A solution is prepared by mixing 30.0 cm $^3$ of 0.100 mol dm $^{-3}$ $H_2SO_4$ with 50.0 cm $^3$ of 0.120 mol dm $^{-3}$ $NaOH$ . Calculate the pH of the resulting solution.

[3 marks]

.......................................................................................................................

17. A student carried out a titration to determine the concentration of a solution of hydrochloric acid using 0.100 mol dm $^{-3}$ sodium hydroxide solution.

(a) Describe how the student would use a pipette to transfer 25.0 cm $^3$ of the hydrochloric acid solution into a conical flask. Include details of rinsing and filling to the mark.

[2 marks]

.......................................................................................................................

(b) The student's titration results are shown below:

Titration	Rough	1	2	3
Final reading / cm $^3$	26.50	25.80	25.70	25.90
Initial reading / cm $^3$	0.50	0.30	0.20	0.40
Volume used / cm $^3$	26.00	25.50	25.50	25.50

(i) Identify any anomalous result and explain your reasoning.

[1 mark]

.......................................................................................................................

(ii) Calculate the mean titre to be used in your calculation. Show clearly how you obtained this value.

[1 mark]

.......................................................................................................................

(c) Using your mean titre from (b)(ii), calculate the concentration of the hydrochloric acid.

[2 marks]

.......................................................................................................................

(d) The student used methyl orange as the indicator. State the colour change observed at the endpoint.

[1 mark]

.......................................................................................................................

(e) Explain why phenolphthalein would also be a suitable indicator for this titration.

[1 mark]

.......................................................................................................................

18. (a) Write an expression for the acid dissociation constant, $K_a$ , for methanoic acid, $HCOOH$ .

[1 mark]

.......................................................................................................................

(b) The $K_a$ of methanoic acid is $1.8 \times 10^{-4}$ mol dm $^{-3}$ at 25 °C. Calculate the pH of a 0.250 mol dm $^{-3}$ solution of methanoic acid. State any assumption you make.

[4 marks]

.......................................................................................................................

(c) A buffer solution is prepared by mixing 100 cm $^3$ of 0.250 mol dm $^{-3}$ $HCOOH$ with 100 cm $^3$ of 0.150 mol dm $^{-3}$ $HCOONa$ .

(i) Calculate the pH of this buffer solution.

[2 marks]

.......................................................................................................................

(ii) Explain, with reference to the equilibrium involved, how this buffer solution resists a change in pH when a small amount of dilute hydrochloric acid is added.

[3 marks]

.......................................................................................................................

Section C: Free Response [15 marks]

19. The solubility product of lead(II) iodide, $PbI_2$ , is $8.7 \times 10^{-9}$ mol $^3$ dm $^{-9}$ at 25 °C.

(a) Write an expression for $K_{sp}$ of $PbI_2$ .

[1 mark]

.......................................................................................................................

(b) Calculate the solubility of $PbI_2$ in pure water at 25 °C, in mol dm $^{-3}$ .

[3 marks]

.......................................................................................................................

(c) Calculate the concentration of $Pb^{2+}$ in a saturated solution of $PbI_2$ that also contains 0.050 mol dm $^{-3}$ $NaI$ .

[2 marks]

.......................................................................................................................

(d) Predict and explain whether more or less $PbI_2$ will dissolve in 0.050 mol dm $^{-3}$ $NaI$ compared to pure water. Use your calculations from (b) and (c) to support your answer.

[2 marks]

.......................................................................................................................

(e) A student adds concentrated $KI$ solution dropwise to a dilute solution of $Pb(NO_3)_2$ . Describe what the student would observe.

[1 mark]

.......................................................................................................................

20. The following curve shows the pH change when 0.100 mol dm $^{-3}$ $NaOH$ is added to 25.0 cm $^3$ of 0.100 mol dm $^{-3}$ ethanoic acid, $CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ mol dm $^{-3}$ ).

Graph for Q20 (ALEVEL Chemistry H2)

Generated graph for Q20.

(a) Using the graph, determine the initial pH of the ethanoic acid solution. Explain why this value is consistent with ethanoic acid being a weak acid.

[2 marks]

.......................................................................................................................

(b) Calculate the initial concentration of $H^+$ ions in the ethanoic acid solution. Verify that this is consistent with the $K_a$ value of $1.8 \times 10^{-5}$ mol dm $^{-3}$ .

[3 marks]

.......................................................................................................................

(c) On the graph, the half-equivalence point occurs at 12.5 cm $^3$ of $NaOH$ added. Explain why the pH at this point equals $pK_a$ of ethanoic acid.

[2 marks]

.......................................................................................................................

(d) At the equivalence point, the pH is approximately 8.7, not 7.0. Explain why the pH at the equivalence point is greater than 7.

[2 marks]

.......................................................................................................................

(e) A student suggests using methyl orange (pH range 3.1–4.4) as the indicator for this titration. Explain whether this is a suitable choice, with reference to the titration curve.

[2 marks]

.......................................................................................................................

(f) Calculate the concentration of $OH^-$ ions at the equivalence point. Hence determine the $K_b$ of the ethanoate ion, $CH_3COO^-$ .

[4 marks]

.......................................................................................................................

END OF PAPER

Data and Constants

Constant	Value
$K_w$ at 25 °C	$1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$
$R$	8.31 J K $^{-1}$ mol $^{-1}$

Acid	$K_a$ / mol dm $^{-3}$
$CH_3COOH$	$1.8 \times 10^{-5}$
$HCOOH$	$1.8 \times 10^{-4}$
$HNO_2$	$4.5 \times 10^{-4}$
$HF$	$6.8 \times 10^{-4}$
$H_2CO_3$	$4.3 \times 10^{-7}$
$H_3PO_4$	$7.5 \times 10^{-3}$

Answers

TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Acids, Bases & Salts

Section A: Multiple Choice [15 marks]

1. B — $SO_4^{2-}$

Explanation: A conjugate base is formed when a Brønsted–Lowry acid donates a proton ( $H^+$ ). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$ . Option A ( $H_2SO_4$ ) is the conjugate acid of $HSO_4^-$ , not its conjugate base.

2. C — $2.51 \times 10^{-11}$ mol dm $^{-3}$

Working: $pH = 3.40 \Rightarrow [H_3O^+] = 10^{-3.40} = 3.98 \times 10^{-4} \text{ mol dm}^{-3}$ $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$ $[OH^-] = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11} \text{ mol dm}^{-3}$

Common mistake: Students often select A, which is the $[H_3O^+]$ value, not $[OH^-]$ .

3. D — Sodium ethanoate, $CH_3COONa$

Explanation: Sodium ethanoate is a salt of a weak acid ( $CH_3COOH$ ) and strong base ( $NaOH$ ). The ethanoate ion ( $CH_3COO^-$ ) hydrolyses in water to produce $OH^-$ ions, making the solution basic (pH > 7). Options A produces an acidic solution (salt of weak base + strong acid). Options B and C are salts of strong acid + strong base, giving neutral solutions.

4. C — 3.83

Working: $K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{x^2}{0.050}$ $x^2 = 4.7 \times 10^{-6} \times 0.050 = 2.35 \times 10^{-7}$ $x = \sqrt{2.35 \times 10^{-7}} = 4.85 \times 10^{-4} \text{ mol dm}^{-3}$ $pH = -\log(4.85 \times 10^{-4}) = 3.31$

Recheck: Using the approximation more carefully: $x = \sqrt{4.7 \times 10^{-6} \times 0.050} = \sqrt{2.35 \times 10^{-7}} = 4.848 \times 10^{-4}$ $pH = -\log(4.848 \times 10^{-4}) = 3.31$

Note: The answer is closest to C (3.83) if we re-examine. Let me recalculate precisely: $x = \sqrt{4.7 \times 10^{-6} \times 0.050} = \sqrt{2.35 \times 10^{-7}} = 4.848 \times 10^{-4}$ $pH = 3.31$

Given the options, the closest answer is B (3.33). However, the question asks for the most appropriate answer. With $K_a = 4.7 \times 10^{-6}$ and $[HA] = 0.050$ : $[H^+] = \sqrt{4.7 \times 10^{-6} \times 0.050} = 4.85 \times 10^{-4}$ $pH = 3.31$

The answer is B (3.33).

Correction: Answer is B (3.33) — the slight difference arises from rounding.

5. C — 7.0

Explanation: $HCl$ is a strong acid and $NaOH$ is a strong base. At the equivalence point, the salt $NaCl$ is formed, which does not hydrolyse (it is the salt of a strong acid and strong base). The solution is neutral, so pH = 7.0.

6. C — It maintains a relatively constant pH by shifting the equilibrium position when small amounts of acid or base are added.

Explanation: A buffer works by Le Chatelier's principle. When acid is added, the conjugate base component reacts with it; when base is added, the weak acid component reacts. The equilibrium shifts to minimise the pH change. Buffers do not neutralise all added acid/base (they have limited capacity), and they do not require strong acids.

7. A — $1.1 \times 10^{-4}$

Working: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ Let solubility = $s$ mol dm $^{-3}$ . Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$ . $K_{sp} = [Mg^{2+}][OH^-]^2 = s(2s)^2 = 4s^3$ $4s^3 = 5.6 \times 10^{-12}$ $s^3 = 1.4 \times 10^{-12}$ $s = \sqrt[3]{1.4 \times 10^{-12}} = 1.12 \times 10^{-4} \text{ mol dm}^{-3}$

Answer: A ( $1.1 \times 10^{-4}$ )

8. B — $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$

Explanation: The ionic product of water is defined as the product of the concentrations of $H_3O^+$ and $OH^-$ ions. At 25 °C, $K_w = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$ . Option A has the wrong value ( $10^{-7}$ is the $[H_3O^+]$ in pure water, not $K_w$ ).

9. A — $1.8 \times 10^{-5}$

Working: $pH = 2.72 \Rightarrow [H^+] = 10^{-2.72} = 1.91 \times 10^{-3} \text{ mol dm}^{-3}$ For a weak monoprotic acid $HA$ : $K_a \approx \frac{[H^+]^2}{[HA]_0} = \frac{(1.91 \times 10^{-3})^2}{0.200}$ $K_a = \frac{3.65 \times 10^{-6}}{0.200} = 1.82 \times 10^{-5} \text{ mol dm}^{-3}$

Answer: A ( $1.8 \times 10^{-5}$ )

10. B — $pK_a$ of ethanoic acid

Explanation: At the half-equivalence point, exactly half the weak acid has been neutralised, so $[CH_3COOH] = [CH_3COO^-]$ . From the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[A^-]}{[HA]} = pK_a + log(1) = pK_a$

11. C — $FeCl_3$

Explanation: $FeCl_3$ is a salt of a strong acid ( $HCl$ ) and weak base ( $Fe(OH)_3$ ). The $Fe^{3+}$ ion is a small, highly charged cation that acts as a Lewis acid, hydrolysing water to produce $H^+$ ions, making the solution acidic. $Na_2CO_3$ and $CH_3COOK$ produce basic solutions (salts of weak acid + strong base). $KNO_3$ is neutral (strong acid + strong base).

12. A — $1.3 \times 10^{-5}$ mol dm $^{-3}$

Working: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ Let solubility = $s$ . Then $[Ag^+] = [Cl^-] = s$ . $K_{sp} = s^2 = 1.8 \times 10^{-10}$ $s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ mol dm}^{-3}$

Answer: A ( $1.3 \times 10^{-5}$ mol dm $^{-3}$ )

13. C — Phenolphthalein

Explanation: The indicator should change colour within the steep rise portion of the titration curve, which includes the equivalence point. For a weak acid–strong base titration, the equivalence point pH ≈ 8.7. Phenolphthalein (pH range 8.2–10.0) changes colour in this region. Methyl orange changes colour at too low a pH, well before the equivalence point.

14. C — 4.74

Working: Moles of $CH_3COOH$ initially = $\frac{50.0 \times 0.400}{1000} = 0.0200$ mol Moles of $NaOH$ added = $\frac{50.0 \times 0.200}{1000} = 0.0100$ mol

$NaOH$ reacts with $CH_3COOH$ in a 1:1 ratio: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Moles of $CH_3COOH$ remaining = $0.0200 - 0.0100 = 0.0100$ mol Moles of $CH_3COO^-$ formed = $0.0100$ mol

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

Since $[CH_3COOH] = [CH_3COO^-]$ (equal moles in the same volume): $pH = pK_a + log\frac{[CH_3COO^-]}{[CH_3COOH]} = pK_a + log(1) = pK_a$ $pK_a = -\log(1.8 \times 10^{-5}) = 4.74$

Answer: C (4.74)

15. C — Adding $NaCl$ to a saturated $AgCl$ solution increases the concentration of $Ag^+$ .

Explanation: This statement is incorrect. Adding $NaCl$ introduces $Cl^-$ ions (a common ion), which shifts the equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ to the left (Le Chatelier's principle). This decreases the concentration of $Ag^+$ , not increases it. The solubility of $AgCl$ decreases in the presence of a common ion.

Section B: Structured Questions [30 marks]

16. (a) [2 marks]

A Brønsted–Lowry acid is a substance that donates a proton ( $H^+$ ). [1 mark]
A Brønsted–Lowry base is a substance that accepts a proton ( $H^+$ ). [1 mark]

(b) [2 marks]

Conjugate acid–base pairs:

$H_2PO_4^-$ / $HPO_4^{2-}$ (acid / conjugate base) [1 mark]
$NH_3$ / $NH_4^+$ (base / conjugate acid) [1 mark]

Explanation: In the forward reaction, $H_2PO_4^-$ donates a proton to become $HPO_4^{2-}$ , and $NH_3$ accepts a proton to become $NH_4^+$ . Each acid–base pair differs by one proton.

(c) [2 marks]

$H_2PO_4^-$ is amphiprotic (or amphoteric). [1 mark] It can act as an acid by donating a proton to form $HPO_4^{2-}$ , or as a base by accepting a proton to form $H_3PO_4$ . When acting as an acid, the species formed is $HPO_4^{2-}$ . [1 mark]

(d) [1 mark]

$HNO_3$ is a strong acid and dissociates completely: $[H^+] = 0.150 \text{ mol dm}^{-3}$ $pH = -\log(0.150) = 0.82$

Answer: pH = 0.82 [1 mark]

(e) [2 marks]

$Ba(OH)_2$ is a strong base and dissociates completely: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$ $[OH^-] = 2 \times 0.150 = 0.300 \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$ $pOH = -\log(0.300) = 0.523$ $pH = 14.00 - 0.523 = 13.48 \quad \textbf{[1 mark]}$

Answer: pH = 13.48

(f) [3 marks]

Moles of $H_2SO_4$ = $\frac{30.0 \times 0.100}{1000} = 0.00300$ mol Moles of $H^+$ from $H_2SO_4$ = $2 \times 0.00300 = 0.00600$ mol [1 mark]

Moles of $NaOH$ = $\frac{50.0 \times 0.120}{1000} = 0.00600$ mol [1 mark]

Since moles of $H^+$ = moles of $OH^-$ , the acid and base neutralise exactly: $H^+ + OH^- \rightarrow H_2O$

The resulting solution contains $Na_2SO_4$ (salt of strong acid + strong base), which does not hydrolyse. The solution is neutral. [1 mark]

Answer: pH = 7.00

17. (a) [2 marks]

Rinse the pipette with the hydrochloric acid solution (not water). [1 mark]
Fill the pipette to just above the 25.0 cm $^3$ mark using a pipette filler. Slowly release the meniscus until the bottom of the meniscus sits exactly on the 25.0 cm $^3$ mark at eye level. Transfer the acid to the conical flask, allowing it to drain freely; touch the tip against the inner wall of the flask but do not blow out the last drop (the pipette is calibrated to deliver 25.0 cm $^3$ with the residual drop remaining). [1 mark]

(b)(i) [1 mark]

The rough titration (26.00 cm $^3$ ) is anomalous. [1 mark] It differs significantly from the other three concordant titres (all 25.50 cm $^3$ ). The rough titration is only used to estimate the approximate volume needed and is not used in the calculation.

(b)(ii) [1 mark]

Titrations 1, 2, and 3 are concordant (all 25.50 cm $^3$ ). $\text{Mean titre} = \frac{25.50 + 25.50 + 25.50}{3} = 25.50 \text{ cm}^3 \quad \textbf{[1 mark]}$

(c) [2 marks]

$HCl + NaOH \rightarrow NaCl + H_2O$

Moles of $NaOH$ used = $\frac{25.50 \times 0.100}{1000} = 2.55 \times 10^{-3}$ mol [1 mark]

From the 1:1 stoichiometry, moles of $HCl$ = $2.55 \times 10^{-3}$ mol $[HCl] = \frac{2.55 \times 10^{-3}}{25.0/1000} = \frac{2.55 \times 10^{-3}}{0.0250} = 0.102 \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

Answer: 0.102 mol dm $^{-3}$

(d) [1 mark]

Methyl orange changes from red to orange (or yellow) at the endpoint. [1 mark]

(e) [1 mark]

For a strong acid–strong base titration, the equivalence point is at pH 7.0 and the pH change is very steep (approximately pH 3–11). Both methyl orange (pH 3.1–4.4) and phenolphthalein (pH 8.2–10.0) change colour within this steep region, so phenolphthalein is also suitable. [1 mark]

18. (a) [1 mark]

$K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} \quad \textbf{[1 mark]}$

(b) [4 marks]

Assumption: The dissociation of $HCOOH$ is small, so $[HCOOH] \approx 0.250$ mol dm $^{-3}$ at equilibrium. [1 mark]

$K_a = \frac{x^2}{0.250} = 1.8 \times 10^{-4}$ $x^2 = 1.8 \times 10^{-4} \times 0.250 = 4.5 \times 10^{-5} \quad \textbf{[1 mark]}$ $x = \sqrt{4.5 \times 10^{-5}} = 6.71 \times 10^{-3} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$ $pH = -\log(6.71 \times 10^{-3}) = 2.17 \quad \textbf{[1 mark]}$

Answer: pH = 2.17

Check assumption: $\frac{6.71 \times 10^{-3}}{0.250} \times 100\% = 2.7\% < 5\%$ , so the assumption is valid.

(c)(i) [2 marks]

Moles of $HCOOH$ = $\frac{100 \times 0.250}{1000} = 0.0250$ mol Moles of $HCOO^-$ (from $HCOONa$ ) = $\frac{100 \times 0.150}{1000} = 0.0150$ mol

Total volume = $100 + 100 = 200$ cm $^3$ = 0.200 dm $^3$

$pH = pK_a + log\frac{[HCOO^-]}{[HCOOH]} \quad \textbf{[1 mark]}$ $pK_a = -\log(1.8 \times 10^{-4}) = 3.74$ $pH = 3.74 + log\frac{0.0150/0.200}{0.0250/0.200} = 3.74 + log\frac{0.0150}{0.0250}$ $pH = 3.74 + log(0.600) = 3.74 + (-0.222) = 3.52 \quad \textbf{[1 mark]}$

Answer: pH = 3.52

(c)(ii) [3 marks]

The buffer contains the equilibrium: $HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq) \quad \textbf{[1 mark]}$

When a small amount of $HCl$ is added, the $H^+$ ions from $HCl$ react with the $HCOO^-$ ions (the conjugate base component of the buffer) to form more $HCOOH$ : [1 mark] $H^+ + HCOO^- \rightarrow HCOOH$

By Le Chatelier's principle, the equilibrium shifts to the left, consuming most of the added $H^+$ ions. The ratio $[HCOO^-]/[HCOOH]$ changes only slightly, so the pH remains relatively constant. [1 mark]

Section C: Free Response [15 marks]

19. (a) [1 mark]

$K_{sp} = [Pb^{2+}][I^-]^2 \quad \textbf{[1 mark]}$

(b) [3 marks]

$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$

Let solubility = $s$ mol dm $^{-3}$ . Then $[Pb^{2+}] = s$ and $[I^-] = 2s$ . [1 mark]

$K_{sp} = s(2s)^2 = 4s^3 = 8.7 \times 10^{-9} \quad \textbf{[1 mark]}$ $s^3 = \frac{8.7 \times 10^{-9}}{4} = 2.175 \times 10^{-9}$ $s = \sqrt[3]{2.175 \times 10^{-9}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

Answer: $1.30 \times 10^{-3}$ mol dm $^{-3}$

(c) [2 marks]

In 0.050 mol dm $^{-3}$ $NaI$ , $[I^-] = 0.050$ mol dm $^{-3}$ (from the fully dissociated $NaI$ ; the contribution from $PbI_2$ dissolution is negligible in comparison). [1 mark]

$K_{sp} = [Pb^{2+}][I^-]^2$ $8.7 \times 10^{-9} = [Pb^{2+}](0.050)^2$ $[Pb^{2+}] = \frac{8.7 \times 10^{-9}}{2.5 \times 10^{-3}} = 3.48 \times 10^{-7} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

Answer: $3.48 \times 10^{-7}$ mol dm $^{-3}$

(d) [2 marks]

Much less $PbI_2$ dissolves in 0.050 mol dm $^{-3}$ $NaI$ compared to pure water. [1 mark]

In pure water, $[Pb^{2+}] = 1.30 \times 10^{-3}$ mol dm $^{-3}$ , whereas in 0.050 mol dm $^{-3}$ $NaI$ , $[Pb^{2+}] = 3.48 \times 10^{-7}$ mol dm $^{-3}$ — a decrease by a factor of approximately 3700. This is because the common ion ( $I^-$ ) from $NaI$ shifts the solubility equilibrium to the left (Le Chatelier's principle), suppressing the dissolution of $PbI_2$ . [1 mark]

(e) [1 mark]

A yellow precipitate of $PbI_2$ is observed. [1 mark]

20. (a) [2 marks]

From the graph, the initial pH (at 0 cm $^3$ $NaOH$ added) is approximately 2.9. [1 mark]

If ethanoic acid were a strong acid at 0.100 mol dm $^{-3}$ , the pH would be $-\log(0.100) = 1.0$ . The observed pH of 2.9 is significantly higher, indicating that ethanoic acid does not dissociate completely — it is a weak acid with only partial dissociation. [1 mark]

(b) [3 marks]

$[H^+] = 10^{-2.9} = 1.26 \times 10^{-3} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

Verification using $K_a$ : $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \approx \frac{(1.26 \times 10^{-3})^2}{0.100 - 1.26 \times 10^{-3}} \quad \textbf{[1 mark]}$ $K_a = \frac{1.59 \times 10^{-6}}{0.0987} = 1.61 \times 10^{-5} \text{ mol dm}^{-3}$

This is reasonably close to the given $K_a$ of $1.8 \times 10^{-5}$ mol dm $^{-3}$ , confirming consistency. The small difference is due to rounding of the initial pH read from the graph. [1 mark]

(c) [2 marks]

At the half-equivalence point, exactly half of the ethanoic acid has been neutralised by $NaOH$ : $[CH_3COOH] = [CH_3COO^-] \quad \textbf{[1 mark]}$

From the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[CH_3COO^-]}{[CH_3COOH]} = pK_a + log(1) = pK_a \quad \textbf{[1 mark]}$

Therefore, the pH at the half-equivalence point equals $pK_a = -\log(1.8 \times 10^{-5}) = 4.74$ .

(d) [2 marks]

At the equivalence point, all the ethanoic acid has been converted to sodium ethanoate ( $CH_3COONa$ ). [1 mark] The ethanoate ion ( $CH_3COO^-$ ) is the conjugate base of a weak acid. It hydrolyses in water: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ This produces $OH^-$ ions, making the solution slightly basic, so pH > 7. [1 mark]

(e) [2 marks]

Methyl orange (pH range 3.1–4.4) is not suitable for this titration. [1 mark]

The steep rise in the titration curve for a weak acid–strong base titration occurs between approximately pH 7 and pH 11. Methyl orange would change colour at pH ~4, which is well before the equivalence point (pH ≈ 8.7), in the buffer region. This would result in a significant titration error. Phenolphthalein (pH 8.2–10.0) is a better choice as it changes colour within the steep rise region. [1 mark]

(f) [4 marks]

At the equivalence point, pH ≈ 8.7: $pOH = 14.0 - 8.7 = 5.3 \quad \textbf{[1 mark]}$ $[OH^-] = 10^{-5.3} = 5.01 \times 10^{-6} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

At the equivalence point, all $CH_3COOH$ has been converted to $CH_3COO^-$ . The total volume is $25.0 + 25.0 = 50.0$ cm $^3$ . $[CH_3COO^-] = \frac{0.100 \times 25.0/1000}{50.0/1000} = 0.0500 \text{ mol dm}^{-3}$

For the hydrolysis equilibrium: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ $K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} = \frac{(5.01 \times 10^{-6})^2}{0.0500} \quad \textbf{[1 mark]}$ $K_b = \frac{2.51 \times 10^{-11}}{0.0500} = 5.02 \times 10^{-10} \text{ mol dm}^{-3} \quad \textbf{[1 mark]}$

Verification: $K_b = K_w / K_a = 1.0 \times 10^{-14} / 1.8 \times 10^{-5} = 5.56 \times 10^{-10}$ mol dm $^{-3}$ , which is consistent.

Answer: $K_b$ of $CH_3COO^- \approx 5.0 \times 10^{-10}$ mol dm $^{-3}$

END OF ANSWER KEY

Mark Summary

Section	Marks
A: Multiple Choice (Q1–15)	15
B: Structured Questions (Q16–18)	30
C: Free Response (Q19–20)	15
Total	60

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