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A Level H2 Chemistry Practice Paper 3

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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 marks for this paper is 60.
You are advised to spend no more than 1 hour 45 minutes on this paper.
A copy of the Data Booklet is provided.
Essential working must be shown for calculation questions to earn full marks.
Give answers to calculations to an appropriate number of significant figures unless otherwise stated.

Section A: Multiple Choice [15 marks]

Questions 1–15: Choose the one correct answer for each question. Each question carries 1 mark.

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

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

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

A. $2.5 \times 10^{-4}$ mol dm $^{-3}$ B. $4.0 \times 10^{-11}$ mol dm $^{-3}$ C. $2.5 \times 10^{-11}$ mol dm $^{-3}$ D. $4.0 \times 10^{-4}$ mol dm $^{-3}$

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

A. $Na_2CO_3$ B. $KNO_3$ C. $NH_4Cl$ D. $CH_3COONa$

4. The $K_a$ of a weak acid $HA$ is $4.0 \times 10^{-6}$ mol dm $^{-3}$ . What is the pH of a 0.10 mol dm $^{-3}$ solution of $HA$ ?

A. 3.20 B. 2.70 C. 3.70 D. 6.00

5. A buffer solution is prepared by mixing 50 cm³ of 0.20 mol dm $^{-3}$ $CH_3COOH$ with 50 cm³ of 0.20 mol dm $^{-3}$ $NaOH$ . Which statement about this mixture is correct?

A. The resulting solution is a buffer because it contains $CH_3COOH$ and $CH_3COO^-$ . B. The resulting solution is not a buffer because all the $CH_3COOH$ is neutralised. C. The resulting solution is a buffer because it contains excess $NaOH$ . D. The resulting solution is not a buffer because only $CH_3COONa$ is present.

6. In the titration of 25.0 cm³ of 0.10 mol dm $^{-3}$ $NaOH$ with 0.10 mol dm $^{-3}$ $HCl$ , what is the pH at the equivalence point?

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

7. The solubility product, $K_{sp}$ , of $PbI_2$ is $8.5 \times 10^{-9}$ mol $^3$ dm $^{-9}$ at 25 °C. What is the solubility of $PbI_2$ in water at this temperature?

A. $1.3 \times 10^{-3}$ mol dm $^{-3}$ B. $2.0 \times 10^{-3}$ mol dm $^{-3}$ C. $1.7 \times 10^{-3}$ mol dm $^{-3}$ D. $2.6 \times 10^{-3}$ mol dm $^{-3}$

8. Which indicator is most suitable for a titration between a weak acid and a strong base?

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

A. Methyl orange B. Bromothymol blue C. Phenolphthalein D. Any of the above

9. Which of the following best explains why a solution of $Na_2CO_3$ is alkaline?

A. $Na^+$ ions react with water to produce $OH^-$ . B. $CO_3^{2-}$ ions react with water to produce $OH^-$ . C. $Na_2CO_3$ is a strong base. D. $CO_3^{2-}$ ions donate protons to water.

10. The pH of a 0.050 mol dm $^{-3}$ solution of a weak acid $HX$ is 3.00. What is the value of $K_a$ for $HX$ ?

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

11. A solution contains 0.10 mol dm $^{-3}$ $CH_3COOH$ and 0.10 mol dm $^{-3}$ $CH_3COONa$ . What happens when a small amount of dilute $HCl$ is added?

A. The pH decreases significantly. B. The pH increases significantly. C. The pH remains almost unchanged. D. The pH first increases then decreases.

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

A. $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$ B. $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-7}$ mol $^2$ dm $^{-6}$ C. $K_w = \frac{[H_3O^+]}{[OH^-]} = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$ D. $K_w = [H_3O^+] + [OH^-] = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$

13. When 20.0 cm³ of 0.10 mol dm $^{-3}$ $H_2SO_4$ is titrated with 0.10 mol dm $^{-3}$ $NaOH$ , what volume of $NaOH$ is required to reach the equivalence point?

A. 10.0 cm³ B. 20.0 cm³ C. 40.0 cm³ D. 80.0 cm³

14. The $K_{sp}$ of $Mg(OH)_2$ is $1.8 \times 10^{-11}$ mol $^3$ dm $^{-9}$ . What is the pH of a saturated solution of $Mg(OH)_2$ ?

A. 10.15 B. 10.52 C. 10.87 D. 11.15

15. Which salt undergoes the most extensive hydrolysis in aqueous solution?

A. $NaCl$ B. $NH_4Cl$ C. $Na_2CO_3$ D. $NH_4CH_3COO$

Section B: Structured Questions [30 marks]

Questions 16–19: Answer all questions. Show essential working for calculations.

16. Buffer Solutions and pH Calculations [8 marks]

(a) Define the term buffer solution. [2]

(b) A buffer solution is prepared by mixing 100 cm³ of 0.50 mol dm $^{-3}$ methanoic acid ( $HCOOH$ ) with 100 cm³ of 0.30 mol dm $^{-3}$ sodium methanoate ( $HCOONa$ ). $K_a$ for methanoic acid is $1.8 \times 10^{-4}$ mol dm $^{-3}$ .

(i) Calculate the pH of this buffer solution. [3]

(ii) Explain, with reference to the relevant equilibria, how this buffer solution resists changes in pH when a small amount of dilute hydrochloric acid is added. [3]

17. Solubility Product and Precipitation [8 marks]

The solubility product, $K_{sp}$ , of barium sulfate ( $BaSO_4$ ) is $1.1 \times 10^{-10}$ mol $^2$ dm $^{-6}$ at 25 °C.

(a) Write an expression for $K_{sp}$ of $BaSO_4$ . [1]

(b) Calculate the solubility of $BaSO_4$ in pure water at 25 °C, giving your answer in mol dm $^{-3}$ . [2]

(c) Calculate the solubility of $BaSO_4$ in a 0.10 mol dm $^{-3}$ solution of $Na_2SO_4$ at 25 °C. Explain why this value differs from your answer in (b). [3]

(d) A student adds 50 cm³ of 0.0010 mol dm $^{-3}$ $BaCl_2$ to 50 cm³ of 0.0010 mol dm $^{-3}$ $Na_2SO_4$ . Determine, by calculation, whether a precipitate of $BaSO_4$ forms. [2]

18. Acid-Base Titrations and Indicators [8 marks]

A student carried out a titration to determine the concentration of a solution of ethanoic acid ( $CH_3COOH$ ) using 0.100 mol dm $^{-3}$ $NaOH$ .

The following titration results were obtained:

Titration	Rough	1	2	3
Final burette reading / cm³	24.80	24.35	24.30	24.40
Initial burette reading / cm³	0.00	0.00	0.00	0.00
Volume used / cm³	24.80	24.35	24.30	24.40

In each titration, 25.0 cm³ of ethanoic acid was used.

(a) Record the titration results in a suitable table and calculate the mean titre to be used in your calculations. Show clearly how you obtained this volume. [3]

(b) Calculate the concentration of the ethanoic acid solution. [2]

(c) Explain why phenolphthalein is a suitable indicator for this titration. In your answer, describe the pH change around the equivalence point. [3]

19. Salt Hydrolysis and pH [6 marks]

(a) Define the term salt hydrolysis. [1]

(b) Predict whether aqueous solutions of the following salts are acidic, basic, or neutral. For each, write an equation to support your answer where hydrolysis occurs.

(i) $AlCl_3$ [2]

(ii) $K_2CO_3$ [2]

(iii) $NaNO_3$ [1]

Section C: Data Interpretation and Application [15 marks]

Questions 20: Answer all parts.

20. Polyprotic Acids and Titration Curves [15 marks]

<image_placeholder> id: Q20-fig1 type: graph linked_question: Q20 description: A titration curve showing the pH change when 0.100 mol dm⁻³ NaOH is added to 25.0 cm³ of 0.100 mol dm⁻³ H₃PO₄ (phosphoric acid). The x-axis is 'Volume of NaOH added / cm³' from 0 to 60 cm³. The y-axis is 'pH' from 0 to 14. The curve starts at approximately pH 1.5, rises gradually, has a first buffer region, then a first equivalence point at approximately pH 4.7 around 25 cm³ NaOH, rises to a second buffer region, then a second equivalence point at approximately pH 9.8 around 50 cm³ NaOH, then rises steeply. There is no third equivalence point shown. Two distinct buffer regions and two steep rises are visible. labels: x-axis: Volume of NaOH added / cm³ (0–60); y-axis: pH (0–14); first equivalence point labelled at ~25 cm³, pH ≈ 4.7; second equivalence point labelled at ~50 cm³, pH ≈ 9.8; starting pH ≈ 1.5; buffer regions indicated between 0–25 cm³ and 25–50 cm³ values: First equivalence point: 25.0 cm³ NaOH, pH ≈ 4.7; Second equivalence point: 50.0 cm³ NaOH, pH ≈ 9.8; Starting pH ≈ 1.5; Half-equivalence points at ~12.5 cm³ (pH ≈ pKa₁) and ~37.5 cm³ (pH ≈ pKa₂) must_show: Two distinct equivalence points, buffer regions, starting pH, axes with labels and scales, curve shape showing two steep rises

The graph above shows the titration curve obtained when 0.100 mol dm $^{-3}$ $NaOH$ is added gradually to 25.0 cm³ of 0.100 mol dm $^{-3}$ phosphoric acid ( $H_3PO_4$ ).

Phosphoric acid is a triprotic acid that ionises in three steps:

$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \quad K_{a1} = 7.5 \times 10^{-3} \text{ mol dm}^{-3}$

$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \quad K_{a2} = 6.2 \times 10^{-8} \text{ mol dm}^{-3}$

$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \quad K_{a3} = 4.8 \times 10^{-13} \text{ mol dm}^{-3}$

(a) Explain why the first equivalence point occurs at approximately pH 4.7 and not at pH 7. [3]

(b) Using the $K_a$ values provided, explain why only two equivalence points are clearly visible on the titration curve, even though $H_3PO_4$ is triprotic. [3]

(c) At the first half-equivalence point (when 12.5 cm³ of $NaOH$ has been added), the pH of the solution equals $pK_{a1}$ . Explain this observation using the Henderson-Hasselbalch equation. [2]

(d) A student suggests that a suitable indicator for the second equivalence point would be phenolphthalein (pH range 8.2–10.0). Evaluate this suggestion. [2]

(e) Calculate the concentration of $HPO_4^{2-}$ ions in the solution at the second equivalence point. Assume the total volume of the solution at this point is 75.0 cm³. [3]

(f) Write an expression for $K_{a2}$ and use it to calculate the pH of a solution containing 0.10 mol dm $^{-3}$ $NaH_2PO_4$ and 0.10 mol dm $^{-3}$ $Na_2HPO_4$ . [2]

Answers

TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Acids, Bases & Salts

Section A: Multiple Choice

1. B — $SO_4^{2-}$

Explanation: A conjugate base is formed when an acid donates a proton ( $H^+$ ). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$ . Option A ( $H_2SO_4$ ) is the conjugate acid (gaining a proton), not the conjugate base. This tests understanding of conjugate acid-base pairs: acid → conjugate base + $H^+$ .

2. B — $4.0 \times 10^{-11}$ mol dm $^{-3}$

Explanation:

$pH = 3.40$ , so $[H_3O^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm $^{-3}$
At 25 °C: $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$
$[OH^-] = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm $^{-3}$

Wait — let me recalculate: $10^{-3.40} = 3.981 \times 10^{-4}$ . Then $[OH^-] = 1.0 \times 10^{-14} / 3.981 \times 10^{-4} = 2.51 \times 10^{-11}$ mol dm $^{-3}$ .

Hmm, that gives $2.5 \times 10^{-11}$ , which is option C. Let me re-examine.

Actually: $10^{-3.40} = 10^{-3} \times 10^{-0.40} = 1.0 \times 10^{-3} \times 0.3981 = 3.981 \times 10^{-4}$

$[OH^-] = 1.0 \times 10^{-14} / 3.981 \times 10^{-4} = 2.512 \times 10^{-11}$ mol dm $^{-3}$

This rounds to $2.5 \times 10^{-11}$ mol dm $^{-3}$ , which is C.

Corrected Answer: C — $2.5 \times 10^{-11}$ mol dm $^{-3}$

Common mistake: Students may confuse $[H_3O^+]$ with $[OH^-]$ and select option A, or miscalculate the antilog.

3. C — $NH_4Cl$

Explanation: $NH_4Cl$ is a salt formed from a weak base ( $NH_3$ ) and a strong acid ( $HCl$ ). The $NH_4^+$ ion undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$ , producing $H_3O^+$ ions and making the solution acidic. $Na_2CO_3$ and $CH_3COONa$ produce basic solutions (salts of strong base + weak acid). $KNO_3$ is neutral (strong acid + strong base).

4. A — 3.20

Explanation:

For a weak acid: $[H^+] \approx \sqrt{K_a \times C} = \sqrt{4.0 \times 10^{-6} \times 0.10} = \sqrt{4.0 \times 10^{-7}} = 6.32 \times 10^{-4}$ mol dm $^{-3}$
$pH = -\log(6.32 \times 10^{-4}) = 3.20$

Common mistake: Students may incorrectly calculate $\sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}$ and get pH = 2.70 (option B), forgetting to multiply by the concentration first.

5. D — The resulting solution is not a buffer because only $CH_3COONa$ is present.

Explanation: Moles of $CH_3COOH = 0.050 \times 0.20 = 0.010$ mol. Moles of $NaOH = 0.050 \times 0.20 = 0.010$ mol. The acid and base react in a 1:1 ratio, so all the $CH_3COOH$ is completely neutralised to form $CH_3COONa$ . The resulting solution contains only $CH_3COONa$ (a salt), with no remaining weak acid to act as a buffer component. A buffer requires a weak acid and its conjugate base in comparable amounts.

6. B — 7.0

Explanation: $NaOH$ is a strong base and $HCl$ is a strong acid. At the equivalence point, the salt formed is $NaCl$ , which is neutral (from strong acid + strong base). The pH at the equivalence point is 7.0. This contrasts with weak acid–strong base titrations where the equivalence point pH > 7 due to hydrolysis of the conjugate base.

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

Explanation:

$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$
$K_{sp} = [Pb^{2+}][I^-]^2$
If solubility = $s$ , then $[Pb^{2+}] = s$ and $[I^-] = 2s$
$K_{sp} = s \times (2s)^2 = 4s^3 = 8.5 \times 10^{-9}$
$s^3 = 2.125 \times 10^{-9}$
$s = \sqrt[3]{2.125 \times 10^{-9}} = 1.29 \times 10^{-3} \approx 1.3 \times 10^{-3}$ mol dm $^{-3}$

Common mistake: Forgetting the stoichiometric coefficient of $I^-$ and writing $K_{sp} = s \times s^2 = s^3$ instead of $4s^3$ .

8. C — Phenolphthalein

Explanation: In a weak acid–strong base titration, the equivalence point occurs at pH > 7 (basic) because the conjugate base of the weak acid hydrolyses. The pH change around the equivalence point falls in the range of approximately 7–10. Phenolphthalein has a pH range of 8.2–10.0, which falls within this steep pH change region. Methyl orange (3.1–4.4) would change colour too early, well before the equivalence point.

9. B — $CO_3^{2-}$ ions react with water to produce $OH^-$ .

Explanation: $Na_2CO_3$ is a salt of a strong base ( $NaOH$ ) and a weak acid ( $H_2CO_3$ ). The $CO_3^{2-}$ ion is the conjugate base of $HCO_3^-$ and undergoes hydrolysis: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$ . This produces $OH^-$ ions, making the solution alkaline. $Na^+$ does not hydrolyse (it is the conjugate acid of a strong base).

10. A — $2.0 \times 10^{-5}$ mol dm $^{-3}$

Explanation:

$pH = 3.00$ , so $[H^+] = 10^{-3.00} = 1.0 \times 10^{-3}$ mol dm $^{-3}$
For $HX \rightleftharpoons H^+ + X^-$ : $[H^+] = [X^-] = 1.0 \times 10^{-3}$ mol dm $^{-3}$
Remaining $[HX] = 0.050 - 0.001 = 0.049 \approx 0.050$ mol dm $^{-3}$ (approximation valid since dissociation is small)
$K_a = \frac{[H^+][X^-]}{[HX]} = \frac{(1.0 \times 10^{-3})^2}{0.050} = \frac{1.0 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5}$ mol dm $^{-3}$

11. C — The pH remains almost unchanged.

Explanation: This is a buffer solution containing a weak acid ( $CH_3COOH$ ) and its conjugate base ( $CH_3COO^-$ ). When a small amount of $HCl$ is added, the $H^+$ ions are consumed by the conjugate base: $CH_3COO^- + H^+ \rightarrow CH_3COOH$ . This shifts the equilibrium but the ratio $[CH_3COO^-]/[CH_3COOH]$ changes only slightly, so the pH remains almost unchanged. This is the defining property of a buffer.

12. A — $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$

Explanation: The ionic product of water is defined as $K_w = [H_3O^+][OH^-]$ . At 25 °C, $K_w = 1.0 \times 10^{-14}$ mol $^2$ dm $^{-6}$ . Option B has the wrong value. Option C incorrectly uses a ratio. Option D incorrectly uses a sum.

13. C — 40.0 cm³

Explanation:

$H_2SO_4$ is diprotic: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
Moles of $H_2SO_4 = 0.0200 \times 0.10 = 0.0020$ mol
Moles of $NaOH$ needed $= 2 \times 0.0020 = 0.0040$ mol
Volume of $NaOH = \frac{0.0040}{0.10} = 0.040$ dm $^3 = 40.0$ cm³

Common mistake: Forgetting that $H_2SO_4$ provides 2 moles of $H^+$ per mole of acid and using a 1:1 ratio, giving 20.0 cm³ (option B).

14. B — 10.52

Explanation:

$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
$K_{sp} = [Mg^{2+}][OH^-]^2 = 4s^3 = 1.8 \times 10^{-11}$
$s^3 = 4.5 \times 10^{-12}$
$s = 1.65 \times 10^{-4}$ mol dm $^{-3}$
$[OH^-] = 2s = 3.30 \times 10^{-4}$ mol dm $^{-3}$
$pOH = -\log(3.30 \times 10^{-4}) = 3.48$
$pH = 14.00 - 3.48 = 10.52$

15. C — $Na_2CO_3$

Explanation: The extent of hydrolysis depends on the strength of the parent acid and base. $NaCl$ is neutral (strong acid + strong base, no hydrolysis). $NH_4Cl$ undergoes cationic hydrolysis (weak base cation). $Na_2CO_3$ undergoes anionic hydrolysis with $CO_3^{2-}$ , which is the conjugate base of the weak acid $HCO_3^-$ — since $H_2CO_3$ is weak, $CO_3^{2-}$ is relatively strongly basic and undergoes extensive hydrolysis. $NH_4CH_3COO$ has both ions hydrolysing but the effects partially cancel ( $K_a \approx K_b$ for the parent acid and base). $CO_3^{2-}$ is the strongest base among the anions listed because it is the conjugate base of a weak acid ( $HCO_3^-$ ) which itself comes from a weak acid ( $H_2CO_3$ ).

Section B: Structured Questions

16. Buffer Solutions and pH Calculations [8 marks]

(a) [2 marks]

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted. It typically consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Marking:

[1] for mentioning resistance to pH change
[1] for identifying the components (weak acid + conjugate base, or weak base + conjugate acid)

(b)(i) [3 marks]

Step 1: Calculate moles of each component after mixing.

Moles of $HCOOH = 0.100 \times 0.50 = 0.050$ mol
Moles of $HCOONa = 0.100 \times 0.30 = 0.030$ mol
Total volume $= 100 + 100 = 200$ cm³ $= 0.200$ dm³

Step 2: Calculate concentrations in the mixture.

$[HCOOH] = \frac{0.050}{0.200} = 0.25$ mol dm $^{-3}$
$[HCOO^-] = \frac{0.030}{0.200} = 0.15$ mol dm $^{-3}$

Step 3: Apply the Henderson-Hasselbalch equation.

$pH = pK_a + \log\frac{[A^-]}{[HA]}$

$pK_a = -\log(1.8 \times 10^{-4}) = 3.74$

$pH = 3.74 + \log\frac{0.15}{0.25} = 3.74 + \log(0.60) = 3.74 + (-0.22) = 3.52$

Marking:

[1] for correct moles/concentrations
[1] for correct $pK_a$ and correct substitution into Henderson-Hasselbalch
[1] for correct final answer (pH = 3.52)

(b)(ii) [3 marks]

When a small amount of dilute $HCl$ is added, the $H^+$ ions from $HCl$ react with the methanoate ions ( $HCOO^-$ ) in the buffer:

$HCOO^-(aq) + H^+(aq) \rightarrow HCOOH(aq)$

This removes the added $H^+$ ions by converting them into un-ionised $HCOOH$ . The equilibrium:

$HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq)$

shifts to the left (Le Chatelier's principle) as $HCOO^-$ is consumed. Since the buffer contains relatively large reservoirs of both $HCOOH$ and $HCOO^-$ , the ratio $[HCOO^-]/[HCOOH]$ changes only slightly, and hence the pH remains nearly constant.

Marking:

[1] for stating that added $H^+$ reacts with $HCOO^-$
[1] for the equation showing the reaction
[1] for explaining that the ratio changes little / reference to Le Chatelier's principle

17. Solubility Product and Precipitation [8 marks]

(a) [1 mark]

$K_{sp} = [Ba^{2+}][SO_4^{2-}]$

Marking:

[1] for correct expression (no state symbols needed, no solids included)

(b) [2 marks]

$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$

Let solubility $= s$ mol dm $^{-3}$

$[Ba^{2+}] = s$ and $[SO_4^{2-}] = s$

$K_{sp} = s \times s = s^2 = 1.1 \times 10^{-10}$

$s = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \approx 1.0 \times 10^{-5} \text{ mol dm}^{-3}$

Marking:

[1] for correct setup ( $s^2 = K_{sp}$ )
[1] for correct answer ( $1.0 \times 10^{-5}$ or $1.05 \times 10^{-5}$ mol dm $^{-3}$ )

(c) [3 marks]

In 0.10 mol dm $^{-3}$ $Na_2SO_4$ , $[SO_4^{2-}] \approx 0.10$ mol dm $^{-3}$ (from the dissolved salt, assuming the contribution from dissolved $BaSO_4$ is negligible).

$K_{sp} = [Ba^{2+}][SO_4^{2-}]$

$1.1 \times 10^{-10} = [Ba^{2+}] \times 0.10$

$[Ba^{2+}] = \frac{1.1 \times 10^{-10}}{0.10} = 1.1 \times 10^{-9} \text{ mol dm}^{-3}$

The solubility is $1.1 \times 10^{-9}$ mol dm $^{-3}$ , which is much lower than in pure water. This is due to the common ion effect: the presence of $SO_4^{2-}$ from $Na_2SO_4$ shifts the equilibrium $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$ to the left (Le Chatelier's principle), suppressing the dissolution of $BaSO_4$ .

Marking:

[1] for correct calculation
[1] for identifying the common ion effect
[1] for explaining the shift in equilibrium

(d) [2 marks]

After mixing equal volumes, the concentrations are halved:

$[Ba^{2+}] = \frac{0.0010}{2} = 5.0 \times 10^{-4}$ mol dm $^{-3}$
$[SO_4^{2-}] = \frac{0.0010}{2} = 5.0 \times 10^{-4}$ mol dm $^{-3}$

Ionic product $= [Ba^{2+}][SO_4^{2-}] = (5.0 \times 10^{-4})^2 = 2.5 \times 10^{-7}$

Since ionic product ( $2.5 \times 10^{-7}$ ) $> K_{sp}$ ( $1.1 \times 10^{-10}$ ), a precipitate of $BaSO_4$ will form.

Marking:

[1] for correct calculation of ionic product
[1] for correct comparison with $K_{sp}$ and conclusion

18. Acid-Base Titrations and Indicators [8 marks]

(a) [3 marks]

Titration	Rough	1	2	3
Final reading / cm³	24.80	24.35	24.30	24.40
Initial reading / cm³	0.00	0.00	0.00	0.00
Titre / cm³	24.80	24.35	24.30	24.40

Titrations 2 and 3 are concordant (within 0.10 cm³ of each other: $|24.35 - 24.30| = 0.05$ cm³ ✓; $|24.35 - 24.40| = 0.05$ cm³ ✓).

All three accurate titrations (1, 2, 3) are within 0.10 cm³ of each other, so all three can be used.

$\text{Mean titre} = \frac{24.35 + 24.30 + 24.40}{3} = \frac{73.05}{3} = 24.35 \text{ cm}^3$

Marking:

[1] for correct table format with all readings
[1] for identifying concordant titres and excluding the rough titration
[1] for correct mean titre calculation (24.35 cm³)

(b) [2 marks]

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Moles of $NaOH = 0.02435 \times 0.100 = 2.435 \times 10^{-3}$ mol

Moles of $CH_3COOH = 2.435 \times 10^{-3}$ mol (1:1 ratio)

$[CH_3COOH] = \frac{2.435 \times 10^{-3}}{0.0250} = 0.0974 \text{ mol dm}^{-3}$

Marking:

[1] for correct moles calculation
[1] for correct concentration (0.0974 mol dm $^{-3}$ )

(c) [3 marks]

This is a weak acid ( $CH_3COOH$ ) – strong base ( $NaOH$ ) titration. At the equivalence point, the solution contains $CH_3COONa$ , which hydrolyses to give a solution with pH > 7 (approximately pH 8.7–9.0). The steep portion of the titration curve around the equivalence point spans approximately pH 7 to pH 10.

Phenolphthalein changes colour in the pH range 8.2–10.0, which falls within the steep pH change region around the equivalence point. Therefore, the colour change from colourless to pink will occur sharply at the equivalence point, giving an accurate result.

Marking:

[1] for identifying this as a weak acid–strong base titration with equivalence point pH > 7
[1] for stating the pH range of phenolphthalein (8.2–10.0)
[1] for explaining that this range falls within the steep region of the curve

19. Salt Hydrolysis and pH [6 marks]

(a) [1 mark]

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).

Marking:

[1] for a clear definition involving reaction of salt ions with water producing $H^+$ or $OH^-$

(b)(i) $AlCl_3$ — Acidic [2 marks]

$AlCl_3$ is a salt of a strong acid ( $HCl$ ) and a weak base ( $Al(OH)_3$ ). The $Al^{3+}$ ion undergoes hydrolysis:

$Al^{3+}(aq) + 3H_2O(l) \rightleftharpoons Al(OH)_3(s) + 3H^+(aq)$

Or more precisely (stepwise):

$[Al(H_2O)_6]^{3+}(aq) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H^+(aq)$

The highly charged $Al^{3+}$ ion polarises the coordinated water molecules, making it easier for a proton to be released. This produces $H^+$ ions, making the solution acidic.

Marking:

[1] for correct prediction (acidic)
[1] for correct hydrolysis equation

(b)(ii) $K_2CO_3$ — Basic [2 marks]

$K_2CO_3$ is a salt of a strong base ( $KOH$ ) and a weak acid ( $H_2CO_3$ ). The $CO_3^{2-}$ ion undergoes hydrolysis:

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

This produces $OH^-$ ions, making the solution basic.

Marking:

[1] for correct prediction (basic)
[1] for correct hydrolysis equation

(b)(iii) $NaNO_3$ — Neutral [1 mark]

$NaNO_3$ is a salt of a strong acid ( $HNO_3$ ) and a strong base ( $NaOH$ ). Neither $Na^+$ nor $NO_3^-$ undergoes hydrolysis, so the solution is neutral (pH = 7).

Marking:

[1] for correct prediction (neutral) with valid reasoning

Section C: Data Interpretation and Application

20. Polyprotic Acids and Titration Curves [15 marks]

(a) [3 marks]

The first equivalence point occurs when all $H_3PO_4$ has been converted to $H_2PO_4^-$ :

$H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$

At this point, the solution contains $NaH_2PO_4$ (the amphoteric species $H_2PO_4^-$ ). The pH is determined by the amphoteric behaviour of $H_2PO_4^-$ , which can act as both an acid and a base. For an amphoteric species, the pH is approximately:

$pH \approx \frac{pK_{a1} + pK_{a2}}{2}$

$pK_{a1} = -\log(7.5 \times 10^{-3}) = 2.12$

$pK_{a2} = -\log(6.2 \times 10^{-8}) = 7.21$

$pH \approx \frac{2.12 + 7.21}{2} = \frac{9.33}{2} = 4.67 \approx 4.7$

The pH is not 7 because $H_2PO_4^-$ is not a neutral salt — it is an amphoteric ion whose pH depends on the average of $pK_{a1}$ and $pK_{a2}$ .

Marking:

[1] for identifying the species present at the first equivalence point ( $H_2PO_4^-$ )
[1] for recognising it is amphoteric
[1] for the calculation using $(pK_{a1} + pK_{a2})/2$

(b) [3 marks]

The third ionisation constant $K_{a3} = 4.8 \times 10^{-13}$ is extremely small, meaning $HPO_4^{2-}$ is an extremely weak acid. The third equivalence point would require:

$HPO_4^{2-} + NaOH \rightarrow Na_3PO_4 + H_2O$

At the third equivalence point, the solution would contain $PO_4^{3-}$ , which is a relatively strong conjugate base (since $K_{a3}$ is very small, $K_b$ for $PO_4^{3-}$ is large). The pH at the third equivalence point would be very high (>12), and the pH change would be very gradual because the very weak third dissociation provides very little buffering capacity. Additionally, at such high pH, the effect of atmospheric $CO_2$ dissolution becomes significant, making the endpoint indistinct. The very small $K_{a3}$ means the third buffer region has very low capacity, and the steep pH rise is not sharp enough to detect with an indicator.

Marking:

[1] for noting that $K_{a3}$ is extremely small
[1] for explaining that the third equivalence point pH would be very high / indistinct
[1] for explaining the lack of a sharp pH change / buffer capacity issue

(c) [2 marks]

At the first half-equivalence point, exactly half of the $H_3PO_4$ has been neutralised:

$[H_3PO_4] = [H_2PO_4^-]$

Using the Henderson-Hasselbalch equation:

$pH = pK_{a1} + \log\frac{[H_2PO_4^-]}{H_3PO_4]}$

Since $[H_2PO_4^-] = [H_3PO_4]$ , the ratio $= 1$ , and $\log(1) = 0$ .

Therefore: $pH = pK_{a1} + 0 = pK_{a1}$

Marking:

[1] for stating that $[H_3PO_4] = [H_2PO_4^-]$ at the half-equivalence point
[1] for applying Henderson-Hasselbalch and showing $\log(1) = 0$

(d) [2 marks]

At the second equivalence point, all $H_2PO_4^-$ has been converted to $HPO_4^{2-}$ . The pH at this point is approximately:

$pH \approx \frac{pK_{a2} + pK_{a3}}{2} = \frac{7.21 + 12.32}{2} = \frac{19.53}{2} = 9.77$

Phenolphthalein changes colour in the range pH 8.2–10.0. Since the equivalence point pH (~9.77) falls within this range, and the steep portion of the titration curve around the second equivalence point also falls within this range, phenolphthalein is a suitable indicator for the second equivalence point.

Marking:

[1] for calculating or estimating the pH at the second equivalence point
[1] for concluding that phenolphthalein is suitable because its range overlaps with the equivalence point pH

(e) [3 marks]

At the second equivalence point, all the original $H_3PO_4$ has been converted to $HPO_4^{2-}$ .

Moles of $H_3PO_4$ initially $= 0.0250 \times 0.100 = 2.50 \times 10^{-3}$ mol

From the stoichiometry:

First equivalence: $H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$ (1:1)
Second equivalence: $NaH_2PO_4 + NaOH \rightarrow Na_2HPO_4 + H_2O$ (1:1)

At the second equivalence point, moles of $HPO_4^{2-} = 2.50 \times 10^{-3}$ mol

Total volume $= 25.0 + 50.0 = 75.0$ cm³ $= 0.0750$ dm³

$[HPO_4^{2-}] = \frac{2.50 \times 10^{-3}}{0.0750} = 0.0333 \text{ mol dm}^{-3}$

Marking:

[1] for correct moles of $HPO_4^{2-}$
[1] for correct total volume
[1] for correct concentration (0.0333 mol dm $^{-3}$ )

(f) [2 marks]

$K_{a2} = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$

For the solution containing 0.10 mol dm $^{-3}$ $NaH_2PO_4$ and 0.10 mol dm $^{-3}$ $Na_2HPO_4$ :

Using the Henderson-Hasselbalch equation:

$pH = pK_{a2} + \log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$

$pK_{a2} = -\log(6.2 \times 10^{-8}) = 7.21$

Since $[HPO_4^{2-}] = [H_2PO_4^-] = 0.10$ mol dm $^{-3}$ :

$pH = 7.21 + \log\frac{0.10}{0.10} = 7.21 + \log(1) = 7.21 + 0 = 7.21$

Marking:

[1] for correct $K_{a2}$ expression and $pK_{a2}$ value
[1] for correct pH calculation (7.21)

Mark Summary:

Section	Marks
A: Q1–15 (MCQ)	15
B: Q16	8
B: Q17	8
B: Q18	8
B: Q19	6
C: Q20	15
Total	60