A Level Chemistry H3 Practice Paper 1

TuitionGoWhere Practice Paper (AI) — Answer Key

Subject: Chemistry Level: A-Level H3 Paper: Practice Paper — Acids, Bases and Salts Version: 1 of 1 Total Marks: 40

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Section A — Structured Questions (30 marks)

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Question 1 (6 marks)

(a) Define the Brønsted–Lowry theory of acids and bases. [2]

Answer: A Brønsted–Lowry acid is a proton ($H^+$) donor. [1] A Brønsted–Lowry base is a proton ($H^+$) acceptor. [1]

Marking notes: Award 1 mark for "proton donor" and 1 mark for "proton acceptor". Do not accept "H⁺ donor/acceptor" without the word "proton" — though in practice, "H⁺ donor" is acceptable. The key idea is the transfer of a proton.

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(b) Classify the following species. For each, state the conjugate partner.

(i) $HSO_4^-$ [2]

Answer: $HSO_4^-$ is amphiprotic. [1]

• As an acid: $HSO_4^- \rightarrow SO_4^{2-} + H^+$; conjugate base is $SO_4^{2-}$.
• As a base: $HSO_4^- + H^+ \rightarrow H_2SO_4$; conjugate acid is $H_2SO_4$.

Accept: "amphiprotic" with either conjugate partner stated correctly for [1] + [1]. [2]

Marking notes: Award 1 mark for identifying as amphiprotic. Award 1 mark for correctly stating both conjugate partners (either direction is acceptable as long as the acid/base role is clear). If the student only gives one conjugate partner, award 1 mark only.

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(ii) $HPO_4^{2-}$ [2]

Answer: $HPO_4^{2-}$ is amphiprotic. [1]

• As an acid: $HPO_4^{2-} \rightarrow PO_4^{3-} + H^+$; conjugate base is $PO_4^{3-}$.
• As a base: $HPO_4^{2-} + H^+ \rightarrow H_2PO_4^-$; conjugate acid is $H_2PO_4^-$.

Accept: "amphiprotic" with either conjugate partner stated correctly for [1] + [1]. [2]

Marking notes: Same as (b)(i).

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Question 2 (7 marks)

(a)(i) Write an expression for $K_w$. [1]

Answer: $K_w = [H^+][OH^-]$ [1]

Marking notes: Accept $K_w = [H_3O^+][OH^-]$. Do not accept $K_w = [H^+]^2$ unless the student has explicitly stated this applies to pure water only.

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(a)(ii) Calculate the pH of $0.050 \text{ mol dm}^{-3}$ NaOH(aq) at 298 K. [2]

Answer: NaOH is a strong base and fully dissociates: $[OH^-] = 0.050 \text{ mol dm}^{-3}$

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.00 \times 10^{-14}}{0.050} = 2.0 \times 10^{-13} \text{ mol dm}^{-3}$

$pH = -\log_{10}(2.0 \times 10^{-13}) = 12.7$ [2]

Marking notes: Award 1 mark for correct calculation of $[H^+]$. Award 1 mark for correct pH value (12.7, accept 12.70). If the student uses $pH = 14 - pOH$ method, this is also acceptable: $pOH = -\log(0.050) = 1.30$, so $pH = 14.00 - 1.30 = 12.70$. Award full marks for either method. Deduct 1 mark if the final answer is given to more than 2 decimal places or lacks appropriate sig figs.

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(b)(i) Calculate the pH of pure water at 310 K. [2]

Answer: In pure water, $[H^+] = [OH^-]$, so: $K_w = [H^+]^2$ $[H^+] = \sqrt{2.52 \times 10^{-14}} = 1.587 \times 10^{-7} \text{ mol dm}^{-3}$

$pH = -\log_{10}(1.587 \times 10^{-7}) = 6.80$ [2]

Marking notes: Award 1 mark for correct calculation of $[H^+]$ (square root step). Award 1 mark for correct pH value (6.80, accept 6.8). The answer should be given to 2 decimal places. Note: the pH is below 7 but the water is still neutral — this is addressed in (b)(ii).

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(b)(ii) Explain whether pure water at 310 K is acidic, acidic, or neutral. [2]

Answer: Pure water at 310 K is neutral. [1]

This is because $[H^+] = [OH^-]$ in pure water. The pH of 6.80 is less than 7 only because $K_w$ has increased with temperature, shifting the position of the self-ionisation equilibrium to the right. Neutrality is defined by equal concentrations of $H^+$ and $OH^-$, not by pH = 7. [1]

Marking notes: Award 1 mark for stating "neutral". Award 1 mark for the explanation that $[H^+] = [OH^-]$ defines neutrality, not pH = 7. A common student error is to say the water is acidic because pH < 7 — this is incorrect and should not be credited.

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Question 3 (8 marks)

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

Answer: $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$ [1]

Marking notes: Accept $H_3O^+$ in place of $H^+$ in the numerator. The expression must be a correct equilibrium expression with products over reactants.

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

Answer: Ethanoic acid is a weak acid that partially dissociates: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

Let $[H^+] = x$ at equilibrium. Then $[CH_3COO^-] = x$ and $[CH_3COOH] \approx 0.100 - x \approx 0.100$ (since $K_a$ is small, $x \ll 0.100$).

$K_a = \frac{x^2}{0.100} = 1.74 \times 10^{-5}$

$x^2 = 1.74 \times 10^{-6}$

$x = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$

$pH = -\log_{10}(1.32 \times 10^{-3}) = 2.88$ [3]

Marking notes:

• Award 1 mark for setting up the $K_a$ expression with the approximation $0.100 - x \approx 0.100$.
• Award 1 mark for correct calculation of $[H^+] = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$.
• Award 1 mark for correct pH = 2.88 (accept 2.88 or 2.9).

Common mistakes:

• Forgetting the approximation and solving the quadratic (this is not wrong but is unnecessary; if done correctly, award full marks).
• Using $[H^+] = 0.100$ (treating the acid as strong) — award 0 marks.
• Calculation errors in the square root — check working.

Validation of approximation: $\frac{1.32 \times 10^{-3}}{0.100} \times 100\% = 1.32\% < 5\%$, so the approximation is valid.

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(c) Calculate the pH when $12.5 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ NaOH has been added. [2]

Answer: Initial moles of $CH_3COOH = 0.0250 \times 0.100 = 2.50 \times 10^{-3}$ mol Moles of NaOH added $= 0.0125 \times 0.100 = 1.25 \times 10^{-3}$ mol

The reaction: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$

Moles of $CH_3COOH$ remaining $= 2.50 \times 10^{-3} - 1.25 \times 10^{-3} = 1.25 \times 10^{-3}$ mol Moles of $CH_3COO^-$ formed $= 1.25 \times 10^{-3}$ mol

Total volume $= 25.0 + 12.5 = 37.5 \text{ cm}^3 = 0.0375 \text{ dm}^3$

This is a buffer solution where $[CH_3COOH] = [CH_3COO^-]$.

Using the Henderson–Hasselbalch equation: $pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$

$pK_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$

$pH = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76$ [2]

Marking notes:

• Award 1 mark for recognising this is a buffer and correctly calculating the moles of acid and conjugate base (or their equal ratio).
• Award 1 mark for the correct pH value of 4.76.

Alternative approach: Using $K_a$ expression directly with the buffer concentrations gives the same result. Accept this method.

Key insight: At the half-equivalence point, $pH = pK_a$. This is a useful result that students should recognise.

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(d) Calculate the pH at the equivalence point. [2]

Answer: At the equivalence point, all $CH_3COOH$ has been converted to $CH_3COO^-$.

Moles of $CH_3COO^-$ formed $= 2.50 \times 10^{-3}$ mol Volume of NaOH needed $= \frac{2.50 \times 10^{-3}}{0.100} = 0.0250 \text{ dm}^3 = 25.0 \text{ cm}^3$ Total volume $= 25.0 + 25.0 = 50.0 \text{ cm}^3 = 0.0500 \text{ dm}^3$

$[CH_3COO^-] = \frac{2.50 \times 10^{-3}}{0.0500} = 0.0500 \text{ mol dm}^{-3}$

$CH_3COO^-$ is the conjugate base of a weak acid and undergoes hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$

$K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$

Let $[OH^-] = y$: $K_b = \frac{y^2}{0.0500} = 5.75 \times 10^{-10}$

$y = \sqrt{5.75 \times 10^{-10} \times 0.0500} = \sqrt{2.875 \times 10^{-11}} = 5.36 \times 10^{-6} \text{ mol dm}^{-3}$

$pOH = -\log_{10}(5.36 \times 10^{-6}) = 5.27$

$pH = 14.00 - 5.27 = 8.73$ [2]

Marking notes:

• Award 1 mark for correctly calculating $K_b$ and setting up the hydrolysis equilibrium.
• Award 1 mark for the correct pH value of 8.73 (accept 8.7, but 2 d.p. is preferred).

Common mistakes:

• Forgetting that the equivalence point of a weak acid–strong base titration has pH > 7.
• Using $K_a$ instead of $K_b$ for the conjugate base hydrolysis.
• Forgetting to account for the total volume when calculating $[CH_3COO^-]$.
• Getting pH < 7 at the equivalence point — this indicates a fundamental misunderstanding.

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Question 4 (5 marks)

(a) Describe how a buffer solution resists changes in acid when a small amount of acid is added. [3]

Answer: A buffer solution contains significant amounts of a weak acid and its conjugate base (e.g., $CH_3COOH$ and $CH_3COO^-$). [1]

When a small amount of acid ($H^+$) is added, the conjugate base ($CH_3COO^-$) reacts with the added $H^+$ ions:

$CH_3COO^- + H^+ \rightarrow CH_3COOH$ [1]

This removes most of the added $H^+$ ions, converting them into the undissociated weak acid. Since the equilibrium position shifts only slightly, the $[H^+]$ (and hence pH) changes very little. [1]

Marking notes:

• Award 1 mark for identifying the buffer composition (weak acid + conjugate base).
• Award 1 mark for the correct reaction equation showing the conjugate base neutralising added $H^+$.
• Award 1 mark for explaining that the equilibrium shift is small and the pH change is minimal.

Note: Students should also mention that when base is added, the weak acid neutralises it. However, the question specifically asks about acid addition, so this is sufficient.

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(b) Calculate the pH of the buffer. [2]

Answer: After mixing:

• Moles of $CH_3COOH = 0.0500 \times 0.200 = 0.0100$ mol
• Moles of $CH_3COO^- = 0.0500 \times 0.100 = 0.00500$ mol
• Total volume $= 100.0 \text{ cm}^3 = 0.100 \text{ dm}^3$

$[CH_3COOH] = \frac{0.0100}{0.100} = 0.100 \text{ mol dm}^{-3}$ $[CH_3COO^-] = \frac{0.00500}{0.100} = 0.0500 \text{ mol dm}^{-3}$

Using the Henderson–Hasselbalch equation: $pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$

$pH = 4.76 + \log_{10}\frac{0.0500}{0.100} = 4.76 + \log_{10}(0.500) = 4.76 + (-0.301) = 4.46$ [2]

Marking notes:

• Award 1 mark for correct calculation of the concentrations (or mole ratio) after mixing.
• Award 1 mark for the correct pH value of 4.46 (accept 4.46 or 4.5).

Common mistake: Forgetting that the total volume is $100 \text{ cm}^3$ (not $50 \text{ cm}^3$) when calculating concentrations. However, if the student uses the mole ratio directly (since the volume cancels), this is acceptable and should be credited.

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Question 5 (4 marks)

(a) Define the term solubility product, $K_{sp}$. [1]

Answer: The solubility product, $K_{sp}$, is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It is the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the dissolution equation. [1]

Marking notes: Award 1 mark for a clear definition that includes the idea of an equilibrium constant for dissolution and the product of ion concentrations raised to their stoichiometric powers. A concise answer such as "the equilibrium constant for the dissolution of a sparingly soluble salt" is sufficient.

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(b)(i) Write an expression for $K_{sp}$ of $PbI_2$. [1]

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

$K_{sp} = [Pb^{2+}][I^-]^2$ [1]

Marking notes: Award 1 mark for the correct expression. The dissolution equation is not required but helps clarify the stoichiometry.

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(b)(ii) Calculate the solubility of $PbI_2$ in pure water. [2]

Answer: Let the solubility of $PbI_2 = s \text{ mol dm}^{-3}$.

From the dissolution equation: $[Pb^{2+}] = s$ $[I^-] = 2s$

$K_{sp} = [Pb^{2+}][I^-]^2 = (s)(2s)^2 = 4s^3$

$4s^3 = 8.5 \times 10^{-9}$

$s^3 = \frac{8.5 \times 10^{-9}}{4} = 2.125 \times 10^{-9}$

$s = \sqrt[3]{2.125 \times 10^{-9}} = 1.29 \times 10^{-3} \text{ mol dm}^{-3}$ [2]

Marking notes:

• Award 1 mark for correctly setting up the relationship $[I^-] = 2s$ and substituting into the $K_{sp}$ expression to get $K_{sp} = 4s^3$.
• Award 1 mark for the correct final answer: $1.29 \times 10^{-3} \text{ mol dm}^{-3}$ (accept $1.3 \times 10^{-3}$ to 2 s.f.).

Common mistakes:

• Forgetting that $[I^-] = 2s$ (not $s$) — this is the most common error. Students who write $K_{sp} = s^2$ or $K_{sp} = s^3$ will get the wrong answer.
• Calculation errors in the cube root.
• Giving the answer for $[Pb^{2+}]$ vs. the solubility — in this case they are the same, but for other salts they may differ.

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Section B — Stimulus-Based Question (10 marks)

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Question 6 (10 marks)

(a) Explain why adding $Na_2CO_3$ removes $Ca^{2+}$ ions from hard water. [3]

Answer: Adding $Na_2CO_3$ introduces $CO_3^{2-}$ ions into the solution. [1]

The dissolution equilibrium for $CaCO_3$ is: $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$

According to Le Chatelier's principle, increasing the concentration of $CO_3^{2-}$ shifts the equilibrium to the left, [1] causing $Ca^{2+}$ ions to precipitate as solid $CaCO_3$, thereby removing them from the water. [1]

Marking notes:

• Award 1 mark for identifying that $Na_2CO_3$ provides $CO_3^{2-}$ ions.
• Award 1 mark for applying Le Chatelier's principle (equilibrium shifts left).
• Award 1 mark for stating that $Ca^{2+}$ precipitates as $CaCO_3$ and is removed.

Alternative acceptable answer: Students may use the $K_{sp}$ argument — when $[CO_3^{2-}]$ increases, the ionic product $[Ca^{2+}][CO_3^{2-}]$ exceeds $K_{sp}$, so precipitation occurs until the product equals $K_{sp}$ again. Award full marks for this approach.

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(b) Calculate the maximum concentration of $CO_3^{2-}$ without precipitating $CaCO_3$. [2]

Answer: Precipitation occurs when $[Ca^{2+}][CO_3^{2-}] > K_{sp}$.

At the point just before precipitation: $[Ca^{2+}][CO_3^{2-}] = K_{sp} = 4.96 \times 10^{-9}$

$[CO_3^{2-}] = \frac{K_{sp}}{[Ca^{2+}]} = \frac{4.96 \10^{-9}}{1.50 \times 10^{-3}} = 3.31 \times 10^{-6} \text{ mol dm}^{-3}$ [2]

Marking notes:

• Award 1 mark for correctly substituting into the $K_{sp}$ expression.
• Award 1 mark for the correct answer: $3.31 \times 10^{-6} \text{ mol dm}^{-3}$ (accept $3.3 \times 10^{-6}$ to 2 s.f.).

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(c) Explain why $EDTA^{4-}$ is effective at sequestering $Ca^{2+}$ ions. [2]

Answer: $EDTA^{4-}$ forms a stable complex with $Ca^{2+}$ ions: $Ca^{2+}(aq) + EDTA^{4-}(aq) \rightleftharpoons [Ca(EDTA)]^{2-}(aq)$

The formation constant $K_f = 5.0 \times 10^{10} \text{ dm}^3 \text{ mol}^{-1}$ is very large, [1] meaning the equilibrium lies far to the right. This means $EDTA^{4-}$ effectively binds $Ca^{2+}$ ions into a stable, soluble complex, preventing them from reacting with soap or forming scale. [1]

Marking notes:

• Award 1 mark for referencing the large $K_f$ value and its implication (equilibrium lies to the right / reaction goes essentially to completion).
• Award 1 mark for explaining the consequence: $Ca^{2+}$ is tied up in the complex and is no longer free to cause hardness.

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(d) Comment on the student's claim. [3]

Answer: The student's claim is incorrect or not necessarily valid. [1]

While $Ca(OH)_2$ may have a lower $K_{sp}$ than $CaCO_3$, this does not automatically make $NaOH$ a better water-softening agent. The key consideration is the solubility and the practicality of the method. [1]

$Ca(OH)_2$ is only sparingly soluble in water (approximately $0.02 \text{ mol dm}^{-2}$ at 298 K), so adding $NaOH$ to hard water will produce only a limited amount of $OH^-$ in solution. The $Ca(OH)_2$ that precipitates will form a suspension rather than being easily removed, and excess $NaOH$ would make the water strongly alkaline, which is undesirable. In contrast, $Na_2CO_3$ is highly soluble, readily provides $CO_3^{2-}$ ions, and the $CaCO_3$ precipitate is easily filtered off. [1]

Marking notes:

• Award 1 mark for stating that the claim is incorrect or questionable.
• Award 1 mark for discussing the limited solubility of $Ca(OH)_2$ or the practical difficulties of using $NaOH$.
• Award 1 mark for a balanced comparison with $Na_2CO_3$ (solubility, ease of removal, pH considerations).

Acceptable alternative arguments:

• Students may argue that the claim has some validity in principle (lower $K_{sp}$ means less $Ca^{2+}$ remains in solution at equilibrium) but is impractical for the reasons above. Award marks for well-reasoned arguments.
• Students may discuss that $NaOH$ would also remove $Mg^{2+}$ (as $Mg(OH)_2$, which has an even lower $K_{sp}$), which could be seen as an advantage. This is a valid point and should be credited.

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Summary of Marks

Question	Marks
1	6
2	7
3	8
4	5
5	4
6	10
Total	40

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Notes for Students

Key concepts tested in this paper:

1. Brønsted–Lowry acid–base theory and conjugate pairs
2. Ionic product of water ($K_w$) and its temperature dependence
3. pH calculations for strong bases, weak acids, buffer solutions, and salt hydrolysis
4. Acid–base titrations: half-equivalence point, equivalence point
5. Buffer action and the Henderson–Hasselbalch equation
6. Solubility product ($K_{sp}$) calculations and precipitation criteria
7. Application of equilibrium principles to real-world contexts (water softening)

Common pitfalls to avoid:

• Confusing pH < 7 with acidity at temperatures other than 298 K
• Forgetting that $[I^-] = 2s$ (not $s$) in $K_{sp}$ calculations for salts like $PbI_2$
• Not accounting for total volume changes in titration calculations
• Assuming the equivalence point of a weak acid–strong base titration is at pH 7
• Using $K_a$ instead of $K_b$ for conjugate base hydrolysis at the equivalence point