A Level H1 Chemistry Practice Paper 2

TuitionGoWhere Practice Paper — Chemistry H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper — Acids, Bases & Salts (Version 2 of 5) Total Marks: 60

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Section A: Multiple Choice [10 marks]

1. (b) $K_w = [H^+][OH^-]$ [1]

Teaching note: The ionic product of water, $K_w$, is the equilibrium constant for the self-ionisation of water: $H_2O \rightleftharpoons H^+ + OH^-$. Since $[H_2O]$ is essentially constant, $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ mol$^2$ dm$^{-6}$ at 25 °C.

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2. (b) $3.16 \times 10^{-11}$ mol dm$^{-3}$ [1]

Working: $[H^+] = 10^{-3.50} = 3.16 \times 10^{-4}$ mol dm$^{-3}$

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{3.16 \times 10^{-4}} = 3.16 \times 10^{-11}$ mol dm$^{-3}$

Common mistake: Students may select (a), which is the $[H^+]$ value, not $[OH^-]$.

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3. (b) $NH_4Cl$ [1]

Teaching note: $NH_4Cl$ is formed from a weak base ($NH_3$) and a strong acid ($HCl$). The $NH_4^+$ ion is the conjugate acid of a weak base and undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$, producing $H^+$ ions and making the solution acidic. The other salts are from strong acid + strong base ($Na_2SO_4$, $KNO_3$) or weak acid + strong base ($CH_3COONa$, which is basic).

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4. (b) The resulting solution is not a buffer because all the acid is neutralised. [1]

Teaching note: 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 ethanoic acid is neutralised, producing only sodium ethanoate ($CH_3COONa$). A buffer requires a mixture of a weak acid and its conjugate base. Since no weak acid remains, this is not a buffer.

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5. (c) 2.80 [1]

Working: $K_a = \frac{[H^+]^2}{[HA]}$ (assuming $[H^+] = [A^-]$ and dissociation is small)

$[H^+] = \sqrt{K_a \times [HA]} = \sqrt{2.5 \times 10^{-5} \times 0.10} = \sqrt{2.5 \times 10^{-6}} = 1.58 \times 10^{-3}$ mol dm$^{-3}$

$pH = -\log(1.58 \times 10^{-3}) = 2.80$

Common mistake: Students may select (a) by assuming pH = –log(0.10) = 1.00, which would be correct for a strong acid but not a weak acid.

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6. (b) 25.0 cm³ [1]

Working: At the equivalence point, moles of acid = moles of base (1:1 stoichiometry for HCl and NaOH).

$moles\ of\ HCl = 0.0250 \times 0.10 = 2.5 \times 10^{-3}$ mol

$Volume\ of\ NaOH = \frac{2.5 \times 10^{-3}}{0.10} = 0.0250$ dm³ $= 25.0$ cm³

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7. (c) Phenolphthalein [1]

Teaching note: In a weak acid–strong base titration, the equivalence point occurs at pH > 7 (due to hydrolysis of the conjugate base). Phenolphthalein changes colour in the pH range 8.2–10.0, which encompasses the steep portion of the titration curve near the equivalence point. Methyl orange (pH 3.1–4.4) would change colour too early, well before the equivalence point.

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8. (d) It completely dissociates in aqueous solution. [1]

Teaching note: A strong acid is defined as one that undergoes complete (100%) dissociation in water. For example: $HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$. This is in contrast to a weak acid, which only partially dissociates and establishes an equilibrium.

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9. (c) The pH remains almost unchanged. [1]

Teaching note: A buffer resists changes in pH when small amounts of acid or base are added. When NaOH is added, the $OH^-$ ions react with the ethanoic acid component of the buffer: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. This converts a small amount of weak acid into its conjugate base, so the ratio $[A^-]/[HA]$ changes only slightly, and the pH remains nearly constant.

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10. (a) $1.6 \times 10^{-4}$ mol dm$^{-3}$ [1]

Working: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$

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

Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$

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

$4s^3 = 1.8 \times 10^{-11}$

$s^3 = 4.5 \times 10^{-12}$

$s = 1.65 \times 10^{-4} \approx 1.6 \times 10^{-4}$ mol dm$^{-3}$

Common mistake: Students may forget to square the $[OH^-]$ term or forget the factor of 4, leading to incorrect answers.

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Section B: Structured Questions [30 marks]

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11. (a) A weak acid is an acid that partially dissociates in aqueous solution. [1]

Marking: Award 1 mark for "partially dissociates" or equivalent wording. Do not accept "does not fully dissociate" without the idea of establishing an equilibrium.

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(b) $HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq)$ [1]

Marking: Award 1 mark for correct equation with reversible arrow and state symbols. Accept $H_3O^+$ in place of $H^+$.

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(c) The student's claim is correct. [1]

HCl is a strong acid and dissociates completely in water, so $[H^+] = 0.01$ mol dm$^{-3}$ and $pH = 2.00$. [½]

Methanoic acid is a weak acid and only partially dissociates, so $[H^+] < 0.01$ mol dm$^{-3}$ and $pH > 2.00$. [½]

Therefore, the HCl solution has a lower pH (more acidic) than the methanoic acid solution of the same concentration. [½]

The student's claim that HCl has a higher pH is incorrect. [½]

Wait — re-reading the question: The student claims HCl has a higher pH than methanoic acid. Since HCl is strong, it has a lower pH. Therefore the student's claim is incorrect.

Corrected answer: The student's claim is incorrect. [1]

HCl is a strong acid and dissociates completely, giving $[H^+] = 0.01$ mol dm$^{-3}$ and $pH = 2.00$. Methanoic acid is weak and partially dissociates, giving $[H^+] < 0.01$ mol dm$^{-3}$ and $pH > 2.00$. [1]

Marking: 1 mark for stating the claim is incorrect. 1 mark for explanation involving complete vs. partial dissociation and comparison of pH values.

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(d) [3]

$K_a = \frac{[H^+][HCOO^-]}{[HCOOH]}$

Assuming $[H^+] = [HCOO^-]$ and dissociation is small so $[HCOOH] \approx 0.050$:

$[H^+]^2 = K_a \times [HCOOH] = 1.6 \times 10^{-4} \times 0.050 = 8.0 \times 10^{-6}$ [1]

$[H^+] = \sqrt{8.0 \times 10^{-6}} = 2.83 \times 10^{-3}$ mol dm$^{-3}$ [1]

$pH = -\log(2.83 \times 10^{-3}) = 2.55$ [1]

Marking: 1 mark for correct expression/substitution. 1 mark for correct $[H^+]$. 1 mark for correct pH (accept 2.54–2.55).

Common mistake: Students may use $[HCOOH] = 0.050$ directly in $K_a = [H^+]/[HA]$ without squaring, or may forget to take the square root.

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12. (a) [2]

Moles of $CH_3COOH = 0.100 \times 0.30 = 0.030$ mol [1]

Moles of $NaOH = 0.100 \times 0.20 = 0.020$ mol [1]

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(b) [2]

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

NaOH is the limiting reagent. [1]

Moles of $CH_3COOH$ remaining $= 0.030 - 0.020 = 0.010$ mol

Moles of $CH_3COONa$ formed $= 0.020$ mol [1]

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(c) [3]

Total volume $= 100 + 100 = 200$ cm³ $= 0.200$ dm³

$[CH_3COOH] = \frac{0.010}{0.200} = 0.050$ mol dm$^{-3}$

$[CH_3COO^-] = \frac{0.020}{0.200} = 0.10$ mol dm$^{-3}$

Using Henderson-Hasselbalch equation:

$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$ [1]

$pH = pK_a + \log\frac{[A^-]}{[HA]} = 4.74 + \log\frac{0.10}{0.050} = 4.74 + \log 2 = 4.74 + 0.30 = 5.04$ [1]

Marking: 1 mark for correct concentrations. 1 mark for correct substitution into Henderson-Hasselbalch. 1 mark for correct final answer (accept 5.04).

Alternative method using $K_a$: $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$ $[H^+] = K_a \times \frac{[CH_3COOH]}{[CH_3COO^-]} = 1.8 \times 10^{-5} \times \frac{0.050}{0.10} = 9.0 \times 10^{-6}$ $pH = -\log(9.0 \times 10^{-6}) = 5.04$

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(d) [2]

When HCl is added, the $H^+$ ions react with the ethanoate ions ($CH_3COO^-$) in the buffer to form ethanoic acid: [1]

$H^+ + CH_3COO^- \rightarrow CH_3COOH$

This converts a small amount of conjugate base into weak acid. The ratio $[A^-]/[HA]$ decreases only slightly, so the pH decreases only very slightly (remains almost unchanged). [1]

Marking: 1 mark for identifying the reaction between $H^+$ and $CH_3COO^-$. 1 mark for explaining that the pH change is very small because the buffer components absorb the added acid.

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13. (a) [3]

Marking scheme for graph:

• Correctly labelled axes with units [1]
• Correct plotting of at least 8 data points [1]
• Smooth curve drawn through the points, showing the characteristic weak acid–strong base titration shape with a steep rise near 25.0 cm³ [1]

Expected features of the graph:

• The curve starts at pH 2.94 and rises gradually.
• There is a relatively flat buffer region between about 5–12.5 cm³.
• The curve rises steeply between 24.9 and 25.1 cm³ (the equivalence point region).
• Beyond 25.0 cm³, the curve flattens again at high pH due to excess NaOH.
• The equivalence point (steepest part of the curve) occurs at 25.0 cm³ NaOH.

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(b) Equivalence point volume $= 25.0$ cm³ [1]

From the graph, the steepest part of the curve (midpoint of the vertical section) occurs at 25.0 cm³.

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(c) [2]

At the equivalence point, all the propanoic acid has been neutralised to form sodium propanoate ($C_2H_5COONa$). [1]

The propanoate ion ($C_2H_5COO^-$) is the conjugate base of a weak acid and undergoes hydrolysis:

$C_2H_5COO^- + H_2O \rightleftharpoons C_2H_5COOH + OH^-$

This produces $OH^-$ ions, making the solution slightly alkaline, so pH > 7. [1]

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(d) [2]

The half-equivalence point occurs when half the acid has been neutralised, i.e. at $\frac{25.0}{2} = 12.5$ cm³ of NaOH. [1]

From the table, at 12.5 cm³, the pH $= 4.87$.

At the half-equivalence point, $[HA] = [A^-]$, so $pH = pK_a = 4.87$. [1]

Marking: 1 mark for identifying the half-equivalence point at 12.5 cm³. 1 mark for stating $pK_a = pH = 4.87$.

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14. (a) The solubility product, $K_{sp}$, is the equilibrium constant for the dissolution of a sparingly soluble salt 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. [1]

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(b)(i) $K_{sp} = [Ca^{2+}][F^-]^2$ [1]

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(b)(ii) [3]

$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$

Let solubility of $CaF_2 = s$ mol dm$^{-3}$

Then $[Ca^{2+}] = s$ and $[F^-] = 2s$

$K_{sp} = [Ca^{2+}][F^-]^2 = s \times (2s)^2 = 4s^3$ [1]

$4s^3 = 3.9 \times 10^{-11}$

$s^3 = 9.75 \times 10^{-12}$

$s = 2.14 \times 10^{-4}$ mol dm$^{-3}$ [1]

Marking: 1 mark for correct expression ($K_{sp} = 4s^3$). 1 mark for correct substitution. 1 mark for correct answer (accept $2.1 \times 10^{-4}$ to $2.14 \times 10^{-4}$ mol dm$^{-3}$).

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(c) [2]

The solubility of $CaF_2$ would decrease in 0.10 mol dm$^{-3}$ NaF solution. [1]

This is due to the common ion effect. NaF provides $F^-$ ions, which are a common ion with $CaF_2$. According to Le Chatelier's principle, the increased $[F^-]$ shifts the equilibrium $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$ to the left, reducing the dissolution of $CaF_2$. [1]

Marking: 1 mark for stating solubility decreases. 1 mark for explanation involving common ion effect and Le Chatelier's principle.

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Section C: Free Response [20 marks]

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15. (a) $2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$ [1]

Accept: $OH^- + H^+ \rightarrow H_2O$ (ionic equation)

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(b)(i) Titration 2 (32.10 cm³) is anomalous. [1]

The other three concordant titres are 25.80, 25.60, and 25.70 cm³ (within 0.20 cm³ of each other), while titration 2 gives 25.60 cm³ as the volume used (32.10 – 6.50 = 25.60 cm³).

Re-evaluation: Titration 2: 32.10 – 6.50 = 25.60 cm³. Titration 1: 25.80 cm³. Titration 3: 25.70 cm³.

All three (25.80, 25.60, 25.70) are within 0.20 cm³ of each other, so none is anomalous. However, the rough titre (26.50 cm³) is not used in the mean calculation as it is only an estimate.

Corrected answer: The rough titre (26.50 cm³) is not used in calculating the mean because it is only an approximate reading. Titrations 1, 2, and 3 are concordant (within 0.20 cm³ of each other). [1]

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(b)(ii) Mean titre $= \frac{25.80 + 25.60 + 25.70}{3} = \frac{77.10}{3} = 25.70$ cm³ [1]

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(b)(iii) [3]

Moles of $H_2SO_4 = 0.02570 \times 0.150 = 3.855 \times 10^{-3}$ mol [1]

From the equation: $2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$

Mole ratio $KOH : H_2SO_4 = 2 : 1$

Moles of $KOH = 2 \times 3.855 \times 10^{-3} = 7.71 \times 10^{-3}$ mol [1]

Concentration of $KOH = \frac{7.71 \times 10^{-3}}{0.0250} = 0.308$ mol dm$^{-3}$ [1]

Marking: 1 mark for moles of $H_2SO_4$. 1 mark for correct mole ratio and moles of KOH. 1 mark for correct concentration (accept 0.308 mol dm$^{-3}$).

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(c) [2]

This is a strong acid–strong base titration. The equivalence point occurs at pH 7. [1]

Methyl orange changes colour in the pH range 3.1–4.4, which is well below the equivalence point pH of 7. It would change colour too early, leading to a significant titration error. Phenolphthalein (pH 8.2–10.0) is also not ideal for strong acid–strong base, but the question context uses it. More precisely: for a strong acid–strong base titration, the pH change is very steep around pH 7, and phenolphthalein's range (8.2–10.0) captures the steep portion, whereas methyl orange (3.1–4.4) changes colour before the equivalence point is reached. [1]

Refined answer: Methyl orange changes colour in the pH range 3.1–4.4. For a strong acid–strong base titration, the equivalence point is at pH 7, and the steep portion of the titration curve spans approximately pH 4–10. Methyl orange would change colour well before the equivalence point, resulting in a large systematic error. Phenolphthalein changes colour in the range 8.2–10.0, which falls within the steep portion of the curve, making it a suitable choice. [1]

Marking: 1 mark for identifying the pH range of methyl orange. 1 mark for explaining that it changes colour before the equivalence point, causing error.

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16. (a) [3]

The common ion effect is the reduction in solubility of a sparingly soluble salt when a soluble compound containing a common ion is added to the solution. [1]

For $AgCl$: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

When $NaCl$ is added, it dissociates completely: $NaCl \rightarrow Na^+ + Cl^-$

This increases the concentration of $Cl^-$ ions in solution. [1]

By Le Chatelier's principle, the equilibrium shifts to the left (towards the solid), reducing the dissolution of $AgCl$ and hence decreasing its solubility. [1]

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(b)(i) [2]

$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

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

$K_{sp} = [Ag^+][Cl^-] = s \times s = s^2$ [1]

$s^2 = 1.8 \times 10^{-10}$

$s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ mol dm$^{-3}$ [1]

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(b)(ii) [2]

In 0.050 mol dm$^{-3}$ NaCl, $[Cl^-] \approx 0.050$ mol dm$^{-3}$ (from NaCl, since the contribution from dissolved AgCl is negligible). [1]

$K_{sp} = [Ag^+][Cl^-]$

$1.8 \times 10^{-10} = [Ag^+] \times 0.050$

$[Ag^+] = \frac{1.8 \times 10^{-10}}{0.050} = 3.6 \times 10^{-9}$ mol dm$^{-3}$

Solubility of $AgCl = 3.6 \times 10^{-9}$ mol dm$^{-3}$ [1]

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(b)(iii) [1]

The solubility of $AgCl$ in 0.050 mol dm$^{-3}$ NaCl ($3.6 \times 10^{-9}$ mol dm$^{-3}$) is much lower than in pure water ($1.34 \times 10^{-5}$ mol dm$^{-3}$). This demonstrates the common ion effect — the presence of $Cl^-$ from NaCl suppresses the dissolution of $AgCl$.

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17. (a) $CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$ [1]

Accept: $CO_2 + H_2O \rightleftharpoons H^+ + HCO_3^-$ (simplified)

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(b)(i) $[H^+] = 10^{-3.8} = 1.58 \times 10^{-4}$ mol dm$^{-3}$ [1]

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(b)(ii) $[H^+]_{unpolluted} = 10^{-5.6} = 2.51 \times 10^{-6}$ mol dm$^{-3}$

Ratio $= \frac{1.58 \times 10^{-4}}{2.51 \times 10^{-6}} = 63.0$ [1]

The polluted rainwater has approximately 63 times the $H^+$ concentration of unpolluted rainwater.

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(c) [3]

Sulfur dioxide dissolves in water to form sulfurous acid: [1]

$SO_2(g) + H_2O(l) \rightleftharpoons H_2SO_3(aq)$

Sulfurous acid is then oxidised to sulfuric acid in the atmosphere (catalysed by particulates or in the presence of oxygen and water): [1]

$2SO_2(g) + O_2(g) + 2H_2O(l) \rightarrow 2H_2SO_4(aq)$

Alternative: $2H_2SO_3(aq) + O_2(g) \rightarrow 2H_2SO_4(aq)$

Sulfuric acid is a strong acid and dissociates completely, releasing $H^+$ ions that lower the pH of rainwater. [1]

Marking: 1 mark for dissolution of $SO_2$. 1 mark for oxidation to $H_2SO_4$. 1 mark for explaining that $H_2SO_4$ dissociates to release $H^+$, lowering pH.

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(d) [2]

Environmental consequence (any one): [1]

• Acidification of lakes and rivers, harming aquatic life (fish, invertebrates)
• Damage to vegetation and forests (damage to leaves, reduced nutrient uptake)
• Corrosion of buildings and statues made of limestone/marble
• Leaching of toxic metals (e.g., aluminium) from soil into water bodies

Method to reduce (any one): [1]

• Use of catalytic converters in vehicles to reduce $NO_x$ emissions
• Flue gas desulfurisation in power plants to remove $SO_2$ from exhaust gases
• Switching to renewable energy sources to reduce fossil fuel combustion
• Using low-sulfur fuels

Marking: 1 mark for a valid consequence. 1 mark for a valid reduction method.

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End of Answer Key

Mark Summary:

Section	Marks
A: Q1–10 (MCQ)	10
B: Q11 (a–d)	7
B: Q12 (a–d)	9
B: Q13 (a–d)	8
B: Q14 (a–c)	6
C: Q15 (a–c)	8
C: Q16 (a–b)	8
C: Q17 (a–d)	8
Total	60 ✓