A Level H2 Chemistry Practice Paper 1

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key — Acids, Bases & Salts (Version 1 of 5)

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Section A: Short-Answer Questions

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1. [2 marks]

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

Example: $HCl$ (or $H_2SO_4$, $HNO_3$, $CH_3COOH$, etc.) [1]

Teaching note: The Brønsted–Lowry definition focuses on the transfer of a proton from the acid to a base. Any species that can donate $H^+$ qualifies. This is broader than the Arrhenius definition, which requires the acid to produce $H^+$ in aqueous solution specifically.

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2. [3 marks total]

(a) [1 mark]

$[H^+] = 10^{-pH} = 10^{-1.20}$ $\boxed{[H^+] = 0.063 \text{ mol dm}^{-3}}$

Teaching note: The relationship $[H^+] = 10^{-pH}$ is fundamental. Students should be able to use this directly from their calculators. The answer should be given to 2 significant figures (matching the 2 decimal places in the pH value).

(b) [2 marks]

$K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{6.31 \times 10^{-2}}$

$\boxed{[OH^-] = 1.58 \times 10^{-13} \text{ mol dm}^{-3}}$

Teaching note: At 25 °C, $K_w$ is always $1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$. Students must rearrange correctly. Common mistake: using $[H^+] = 0.063$ without sufficient precision, which introduces rounding errors. Using $10^{-1.20}$ directly is more accurate.

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3. [3 marks]

Answer: Hydrochloric acid is a strong acid and dissociates completely in water, producing a high concentration of $H^+$ ions. [1]

Ethanoic acid is a weak acid and only partially dissociates in water, so the concentration of $H^+$ ions is much lower than the nominal concentration of the acid. [1]

Since pH = $-\log[H^+]$, the lower $[H^+]$ in ethanoic acid gives a higher pH. [1]

Teaching note: Key terms required: "strong/weak acid," "complete/partial dissociation," and the link between $[H^+]$ and pH. Students often state that weak acids have "fewer $H^+$ ions" without explaining why (partial dissociation), which would not earn full marks.

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4. [2 marks]

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

Units: $\frac{\text{mol dm}^{-3} \times \text{mol dm}^{-3}}{\text{mol dm}^{-3}} = \boxed{\text{mol dm}^{-3}}$ [1]

Teaching note: $K_a$ expressions always have products (right-hand side of equilibrium) over reactants (left-hand side). Pure solids and solvents are omitted. For a monobasic acid, the units of $K_a$ are always mol dm$^{-3}$.

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5. [5 marks total]

(a) [3 marks]

$[H^+] = 10^{-3.10} = 7.94 \times 10^{-4} \text{ mol dm}^{-3}$

For the equilibrium: $HX \rightleftharpoons H^+ + X^-$

At equilibrium: $[H^+] = [X^-] = 7.94 \times 10^{-4}$ mol dm$^{-3}$

$[HX] = 0.025 - 7.94 \times 10^{-4} \approx 0.0242 \text{ mol dm}^{-3}$

$K_a = \frac{(7.94 \times 10^{-4})^2}{0.0242} = \frac{6.31 \times 10^{-7}}{0.0242}$

$\boxed{K_a = 2.6 \times 10^{-5} \text{ mol dm}^{-3}}$

Marking: 1 mark for $[H^+]$ calculation, 1 mark for correct substitution, 1 mark for final answer.

Teaching note: Since the acid is weak and the degree of dissociation is small (~3.2%), the approximation $[HX] \approx$ initial concentration is valid. However, showing the subtraction demonstrates understanding. Students should check that dissociation is <5% to justify the approximation.

(b) [2 marks]

NaX is the salt of a weak acid and strong base, so $X^-$ undergoes hydrolysis:

$X^- + H_2O \rightleftharpoons HX + OH^-$

The solution will be alkaline (pH > 7). [1]

Using $K_b = K_w / K_a = 1.00 \times 10^{-14} / 2.61 \times 10^{-5} = 3.83 \times 10^{-10}$

Since $K_b$ is very small, the pH will be only slightly above 7, approximately pH ≈ 8–9. [1]

Teaching note: The conjugate base of a weak acid is itself a weak base, so the salt solution is alkaline. Students should recognise the pattern: weak acid + strong base → alkaline salt solution.

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6. [3 marks]

Answer: The pH of the ethanoic acid solution increases. [1]

$CH_3COONa$ dissociates completely to provide $CH_3COO^-$ ions, which is the common ion. [1]

By Le Chatelier's principle, the increased $[CH_3COO^-]$ shifts the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ to the left, reducing $[H^+]$ and hence increasing the pH. [1]

Teaching note: This is a classic buffer/common ion effect question. Students must identify the common ion ($CH_3COO^-$), state the direction of equilibrium shift, and link this to a change in $[H^+]$ and therefore pH. Simply stating "pH increases" without explanation earns only 1 mark.

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7. [5 marks total]

(a) [3 marks]

Moles of $CH_3COOH$ = $0.050 \times 0.20 = 0.010$ mol

Moles of $NaOH$ = $0.050 \times 0.20 = 0.010$ mol

Since moles of acid = moles of base, all the $CH_3COOH$ is converted to $CH_3COONa$. [1]

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

$[CH_3COO^-] = \frac{0.010}{0.100} = 0.10 \text{ mol dm}^{-3}$

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

$[OH^-] = \sqrt{K_b \times [CH_3COO^-]} = \sqrt{5.75 \times 10^{-10} \times 0.10} = \sqrt{5.75 \times 10^{-11}} = 7.58 \times 10^{-6}$

$pOH = -\log(7.58 \times 10^{-6}) = 5.12$

$\boxed{pH = 14.00 - 5.12 = 8.88}$

Marking: 1 mark for identifying complete neutralisation to salt, 1 mark for correct $K_b$ and $[OH^-]$ calculation, 1 mark for final pH.

(b) [2 marks]

The resulting solution is not a buffer. [1]

A buffer requires significant amounts of both a weak acid and its conjugate base. Here, all the $CH_3COOH$ has been converted to $CH_3COO^-$, so there is no remaining weak acid to act as the buffer component. [1]

Teaching note: This is a common exam trap. Students often assume that mixing a weak acid and strong base always produces a buffer. A buffer is only formed when the acid is in excess so that both the weak acid and its salt coexist in significant amounts.

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8. [3 marks]

Answer: Add a small piece of magnesium ribbon (or zinc, or $Na_2CO_3$) to equal volumes of each acid at the same concentration. [1]

With the strong acid, the reaction is vigorous/faster, producing hydrogen gas (effervescence) rapidly. [1]

With the weak acid, the reaction is slower/less vigorous because the $[H^+]$ is lower (the acid is only partially dissociated). [1]

Alternative acceptable answers:

• Measure the pH of each solution: the strong acid has a lower pH.
• Measure electrical conductivity: the strong acid has higher conductivity due to greater $[ions]$.
• Measure initial rate of reaction with a carbonate: strong acid produces $CO_2$ faster.

Teaching note: The key principle is that at the same concentration, a strong acid has a higher $[H^+]$ than a weak acid, leading to faster reaction rates. Any experiment that exploits this difference is acceptable.

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9. [4 marks total]

(a) [1 mark]

$K_{sp} = [Mg^{2+}][OH^-]^2$

(b) [3 marks]

Let the solubility of $Mg(OH)_2$ = $s$ mol dm$^{-3}$

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

$[Mg^{2+}] = s, \quad [OH^-] = 2s$

$K_{sp} = s \times (2s)^2 = 4s^3 = 5.61 \times 10^{-12}$

$s^3 = \frac{5.61 \times 10^{-12}}{4} = 1.4025 \times 10^{-12}$

$\boxed{s = \sqrt[3]{1.4025 \times 10^{-12}} = 1.12 \times 10^{-4} \text{ mol dm}^{-3}}$

Marking: 1 mark for correct relationship between ion concentrations and $s$, 1 mark for correct substitution into $K_{sp}$ expression, 1 mark for final answer.

Teaching note: The stoichiometry of dissolution is critical. For every 1 mole of $Mg(OH)_2$ that dissolves, 2 moles of $OH^-$ are produced, hence $[OH^-] = 2s$. This is the most common source of error — students often write $[OH^-] = s$.

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10. [3 marks]

Answer: The common ion effect is the reduction in solubility of a sparingly soluble salt when a solution already contains an ion that is common to the salt. [1]

$AgCl$ dissolves according to: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

$NaCl$ dissociates completely: $NaCl \rightarrow Na^+ + Cl^-$, providing $Cl^-$ ions. [1]

By Le Chatelier's principle, the increased $[Cl^-]$ shifts the equilibrium to the left, reducing the solubility of $AgCl$. [1]

Teaching note: Students must define the term, write relevant equations, and apply Le Chatelier's principle. The common ion ($Cl^-$) suppresses dissolution. This is a direct application of equilibrium principles to solubility.

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Section B: Structured Questions

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11. [9 marks total]

(a) [1 mark]

The rough titration (24.80 cm$^3$) is anomalous. [1]

It differs significantly from the other three results (24.35, 24.30, 24.40), which are all within 0.10 cm$^3$ of each other.

(b) [2 marks]

The rough titration is excluded. Titrations 1, 2, and 3 are concordant (within 0.10 cm$^3$).

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

$\boxed{\text{Mean titre} = 24.35 \text{ cm}^3}$

Marking: 1 mark for excluding the rough/anomalous result, 1 mark for correct mean.

(c) [1 mark]

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

(d) [3 marks]

Moles of $NaOH$ used = $0.100 \times \frac{24.35}{1000} = 2.435 \times 10^{-3}$ mol [1]

From the equation, mole ratio $H_2SO_4 : NaOH = 1 : 2$

Moles of $H_2SO_4 = \frac{2.435 \times 10^{-3}}{2} = 1.2175 \times 10^{-3}$ mol [1]

$[H_2SO_4] = \frac{1.2175 \times 10^{-3}}{\frac{25.0}{1000}} = \frac{1.2175 \times 10^{-3}}{0.025}$

$\boxed{[H_2SO_4] = 0.0487 \text{ mol dm}^{-3}}$

(e) [2 marks]

There is no effect on the calculated concentration. [1]

The conical flask contains a known volume of $H_2SO_4$. Rinsing with distilled water adds more water but does not change the number of moles of $H_2SO_4$ in the flask. The same amount of $NaOH$ is still required for neutralisation, so the titre volume and calculated concentration remain unchanged. [1]

Teaching note: This is a classic conceptual question. Students often confuse rinsing the conical flask (no effect) with rinsing the burette (would cause error). The burette must be rinsed with the titrant, not water.

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12. [8 marks total]

(a) [2 marks]

From the graph, initial pH ≈ 1.0 (at 0 cm$^3$ NaOH added). [1]

$[H^+] = 10^{-1.0} = 0.10 \text{ mol dm}^{-3}$

Since $HCl$ is a strong monoprotic acid, $[HCl] = [H^+] = \boxed{0.10 \text{ mol dm}^{-3}}$ [1]

(b) [1 mark]

$HCl$ is a strong acid and $NaOH$ is a strong base. The salt $NaCl$ formed does not hydrolyse, so the solution is neutral at pH 7.0. [1]

(c) [3 marks]

The sketch should show:

• Starting pH higher than 1.0 (approximately pH ≈ 2–3 for 0.10 mol dm$^{-3}$ weak acid) [1]
• A buffer region before the equivalence point (gentle rise) [1]
• Equivalence point at 25.0 cm$^3$ NaOH but at pH > 7 (approximately pH ≈ 8–9), with a less steep vertical rise compared to the strong acid curve [1]

Teaching note: Key differences for weak acid vs. strong base titration: (1) higher initial pH, (2) buffer region with gradual pH change, (3) equivalence point above pH 7 due to hydrolysis of the conjugate base, (4) less steep pH jump at equivalence.

(d) [2 marks]

At the equivalence point, all the ethanoic acid has been converted to sodium ethanoate ($CH_3COONa$). [1]

The $CH_3COO^-$ ion is the conjugate base of a weak acid and undergoes hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$, producing $OH^-$ ions and making the solution alkaline (pH > 7). [1]

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13. [10 marks total]

(a) [3 marks]

Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

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

$5.00 = 4.76 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$\log\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.24$

$\boxed{\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.24} = 1.74}$

Marking: 1 mark for $pK_a$ calculation, 1 mark for correct substitution, 1 mark for final ratio.

(b) [4 marks]

To form a buffer, some $CH_3COOH$ must react with $NaOH$ to produce $CH_3COO^-$, with excess $CH_3COOH$ remaining.

Let volume of $NaOH$ = $V$ dm$^3$, volume of $CH_3COOH$ = $(1.00 - V)$ dm$^3$

Moles of $NaOH$ = $0.50V$ → produces $0.50V$ mol $CH_3COO^-$

Moles of $CH_3COOH$ remaining = $0.50(1.00 - V) - 0.50V = 0.50 - 1.00V$

$\frac{[CH_3COO^-]}{[CH_3COOH]} = \frac{0.50V}{0.50 - 1.00V} = 1.74$

$0.50V = 1.74(0.50 - 1.00V) = 0.87 - 1.74V$

$2.24V = 0.87$

$V = 0.388 \text{ dm}^3 = 388 \text{ cm}^3$

Volume of 0.50 mol dm$^{-3}$ $NaOH$ = $\boxed{388 \text{ cm}^3}$ [1]

Volume of 0.50 mol dm$^{-3}$ $CH_3COOH$ = $1000 - 388 = \boxed{612 \text{ cm}^3}$ [1]

Marking: 1 mark for correct setup of ratio equation, 1 mark for each volume.

(c) [3 marks]

In 1.00 dm$^3$ of the buffer:

Moles of $CH_3COO^-$ = $0.50 \times 0.388 = 0.194$ mol

Moles of $CH_3COOH$ = $0.50 - 0.388 = 0.112$ mol

Adding 0.005 mol $NaOH$:

• $CH_3COOH$ reacts with $NaOH$: moles of $CH_3COOH$ = $0.112 - 0.005 = 0.107$ mol [1]
• Moles of $CH_3COO^-$ = $0.194 + 0.005 = 0.199$ mol [1]

$pH = 4.76 + \log\frac{0.199}{0.107} = 4.76 + \log(1.860) = 4.76 + 0.270$

$\boxed{pH = 5.03}$

Teaching note: Adding a small amount of strong base to a buffer causes only a small pH change — this is the key property of buffers. The pH changes from 5.00 to 5.03, demonstrating buffer action.

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14. [5 marks total]

(a) [2 marks]

$HCl$ is a strong acid and dissociates completely, providing $[H^+] = 0.10$ mol dm$^{-3}$. [1]

The high $[H^+]$ from $HCl$ suppresses the dissociation of $CH_3COOH$ (common ion effect), so the additional $[H^+]$ contributed by $CH_3COOH$ is negligible. [1]

(b) [1 mark]

$pH = -\log(0.10) = \boxed{1.00}$

(c) [2 marks]

From the $CH_3COOH$ equilibrium:

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

$1.74 \times 10^{-5} = \frac{[CH_3COO^-](0.10)}{0.10}$

$\boxed{[CH_3COO^-] = 1.74 \times 10^{-5} \text{ mol dm}^{-3}}$

Teaching note: In a mixture of strong and weak acids, the strong acid dominates the pH. The weak acid's dissociation is suppressed, and $[CH_3COO^-]$ is determined solely by $K_a$ with $[H^+]$ fixed by the strong acid.

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15. [4 marks total]

(a) [2 marks]

An indicator changes colour over the approximate range $pK_{in} \pm 1$. [1]

$pK_{in} = -\log(3.5 \times 10^{-4}) = 3.46$

Colour change range: pH 2.46 to 4.46 (approximately pH 2.5–4.5). [1]

(b) [2 marks]

For a strong acid–strong base titration, the equivalence point is at pH 7.0, and the pH jump spans approximately pH 4–10. [1]

Either indicator is suitable since the steep portion of the titration curve spans both their colour-change ranges. However, phenolphthalein (pH 8.2–10.0) is more commonly used because its colour change (colourless → pink) is easier to detect than methyl orange's (red → yellow). [1]

Teaching note: For strong acid–strong base titrations, both indicators work because the vertical portion of the pH curve covers a wide range. The choice is often based on ease of observing the colour change.

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Section C: Data Interpretation & Application

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16. [7 marks total]

(a) [2 marks]

Order of increasing acid strength: $\boxed{HCOOH < HNO_2 < HF}$ [1]

A larger $K_a$ value indicates a greater degree of dissociation and therefore a stronger acid. [1]

(b) [3 marks]

$K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} = \frac{x^2}{0.20 - x} \approx \frac{x^2}{0.20}$

$x^2 = 1.78 \times 10^{-4} \times 0.20 = 3.56 \times 10^{-5}$

$x = [H^+] = 5.97 \times 10^{-3} \text{ mol dm}^{-1}$ [1]

$pH = -\log(5.97 \times 10^{-3}) = \boxed{2.22}$ [1]

Marking: 1 mark for correct setup, 1 mark for $[H^+]$, 1 mark for pH. Check: dissociation = 5.97 × 10⁻³/0.20 = 3.0% < 5%, so approximation is valid.

(c) [2 marks]

$\boxed{HCOONa}$ would produce the most alkaline solution. [1]

$HCOOH$ has the smallest $K_a$, meaning $HCOO^-$ is the strongest conjugate base (of the three). A stronger conjugate base undergoes more hydrolysis, producing more $OH^-$ and hence a higher pH. [1]

Teaching note: The weaker the acid, the stronger its conjugate base. This inverse relationship is key: smallest $K_a$ → strongest conjugate base → most alkaline salt solution.

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17. [7 marks total]

(a) [2 marks]

Bubble the gas through limewater (calcium hydroxide solution). [1]

The limewater turns milky/cloudy due to the formation of white precipitate $CaCO_3$. [1]

(b) [2 marks]

The solution is calcium hydroxide solution (limewater). [1]

$CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq)$

(c) (i) [1 mark]

$Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$

(ii) [2 marks]

Moles of $HCl$ = $0.100 \times \frac{30.0}{1000} = 3.00 \times 10^{-3}$ mol [1]

From the equation, mole ratio $Ca(OH)_2 : HCl = 1 : 2$

Moles of $Ca(OH)_2 = \frac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}$ mol

$[Ca(OH)_2] = \frac{1.50 \times 10^{-3}}{\frac{25.0}{1000}} = \boxed{0.0600 \text{ mol dm}^{-3}}$

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18. [6 marks total]

(a) [1 mark]

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

(b) [1 mark]

From the stoichiometry, $[Pb^{2+}] = \frac{1}{2}[I^-] = \frac{2.40 \times 10^{-3}}{2} = \boxed{1.20 \times 10^{-3} \text{ mol dm}^{-3}}$

(c) [2 marks]

$K_{sp} = [Pb^{2+}][I^-]^2 = (1.20 \times 10^{-3})(2.40 \times 10^{-3})^2$

$= 1.20 \times 10^{-3} \times 5.76 \times 10^{-6} = 6.91 \times 10^{-9}$

$\boxed{K_{sp} = 6.91 \times 10^{-9} \text{ mol}^3 \text{ dm}^{-9}}$

Marking: 1 mark for correct substitution, 1 mark for final answer with units.

(d) [2 marks]

The solubility of $PbI_2$ would decrease. [1]

Adding $KI$ increases $[I^-]$ (common ion effect). By Le Chatelier's principle, the equilibrium shifts to the left, reducing the dissolution of $PbI_2$. [1]

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19. [5 marks total]

(a) [3 marks]

Moles of $H_2SO_4$ = $0.15 \times \frac{20.0}{1000} = 3.00 \times 10^{-3}$ mol

Moles of $KOH$ = $0.10 \times \frac{30.0}{1000} = 3.00 \times 10^{-3}$ mol

Reaction: $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$

Moles of $KOH$ needed to neutralise all $H_2SO_4$ = $2 \times 3.00 \times 10^{-3} = 6.00 \times 10^{-3}$ mol

Since only $3.00 \times 10^{-3}$ mol $KOH$ is available, $KOH$ is the limiting reagent. [1]

Moles of $H_2SO_4$ reacted = $\frac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}$ mol

Moles of excess $H_2SO_4$ = $3.00 \times 10^{-3} - 1.50 \times 10^{-3} = \boxed{1.50 \times 10^{-3} \text{ mol}}$ [1]

Marking: 1 mark for identifying limiting reagent, 1 mark for moles of excess reagent, 1 mark for correct calculation.

(b) [2 marks]

Total volume = $20.0 + 30.0 = 50.0$ cm$^3$ = 0.050 dm$^3$

Excess $[H^+]$: Each mole of $H_2SO_4$ provides 2 moles of $H^+$:

$[H^+] = \frac{2 \times 1.50 \times 10^{-3}}{0.050} = \frac{3.00 \times 10^{-3}}{0.050} = 0.060 \text{ mol dm}^{-3}$

$pH = -\log(0.060) = \boxed{1.22}$

Teaching note: $H_2SO_4$ is diprotic, so each mole produces 2 moles of $H^+$. This is a common source of error — students sometimes forget the factor of 2.

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20. [6 marks total]

(a) [2 marks]

$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$ [1]

$NaHCO_3(s) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) + CO_2(g)$ [1]

(b) [3 marks]

Moles of $CaCO_3 = \frac{0.50}{100.1} = 4.995 \times 10^{-3}$ mol [1]

From the equation, mole ratio $CaCO_3 : HCl = 1 : 2$

Moles of $HCl$ neutralised = $2 \times 4.995 \times 10^{-3} = 9.99 \times 10^{-3}$ mol [1]

$V = \frac{9.99 \times 10^{-3}}{0.50} = 1.998 \times 10^{-2} \text{ dm}^3 = \boxed{20.0 \text{ cm}^3}$

(c) [1 mark]

$NaHCO_3$ is preferred because it is milder/less corrosive (or: it reacts more gently / produces less gas per mole / has a lower alkalinity so is less likely to cause alkalosis). [1]

Alternative acceptable answers:

• $NaHCO_3$ is more soluble and acts faster.
• $NaHCO_3$ produces less $CO_2$ per mole of acid neutralised (1:1 vs 1:2 ratio with $CaCO_3$).

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Total: 60 marks

Mark Distribution Summary:

Section	Marks
A: Questions 1–10	20
B: Questions 11–15	25
C: Questions 16–20	15
Total	60