A Level H2 Chemistry Practice Paper 5

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

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

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Question 1 [2]

Definition: A Brønsted–Lowry acid is a proton ($H^+$) donor.

Example: $HCl$ (or any valid acid such as $H_2SO_4$, $CH_3COOH$, $NH_4^+$)

Equation: $HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

or $HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$

Marking:

• 1 mark for correct definition (proton donor)
• 1 mark for a valid example with a correct equation showing proton donation

Teaching note: A Brønsted–Lowry acid donates a proton to another species (the base). The key idea is proton transfer. Students sometimes confuse this with the Lewis acid definition (electron pair acceptor).

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Question 2 [2]

$HCl$ is a strong acid and dissociates completely:

$HCl \rightarrow H^+ + Cl^-$

$[H^+] = 0.150 \text{ mol dm}^{-3}$

$pH = -\log_{10}[H^+] = -\log_{10}(0.150)$

$pH = 0.82 \text{ (2 d.p.)}$

Marking:

• 1 mark for correct $[H^+] = 0.150 \text{ mol dm}^{-3}$ (complete dissociation)
• 1 mark for correct pH = 0.82

Common mistake: Students may forget that $HCl$ is monoprotic and strong, so $[H^+] = [HCl]$. For diprotic strong acids like $H_2SO_4$, $[H^+] = 2 \times [H_2SO_4]$ (for the first complete dissociation).

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Question 3

(a) [1]

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

Marking: 1 mark for correct expression.

(b) [2]

$[H^+] = \sqrt{K_a \times c} = \sqrt{1.74 \times 10^{-5} \times 0.100}$

$[H^+] = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$

$pH = -\log_{10}(1.32 \times 10^{-3}) = 2.88 \text{ (2 d.p.)}$

Marking:

• 1 mark for correct substitution
• 1 mark for correct pH = 2.88

Teaching note: For weak acids, we use the approximation $[H^+] = \sqrt{K_a \cdot c}$ because the degree of dissociation is small. This is derived from $K_a = \frac{x^2}{c-x} \approx \frac{x^2}{c}$ when $x \ll c$.

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Question 4 [3]

$NaCl$: Formed from a strong acid ($HCl$) and a strong base ($NaOH$). Neither ion hydrolyses in water, so the solution is neutral, pH ≈ 7.

$CH_3COONa$: Formed from a weak acid ($CH_3COOH$) and a strong base ($NaOH$). The ethanoate ion ($CH_3COO^-$) is the conjugate base of a weak acid and undergoes hydrolysis:

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$

This produces $OH^-$ ions, making the solution alkaline (pH > 7).

Marking:

• 1 mark: $NaCl$ from strong acid + strong base → neutral
• 1 mark: $CH_3COONa$ from weak acid + strong base → alkaline
• 1 mark: Correct explanation involving hydrolysis of $CH_3COO^-$ producing $OH^-$

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Question 5

(a) [1]

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

Marking: 1 mark for correct balanced equation.

(b) [2]

Moles of $NaOH$ used:

$n(NaOH) = 0.100 \times \frac{24.80}{1000} = 2.48 \times 10^{-3} \text{ mol}$

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

$n(H_2SO_4) = \frac{2.48 \times 10^{-3}}{2} = 1.24 \times 10^{-3} \text{ mol}$

$c(H_2SO_4) = \frac{1.24 \times 10^{-3}}{25.0/1000} = 0.0496 \text{ mol dm}^{-3}$

Marking:

• 1 mark for correct moles of NaOH and use of 1:2 ratio
• 1 mark for correct concentration = 0.0496 mol dm⁻³

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Question 6

(a) [2]

Adding sodium ethanoate increases the concentration of the common ion $CH_3COO^-$. By Le Chatelier's principle, the equilibrium:

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

shifts to the left, reducing $[H^+]$. Therefore, the pH increases.

Marking:

• 1 mark: Common ion effect / equilibrium shifts left
• 1 mark: pH increases

(b) [2]

Dilution reduces the concentration of all species. For a weak acid, dilution increases the degree of dissociation (more molecules dissociate), but the overall $[H^+]$ decreases because the solution is more dilute. Therefore, the pH increases (moves closer to 7).

Marking:

• 1 mark: Degree of dissociation increases on dilution
• 1 mark: pH increases (but remains below 7)

Common mistake: Students may think dilution of a weak acid decreases pH. While more molecules dissociate, the total $[H^+]$ still decreases because the volume increase dominates.

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Question 7

(a) [1]

$n(CH_3COOH) = 0.200 \times \frac{50.0}{1000} = 0.0100 \text{ mol}$

$n(NaOH) = 0.100 \times \frac{50.0}{1000} = 0.00500 \text{ mol}$

Marking: 1 mark for both correct values.

(b) [2]

The reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

$NaOH$ is the limiting reagent.

$n(CH_3COOH)_{\text{remaining}} = 0.0100 - 0.00500 = 0.00500 \text{ mol}$

$n(CH_3COONa)_{\text{formed}} = 0.00500 \text{ mol}$

Marking:

• 1 mark for identifying NaOH as limiting reagent and calculating remaining acid
• 1 mark for moles of sodium ethanoate formed

(c) [2]

Total volume = $50.0 + 50.0 = 100.0 \text{ cm}^3 = 0.100 \text{ dm}^3$

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

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

Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$

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

$pH = 4.76 + \log_{10}\frac{0.0500}{0.0500} = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76$

Marking:

• 1 mark for correct concentrations or correct use of Henderson–Hasselbalch
• 1 mark for correct pH = 4.76

Teaching note: When $[acid] = [salt]$, $pH = pK_a$. This is a useful shortcut for buffer problems.

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Question 8 [2]

A strong acid dissociates completely in aqueous solution (e.g., $HCl$, $H_2SO_4$, $HNO_3$). The degree of dissociation is essentially 100%. The solution contains only $H^+$ ions and the conjugate base anions — no undissociated acid molecules remain.

A weak acid only partially dissociates in aqueous solution (e.g., $CH_3COOH$, $H_2CO_3$). The degree of dissociation is small (typically < 5%). The solution contains $H^+$ ions, conjugate base anions, and undissociated acid molecules in equilibrium.

Marking:

• 1 mark: Strong acid = complete dissociation; weak acid = partial dissociation
• 1 mark: Reference to species present (undissociated molecules exist in weak acid solution)

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Question 9

(a) [1]

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$

Marking: 1 mark for correct expression.

(b) [3]

$[OH^-] = \sqrt{K_b \times c} = \sqrt{1.78 \times 10^{-5} \times 0.050}$

$[OH^-] = \sqrt{8.90 \times 10^{-7}} = 9.43 \times 10^{-4} \text{ mol dm}^{-3}$

$pOH = -\log_{10}(9.43 \times 10^{-4}) = 3.03$

$pH = 14.00 - 3.03 = 10.97$

Marking:

• 1 mark for correct $[OH^-]$ calculation
• 1 mark for correct pOH
• 1 mark for correct pH = 10.97

Teaching note: For weak bases, we use the same approximation method as weak acids but calculate $[OH^-]$ first, then convert to pH using $pH = 14 - pOH$.

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Question 10 [3]

The equivalence point is the point in an acid–base titration at which the amount (in moles) of acid is exactly neutralised by the amount of base, according to the stoichiometric ratio in the balanced equation. At this point, neither reactant is in excess.

Indicator choice: The indicator must have a colour-change range (transition range) that falls within the steep portion of the titration curve near the equivalence point.

• For a strong acid–strong base titration, the equivalence point is at pH 7 and the pH change is very steep. Many indicators work (e.g., phenolphthalein, methyl orange).
• For a weak acid–strong base titration, the equivalence point is above pH 7 (due to hydrolysis of the conjugate base). Phenolphthalein (range 8.2–10.0) is suitable.
• For a strong acid–weak base titration, the equivalence point is below pH 7. Methyl orange (range 3.1–4.4) is suitable.

Marking:

• 1 mark: Correct definition of equivalence point
• 1 mark: Indicator range must fall within the steep pH change region
• 1 mark: Specific example linking acid/base strength to indicator choice

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Question 11

(a) [1]

Titration	Rough	1	2	3
Volume used / $\text{cm}^3$	25.50	24.90	24.85	25.10

Marking: 1 mark for all three values correctly calculated.

(b) [2]

Titrations 1, 2, and 3 are concordant (within 0.10 cm³ of each other). The rough titration is not used in the mean.

$\text{Mean titre} = \frac{24.90 + 24.85 + 25.10}{3} = \frac{74.85}{3} = 24.95 \text{ cm}^3$

Marking:

• 1 mark for identifying concordant titres (1, 2, 3) and excluding the rough
• 1 mark for correct mean = 24.95 cm³

(c) [2]

$n(NaOH) = 0.100 \times \frac{24.95}{1000} = 2.495 \times 10^{-3} \text{ mol}$

HA is monoprotic, so mole ratio HA : NaOH = 1 : 1

$n(HA) = 2.495 \times 10^{-3} \text{ mol}$

$c(HA) = \frac{2.495 \times 10^{-3}}{25.0/1000} = 0.0998 \text{ mol dm}^{-3}$

Marking:

• 1 mark for correct moles of NaOH and 1:1 ratio
• 1 mark for correct concentration = 0.0999 mol dm⁻³ (accept 0.0998–0.100)

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Question 12

(a) [1]

Equivalence point at 25.0 cm³ NaOH (read from the graph where the steep rise passes through pH 7).

Marking: 1 mark for reading 25.0 cm³ from the graph.

(b) [2]

$HCl$ is a strong acid and $NaOH$ is a strong base. At the equivalence point, the salt formed is $NaCl$, which is derived from a strong acid and a strong base. Neither $Na^+$ nor $Cl^-$ undergoes hydrolysis, so the solution is neutral and the pH is 7.

Marking:

• 1 mark: NaCl formed from strong acid + strong base
• 1 mark: No hydrolysis → pH = 7

(c) [3]

Key differences for ethanoic acid (weak acid) vs. NaOH titration:

1. Starting pH is higher (around 2.9, not 1) because ethanoic acid is weak and only partially dissociates.
2. Equivalence point pH is above 7 (approximately 8.7–9) because the salt $CH_3COONa$ is formed from a weak acid and strong base; the ethanoate ion hydrolyses to produce an alkaline solution.
3. The curve has a buffer region before the equivalence point where the pH rises gradually (the "S" shape is less steep initially).
4. The steep rise is shorter in vertical extent compared to the strong acid titration.

Marking:

• 1 mark: Starting pH is higher
• 1 mark: Equivalence point pH > 7
• 1 mark: Any other valid difference (buffer region, shorter steep section, etc.)

Image note: The graph should show a curve starting at pH ≈ 2.9, with a buffer region from about 0–20 cm³, a steep rise between approximately 23–26 cm³ passing through pH ≈ 8.7 at 25.0 cm³, and levelling off at pH ≈ 11–12.

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Question 13

(a) [2]

Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$

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

$4.75 = 4.76 + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$

$\log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.75 - 4.76 = -0.01$

$\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{-0.01} = 0.977$

Marking:

• 1 mark for correct pKa calculation and substitution
• 1 mark for correct ratio = 0.977 (or 0.98 to 2 s.f.)

(b) [2]

Moles of $CH_3COOH$ in 100 cm³ of 0.500 mol dm⁻³:

$n(CH_3COOH) = 0.500 \times \frac{100}{1000} = 0.0500 \text{ mol}$

From the ratio:

$\frac{n(CH_3COO^-)}{n(CH_3COOH)} = 0.977$

$n(CH_3COO^-) = 0.977 \times 0.0500 = 0.04885 \text{ mol}$

Mass of $CH_3COONa$:

$m = n \times M_r = 0.04885 \times 82.0 = 4.01 \text{ g}$

Marking:

• 1 mark for correct moles of ethanoic acid and use of ratio
• 1 mark for correct mass = 4.01 g

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Question 14

(a) [1]

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

Marking: 1 mark for correct expression.

(b) [2]

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

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

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

$K_{sp} = s \times (2s)^2 = 4s^3$

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

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

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

Marking:

• 1 mark for correct relationship $[OH^-] = 2s$ and $K_{sp} = 4s^3$
• 1 mark for correct solubility = $1.12 \times 10^{-4}$ mol dm⁻³

(c) [2]

$[OH^-] = 2s = 2 \times 1.12 \times 10^{-4} = 2.24 \times 10^{-4} \text{ mol dm}^{-3}$

$pOH = -\log_{10}(2.24 \times 10^{-4}) = 3.65$

$pH = 14.00 - 3.65 = 10.35$

Marking:

• 1 mark for correct $[OH^-] = 2.24 \times 10^{-4}$
• 1 mark for correct pH = 10.35

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Question 15

(a) [2]

$HCl$ is a strong acid and dissociates completely:

$[H^+] = 0.100 \text{ mol dm}^{-3}$

$pH = -\log_{10}(0.100) = 1.00$

Marking:

• 1 mark for $[H^+] = 0.100$ from HCl
• 1 mark for pH = 1.00

(b) [2]

The $H^+$ from $HCl$ suppresses the dissociation of ethanoic acid due to the common ion effect. The equilibrium:

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

is shifted far to the left by the high $[H^+]$ from $HCl$. The additional $[H^+]$ from ethanoic acid is negligible compared to 0.100 mol dm⁻³ from $HCl$.

Marking:

• 1 mark: Common ion effect suppresses ethanoic acid dissociation
• 1 mark: $[H^+]$ from ethanoic acid is negligible compared to 0.100 mol dm⁻³

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Question 16

(a) [2]

Methyl orange changes colour over pH 3.1–4.4. For a weak acid–strong base titration, the equivalence point occurs at approximately pH 8.7–9, and the steep pH change occurs over approximately pH 7–10. Since the colour change range of methyl orange (3.1–4.4) does not fall within the steep region of the titration curve, methyl orange is not suitable. The colour change would occur long before the equivalence point is reached.

Marking:

• 1 mark: Methyl orange is not suitable
• 1 mark: Its range (3.1–4.4) does not fall within the steep pH change region (pH 7–10)

(b) [2]

Phenolphthalein is more suitable. Its colour change range is pH 8.2–10.0, which falls within the steep portion of the titration curve for a weak acid–strong base titration (pH 7–10). The colour change (colourless to pink) would occur very close to the equivalence point.

Marking:

• 1 mark: Phenolphthalein
• 1 mark: Its range (8.2–10.0) falls within the steep region near the equivalence point

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Question 17

(a) [1]

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

Marking: 1 mark for correct equation.

(b) [2]

$Na_2CO_3$ is formed from a strong base ($NaOH$) and a weak acid ($H_2CO_3$). The carbonate ion ($CO_3^{2-}$) is the conjugate base of the weak acid $HCO_3^-$. It hydrolyses with water to produce $OH^-$ ions, making the solution alkaline.

Marking:

• 1 mark: Strong base + weak acid
• 1 mark: $CO_3^{2-}$ hydrolyses to produce $OH^-$

(c) [2]

$NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$

$NH_4Cl$ is formed from a weak base ($NH_3$) and a strong acid ($HCl$). The ammonium ion ($NH_4^+$) is the conjugate acid of the weak base $NH_3$. It hydrolyses with water to produce $H_3O^+$ (or $H^+$) ions, making the solution acidic.

Marking:

• 1 mark: Correct hydrolysis equation
• 1 mark: $NH_4^+$ is the conjugate acid of a weak base; produces $H^+$ → acidic

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Question 18

(a) [1]

$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Marking: 1 mark for correct balanced equation.

(b) [1]

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

Marking: 1 mark for correct balanced equation.

(c) [2]

$Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O$

Mole ratio: $Ca(OH)_2 : H_2SO_4 = 1 : 1$

$n(Ca(OH)_2) = n(H_2SO_4) = 0.025 \text{ mol}$

$m(Ca(OH)_2) = 0.025 \times 74.1 = 1.85 \text{ g}$

Marking:

• 1 mark for correct 1:1 mole ratio and moles of Ca(OH)₂
• 1 mark for correct mass = 1.85 g

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Question 19

(a) [2]

Total volume = $50.0 + 50.0 = 100.0 \text{ cm}^3$

Mass of solution = $100.0 \times 1.00 = 100.0 \text{ g}$

$q = mc\Delta T = 100.0 \times 4.18 \times 6.8 = 2842.4 \text{ J} = 2.84 \text{ kJ}$

Marking:

• 1 mark for correct mass and substitution
• 1 mark for correct q = 2.84 kJ

(b) [2]

$n(HCl) = 1.00 \times \frac{50.0}{1000} = 0.0500 \text{ mol}$

$n(NaOH) = 1.00 \times \frac{50.0}{1000} = 0.0500 \text{ mol}$

The reaction is 1:1, so 0.0500 mol of water is formed.

$\Delta H = \frac{-q}{n} = \frac{-2.84}{0.0500} = -56.8 \text{ kJ mol}^{-1}$

Marking:

• 1 mark for correct moles of reaction
• 1 mark for correct $\Delta H = -56.8 \text{ kJ mol}^{-1}$ (negative sign required)

(c) [1]

Any one of:

• Heat loss to the surroundings (polystyrene cup is not a perfect insulator)
• Heat absorbed by the calorimeter/cup
• Assumption that the specific heat capacity and density of the solution are the same as water

Marking: 1 mark for any valid reason.

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Question 20

(a) [2]

$K_{a1} \gg K_{a2}$ because:

The first proton is removed from the neutral molecule $H_2A$, which is relatively easy. The second proton must be removed from the already negatively charged species $HA^-$. Removing a positively charged proton ($H^+$) from a negatively charged ion is energetically more difficult because of the stronger electrostatic attraction between the negative ion and the proton. Additionally, the negative charge on $HA^-$ makes it harder to lose another proton.

Marking:

• 1 mark: Second dissociation is from a negatively charged species
• 1 mark: Electrostatic attraction makes removal of H⁺ harder / energetically less favourable

(b) [3]

Considering only the first dissociation:

$H_2A \rightleftharpoons H^+ + HA^- \quad K_{a1} = 4.50 \times 10^{-3}$

$K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} = \frac{x^2}{0.050 - x} = 4.50 \times 10^{-3}$

Since $K_{a1}$ is not very small compared to $c$, we should solve the quadratic:

$x^2 = 4.50 \times 10^{-3}(0.050 - x)$

$x^2 = 2.25 \times 10^{-4} - 4.50 \times 10^{-3}x$

$x^2 + 4.50 \times 10^{-3}x - 2.25 \times 10^{-4} = 0$

Using the quadratic formula:

$x = \frac{-4.50 \times 10^{-3} + \sqrt{(4.50 \times 10^{-3})^2 + 4 \times 2.25 \times 10^{-4}}}{2}$

$x = \frac{-4.50 \times 10^{-3} + \sqrt{2.025 \times 10^{-5} + 9.00 \times 10^{-4}}}{2}$

$x = \frac{-4.50 \times 10^{-3} + \sqrt{9.2025 \times 10^{-4}}}{2}$

$x = \frac{-4.50 \times 10^{-3} + 3.034 \times 10^{-2}}{2} = \frac{2.584 \times 10^{-2}}{2} = 1.29 \times 10^{-2}$

$[H^+] = 1.29 \times 10^{-2} \text{ mol dm}^{-3}$

$pH = -\log_{10}(1.29 \times 10^{-2}) = 1.89$

Marking:

• 1 mark for setting up the $K_{a1}$ expression correctly
• 1 mark for solving the quadratic (or valid attempt)
• 1 mark for correct pH = 1.89

(c) [2]

For the second dissociation:

$HA^- \rightleftharpoons H^+ + A^{2-} \quad K_{a2} = 6.20 \times 10^{-8}$

Since $K_{a2} \ll K_{a1}$, the second dissociation contributes negligibly to $[H^+]$, so:

$[H^+] \approx [HA^-] \approx 1.29 \times 10^{-2} \text{ mol dm}^{-3}$

$K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$

$6.20 \times 10^{-8} = \frac{(1.29 \times 10^{-2})[A^{2-}]}{1.29 \times 10^{-2}}$

$[A^{2-}] = K_{a2} = 6.20 \times 10^{-8} \text{ mol dm}^{-3}$

Marking:

• 1 mark for recognising that $[H^+] \approx [HA^-]$ from first dissociation
• 1 mark for $[A^{2-}] = K_{a2} = 6.20 \times 10^{-8}$ mol dm⁻³

Teaching note: For diprotic acids where $K_{a1} \gg K_{a2}$, the concentration of the dianion $A^{2-}$ is approximately equal to $K_{a2}$. This is a useful result that students should remember.

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End of Answer Key

Total: 60 marks