A Level H2 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H2 Quiz - Acids Bases Salts: Answer Key

Section A: Multiple Choice & Short Answer (Questions 1–8)

1. Answer: B — $SO_4^{2-}$

Explanation: A conjugate base is formed when an acid donates a proton ($H^+$). $HSO_4^-$ acts as an acid by losing one $H^+$ to become $SO_4^{2-}$. Option A ($H_2SO_4$) is the conjugate acid (gaining a proton), not the conjugate base. Option C is the conjugate acid of water. Option D is unrelated.

[1 mark]

2. Answer: C — $2.51 \times 10^{-11}$ mol dm$^{-3}$

Explanation:

• $pH = 3.40$, so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$
• At 25 °C, $K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$
• $[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm$^{-3}$

Common mistake: Students often select option A, which is the $[H^+]$ value, not $[OH^-]$. Always check what the question is asking for.

[1 mark]

3. Answer: 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:

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

This produces $H_3O^+$ ions, making the solution acidic. $Na_2CO_3$ and $CH_3COONa$ produce alkaline solutions (salts of strong base + weak acid). $KNO_3$ is neutral (strong acid + strong base).

[1 mark]

4. Answer: C — 3.85

Explanation:

• For a weak acid: $K_a = \frac{[H^+]^2}{[HA]}$ (assuming $[H^+] = [A^-]$ and dissociation is small)
• $[H^+] = \sqrt{K_a \times [HA]} = \sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.47 \times 10^{-4}$ mol dm$^{-3}$
• $pH = -\log(4.47 \times 10^{-4}) = 3.35$

Wait — let me recalculate:

• $[H^+] = \sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.472 \times 10^{-4}$
• $pH = -\log(4.472 \times 10^{-4}) = 3.35$

Hmm, that gives 3.35 which is option B. Let me recheck: $\sqrt{2.0 \times 10^{-7}} = \sqrt{20 \times 10^{-8}} = 4.472 \times 10^{-4}$. $-\log(4.472 \times 10^{-4}) = 4 - \log(4.472) = 4 - 0.650 = 3.35$.

Corrected Answer: B — 3.35

Explanation:

• For a weak acid $HA$: $K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+]^2}{[HA]_0}$ (since dissociation is small)
• $[H^+] = \sqrt{K_a \times [HA]_0} = \sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.47 \times 10^{-4}$ mol dm$^{-3}$
• $pH = -\log(4.47 \times 10^{-4}) = 3.35$

Common mistake: Using $[HA] = 0.050$ directly in $K_a = [H^+]^2/[HA]$ without the square root, or confusing $K_a$ with $pK_a$.

[1 mark]

5. Answer: C — The pH at the equivalence point is greater than 7.0.

Explanation: $CH_3COOH$ is a weak acid and $NaOH$ is a strong base. At the equivalence point, all the $CH_3COOH$ has been converted to $CH_3COO^-$ (the conjugate base), which hydrolyses in water to produce $OH^-$ ions:

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

This makes the solution slightly alkaline, so pH > 7. The moles of $NaOH$ required equal the moles of $CH_3COOH$ (1:1 stoichiometry), so D is incorrect.

[1 mark]

6. Answer: 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).

Teaching note: A buffer typically consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). The key feature is its ability to maintain a relatively constant pH. Students should mention both the resistance to pH change and the small quantity of acid/base added.

[2 marks] — 1 mark for "resists change in pH", 1 mark for "when small amounts of acid or base are added"

7. Answer: Hydrochloric acid is a strong acid and dissociates completely in water, so a 0.10 mol dm$^{-3}$ solution gives $[H^+] = 0.10$ mol dm$^{-3}$ and pH = 1.0. Ethanoic acid is a weak acid and only partially dissociates. For 0.10 mol dm$^{-3}$ $CH_3COOH$:

$[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.8 \times 10^{-5} \times 0.10} = 1.34 \times 10^{-3}$ mol dm$^{-3}$

$pH = -\log(1.34 \times 10^{-3}) = 2.87$

Since $[H^+]$ is much lower for ethanoic acid, its pH is higher.

Teaching note: The key distinction is complete dissociation (strong acid) vs. partial dissociation (weak acid). At the same concentration, a strong acid always has a lower pH.

[2 marks] — 1 mark for identifying HCl as strong (complete dissociation) and CH₃COOH as weak (partial dissociation), 1 mark for calculation or comparison showing higher [H⁺] for HCl

8. Answer:

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

Units: mol$^3$ dm$^{-9}$

Teaching note: $PbI_2$ dissociates as $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$. The $K_{sp}$ expression raises each ion concentration to the power of its stoichiometric coefficient. Since the units are (mol dm$^{-3}$)(mol dm$^{-3}$)$^2$ = mol$^3$ dm$^{-9}$.

[2 marks] — 1 mark for correct expression, 1 mark for correct units

Section B: Structured Questions (Questions 9–15)

9.

(a) Answer: Titration 3 (35.10 → 10.50 = 24.60 cm$^3$) is not anomalous. All three accurate titres (24.30, 24.60, 24.40) are concordant (within 0.20 cm$^3$ of each other). There is no anomalous result.

Wait — let me re-examine. Titration 1: 24.30, Titration 2: 24.60, Titration 3: 24.40. The range is 24.60 − 24.30 = 0.30 cm$^3$. This is slightly more than 0.20 cm$^3$. However, in many exam mark schemes, titres within 0.30 cm$^3$ are still considered acceptable. Titration 2 (24.60) could be considered slightly anomalous as it is 0.30 cm$^3$ from titration 1.

Revised Answer: Titration 2 (24.60 cm$^3$) could be considered anomalous as it differs by 0.30 cm$^3$ from the closest titre (24.30 cm$^3$), which exceeds the usual 0.20 cm$^3$ concordancy criterion.

Actually, let me reconsider the data more carefully:

• Titration 1: 24.30 cm$^3$
• Titration 2: 24.60 cm$^3$ (difference from T1 = 0.30)
• Titration 3: 24.40 cm$^3$ (difference from T1 = 0.10, from T2 = 0.20)

Titration 2 is 0.30 cm$^3$ from Titration 1, which is outside the 0.20 cm$^3$ concordancy range. So Titration 2 is anomalous.

Answer: Titration 2 (24.60 cm$^3$) is anomalous because it differs by more than 0.20 cm$^3$ from the closest titre (24.30 cm$^3$).

[1 mark]

(b) Answer: Using concordant titres (Titrations 1 and 3):

$\text{Mean volume} = \frac{24.30 + 24.40}{2} = 24.35 \text{ cm}^3$

[1 mark]

(c) Answer:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$

[1 mark]

(d) Answer:

• Moles of $NaOH$ used = $0.100 \times \frac{24.35}{1000} = 2.435 \times 10^{-3}$ mol
• 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
• Concentration of $H_2SO_4 = \frac{1.2175 \times 10^{-3}}{25.0/1000} = \frac{1.2175 \times 10^{-3}}{0.0250} = 0.0487$ mol dm$^{-3}$

[3 marks] — 1 mark for moles of NaOH, 1 mark for mole ratio and moles of H₂SO₄, 1 mark for final concentration

10.

(a) Answer:

• After mixing, total volume = 100.0 cm$^3$
• $[CH_3COOH] = \frac{0.200 \times 50.0}{100.0} = 0.100$ mol dm$^{-3}$
• $[CH_3COO^-] = \frac{0.300 \times 50.0}{100.0} = 0.150$ mol dm$^{-3}$
• Using Henderson-Hasselbalch equation:

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

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

$pH = 4.74 + \log\frac{0.150}{0.100} = 4.74 + \log(1.50) = 4.74 + 0.18 = 4.92$

[3 marks] — 1 mark for calculating diluted concentrations, 1 mark for correct pKa, 1 mark for correct pH

(b) Answer: When $HCl$ is added, the $H^+$ ions react with the conjugate base $CH_3COO^-$ in the buffer:

$CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$

By Le Chatelier's principle, the equilibrium $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$ shifts to the left to consume the added $H^+$. The $CH_3COO^-$ ions act as a "sink" for the added $H^+$, converting them into undissociated $CH_3COOH$. Since most of the added $H^+$ is consumed, the change in $[H^+]$ and hence pH is very small.

[3 marks] — 1 mark for equation showing H⁺ reacting with CH₃COO⁻, 1 mark for Le Chatelier's principle / equilibrium shift, 1 mark for explaining minimal pH change

(c) Answer: The pH would increase slightly. The $OH^-$ ions from $NaOH$ react with the $CH_3COOH$ component of the buffer:

$CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$

This converts $CH_3COOH$ into $CH_3COO^-$, increasing the ratio $[CH_3COO^-]/[CH_3COOH]$, which increases the pH slightly. However, the buffer minimises the pH change.

[2 marks] — 1 mark for stating pH increases, 1 mark for correct explanation involving reaction with CH₃COOH

11.

(a) Answer:

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

[1 mark]

(b) Answer:

• Let the solubility of $Mg(OH)_2 = s$ mol dm$^{-3}$
• $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
• $[Mg^{2+}] = s$, $[OH^-] = 2s$
• $K_{sp} = s \times (2s)^2 = 4s^3$
• $4s^3 = 1.8 \times 10^{-11}$
• $s^3 = 4.5 \times 10^{-12}$
• $s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4}$ mol dm$^{-3}$

[3 marks] — 1 mark for correct Ksp substitution with 2s, 1 mark for algebraic manipulation, 1 mark for correct answer

(c) Answer:

• $[OH^-] = 2s = 2 \times 1.65 \times 10^{-4} = 3.30 \times 10^{-4}$ mol dm$^{-3}$
• $pOH = -\log(3.30 \times 10^{-4}) = 3.48$
• $pH = 14.00 - 3.48 = 10.52$

[3 marks] — 1 mark for [OH⁻] = 2s, 1 mark for pOH calculation, 1 mark for pH

12.

(a) Answer: Hydrochloric acid gives the highest conductivity reading (4.20 mA). $HCl$ is a strong acid and dissociates completely in water, producing the highest concentration of ions ($H^+$ and $Cl^-$) available to carry current.

[2 marks] — 1 mark for identifying HCl, 1 mark for explanation involving complete dissociation / highest ion concentration

(b) Answer: Citric acid is a weak acid with the smallest $K_a$ value among the three acids. It dissociates only partially in water, producing the lowest concentration of ions. Since electrical conductivity depends on the concentration of ions in solution, the citric acid solution has the lowest conductivity.

[2 marks] — 1 mark for identifying citric acid as weakest/least dissociated, 1 mark for linking to lowest ion concentration and conductivity

(c) Answer: Although both are acids, $HCl$ is a strong acid that dissociates completely, while $H_3PO_4$ is a weak acid that only partially dissociates. Therefore, the concentration of $H^+$ ions (and hence total ion concentration) is lower in the phosphoric acid solution, resulting in lower conductivity.

[2 marks] — 1 mark for identifying HCl as strong and H₃PO₄ as weak, 1 mark for linking to ion concentration and conductivity

13.

(a) Answer:

• $C_2H_5COOH(aq) \rightleftharpoons C_2H_5COO^-(aq) + H^+(aq)$
• $K_a = \frac{[H^+]^2}{[HA]_0} = 1.3 \times 10^{-5}$
• $[H^+] = \sqrt{1.3 \times 10^{-5} \times 0.200} = \sqrt{2.6 \times 10^{-6}} = 1.61 \times 10^{-3}$ mol dm$^{-3}$
• $pH = -\log(1.61 \times 10^{-3}) = 2.79$

[3 marks] — 1 mark for Ka expression, 1 mark for [H⁺] calculation, 1 mark for pH

(b) Answer:

• At equivalence point, all propanoic acid is converted to sodium propanoate ($C_2H_5COONa$).
• Moles of $C_2H_5COOH = 0.200 \times 0.0250 = 5.0 \times 10^{-3}$ mol
• Volume of $NaOH$ needed = $\frac{5.0 \times 10^{-3}}{0.200} = 0.0250$ dm$^3$ = 25.0 cm$^3$
• Total volume = 25.0 + 25.0 = 50.0 cm$^3$
• $[C_2H_5COO^-] = \frac{5.0 \times 10^{-3}}{0.0500} = 0.100$ mol dm$^{-3}$

The propanoate ion hydrolyses: $C_2H_5COO^-(aq) + H_2O(l) \rightleftharpoons C_2H_5COOH(aq) + OH^-(aq)$

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.69 \times 10^{-10}$

$[OH^-] = \sqrt{K_b \times [C_2H_5COO^-]} = \sqrt{7.69 \times 10^{-10} \times 0.100} = \sqrt{7.69 \times 10^{-11}} = 8.77 \times 10^{-6} \text{ mol dm}^{-3}$

$pOH = -\log(8.77 \times 10^{-6}) = 5.06$

$pH = 14.00 - 5.06 = 8.94$

[4 marks] — 1 mark for calculating concentration of salt at equivalence point, 1 mark for Kb calculation, 1 mark for [OH⁻] calculation, 1 mark for pH

(c) Answer: The titration curve should show:

• Starting pH ≈ 2.79 (weak acid)
• A gradually rising curve with a buffer region (relatively flat section) centred around half-equivalence point (12.5 cm$^3$ NaOH) where pH = $pK_a$ = 4.89
• A steep rise near the equivalence point (25.0 cm$^3$)
• Equivalence point at pH ≈ 8.94 (> 7, since salt of weak acid + strong base)
• Curve levels off at high pH after equivalence point

[2 marks] — 1 mark for correct shape with buffer region and steep rise, 1 mark for correct labelling of key points (initial pH, equivalence point pH > 7, buffer region)

14.

(a) Answer: $HCl$ is a strong acid and dissociates completely, giving $[H^+] = 0.10$ mol dm$^{-3}$. Ethanoic acid is a weak acid with $K_a = 1.8 \times 10^{-5}$. In the presence of $HCl$, the high $[H^+]$ from $HCl$ suppresses the dissociation of $CH_3COOH$ (common ion effect / Le Chatelier's principle). The contribution of $H^+$ from $CH_3COOH$ is negligible compared to 0.10 mol dm$^{-3}$ from $HCl$.

[2 marks] — 1 mark for identifying complete dissociation of HCl giving [H⁺] = 0.10, 1 mark for common ion effect suppressing CH₃COOH dissociation

(b) Answer:

• $[H^+] \approx 0.10$ mol dm$^{-3}$ (from $HCl$ only; contribution from $CH_3COOH$ is negligible)
• $pH = -\log(0.10) = 1.00$

[2 marks] — 1 mark for stating [H⁺] ≈ 0.10, 1 mark for pH = 1.00

(c) Answer: The dissociation equilibrium of ethanoic acid is:

$CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$

$HCl$ dissociates completely, increasing $[H^+]$. By Le Chatelier's principle, the equilibrium shifts to the left to counteract the increase in $[H^+]$. This suppresses the dissociation of $CH_3COOH$, so very little $CH_3COO^-$ and additional $H^+$ is produced.

[2 marks] — 1 mark for identifying the equilibrium, 1 mark for stating shift to the left with Le Chatelier's explanation

15.

(a) Answer:

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

[1 mark]

(b) Answer: Order of increasing solubility: Z < Y < X

Explanation: The solubility of $AgCl$ is reduced in solutions containing a common ion (either $Ag^+$ or $Cl^-$) due to the common ion effect.

• Solution Z (0.10 mol dm$^{-3}$ $AgNO_3$): Contains $Ag^+$ ions — common ion suppresses dissolution. Lowest solubility.
• Solution Y (0.10 mol dm$^{-3}$ $NaCl$): Contains $Cl^-$ ions — common ion suppresses dissolution. Intermediate solubility.
• Solution X (pure water): No common ions. Highest solubility.

Both Y and Z have the same concentration of common ion (0.10 mol dm$^{-3}$), so their solubilities would be similar but not identical (depending on which ion's concentration is held constant in the Ksp expression). However, since $K_{sp} = [Ag^+][Cl^-]$ and both common ion concentrations are equal, the solubilities are the same. The expected answer is Z ≈ Y < X.

Revised: Since both NaCl and AgNO₃ provide the same concentration (0.10 mol dm⁻³) of common ion, the solubility of AgCl in Y and Z would be approximately equal. The order is: Z ≈ Y < X (or Z = Y < X).

[3 marks] — 1 mark for correct order, 1 mark for identifying common ion effect, 1 mark for explaining that pure water has no common ion

(c) Answer:

• Let solubility of $AgCl = s$ mol dm$^{-3}$
• $K_{sp} = s^2 = 1.8 \times 10^{-10}$
• $s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ mol dm$^{-3}$
• Solubility in g dm$^{-3} = 1.34 \times 10^{-5} \times 143.4 = 1.92 \times 10^{-3}$ g dm$^{-3}$

[3 marks] — 1 mark for s² = Ksp, 1 mark for s value, 1 mark for conversion to g dm⁻³

Section C: Data-Based & Application Questions (Questions 16–20)

16.

(a) Answer: As pH increases, $[H^+]$ decreases. By Le Chatelier's principle, the equilibrium:

$HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$

shifts to the right to produce more $H^+$ (counteracting the decrease). This increases the concentration of $In^-$ (colour B), causing the colour to change from colour A to colour B.

[2 marks] — 1 mark for decrease in [H⁺] shifting equilibrium right, 1 mark for linking to colour change

(b) Answer:

• $pK_{In} = -\log(5.0 \times 10^{-7}) = 6.30$
• pH range = $6.30 \pm 1$ = 5.30 to 7.30

[2 marks] — 1 mark for pKIn calculation, 1 mark for correct range

(c) Answer: In a strong acid–strong base titration, the equivalence point is at pH 7.0, and there is a steep pH jump from approximately pH 4 to pH 10. The indicator's range (5.30–7.30) falls within this steep region, so the colour change accurately signals the equivalence point.

In a weak acid–strong base titration, the equivalence point is at pH > 7 (typically pH 8–9). The steep portion of the titration curve occurs at higher pH values. The indicator's range (5.30–7.30) falls before the equivalence point, so the colour change would occur too early, leading to a significant titration error.

[2 marks] — 1 mark for explaining suitability for strong-strong (range within steep region at pH 7), 1 mark for explaining unsuitability for weak-strong (range below equivalence point pH)

17.

(a) Answer:

• $[H^+] = 10^{-4.2} = 6.31 \times 10^{-5}$ mol dm$^{-3}$

[1 mark]

(b) Answer:

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

(Alternatively: $SO_2 + H_2O \rightarrow H_2SO_3$, then $2H_2SO_3 + O_2 \rightarrow 2H_2SO_4$)

[1 mark]

(c) Answer: Sulfuric acid is diprotic. The first dissociation is complete ($H_2SO_4 \rightarrow H^+ + HSO_4^-$), but the second dissociation ($HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$) is not complete ($K_{a2} = 1.2 \times 10^{-2}$). Therefore, not all $H_2SO_4$ molecules release both protons, so the $[H^+]$ is less than twice the $H_2SO_4$ concentration, resulting in a pH that is higher than expected if both dissociations were complete.

[2 marks] — 1 mark for identifying incomplete second dissociation, 1 mark for explaining effect on [H⁺]

(d) Answer:

• Environmental effect: Acid rain damages aquatic ecosystems by lowering the pH of lakes and rivers, killing fish and other aquatic organisms. It also corrodes buildings and statues made of limestone/marble, and damages soil by leaching essential nutrients.
• Method to reduce: Use flue gas desulfurisation (scrubbers) in power plants to remove $SO_2$ from exhaust gases before release. Alternatively, use catalytic converters, switch to low-sulfur fuels, or use renewable energy sources.

[2 marks] — 1 mark for environmental effect, 1 mark for reduction method

18.

(a) Answer:

• Using Henderson-Hasselbalch: $pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$
• $5.00 = 4.74 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$
• $\log\frac{[CH_3COO^-]}{[CH_3COOH]} = 5.00 - 4.74 = 0.26$
• $\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.26} = 1.82$

[3 marks] — 1 mark for correct substitution, 1 mark for log calculation, 1 mark for ratio

(b) Answer:

• Moles of $CH_3COOH = 0.500 \times 0.100 = 0.0500$ mol
• From part (a): $\frac{[CH_3COO^-]}{[CH_3COOH]} = 1.82$
• Since both are in the same volume: $\frac{n_{CH_3COO^-}}{n_{CH_3COOH}} = 1.82$
• $n_{CH_3COO^-} = 1.82 \times 0.0500 = 0.0910$ mol
• Mass of $CH_3COONa = 0.0910 \times 82.0 = 7.46$ g

[3 marks] — 1 mark for moles of CH₃COOH, 1 mark for using ratio to find moles of CH₃COONa, 1 mark for mass calculation

19.

(a) Answer: $Na_2SO_4$ is formed from a strong base ($NaOH$) and a strong acid ($H_2SO_4$). Neither the $Na^+$ cation nor the $SO_4^{2-}$ anion undergoes hydrolysis in water, so the solution remains neutral (pH = 7.0).

[1 mark]

(b) Answer: $Na_2CO_3$ is formed from a strong base ($NaOH$) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion is the conjugate base of the weak acid $HCO_3^-$ and undergoes hydrolysis:

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

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

[2 marks] — 1 mark for identifying salt of strong base + weak acid, 1 mark for hydrolysis equation producing OH⁻

(c) Answer: $AlCl_3$ is formed from a weak base ($Al(OH)_3$) and a strong acid ($HCl$). The $Al^{3+}$ ion is a small, highly charged cation that acts as a Lewis acid and undergoes hydrolysis:

$[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)$

The high charge density of $Al^{3+}$ polarises the O–H bonds in the coordinated water molecules, making it easier for a proton to be released. This produces $H_3O^+$ ions, making the solution acidic.

[2 marks] — 1 mark for identifying salt of weak base + strong acid, 1 mark for hydrolysis equation producing H₃O⁺

(d) Answer: The solution would be neutral. Ammonium ethanoate is a salt of a weak acid ($CH_3COOH$, $K_a = 1.8 \times 10^{-5}$) and a weak base ($NH_3$, $K_b = 1.8 \times 10^{-5}$). Since $K_a = K_b$, the extent of hydrolysis of $CH_3COO^-$ (producing $OH^-$) is equal to the extent of hydrolysis of $NH_4^+$ (producing $H^+$). The $[H^+]$ and $[OH^-]$ produced are equal, so the solution is neutral (pH ≈ 7).

[2 marks] — 1 mark for stating neutral, 1 mark for explaining that Ka = Kb so hydrolysis is equal

20.

(a) Answer:

• Moles of $OH^- = 0.150 \times \frac{30.0}{1000} = 4.50 \times 10^{-3}$ mol
• Moles of $H^+ = 2 \times 0.100 \times \frac{20.0}{1000} = 4.00 \times 10^{-3}$ mol (Note: $H_2SO_4$ provides 2 moles of $H^+$ per mole of acid)

[2 marks] — 1 mark for OH⁻ moles, 1 mark for H⁺ moles (including factor of 2)

(b) Answer: The solution is basic.

$H^+ + OH^- \rightarrow H_2O$

Moles of $OH^-$ remaining = $4.50 \times 10^{-3} - 4.00 \times 10^{-3} = 0.50 \times 10^{-3}$ mol

Since excess $OH^-$ remains after neutralisation, the solution is basic.

[2 marks] — 1 mark for identifying excess OH⁻, 1 mark for stating basic

(c) Answer:

• Total volume = 30.0 + 20.0 = 50.0 cm$^3$ = 0.0500 dm$^3$
• $[OH^-] = \frac{0.50 \times 10^{-3}}{0.0500} = 0.010$ mol dm$^{-3}$
• $pOH = -\log(0.010) = 2.00$
• $pH = 14.00 - 2.00 = 12.00$

[2 marks] — 1 mark for [OH⁻] calculation, 1 mark for pH

(d) Answer:

• Additional moles of $H^+ = 2 \times 0.100 \times \frac{10.0}{1000} = 2.00 \times 10^{-3}$ mol
• Moles of $OH^-$ before addition = $0.50 \times 10^{-3}$ mol
• After reaction: moles of $H^+$ remaining = $2.00 \times 10^{-3} - 0.50 \times 10^{-3} = 1.50 \times 10^{-3}$ mol
• New total volume = 50.0 + 10.0 = 60.0 cm$^3$ = 0.0600 dm$^3$
• $[H^+] = \frac{1.50 \times 10^{-3}}{0.0600} = 0.0250$ mol dm$^{-3}$
• $pH = -\log(0.0250) = 1.60$

[3 marks] — 1 mark for additional H⁺ moles, 1 mark for excess H⁺ and new volume, 1 mark for pH

End of Answer Key

Total: 60 marks