A Level H1 Chemistry Practice Paper 2

A-Level Chemistry H1 Quiz - Acids Bases Salts

Answer Key

Section A: Multiple Choice Questions (10 marks)

1. Answer: B
[1 mark]
A weak acid is defined as an acid that partially dissociates in aqueous solution. Option A confuses concentration with strength. Option C is incorrect because weak acids do produce $\text{H}^+$ ions, just not completely. Option D is incorrect because acids have pH less than 7.

2. Answer: B
[1 mark]
For a strong monoprotic acid, $[\text{H}^+] = 0.050 \text{ mol dm}^{-3}$.
$\text{pH} = -\log_{10}(0.050) = -\log_{10}(5.0 \times 10^{-2}) = 2 - \log_{10}5 = 2 - 0.70 = 1.30$
Common mistake: Students may incorrectly calculate $\text{pH} = -\log(0.05) = 1.3$ but choose option A (1.0) by rounding incorrectly.

3. Answer: B
[1 mark]
$\text{NH}_4\text{Cl}$ is a salt of a weak base ($\text{NH}_3$) and a strong acid ($\text{HCl}$). The $\text{NH}_4^+$ ion undergoes hydrolysis to produce $\text{H}^+$ ions, making the solution acidic. $\text{Na}_2\text{CO}_3$ and $\text{CH}_3\text{COONa}$ produce basic solutions (salts of strong base + weak acid). $\text{KNO}_3$ is neutral (strong acid + strong base).

4. Answer: B
[1 mark]
In pure water, $[\text{H}^+] = [\text{OH}^-]$. Since $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$, we have $[\text{H}^+]^2 = 1.0 \times 10^{-14}$, so $[\text{H}^+] = 1.0 \times 10^{-7} \text{ mol dm}^{-3}$.
$\text{pH} = -\log_{10}(1.0 \times 10^{-7}) = 7$

5. Answer: A
[1 mark]
When equal volumes of equimolar weak acid and its conjugate base are mixed, the concentrations of acid and conjugate base become equal. Using the Henderson–Hasselbalch equation: $\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = \text{p}K_a + \log(1) = \text{p}K_a$. The pH is less than 7 (ethanoic acid has $\text{p}K_a \approx 4.77$). The buffer CAN resist pH changes. The concentrations are equal, not unequal.

6. Answer: B
[1 mark]
At the equivalence point of a strong acid–strong base titration, the salt formed is neutral (e.g., $\text{NaCl}$), so the pH is 7. The indicator does not determine the pH at the equivalence point; it only signals when the endpoint is reached.

7. Answer: B
[1 mark]
$\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(1.6 \times 10^{-4}) = 4 - \log_{10}1.6 = 4 - 0.20 = 3.80$

8. Answer: C
[1 mark]
A buffer does not necessarily maintain a pH of 7. The pH of a buffer depends on the $\text{p}K_a$ of the weak acid and the ratio of conjugate base to acid. Buffers can be acidic (pH < 7) or basic (pH > 7). The other options are all correct properties of buffers.

9. Answer: B
[1 mark]
Moles of $\text{HCl} = 0.0250 \times 0.100 = 0.00250 \text{ mol}$
Since the reaction is 1:1 ($\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$), moles of $\text{NaOH} = 0.00250 \text{ mol}$
Volume of $\text{NaOH} = \frac{0.00250}{0.100} = 0.0250 \text{ dm}^3 = 25.0 \text{ cm}^3$

10. Answer: C
[1 mark]
For a weak acid: $[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{4.0 \times 10^{-6} \times 0.010} = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(2.0 \times 10^{-4}) = 4 - \log_{10}2 = 4 - 0.30 = 3.70$

Section B: Structured Questions (30 marks)

11. (a) A Brønsted–Lowry acid is a proton ($\text{H}^+$) donor.
[1 mark]
Marking note: Must mention "proton" or "H⁺" and "donor" or "donates".

(b) Acid: $\text{H}_2\text{O}$
Base: $\text{NH}_3$
[2 marks – 1 mark each]
Water donates a proton to ammonia, forming $\text{NH}_4^+$ and $\text{OH}^-$. Therefore, water acts as the acid and ammonia acts as the base.

(c) Water is amphoteric because it can act as both an acid (proton donor) and a base (proton acceptor).
Acting as an acid: $\text{H}_2\text{O} + \text{NH}_3 \rightleftharpoons \text{OH}^- + \text{NH}_4^+$
Acting as a base: $\text{H}_2\text{O} + \text{HCl} \rightleftharpoons \text{H}_3\text{O}^+ + \text{Cl}^-$
(or $\text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_3\text{O}^+ + \text{HSO}_4^-$)
[3 marks – 1 mark for definition, 1 mark for each equation]
Marking note: The definition must state that water can act as both acid and base. Equations must show water donating H⁺ (acid) and accepting H⁺ (base).

12. (a) Molar mass of $\text{HCl} = 1.0 + 35.5 = 36.5 \text{ g mol}^{-1}$
Moles of $\text{HCl} = \frac{0.365}{36.5} = 0.0100 \text{ mol}$
Concentration = $\frac{0.0100}{0.250} = 0.0400 \text{ mol dm}^{-3}$
[2 marks – 1 mark for moles, 1 mark for concentration]

(b) $\text{HCl}$ is a strong acid, so $[\text{H}^+] = 0.0400 \text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(0.0400) = 1.40$
[1 mark]

(c) On dilution by a factor of 10, the new concentration = $\frac{0.0400}{10} = 0.00400 \text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(0.00400) = 2.40$
[2 marks – 1 mark for new concentration, 1 mark for pH]
Teaching note: Diluting a strong acid by a factor of 10 increases the pH by 1 unit.

13. (a) $K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$
[1 mark]
Marking note: Must be an expression with correct species. State symbols are not required in the $K_a$ expression.

(b) For a weak acid: $[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.7 \times 10^{-5} \times 0.10} = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(1.30 \times 10^{-3}) = 2.89$
[3 marks – 1 mark for [H⁺] expression/substitution, 1 mark for [H⁺] value, 1 mark for pH]
Common mistake: Students may use $[\text{H}^+] = K_a \times c$ instead of $\sqrt{K_a \times c}$. The approximation is valid when $c \gg K_a$ (here $0.10 \gg 1.7 \times 10^{-5}$, so the approximation is valid).

(c) After mixing equal volumes, concentrations are halved:
$[\text{CH}_3\text{COOH}] = \frac{0.10}{2} = 0.050 \text{ mol dm}^{-3}$
$[\text{CH}_3\text{COO}^-] = \frac{0.20}{2} = 0.10 \text{ mol dm}^{-3}$
Using Henderson–Hasselbalch:
$\text{pH} = \text{p}K_a + \log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = -\log(1.7 \times 10^{-5}) + \log\frac{0.10}{0.050}$
$= 4.77 + \log(2.0) = 4.77 + 0.30 = 5.07$
[3 marks – 1 mark for calculating diluted concentrations, 1 mark for pKₐ or Kₐ expression, 1 mark for final pH]
Alternative method using $K_a$: $[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.7 \times 10^{-5} \times \frac{0.050}{0.10} = 8.5 \times 10^{-6}$; $\text{pH} = -\log(8.5 \times 10^{-6}) = 5.07$

14. (a) pH at equivalence point ≈ 8.8 (accept 8.5–9.0)
[1 mark]

(b) At the equivalence point, all the ethanoic acid has reacted with $\text{NaOH}$ to form sodium ethanoate ($\text{CH}_3\text{COONa}$). The ethanoate ion ($\text{CH}_3\text{COO}^-$) is the conjugate base of a weak acid and undergoes hydrolysis:
$\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$
This produces $\text{OH}^-$ ions, making the solution slightly alkaline (pH > 7).
[2 marks – 1 mark for identifying the salt formed, 1 mark for explaining hydrolysis producing OH⁻]

(c) Phenolphthalein. The equivalence point occurs at pH ≈ 8.8, which falls within the pH range of phenolphthalein (8.2–10.0).
[2 marks – 1 mark for indicator, 1 mark for reason]
Note: Methyl orange (3.1–4.4) would not be suitable as its colour change occurs well before the equivalence point.

(d) $\text{pH} = 8.8$, so $[\text{H}^+] = 10^{-8.8} = 1.58 \times 10^{-9} \text{ mol dm}^{-3}$
$[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{1.58 \times 10^{-9}} = 6.3 \times 10^{-6} \text{ mol dm}^{-3}$
[1 mark]
Alternative: $\text{pOH} = 14 - 8.8 = 5.2$; $[\text{OH}^-] = 10^{-5.2} = 6.3 \times 10^{-6} \text{ mol dm}^{-3}$

15. (a) A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It typically contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in approximately equal concentrations.
[2 marks – 1 mark for resistance to pH change, 1 mark for composition]

(b) (i) Using the Henderson–Hasselbalch equation:
$\text{p}K_a = -\log(6.3 \times 10^{-5}) = 4.20$
$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = 4.20 + \log\frac{0.30}{0.20} = 4.20 + \log(1.5) = 4.20 + 0.18 = 4.38$
[2 marks – 1 mark for pKₐ, 1 mark for pH]

(ii) When $\text{HCl}$ is added, the $\text{H}^+$ ions react with the conjugate base $\text{A}^-$:
$\text{A}^- + \text{H}^+ \rightarrow \text{HA}$
This removes the added $\text{H}^+$ ions by converting them into the weak acid HA, which only partially dissociates. The ratio $[\text{A}^-]/[\text{HA}]$ changes only slightly, so the pH remains relatively constant.
[2 marks – 1 mark for equation, 1 mark for explanation]
Teaching note: The buffer works because the added H⁺ is consumed by A⁻, and the resulting HA does not significantly increase [H⁺] because it is a weak acid.

Section C: Data-Based and Application Questions (20 marks)

16. (a) Solution P: Acidic
Solution Q: Neutral
Solution R: Alkaline
[3 marks – 1 mark each]

(b) $\text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$
(or $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$)
[1 mark]

(c) A strong acid is an acid that completely dissociates in aqueous solution.
Equation: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$
(or $\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$ with a single arrow)
[2 marks – 1 mark for definition, 1 mark for equation with single arrow]
Key distinction: Strong acids use a single arrow (→) showing complete dissociation; weak acids use an equilibrium arrow (⇌) showing partial dissociation.

(d) Solution Q has a pH of 7.0 and very low electrical conductivity, indicating it is pure water. Pure water has very few ions ($\text{H}^+$ and $\text{OH}^-$ at $10^{-7} \text{ mol dm}^{-3}$ each), so it conducts electricity very poorly.
[1 mark]

17. (a) Ethanoic acid < Benzoic acid < Methanoic acid
[1 mark]

(b) A larger $K_a$ value indicates a greater degree of dissociation, meaning the acid is stronger. Methanoic acid has the largest $K_a$ ($1.6 \times 10^{-4}$), so it is the strongest. Ethanoic acid has the smallest $K_a$ ($1.7 \times 10^{-5}$), so it is the weakest.
[1 mark]

(c) $[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.6 \times 10^{-4} \times 0.050} = \sqrt{8.0 \times 10^{-6}} = 2.83 \times 10^{-3} \text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(2.83 \times 10^{-3}) = 2.55$
[2 marks – 1 mark for [H⁺], 1 mark for pH]

(d) $\text{pH} = 2.50$, so $[\text{H}^+] = 10^{-2.50} = 3.16 \times 10^{-3} \text{ mol dm}^{-3}$
For a weak acid: $K_a = \frac{[\text{H}^+]^2}{c - [\text{H}^+]}$ (using the exact expression)
$1.6 \times 10^{-4} = \frac{(3.16 \times 10^{-3})^2}{c - 3.16 \times 10^{-3}}$
$c - 3.16 \times 10^{-3} = \frac{1.0 \times 10^{-5}}{1.6 \times 10^{-4}} = 0.0625$
$c = 0.0625 + 0.00316 = 0.0657 \text{ mol dm}^{-3}$
Alternatively, using the approximation $[\text{H}^+] = \sqrt{K_a \times c}$:
$c = \frac{[\text{H}^+]^2}{K_a} = \frac{(3.16 \times 10^{-3})^2}{1.6 \times 10^{-4}} = \frac{1.0 \times 10^{-5}}{1.6 \times 10^{-4}} = 0.0625 \text{ mol dm}^{-3}$
[3 marks – 1 mark for [H⁺], 1 mark for Kₐ expression, 1 mark for concentration]
Note: The approximation gives 0.0625 mol dm⁻³ while the exact calculation gives 0.0657 mol dm⁻³. Both are acceptable, but the exact method is preferred when [H⁺] is not negligible compared to c.

18. (a) $\text{p}K_a = -\log(1.7 \times 10^{-5}) = 4.77$
Using Henderson–Hasselbalch:
$\text{pH} = \text{p}K_a + \log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$
$4.50 = 4.77 + \log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$
$\log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 4.50 - 4.77 = -0.27$
$\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 10^{-0.27} = 0.54$
[3 marks – 1 mark for pKₐ, 1 mark for substitution, 1 mark for ratio]

(b) From (a), $\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 0.54$
$[\text{CH}_3\text{COOH}] = 0.50 \text{ mol dm}^{-3}$ (in the final mixture, assuming volume change is negligible)
$[\text{CH}_3\text{COO}^-] = 0.54 \times 0.50 = 0.27 \text{ mol dm}^{-3}$
Moles of $\text{CH}_3\text{COONa}$ needed in $100 \text{ cm}^3 = 0.27 \times 0.100 = 0.027 \text{ mol}$
Molar mass of $\text{CH}_3\text{COONa} = 2(12.0) + 3(1.0) + 2(16.0) + 23.0 = 82.0 \text{ g mol}^{-1}$
Mass of $\text{CH}_3\text{COONa} = 0.027 \times 82.0 = 2.21 \text{ g}$
[3 marks – 1 mark for [CH₃COO⁻], 1 mark for moles, 1 mark for mass]

19. (a) The mixture contains two acids: $\text{HCl}$ (a strong monoprotic acid) and $\text{H}_2\text{SO}_4$ (a strong diprotic acid). The first equivalence point corresponds to the neutralisation of $\text{HCl}$ and the first proton of $\text{H}_2\text{SO}_4$. The second equivalence point corresponds to the neutralisation of the second proton of $\text{H}_2\text{SO}_4$.
[2 marks – 1 mark for identifying two acids, 1 mark for explaining two protons/stages]

(b) Total volume of $\text{NaOH} = 40.0 \text{ cm}^3$
[1 mark]

(c) At the first equivalence point (15.0 cm³), both $\text{HCl}$ and the first proton of $\text{H}_2\text{SO}_4$ have been neutralised.
At the second equivalence point (40.0 cm³ total), the second proton of $\text{H}_2\text{SO}_4$ is neutralised.
Volume of $\text{NaOH}$ for the second proton of $\text{H}_2\text{SO}_4 = 40.0 - 15.0 = 25.0 \text{ cm}^3$
Since $\text{H}_2\text{SO}_4$ is diprotic, the volume for the first proton = volume for the second proton = $25.0 \text{ cm}^3$
Volume of $\text{NaOH}$ for $\text{HCl} = 15.0 - 25.0 = \text{Wait—this gives a negative value. Let me reconsider.}$

Re-analysis: The first equivalence point at 15.0 cm³ represents the neutralisation of HCl only (since HCl is a strong acid and reacts first). The second equivalence point at 40.0 cm³ represents the neutralisation of both protons of H₂SO₄.
Volume for $\text{HCl} = 15.0 \text{ cm}^3$
Moles of $\text{NaOH}$ for $\text{HCl} = 0.0150 \times 0.100 = 0.00150 \text{ mol}$
Concentration of $\text{HCl} = \frac{0.00150}{0.0250} = 0.060 \text{ mol dm}^{-3}$
[2 marks – 1 mark for moles of NaOH, 1 mark for concentration]
Note: The interpretation depends on the relative acid strengths. Since both HCl and H₂SO₄ are strong acids, the first equivalence point likely corresponds to HCl (monoprotic) and the second to H₂SO₄ (diprotic, requiring twice the volume). However, the data shows 15.0 cm³ for the first and 25.0 cm³ more for the second, suggesting HCl = 15.0 cm³ and H₂SO₄ = 25.0 cm³ (for both protons).

20. (a) As temperature increases, $K_w$ increases, meaning the dissociation of water increases. This produces more $\text{H}^+$ and $\text{OH}^-$ ions in equal amounts. Since $[\text{H}^+]$ increases, the pH decreases. However, because $[\text{H}^+] = [\text{OH}^-]$ at all temperatures, pure water remains neutral.
[2 marks – 1 mark for explaining increased dissociation, 1 mark for stating water remains neutral because [H⁺] = [OH⁻]]

(b) At $40^\circ\text{C}$: $K_w = 2.9 \times 10^{-14}$
$[\text{H}^+] = \sqrt{K_w} = \sqrt{2.9 \times 10^{-14}} = 1.70 \times 10^{-7} \text{ mol dm}^{-3}$
[1 mark]

(c) At $60^\circ\text{C}$, the pH of neutral water is 6.51 (from the table). A solution with pH 5.50 has a pH lower than 6.51, meaning $[\text{H}^+] > [\text{OH}^-]$. Therefore, the solution is acidic at this temperature.
[2 marks – 1 mark for identifying the neutral pH at 60°C, 1 mark for comparison and conclusion]

(d) The dissociation of water is an endothermic process. As temperature increases, $K_w$ increases, meaning the equilibrium shifts to the right (towards more dissociation). By Le Chatelier's principle, increasing temperature favours the endothermic direction.
[1 mark]