A Level H1 Chemistry Practice Paper 5

TuitionGoWhere Exam Practice (AI) — Chemistry H1 A-Level

Practice Paper Answers: Acids, Bases & Salts (Version 5 of 5)

Total Marks: 60

Section A

1. [2 marks]
A weak acid is one that only partially dissociates/ionises in water (1 mark).
Equation: $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)$ (1 mark for reversible arrow and state symbols).
Teaching note: Weak ≠ dilute. Use ⇌ not →. Common error: writing (l) or omitting (aq).

2. [1 mark]
A Brønsted–Lowry base is a proton ($\text{H}^+$) acceptor.
Teaching note: Contrast with Arrhenius (produces OH⁻ in water).

3. [2 marks]
Conjugate acid–base pairs: $\text{NH}_3/\text{NH}_4^+$ and $\text{H}_2\text{O}/\text{OH}^-$.
(1 mark each pair).
Teaching note: Conjugate acid has one more H⁺ than base.

4. [1 mark]
High acidity (low pH) denatures enzymes, changing active site shape so substrate cannot bind.
Teaching note: Link H⁺ to disruption of H-bonds/ionic bonds.

5. [1 mark]
$K_w = [\text{H}^+(aq)][\text{OH}^-(aq)] = 1.0 \times 10^{-14}$ mol² dm⁻⁶ at 25 °C.

6. [1 mark]
$[\text{H}^+] = 10^{-3} = 1.0 \times 10^{-3}$ mol dm⁻³.

7. [2 marks]
Phenolphthalein (1 mark), pH range 8.3–10.0 (1 mark). Or methyl orange (3.1–4.4) acceptable if strong acid–strong base but less ideal.
Teaching note: Strong–strong titration has steep midpoint near 7; either indicator with range covering 7 works.

8. [1 mark]
A buffer solution resists changes in pH when small amounts of acid or base are added.

9. [2 marks]
$\text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{HCO}_3^-(aq) + \text{H}^+(aq)$ (2 marks: equation + state symbols + reversible).

10. [2 marks]
$K_a$ is the acid dissociation constant (1 mark); it measures the extent of dissociation of a weak acid in water (1 mark).
Expression: $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$.

Section B

11. [3 marks]
$n(\text{NaOH}) = c \times V = 0.100 \times (20.0/1000) = 0.00200$ mol (1 mark)
1:1 ratio ⇒ $n(\text{CH}_3\text{COOH}) = 0.00200$ mol (1 mark)
$c = n/V = 0.00200 / (25.0/1000) = 0.0800$ mol dm⁻³ (1 mark)
Common mistake: Not converting cm³ to dm³.

12. [2 marks]
HCl is strong: $[\text{H}^+] = 0.010$ mol dm⁻³ (1 mark)
$\text{pH} = -\log(0.010) = 2.00$ (1 mark)

13. [3 marks]
$\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-$ so $[\text{OH}^-] = 2 \times 0.050 = 0.100$ mol dm⁻³ (1 mark)
$\text{pOH} = -\log(0.100) = 1.00$ (1 mark)
$\text{pH} = 14.00 - 1.00 = 13.00$ (1 mark)

14. [3 marks]
$\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}$
$\text{p}K_a = -\log(1.8\times10^{-4}) = 3.74$ (1 mark)
$\log(0.10/0.20) = \log(0.5) = -0.30$ (1 mark)
$\text{pH} = 3.74 - 0.30 = 3.44$ (1 mark)

15. [3 marks total]
(a) $K_a = \frac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}$ [1 mark]
(b) $[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}$; assuming $[\text{acid}] \approx [\text{C}_6\text{H}_5\text{COOH}]$ and using given $[\text{C}_6\text{H}_5\text{COO}^-]=0.020$:
If $[\text{acid}] = 0.020$ (typical), $[\text{H}^+] = 6.3\times10^{-5} \times (0.020/0.020) = 6.3\times10^{-5}$ mol dm⁻³ [2 marks for method and answer].
Note: If only $[\text{C}_6\text{H}_5\text{COO}^-]$ given, assume acid concentration equal for illustration.

16. [2 marks]
$n(\text{H}_2\text{SO}_4) = 0.080 \times (30.0/1000) = 0.00240$ mol (1 mark)
From eq: 1 mol $\text{H}_2\text{SO}_4$ needs 2 mol KOH ⇒ $n(\text{KOH}) = 0.00480$ mol (1 mark)

17. [3 marks]
$\text{CO}_3^{2-} + \text{H}^+ \rightleftharpoons \text{HCO}_3^-$ (1 mark) and $\text{HCO}_3^- + \text{H}^+ \rightleftharpoons \text{H}_2\text{CO}_3$ (1 mark). Added acid is consumed by $\text{CO}_3^{2-}$, resisting pH fall (1 mark).
Teaching: Buffer pair $\text{HCO}_3^-/\text{CO}_3^{2-}$.

18. [3 marks]
Initial $n(\text{acid}) = 0.050 \times 0.100 = 0.0050$ mol; add 0.0050 mol salt.
$[\text{acid}] = 0.0050/0.100 = 0.050$, $[\text{salt}] = 0.0050/0.100 = 0.050$ (1 mark)
$\text{p}K_a = -\log(1.8\times10^{-5}) = 4.74$ (1 mark)
$\text{pH} = 4.74 + \log(0.050/0.050) = 4.74$ (1 mark)

19. [2 marks]
HCl reacts: $\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+$ (1 mark). pH decreases slightly due to buffer action (1 mark).

20. [4 marks]
$n(\text{OH}^-) = 0.040 \times 0.0500 = 0.00200$ mol (1 mark)
$n(\text{H}^+) = 0.020 \times 0.0500 = 0.00100$ mol (1 mark)
Excess $\text{OH}^- = 0.00100$ mol in 0.100 dm³ ⇒ $[\text{OH}^-] = 0.0100$ mol dm⁻³ (1 mark)
$\text{pOH} = 2.00$, $\text{pH} = 12.00$ (1 mark)

Section C

21. [3 marks total]
(a) HA and HD (pH 1.0 ⇒ strong) [1 mark]
(b) HB and HC are weak acids with different $K_a$; partial dissociation gives different $[\text{H}^+]$ [2 marks: weak acid idea + $K_a$ difference].

22. [2 marks]
Equivalence volume = 50.0 cm³ (1 mark). Half-neutralisation pH = p$K_a$ ≈ 1.0? Actually for strong acid, half = pH ~1.3? From graph, at 25 cm³ pH ≈ 1.3 (1 mark for stating ~1.3 or reading from curve).
Image must show curve with equivalence at 50 cm³, pH 7.

23. [4 marks]
Lower pH ⇒ more $\text{H}^+$ shifts $\text{CO}_3^{2-} + \text{H}^+ \rightleftharpoons \text{HCO}_3^-$ right (2 marks: Le Chatelier + equation). Reduced $\text{CO}_3^{2-}$ impairs shell formation in corals/molluscs (2 marks: biological effect).

24. [3 marks]
(a) $\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$ (1 mark)
(b) $[\text{OH}^-] = \sqrt{K_b \times c} = \sqrt{4.0\times10^{-5} \times 0.010} = \sqrt{4.0\times10^{-7}} = 6.3\times10^{-4}$ mol dm⁻³ (2 marks method + answer).

25. [3 marks]
Methyl orange changes at pH 3.1–4.4 (1 mark). Weak acid–strong base equivalence near pH 8–9 (1 mark). Indicator changes before equivalence ⇒ inaccurate endpoint (1 mark). Not suitable; use phenolphthalein.

End of Answer Key