Acids and basesAQA A-Level Chemistry: Topic test

20 questions, 54 marks

AQA A-Level Chemistry

Acids and bases topic test

Total 54 marks

Name

Class

Date

1
In the nitration of benzene, concentrated nitric acid is mixed with concentrated sulfuric acid. The first step is an acid–base equilibrium: HNO₃ + H₂SO₄ ⇌ H₂NO₃⁺ + HSO₄⁻.
(a)
Which species is the Brønsted–Lowry base in the forward reaction?
[1 mark]
AH₂SO₄
BHNO₃
CH₂NO₃⁺
DHSO₄⁻
(b)
Which pair is a conjugate acid–base pair?
[1 mark]
AHNO₃ and H₂SO₄
BH₂SO₄ and H₂NO₃⁺
CHNO₃ and HSO₄⁻
DH₂NO₃⁺ and HNO₃
(c)
Nitric acid is a strong acid in water. Explain why it acts as a base in this mixture.
[2 marks]
Total for question 1: 4 marks
2
A technician has a solution of nitric acid, HNO₃, a strong acid, of concentration 0.0350 mol dm⁻³ and a solution of potassium hydroxide of concentration 0.0400 mol dm⁻³. The ionic product of water, Kw, is 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K and 5.48 × 10⁻¹⁴ mol² dm⁻⁶ at 323 K.
(a)
What is the pH of the nitric acid solution at 298 K?
[1 mark]
A1.46
B1.15
C3.35
D12.54
(b)
10.0 cm³ of the nitric acid is diluted with water to a total volume of 1000 cm³. What is the pH of the diluted solution?
[1 mark]
A1.46
B2.46
C3.46
D4.46
(c)
Calculate the pH of the potassium hydroxide solution at 323 K.
[2 marks]
Total for question 2: 4 marks
3
Benzoic acid, C₆H₅COOH, is a weak monoprotic acid used as a food preservative. At 298 K its acid dissociation constant, Ka, is 6.28 × 10⁻⁵ mol dm⁻³.
(a)
Calculate the pH of a 0.0500 mol dm⁻³ solution of benzoic acid at 298 K.
[3 marks]
(b)
Calculate pKa for benzoic acid. A solution of benzoic acid has a pH of 2.60. Calculate the concentration of the benzoic acid in this solution.
[4 marks]
Total for question 3: 7 marks
4
Propanoic acid, CH₃CH₂COOH, is a weak monoprotic acid with Ka = 1.35 × 10⁻⁵ mol dm⁻³ at 298 K.
(a)
A student titrates 25.0 cm³ of 0.100 mol dm⁻³ propanoic acid with 0.100 mol dm⁻³ sodium hydroxide. Indicator pH ranges: methyl orange 3.2–4.4; phenolphthalein 8.3–10.0. Calculate the initial pH, the volume of sodium hydroxide at the equivalence point, and the pH after 12.5 cm³ of sodium hydroxide has been added. Explain why the pH at the equivalence point is above 7 and select a suitable indicator.
[6 marks]
(b)
A buffer solution is prepared by mixing 40.0 cm³ of 0.200 mol dm⁻³ propanoic acid with 20.0 cm³ of 0.150 mol dm⁻³ sodium hydroxide. Calculate the pH of the buffer. Explain how the buffer resists a change in pH when a small amount of acid is added.
[6 marks]
Total for question 4: 12 marks
5
A student titrates 25.0 cm³ of 0.100 mol dm⁻³ methylamine solution, CH₃NH₂, a weak base, with 0.100 mol dm⁻³ hydrochloric acid. Indicator pH ranges: methyl orange 3.2–4.4; bromothymol blue 6.0–7.6; phenolphthalein 8.3–10.0.
(a)
What is the pH at the equivalence point?
[1 mark]
AExactly 7, because the acid and base have the same concentration
BAbove 7, because methylamine is a base
CExactly 7, because chloride ions are neutral
DBelow 7, because the methylammonium ion is a weak acid
(b)
Which indicator is suitable for this titration?
[1 mark]
AMethyl orange
BPhenolphthalein
CBromothymol blue
DNone, because indicators cannot be used for weak bases
(c)
Explain why phenolphthalein is not suitable for this titration.
[2 marks]
Total for question 5: 4 marks
6
A cell culture medium is buffered at pH 7.20 using a mixture of sodium dihydrogenphosphate, NaH₂PO₄, and disodium hydrogenphosphate, Na₂HPO₄. The weak acid in the buffer is H₂PO₄⁻, which has Ka = 6.2 × 10⁻⁸ mol dm⁻³ at 298 K.
(a)
What must be the ratio [HPO₄²⁻] / [H₂PO₄⁻] in this buffer?
[1 mark]
A0.10
B0.50
C1.0
D10
(b)
A small amount of sodium hydroxide is added to the buffer. Which reaction removes the hydroxide ions?
[1 mark]
AHPO₄²⁻ + OH⁻ → PO₄³⁻ + H₂O
BH₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O
CHPO₄²⁻ + H⁺ → H₂PO₄⁻
DH₂PO₄⁻ + H⁺ → H₃PO₄
(c)
The buffer is diluted ten-fold with water. Explain why the pH of the buffer hardly changes.
[2 marks]
Total for question 6: 4 marks
7
A student titrates 25.0 cm³ of a solution of an unknown weak monoprotic acid, HA, with 0.100 mol dm⁻³ sodium hydroxide. The end point is reached after 21.60 cm³ of sodium hydroxide has been added. The pH of the mixture after 10.80 cm³ of sodium hydroxide has been added is 4.35.
(a)
Calculate the concentration of the weak acid HA in the original solution.
[3 marks]
(b)
Use the pH after 10.80 cm³ of sodium hydroxide has been added to calculate Ka for HA, and hence calculate the pH of the original solution of HA.
[4 marks]
Total for question 7: 7 marks
8
Chlorine is used to disinfect swimming pools. It forms hypochlorous acid, HOCl, a weak monoprotic acid with Ka = 3.0 × 10⁻⁸ mol dm⁻³ at 298 K.
(a)
Write the expression for Ka for HOCl. Calculate the pH of a 0.0200 mol dm⁻³ solution of HOCl and explain why it is higher than the pH of 0.0200 mol dm⁻³ hydrochloric acid, which is 1.70.
[6 marks]
(b)
A buffer is made by mixing 50.0 cm³ of 0.0200 mol dm⁻³ HOCl with 25.0 cm³ of 0.0100 mol dm⁻³ sodium hypochlorite, NaOCl. Calculate the pH of the buffer. Explain how the buffer responds to small additions of acid and of alkali.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).