Skip to main content
A-LevelChemistryEquilibria9701/42 · Oct/Nov 2025 · Q315 Marks

Define conjugate acid-base pair.

  1. (a)(i)Define conjugate acid-base pair.[1]
  2. (a)(ii)Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO42–. conjugate acid of HPO42– conjugate base of HPO42–[1]
  3. (b)(i)The Ka of propanoic acid, CH3CH2COOH, is 1.35 × 10–5 mol dm–3 at 298 K. Solution C is a solution of CH3CH2COOH with a pH of 3.60 at 298 K. Calculate the concentration of CH3CH2COOH in solution C. [CH3CH2COOH] = mol dm–3[2]
  4. (b)(ii)Calculate the concentration of hydroxide ions in solution C. [OH–] = moldm–3[1]
  5. (b)(iii)Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C. concentration = .moldm–3[1]
  6. (b)(iv)Table 3.1 shows three possible values of the Ka of dimethylpropanoic acid, (CH3)3CCOOH. Place a tick in Table 3.1 to show the correct value. Explain your answer. Table 3.1 value of Ka /moldm–3 place one tick (✔) in this column 9.33 x 10–6 1.35 x 10–5 3.35 x 10–5 explanation[3]
  7. (c)(i)Solution D is made by mixing 100cm³ of 0.100 mol dm–3 CH3CH2COOH and 100 cm³ of 0.100 mol dm–3 NaCl. The pH of solution D is measured as small amounts of H2SO4(aq) are added to it, and when small amounts of NaOH(aq) are added to it. Solution D only acts as a buffer solution when one of these solutions is added to it. Complete the sentence and write an equation for the reaction that occurs. Solution D acts as a buffer when is added to it. equation[1]
  8. (c)(ii)Complete the sentence and explain why solution D does not act as a buffer when the other solution is added. Solution D does not act as a buffer when explanation[1]
  9. (d)(i)Manganese(II) hydroxide, Mn(OH)2, is only slightly soluble in water. The solubility of Mn(OH)2 in water is 3.28 × 10–3 g dm–3 at 298 K. Calculate the concentration of a saturated solution of Mn(OH)2 at 298 K. [Mn(OH)2] = moldm–3[1]
  10. (d)(ii)Write an expression for the Ksp of Mn(OH)2. Give the units of Ksp. Ksp = units[2]
  11. (d)(iii)Use your answers to (d)(i) and (d)(ii) to calculate the value of Ksp of Mn(OH)2 at 298 K. Ksp =[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

  1. [1] two species that differ by one proton / H+
  2. [1] H2PO4- AND PO43-
  3. [1] M1: [H+] = 2.51 × 10–4
  4. [1] M2: [CH3CH2COOH] = 4.67 × 10-3
  5. [1] 3.98 x 10-11
  6. [1] [HCl] = 2.51 × 10–4
  7. [1] M1: top box ticked AND weaker acid
  8. [1] M2: alkyl group is electron donating
  9. [1] M3: O-H bond strengthened / anion is destabilised
  10. [1] NaOH AND CH3CH2COOH + OH– → CH3CH2COO– + H2O
  11. [1] H2SO4 AND conjugate base of CH3CH2COOH is not present
  12. [1] 3.69 x 10-5
  13. [1] M1: Ksp = [Mn²+][OH-]2
  14. [1] M2: units = mol³ dm-9
  15. [1] 2.0 × 10-13

Write your answer and get it marked

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Sign up free →

About this question

Topic
Equilibria, A-Level Chemistry (syllabus 9701). Structured question, 15 marks.
Source
A-Level Chemistry, Oct/Nov 2025 Paper 4 Variant 2, question 3.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content

Related Chemistry questions

© 2026 Oracle Prep