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A-LevelChemistryEquilibria9701/42 · May/June 2017 · Q610 Marks

The table lists some organic acids and their pKa values. acid formula pKa ethanoic acid CH3CO₂H 4.76 chloroethanoic acid CICH₂CO₂H 2.86 aminoethanoic acid (glycine) H2NCH2CO2H 9.87

  1. (a)(i)State the relationship between pKa and the strength of an acid.[1]
  2. (a)(ii)State the mathematical relationship between pKa and the acidity constant Ka.[1]
  3. (a)(iii)Give reasons for why the pKa value for chloroethanoic acid is smaller than that for ethanoic acid.[2]
  4. (b)(i)Use the zwitterionic structure for aminoethanoic acid (glycine) in aqueous solution to write an equation for its dissociation giving H+(aq) ions.[1]
  5. (b)(ii)Calculate the pH of a 0.100 moldm-³ solution of aminoethanoic acid. pH = ...[2]
  6. (b)(iii)A 10.0 cm³ sample of 0.100 mol dm-3 aminoethanoic acid (glycine) was titrated with 0.100 mol dm-3 NaOH. After 20.0 cm³ of NaOH, an excess, had been added, the pH was found to be 12.5. Using the following axes, sketch a graph showing how the pH changes during this titration.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

  1. [1] the lower / smaller the pKa, the stronger the acid
  2. [1] pKa = -log(Ka) or pKa = -lg(Ka) or Ka = 10⁻pKa
  3. [1] (stronger than ethanoic acid because) Cl is electron-withdrawing
  4. [1] and so stabilises the RCO2⁻ anion / conjugate base or weakens O-H bond (so H⁺ is more easily released)
  5. [1] NH3⁺CH2CO2⁻ → NH2CH2CO2⁻ + H⁺ OR NH3⁺CH2CO2⁻ + H2O → NH2CH2CO2⁻ + H3O⁺
  6. [1] Ka = 10⁻9.87 = 1.35 × 10⁻¹⁰
  7. [1] [H⁺] = √(Ka.c) = 3.67 × 10⁻⁶ pH = 5.4 (5.43–5.44) min 2sf
  8. [1] curve starts at 5.4 and continuous
  9. [1] vertical portion (end point) at vol added = 10.0 cm³
  10. [1] finishes at pH = 12.5 at 20 cm³ (and does not increase in pH)

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About this question

Topic
Equilibria, A-Level Chemistry (syllabus 9701). Structured question, 10 marks.
Source
A-Level Chemistry, May/June 2017 Paper 4 Variant 2, question 6.

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