| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.hl.TZ1.5 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Determine | Question number | 5 | Adapted from | N/A |
Question
Limescale, CaCO3(s), can be removed from water kettles by using vinegar, a dilute solution of ethanoic acid, CH3COOH(aq).
Predict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.
Dissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.
Write an equation to show ammonia, NH3, acting as a Brønsted–Lowry base and a different equation to show it acting as a Lewis base.
Determine the pH of 0.010 mol dm−3 2,2-dimethylpropanoic acid solution.
Ka (2,2-dimethylpropanoic acid) = 9.333 × 10−6
Explain, using appropriate equations, how a suitably concentrated solution formed by the partial neutralization of 2,2-dimethylpropanoic acid with sodium hydroxide acts as a buffer solution.
Markscheme
slower rate with ethanoic acid
OR
smaller temperature rise with ethanoic acid
[H+] lower
OR
ethanoic acid is weak
OR
ethanoic acid is partially dissociated
Accept experimental observations such as “slower bubbling” or “feels less warm”.
[2 marks]
Any one of:
corrosion of materials/metals/carbonate materials
destruction of plant/aquatic life
«indirect» effect on human health
Accept “lowering pH of oceans/lakes/waterways”.
[1 mark]
Brønsted–Lowry base:
NH3 + H+ → NH4+
Lewis base:
NH3 + BF3 → H3NBF3
Accept “AlCl3 as an example of Lewis acid”.
Accept other valid equations such as Cu2+ + 4NH3 → [Cu(NH3)4]2+.
[2 marks]
[H+] «\( = \sqrt {{{\text{K}}_{\text{a}}} \times \left[ {{{\text{C}}_5}{{\text{H}}_{10}}{{\text{O}}_2}} \right]} = \sqrt {9.333 \times {{10}^{ - 6}} \times 0.010} {\text{ }}\)» = 3.055 × 10–4 «mol dm–3»
«pH =» 3.51
Accept “pH = 3.52”.
Award [2] for correct final answer.
Accept other calculation methods.
[2 marks]
(CH3)3CCOOH(aq) + OH–(aq) → (CH3)3CCOO–(aq) + H2O(l)
OR
(CH3)3CCOOH(aq) + OH–(aq) \( \rightleftharpoons \) (CH3)3CCOO–(aq) + H2O(l) AND addition of alkali causes equilibrium to move to right
(CH3)3CCOO–(aq) + H+(aq) → (CH3)3CCOOH(aq)
OR
(CH3)3CCOO–(aq) + H+(aq) \( \rightleftharpoons \) (CH3)3CCOOH(aq) AND addition of acid causes equilibrium to move to right
Accept “HA” for the acid.
Award [1 max] for correct explanations of buffering with addition of acid AND base without equilibrium equations.
[2 marks]
