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Question 18M.2.hl.TZ1.5

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Date May 2018 Marks available [Maximum mark: 9] Reference code 18M.2.hl.TZ1.5
Level hl Paper 2 Time zone TZ1
Command term Apply, Determine, Explain, Formulate, Justify, Predict, State Question number 5 Adapted from N/A
5.
[Maximum mark: 9]
18M.2.hl.TZ1.5

Limescale, CaCO3(s), can be removed from water kettles by using vinegar, a dilute solution of ethanoic acid, CH3COOH(aq).

(a)

Predict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.

[2]

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]

(b)

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.

[1]

Markscheme

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]

(c)

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.

 

[2]

Markscheme

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]

(d)

Determine the pH of 0.010 mol dm−3 2,2-dimethylpropanoic acid solution.

Ka (2,2-dimethylpropanoic acid) = 9.333 × 10−6

[2]

Markscheme

[H+] « = K a × [ C 5 H 10 O 2 ] = 9.333 × 10 − 6 × 0.010   » = 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]

(e)

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.

[2]

Markscheme

(CH3)3CCOOH(aq) + OH–(aq) → (CH3)3CCOO–(aq) + H2O(l)

OR

(CH3)3CCOOH(aq) + OH–(aq) ⇌  (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) ⇌ (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]

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