A-LevelChemistryEquilibria9701/42 · Oct/Nov 2019 · Q38 Marks

Ammonium chloride, NH4Cl(s), dissolves in water to form an acidic solution. This is due to the dissociation of the ammonium ions. NH4+(aq) ⇌ H+(aq) + NH3(aq)

(a)(i)Calculate the [H+] in a 0.300 mol dm³ solution of ammonium chloride. [H+] =[1]
(a)(ii)Calculate the value of pKa of the ammonium ion. pKa =[2]
(b)(i)Explain, with the aid of an equation, how this solution can behave as a buffer when a small amount of a strong acid is added.[1]
(b)(ii)Explain, with the aid of an equation, how this solution can behave as a buffer when a small amount of a strong base is added.[1]
(c)(i)Use the value of Kw to calculate [H+] in pure water under standard conditions. Show your working. [H+] =[1]
(c)(ii)The pH of pure water at 50°C is 6.64. Calculate the numerical value of Kw at 50°C. Kw =[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] 1.3 x 10-5
[2] M1: Ka expression used correctly and Ka = 5.5(3) × 10–10
[2] M2: pKa = 9.26
[2] Award 2 marks for correct answer
[1] NH3 + H+ → NH4+
[1] NH4+ + OH- → NH3 + H2O or H+ + OH- → H2O and reference to expression in Q shifting R
[1] quotes Kw = 1 × 10-14 or 1 × 10-1 [H+][OH-]
[1] [H+] = 1 x 10-7
[2] M1: [H+] = 2.3 × 10-7 (calculator value 2.290867 × 10-7) and Kw = [2.3 × 10-7]2
[2] M2: Kw = 5.2 × 10-14 calculator 5.248074 × 10-14
[2] Award 2 marks for correct answer

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