DP Chemistry (last assessment 2024)

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Question 22M.2.hl.TZ2.a(ii)

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Date May 2022 Marks available [Maximum mark: 3] Reference code 22M.2.hl.TZ2.a(ii)
Level hl Paper 2 Time zone TZ2
Command term Calculate Question number a(ii) Adapted from N/A
a(ii).
[Maximum mark: 3]
22M.2.hl.TZ2.a(ii)

The overall equation for the production of hydrogen cyanide, HCN, is shown below.

CH4 (g) + NH3 (g) +32O2 (g) → HCN (g) + 3H2O (g)

Calculate the pH of a 1.00 × 10−2 mol dm−3 aqueous solution of ammonia.

pKb = 4.75 at 298 K.

[3]

Markscheme

Kb = 10-4.75 /1.78 x 10-5
OR
Kb = OH-2NH3 ✔

 

[OH–] = « 1.00×10-2×10-4.75 =» 4.22 × 10–4 «(mol dm–3)» ✔

 

pOH« = –log10 (4.22 × 10–4)» = 3.37
AND
pH = «14 – 3.37» = 10.6

OR

[H+]« =1.00×10-144.22×10-4» = 2.37 × 10–11
AND
pH« = –log10 2.37 × 10–11» = 10.6 ✔

 

Award [3] for correct final answer.

Examiners report

Rather surprisingly, many students got full marks for this multi-step calculation; others went straight to the pH/pKa acid/base equation so lost at least one of the marks: students often seem less prepared for base calculations, as opposed to acid calculations.

Syllabus sections