DP Chemistry (last assessment 2024)

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Question 21M.2.hl.TZ1.8

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Date May 2021 Marks available [Maximum mark: 6] Reference code 21M.2.hl.TZ1.8
Level hl Paper 2 Time zone TZ1
Command term Calculate, Sketch Question number 8 Adapted from N/A
8.
[Maximum mark: 6]
21M.2.hl.TZ1.8

Propanoic acid, CH3CH2COOH, is a weak organic acid.

(a)

Calculate the pH of 0.00100 mol dm–3 propanoic acid solution. Use section 21 of the data booklet.

[3]

Markscheme

Ka = 10−4.87 / 1.35 × 10−5 ✔

[H+] = «1.35×10-5×0.001=1.35×10-8=» 1.16 × 10−4 «mol dm−3» ✔

pH = 3.94 ✔


Accept alternative methods of calculation.

Award [3] for correct final answer.

Award [3] for 3.96 {answer if solved by quadratic}.

(b)

Sketch the general shape of the variation of pH when 50 cm3 of 0.001 mol dm–3 NaOH (aq) is gradually added to 25 cm3 of 0.001 mol dm–3 CH3CH2COOH (aq).

[3]

Markscheme

Any three of:

correct “S” shape ✔

equivalence point at 25 cm3 ✔

final pH tends to 11 ✔

pH at equivalence point >7 ✔

starting pH between 3.8 - 4 ✔

pH at half equivalence approx. 5 ✔


Do not penalize for incorrect points.
Award any 3 correct.

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