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Date November 2021 Marks available 2 Reference code 21N.2.hl.TZ0.11
Level HL Paper 2 Time zone TZ0
Command term Calculate Question number 11 Adapted from N/A

Question

50.00 cm3 of 0.75 mol dm−3 sodium hydroxide was added in 1.00 cm3 portions to 22.50 cm3 of 0.50 mol dm−3 chloroethanoic acid.

Calculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.

[2]
a.

The concentration of excess sodium hydroxide was 0.362 mol dm−3. Calculate the pH of the solution at the end of the experiment.

[1]
b.

Sketch the neutralisation curve obtained and label the equivalence point.

[3]
c.

Markscheme

«Ka = 10–2.87 = 1.35 × 10–3 »

«1.35 × 10–3chloroethanoate×H+0.50moldm-3=x20.50moldm-3 »

«x = [H+] =1.4×10-3×0.50=» 2.6 × 10–2 «mol dm–3» ✔


«pH = –log[H+] = –log(2.6 × 10–2) =» 1.59 ✔

 

Accept final answer in range 1.58–1.60.

Award [2] for correct final answer.

a.

«pOH = –log(0.362) = 0.441»

«pH = 14.000 – 0.441 =» 13.559 ✔

b.

OR

starts at 1.6 AND finishes at 13.6 ✔

approximately vertical at the correct volume of alkali added ✔

equivalence point labelled AND above pH 7 ✔

 

Accept any range from 1.1-1.9 AND 13.1-13.9 for M1 or ECF from 11c(i) and 11c(ii).

Award M2 for vertical climb at 28 cm3 OR 15 cm3.

Equivalence point must be labelled for M3.

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Additional higher level (AHL) » Topic 18: Acids and bases » 18.2 Calculations involving acids and bases
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Additional higher level (AHL) » Topic 18: Acids and bases
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