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.
The concentration of excess sodium hydroxide was 0.362 mol dm−3. Calculate the pH of the solution at the end of the experiment.
Sketch the neutralisation curve obtained and label the equivalence point.
Markscheme
«Ka = 10–2.87 = 1.35 × 10–3 »
«1.35 × 10–3 = »
«x = [H+] ==» 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.
«pOH = –log(0.362) = 0.441»
«pH = 14.000 – 0.441 =» 13.559 ✔
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.