Date | May 2009 | Marks available | 3 | Reference code | 09M.2.hl.TZ1.2 |
Level | HL | Paper | 2 | Time zone | TZ1 |
Command term | Sketch | Question number | 2 | Adapted from | N/A |
Question
The pKa value for propanoic acid is given in Table 15 of the Data Booklet.
State the equation for the reaction of propanoic acid with water.
Calculate the hydrogen ion concentration (in moldm−3) of an aqueous solution of 0.100 moldm−3 propanoic acid.
The graph below shows a computer simulation of a titration of 25.0 cm3 of 0.100 moldm−3 hydrochloric acid with 0.100 moldm−3 sodium hydroxide and the pH range of phenol red indicator.
Sketch the graph that would be obtained for the titration of 25.0 cm3 of 0.100 moldm−3 propanoic acid with 0.100 moldm−3 potassium hydroxide using bromophenol blue as an indicator. (The pH range of bromophenol blue can be found in Table 16 of the Data Booklet).
Markscheme
CH3CH2COOH+H2O⇌CH3CH2COO−+H3O+ /
CH3CH2COOH⇌CH3CH2COOH−+H+;
⇌ required for mark.
(pKa for propanoic acid = 4.87)
[H+]2=0.100×Ka;
[H+]=1.16×10−3 (moldm−3);
sketch to show:
indicator range between pH 3.0 and pH 4.6 (with “yellow” at pH 3.0 and “blue” at pH 4.6);
initial pH of acid at 2.9 ± 1.0 (when no KOH has been added);
half-equivalence point (does not need to be named) at pH 4.9 when 12.5 cm3 of KOH have been added;
equivalence point at approx pH 8.5 – 9.0 when 25.0 cm3 of KOH(aq) added;
upper part of curve from 25.0 – 50.0 cm3 added identical to original curve;
Award [1] each for any three points.
Examiners report
The equation of propanoic acid with water was problematic for many candidates who omitted the equilibrium arrow (⇌) in part (a)(i). Although candidates were referred to the Data Booklet, some candidates did not know the formula of propanoic acid.
Part (a)(ii) was answered well by about half the candidates.
Part (b) also caused difficulties, with many candidates scoring only the mark for showing the pH range of bromophenol blue. Some candidates were thrown by the choice of indicator and selected a more appropriate indicator for these reagents. It is important to answer the question on the paper as the indicator was deliberately chosen to be different to the indicator used in the example. Graphs were generally badly and roughly drawn. Even candidates who had correctly calculated [H+] in part (a) often did not start the graph at the correct pH. Most graphs finished too low at a pH of 10 or less, and the vertical part of the graph was frequently at a volume less than 25 cm3. Rarely did a candidate get the half-equivalence value correct.