The question was attempted by most candidates and the performance was good.
In part (a)(i) and (ii), candidates correctly calculated volume of acid used for each titre and the average titre value as follows:
(i) Titre I. 24.60 – 0.00 = 24.60 cm3
II. 48.80 – 24.60 = 24.20 cm3
III. 25.55 – 1.25 = 24.30 cm3
- Average titre = 24.20 + 24.30
2
= 24.25 cm3
However, some of the candidates lost marks because they either used the three titre values or two non concordant titre values with wrong units to calculate the average titre.
In part (b) (i) – (iii), candidates correctly calculated the concentration of XOH(aq) in
moldm-3, molar mass of XOH and atomic mass of X respectively. Some of the candidates could not however, express their final answers in the required number of significant figures with the correct unit hence, lost some marks.
The expected answers from candidates were as follows:
(i) CA VA = 1
CB VB 2
CB = 2 CA VA
VB
= 0.06 x 24.25x 2
25
= 0.116 mol dm-3 (2 or 3 sig. figures)
ALTERNATIVE METHOD
1000 cm3 of H2SO4 contain 0.06 mole
∴ 24.25 cm3 H2SO4 will contain .06 x 24.25
1000
= 0.001455 mole
From equation for the reaction,
1 mole H2SO4 reacts with 2 moles KOH
0.001455 mole H2SO4 will react with 2 x 0.001455 mole XOH
= 0.00291 mole XOH
25 cm3 of XOH contains 0.00291mole
1000 cm3 of XOH will contain 0.00291 x 1000
25
= 0.1164 mole
C XOH = 0.116 mol dm-3
- molar mass of XOH = 6.5 g dm-3
0.116mol dm-3
= 56.0 g mol-1
- XOH = 56
X + 16 +1= 56
X = 56 -17 = 39
In question (c), candidates correctly named any of methyl orange, methyl red and phenolphthalein as a suitable indicator for the titration. Some lost the mark because they wrote “any indicator”.
2XOH(aq) + H2SO4(aq) X2SO4(aq) + 2H2O(L)