Part (a): This question was well attempted by majority of the candidates and the performance was quite good. Candidates correctly tabulated their burette readings and calculated the average volume of A used. Some candidates lost marks as a result of deliberate alteration/cancellation of titre value to agree with those of the supervisor and for using non concordant readings to calculate the average titre.
In question (b)(i), (ii) and (iii), candidates calculates the concentration of B in moldm-3, concentration of A in moldm-3 and the molar mass of H2Y. Some candidates however could not express their answers to three significant figures.
Majority of the candidates did not express the molar mass of H2Y in g mol-1
The expected answers were:
(ii) Concentration of A in mol dm-3
From the reaction equation
CA VA = 1
CB VB 2
CA = CB VB
2 VA
CA = 0.0975 x 25
2 VA
= P mol dm-3
Alternative Method:
Amount of acid used = 0.025 x 0.0975 = 0.00244 mol
From the equation of reaction:
2 mol of B require 1 mol of A
0.00244 mol of B requires ½ x 0.00244 mol of A
= 0.00122 mol of A.
i.e VA cm3 contain 0.00122 mol.
1000 cm3 contain 0.00122 x 1000/VA mol.
i.e. Conc of A = 1.22/VA = P mol dm-3
(iii) Molar mass of H2Y
Molar mass of H2Y = 6.22
P
= Q g mol-1
Part (C): Candidates attempted this question very well by stating that the pH of solution A before titration is less than 7 while that of solution B before titration is greater than 7
