Part (a): this question was well atempted by most candidates and the performance was quite good. The table and average titre value were correctly done by majority of the responding candidates. Few candidates however, had no table of titres while majority of others altered and/or cancelled their tables to make their titre values correspond with the teacher’s titre.
In part (b), most candidates were unable to calculate neither the number of moles of the acid nor the number of mole of the base in 25.0cm3 and hence they were unable to determine correctly the mole ratio of acid to base.
The expected answers are:
(b)(i) number of moles of acid = 0.100 x VA
1000
= X mole(s) [3sig. Fig to score]
1000cm3 contains 0.100 mole(s)
VA will contain 0.100 x VA
1000
= X moles [3 Sig. Fig. to score]
(ii) Number of moles of KOH in B
500cm3 of B contains 2.8g of KOH
1000cm3 of B will contain 2.8 x 1000 = 5.6 KOH
500
Molar mass of KOH = 39 + 16 + 1 or 56 gmol-1
Conc of B = 5.6 = 0.100 mol dm-3
56
OR
Molar mass of KOH = 39 + 16 + 1 or 56 gmol-1
Number of moles of KOH = 2.8 = 0.05 mole(s)
56
500cm3 contains 0.05 mole(s)
1000cm3 will contain 0.05 x 1000
500
= 0.100 moldm-3
Number of moles of KOH in B = 0.100 x 20/25 = 0.0025 mole(s)
1000
= Y mole(s)
(iii) Mole ratio of acid to base = X: Y to nearest whole number ratio
