TThis question was attempted by majority of the candidates and the performance was good.
In (a)(i) candidates were required to copy and complete a table of titre values. Majority of the candidates performed well, but a few of them could not still subtract the initial from final readings.
In (a)(i) and (ii), candidates correctly gave the volumes of B used and went on to use consistent titre values in averaging. However, a few candidates lost marks for using non-concordant titres.
in (a)(iii), majority or the candidates correctly balanced the chemical equation. Some candidates lost marks as a result of writing wrong physical state. Candidates should note that balanced chemical equation would be marked even without physical state, but writing the wrong physical state will result to forfeiture of some marks.
In (b), most candidates performed very well in this question but quite a number of them lost marks for wrong mole ratio and for not correctly giving their answers to 3 significant figures.
The expected answers include:
1.(a)(i)
Burette reading |
Rough |
I |
II |
Final reading (cm3) |
23.50 |
44.20 |
25.20 |
Initial reading (cm3) |
2.00 |
23.50 |
4.30 |
Volume of B used (cm3) |
21.50 |
20.70 |
20.90 |
(ii) Average titre value = 20.70 + 20.90
2
= 20.80 cm3
(iii) Na2CO3 + 2HCl → 2NaCl + H2O + CO2
(b)(i) 250 cm3 contains 1.30 g of Na2CO3
\ 1000 cm3 will contain 1.30 x 1000
250
= 5.20 gdm-3
Molar mass of Na2CO3 = (23 x 2) + 12 +(16 x 3)
= 106 gmol-1
\ Concentration in moldm-3 = 5.20
106
= 0.0491 moldm-3
(ii) I CaVa = 2
CbVb 1
Ca x 20.80 = 2
0.0491 x 25 1
Ca = 2 x 0.0491 x 25
20.80
= 0.118 moldm-3
II. Concentration in gdm-3 ρ = Conc. x Molar mass
Molar mass of HCl = 1 + 35.5 = 36.5gmol-1
Conc. gdm-3 = 0.118 x 36.5
= 4.31gdm-3
