This question was based on acid-base titration. It was attempted by majority of the candidates and their performance was good.
In part (a), candidates correctly tabulated their burette readings and calculated the average volume of acid used using concordant tire values. However, some candidates
lost marks due to alteration of their titre values and wrong averaging of titres.
In the part (b), majority of the candidates correctly calculated the concentration of
A and B in moldm-3. However, some candidates lost marks due to wrong substitution and units. In part (b) (iv) only a few candidates correctly calculated the volume of CO2
evolved in the reaction at s.t.p.
The expected answers include:
Question 1
(a) (i) Titrations: 4 x 2 sets
Averaging(Vcm3)
Na2CO3(aq) + 2HC1(aq) 2NaC1(aq) + H2O + CO2(g)
(b) (i) Conc. of B in moldm-3
Molar mass (Na2CO3) = (2 x 23) + 12 + (3 x 16) + 106 gmol-1
Conc. of B (Na2CO3) = 2.45 = 0.0925 mol dm-3
106 x 0.25
Alt. Method
250 cm 3 of solution contained 2.45g of Na2CO3
1000 cm 3 of solution contained 2.45 x 1000
250
= 2.45x4
= 9.80g
M(Na2CO3) = 106gmol-1
Conc. B(Na2 CO3) = 9.80 =0.0925 moldm-3
106
Conc. of A in moldm-3
From reaction equation
CA VA = 2 CA = 2 x CB x VB
CBVB 1 VB
CA = 2 x 0.0925 x 20/25 = a moldm-3 (correct evaluation to 3 sig. figs)
VA
Conc. of A in gdm-3
M(HC) = 1 + 35.5 = 36.5gmol-1
Conc. of A in (gdm-3) = a x molar mass (HC1)
= a x 36.5
= b gdm-3
(iv) Volume of the gas evolved
From reaction equation
1 mole of Na2CO3 = 1 mole of CO2 at stp
n(CO2) = n (Na2CO3) = 0.0925 mole
… Vco2 = n x Vm
= 0.0925 x 22.4dm3
= 2.07dm3 at stp
… Vco2 liberated in 25 cm3 of Na2CO3 will be 2,07 x 25 x 25
1000
= 0.0518 dm3/51.80 cm3 at stp