Question 1A
1.
A is 0.100 mol dm-3 HNO3.
B is a solution containing 2.50 g of a mixture of Na2CO3 and Na2SO4 in 25.0cm3 of solution.
(a)
Put A into the burette and titrate it against 20.0cm3 or 25.0cm3 portion of
B using methyl orange as indicator. Repeat the exercise to obtain consistent titres values. Tabulate your burette readings and calculate the
average volume of A used.
The equation of reaction is involved in the titration is:
Na2CO3(aq) + 2HC1(aq) 
2NaC1(aq)+ H2O(1)+ CO2(g)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm-3;
(ii) concentration of Na2CO3B in gdm-3;
(iii) percentage of Na2CO3 in the mixture. [Na2CO3 = 106] [20 marks]
Observation
This question was attempted by nearly all the candidates and the performance was good. In part (a), the table and average titre value were correctly done by majority of the candidates.
In the part (b), majority of the candidates were able to calculate the concentration of Na2CO3 in B in moldm-3 and in gdm-3. Also, majority of the candidates were able to calculate the percentage of Na2CO3 in the mixture. However, few candidates lost marks due to wrong units, wrong significant figures, wrong substitution, and calculated percentage impurity instead of percentage purity.
The expected answers include:
1(a) Two consistent/concordant titres
Averaging
(b) (i) Conc. of B in moldm-3
CAVA = 2
CBVB 1
CB = CAVA
2VB
=0.100 x VA
2 x 20/25
= q moldm-3
Alternative Method
Amount of acid in A used = 0.100 x VAmol
1000
= 0.0001VAmol
From balanced equation in reaction,
2 moles of A ≡ 1 mole of B
0.0001VAmol of A =½0.0001 VA mol of B
= 0.00005VAmol.
If 25 cm3 of B ≡ 0.00005VAmol
100cm3 of B = 0.00005VA × 1000mol
25
= 0.002VAmol
Hence conc. of B = q modm-3 say
(ii) Conc. of Na2CO3 in B gdm-3
Conc. of Na2CO3 = CB x M(Na2CO3)
= q x 106
= r say
(iii) Percentage of Na2CO3in the mixture
1000cm3 of B contains rg
250cm3 of B contains = r x 250
1000
= s say
% Na2CO3 = 
x 100
= z say
Alternative Method
250 cm3 of B = 2.50g
1000 cm3of B ≡ 2.50 x 1000g
250
= 10.0g
Mass conc. of B = 10 gldms
% Na2CO3 = r ×100
10
= z say