This question was attempted by most candidates and the performance was good In (a), majority of the candidates correctly tabulated the burette readings and used concordant values to calculate the average volume of acid used: . However, some candidates lost marks because they either altered their burette readings to agree with the supervisor's value or outrightly cancelled their table and retabulated.
In b(i) and (ii), a good number of candidates correctly calculated the concentrations of A and B
in moldm:3 respectively thus:
(i) Concentration of A in moldm-3;
Concentration A in gdm-3 = 1.04 x 1000
500
= 2.08gdm-3
Molar mass of HCl = 1 + 35.5
= 36.5 mol-1
Concentration of A in = 2.08
36.5
= 0.057 moldm-3 (2 or 3 sig. Fig.)
(ii) Concentration of solution B in moldm-3
CAVA = 2
CBVB 1
CB = CAVA
2VB
= 0-057 x VA
2 x 20/25
= X moldm-3 (3 sig. Fig)
Alternative Method:
(i) Molar mass' of HCI = I = 35.5 = 36.5 gmol-1 Moles of A = 1.04 = 0.0285
36.5
cm3 contains 0.00285 moles of A 1000cm3 will contain 0.0285 x 1000
500
= 0.057 moldm-3.
1000cm3 will contain 0.057 moles HCI
:. V A will contain 0.057 x VA moles = w moles
1000
From the equation of the reaction,
2 moles of HCI = I mole of Na2C03 Hence, w moles of HCI = w of Na2C03
2
20/25cm3 Na2C03 contain w/2 moles
:. 1000cm3 Na2C03 contains w x 1000 moles
2 20/25
= x moldm-3
Some candidates however lost marks due to their inability to express their answers in the required number of significant figures.
In (b)(iii) most candidates could not calculate the solubility of Na2C03 in moldm-3 because of their poor knowledge of the concept of solubility and dilution factor. The required answer to the question was as follows:
Solubility of Na2C03 in moldm-3:
The dilution factor of B =50 to 1000
= 1.20
The solubility = X x 20
= ymoldm-3
In b(iv), candidates could not calculate the volume of CO2 liberated from 1 dm3 of B if the titration were carried out at s.t.p. The expected answer from candidates were:
The volume of C02(g) liberated from 1 dm3 at stp:
From the equation of the reaction
X mole of Na2C03 == X moles of CO2 1 mole of C02 at stp == 22.4 elm3
X mole of C02 = 22.4 x X dm3
= Z dm3 say
2Nacl(aq) + H2O + CO2(g)