This question was attempted by most candidates and the performance was poor.
In (a), most candidates correctly tabulated their readings and calculated the average volume of A used. However, some of them lost marks because they manipulated their readings to agree with those of the supervisor and used non concordant titres to calculate the average volume of A used.
In (b)(i)-(iii), most candidates could not calculate the amount of acid in the average titre, amount of base in 20.00 cm3 or 25.00 cm3 and the mole ratio. The expected answers to part (b) were as follows:
(i) Amount of acid in average titre
0.05 x Va = a moles
1000
OR
1000 cm3 contains 0.05 moles
∴ Va will contain 0.05 x Va
1000 = a moles
(ii) concentration of NaOH in B
0.025 moles = 0.100 moldm-3
0.25dm3
OR
0.025 x 1000 = 0.100 moldm-3
250
Amount of base (NaOH) in 20/25 cm3
20/25 cm3 x 0.100 moldm3 = b moles say
1000 OR
250 cm3 of B contains 0.025moles of NaOH
∴ 20/25 cm3 of the solution will contain 0.025 x 20/25
250
= b moles say
(iii) moles ratio of acid (HyX) to base (NaOH)
a = z say
b 1
OR
a : b = a : b
b b
= z say : 1
In (c), only few candidates who got (b)(i)-(iii) were able to correctly write a balanced chemical equation for the reaction as follows:
zNaOH + HyX → NazX + yH2O
In (d), candidates could not correctly deduce the basicity of the acid which was z.
