The question was attempted by most candidates and the performance was good.
In part (a), candidates were able to correctly name a suitable indicator for the titration with logical reason for their choice, gave the colour of the indicator in the base and at end-point and stated the type of reaction involved in the experiment as follows:

(a) (i) Methyl orange/methl red/phenolphthalein

Because the end point will concide with the pl+/colour change range of the indicator.Zit is a reaction between strong acid and a strong base.

(ii)

	Colour in Base	Colour at the End Point
*Methyl orange	Yellow	Orange
* Methyl red	Yellow	Orange
* Phenolphtalein	Pink	Colourless

(iii) Neutralization
In part(b) (i) and (ii), candidates correctly wrote a balanced equation for the reaction and determined the average volume of acid used thus:

(i) NaOH(aq) + HCI(aq) → NaCI(aq) + H2O(l);
(ii) Average titre = 23.50 + 23.40
                                            2
                                 = 23.45cm³

In part (c), most candidates were able to correctly calculate the concentration of the acid in moldrrr? and gdrrr" in (i) and (ii) respectively. However, some of them lost marks because of omission/wrong units and lack of understanding of the significant figures. The expected answers from candidates were as follows:

(i)        CAVA/CBVB =  1 mole ratio
CA CAVA/CBVB making CA the subject

CA = 1 x O. 065 x 20 correct substitution
                   23.45
       = 0.0554 moldm^-3

Alternative Method

(i) Amount of NaoH used = 0.065 x 20
                                                          1000
                                                 = 0.0013 mol

From the balanced equation of reaction:

1 mol ofNaOH required 1 mol ofHCL
•• 0.0013 mol of NaOH required 0.0013 mol ofHCL
i.e 23.45cm³of solution contained 0.0013 mol of HCl

:. 1000cm3 of solution contained 0.0013 x 1000 of HCl
                                                                      23.45
                                                               = 0.0554 mol

Hence concentration of acid = 0.0554 mol dm^-3

(ii) Cone. in gdm ^-3 = cone in moldm^-3 x molar mass
       Molar mass ofHCI = 1 + 35.5
       = 36.5 gmol^-1

       Conc. in gdm^-3     = 0.0554 moldm^-3 x 36.5g mol^-1
       = 2.02 g dm^-3

In (c)(iii), candidates could not calculate the mass of HCI in 20cm3 of solution. The expected solution from candidates was a follows:

1000cm³   contains 2.02 of acid solution
                :. 20cm³   will contain               2.02 x 20
                                                                        1000
                                                                = 0.0404g
Alternative method
Amount of acid = 0.554x 20
                                           1000

Mass of acid = amount x molar mass
                = 0.0554 x 20 x 36.5
                                1000
                                = 0.040g