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