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Chemistry Paper 3,Nov/Dec 2010  
Questions:         1 2 3       Main
General Comments
Weakness/Remedies
Strength

























































Question 1

In an experiment, 20.0cm3 portions of 0.065 mol dm-3NaOH were titrated against dilute HCl. The table below shows the results of the titration

Burette readings {cmt]

1st

2nd

3rd

Final reading [cm-]

23.50

46.60

47.40

Initial reading [cmt)

0.00

23.50

24.00

 

 

 

--

Volume of acid used

23.50

23.10

23.40

(a) (i) Name a suitable indicator for the titration. Give a reason for your answer.
    (ii) Give the colour of the indicator in the base and at the end point.
    (iii) What type of reaction is demonstrated by the experiment? [ 5 marks]

(b) (i) Write a balanced equation for the reaction
    (ii) Determine the average volume of acid used. [ 3 marks ]

(c) Calculate the
    (i) concentration of the acid in mol dm-3
    (ii) concentration of the acid in g dm-3
   (iii) mass of HCl in 20cm3 of solution
   [H = 1.00, Cl = 35.5 ]

                                                                                                                   

_____________________________________________________________________________________________________
OBSERVATION

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





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