Chemistry Paper 3 WASSCE (PC), 2018

Question 1

 

  1. A was prepared by dissolving 6.50 g of an impure solid Y(OH)2.XH2O in 250 cm3 of solution.

B is a solution of HCl containing 0.025 mol in 250 cm3.
Titration of 25 cm3 portions of A with B using methyl orange as indicator gave the following results:

Burette readings / cm3

1

2

3

Final reading

33.20

30.00

 

Initial reading

16.80

7.50

1.50

Volume of B used/cm3

 

 

16.65

(a) (i)   Complete the table.

(ii)  Calculate the average volume of B used.
Equation of the reaction is:

(b) From the information provided, calculate the:

concentration of

I. Y(OH)2. XH2O in moldm-3;

II. Y2+ ions in moldm-3;

III. OH- ions in moldm-3.

(ii)  mass of pure Y(OH)2. XH2O in 250 cm3 of A.

(iii) percentage purity of Y(OH)2. XH2O.
[Y(OH)2. XH2O = 150 g mol-I]


Observation

In part (a), majority of the candidates could not complete the table and calculate    the average volume of B used.

In part (b), majority of the candidates could not calculate the concentration of the base in moldm-3. Also, only few of them could calculate the concentration of Y2+  ions in moldm-3. In addition to this, majority of them could not calculate the mass           of pure Y(OH)2. XH2O in 250 cm3 of A.

The expected answers include:

(a)        (i)


Burette reading / cm3

1

2

3

Final reading

33.20

49.60

18.15

Initial reading

16.80

33.20

1.50

Volume of B used / cm3

16.40

16.40

16.65

(ii)        Average volume of B used
VB = 16.40 + 16.40

           2                                                                            

= 16.40 cm3 

(b)        (i)
I.          Concentration of Y(OH)2.XH2O in moldm-3
            From the equation of reaction
            nY(OH)2.XH2O = 1                                                                               n(HCl)      2

            CHCl = 0.025 mol              

                 0.25 dm3
=0.100moldm3                                                                                               
            CY.XH2O = CHCl xVHCl                                                     

                2 x VY(OH)2.XH2O        
= 0.100 x 16.40                                                      

    2 x 25.0
= 0.0328 moldm-3                  

   

OR

(b)  (i)        
Concentration of Y(OH)2.XH2O

CHCI in moldm-3
250 cm3≡ 0.025 mol
1000 cm3 = 0.025 x 1000= 0.100 mol                                                  

           250
CHCI=0.100moldm3=CB                                                                                                                  
CAVA = 1 mole ratio                                                                                       
CBVB   2

= 0.0328 moldm-3              
                                                                           
OR

 

 

II.        Y2+ ions in moldm-3

  Y(OH)2.XH2O → Y2+ + 2OH+XH2O                                                   
n(Y2+)  = 1

n(Y.XH2O) 1

C(Y2+) = C(Y(OH)2.XH2O)
=  0.0328 moldm-3       

                                                                 

III.  OH- ions in moldm-3
 n(OH-)      = 2                                                                                   
n(Y(OH)2.XH2O)  1

  C(OH-) = 2 x 0.0328
=  0.0656 moldm-3     

                       

 

(iii)  Percentage purity of Y(OH)2.XH2O

% purity = mass of pure   x 100

Mass of impure
= 1.30  x 100                                                                                   
6.50

  = 20.0