Question 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