Question 1B
           E is a  solution containing 2.92 g of HCl per dm3.
              F is a  solution obtained by diluting 20.0 cm3 of a saturated solution of  Y(OH)2 at 25 ºC 
           per dm3 of solution.
- Put E into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portion of F using phenolphthalein as indicator.
 
Repeat the titration to obtain concordant titre values.
Tabulate your results and calculate the average volume of acid used.
The  equation for the reaction is:
   Y(OH)2  + 2HCl  -> YCl2   +   2H2O 
- From your results and the information provided, calculate the:
 
(i)            Concentration  of HCl in E mol dm-3;
     (ii)           concentration of  Y(OH)2 in F in mol dm-3-;
     (iii)          solubility  of the substance Y(OH)2 in mol dm-3;
     (iv)          mass  of Y(OH)2 that would be deposited if 1 dm3 of saturated  solution is  
     evaporated  to dryness.
[H=1.0; O =16.0; Cl = 35.5; Y(OH)2 = 74.0]
Observation
This question was on titration experiment. Majority of the candidates that responded to this question performed above average.
In part (a), majority of the candidates obtained concordant values from the titration experiment;
In part (b), majority of the candidates were able to calculate the concentration of HCl in E in mol dm-3 However, they could not calculate the solubility of the substance Y(OH)2 in mol dm-3.
The expected answers include:
(a) Two concordant  titres 
    Averaging 
    (b) (i) Molar mass  of HCl = 1 + 35.5 
    = 36.5 g mol--1
    Concentration of E = 2.92 
    36.5 x 1 
    = 0.080 mol dm-3 
  (ii) From the equation 
    CEVE = 2  
    CFVF    1
  0.080 x VE = 2 
    CF  x VF            1
CF =   0.080 x VE 
                    2 x VF 
    = Say W mol dm-3
(iii) C1V1(Saturated  solution) = C2V2 (dilute solution) 
    C1  x 20 = W x1000 
C1 = W x 1000 
                    20 
    =  Say X mold m-3 
  Alternative  
    20cm3 of saturated solution contains W moles 
    Hence 1000 cm3 of saturated solution = W x 1000 
                                                                        20 
    = Say X mol dm-3
    (iv)      Mass of Y(OH)2 deposited 
    = X x molar mass (Y(OH)2) 
    = X x 74 g 
    = Z g say