This question was based on acid-base titration.  It was attempted by majority of the           candidates and their performance was good.

            In part (a), candidates correctly tabulated their burette readings and calculated the             average volume of acid used using concordant tire values.  However, some candidates
            lost marks due to alteration of their titre values and wrong averaging of titres. 

 

 

            In the part (b), majority of the candidates correctly calculated the concentration of
            A and B in moldm-3.  However, some candidates lost marks due to wrong substitution      and units.   In part (b) (iv) only a few candidates correctly calculated the volume of CO2
            evolved in the reaction at s.t.p.

            The expected answers include:
            Question 1

            (a)        (i)         Titrations:  4 x 2 sets
                                    Averaging(Vcm3)

                                    Na2CO3(aq) + 2HC1(aq)            2NaC1(aq) + H2O + CO2(g)      

            (b)        (i)         Conc. of B in moldm-3           
                                                Molar mass (Na2CO3) = (2 x 23) + 12 + (3 x 16) + 106 gmol-1
                                                Conc. of B (Na2CO3)  =         2.45           =   0.0925 mol dm-3 
                                                                               106 x 0.25
                        Alt. Method
                                                250 cm 3 of solution contained 2.45g of Na2CO3
                                    1000 cm 3 of solution contained  2.45 x 1000
                                                                                            250

                                                                                                                =              2.45x4
                                                                                    =          9.80g
                                    M(Na2CO3)  =  106gmol-1
                                    Conc. B(Na2 CO3)   =   9.80      =0.0925 moldm-3
                                                                                                                106
                                                Conc. of A in moldm-3
                                    From reaction equation
                                           CA VA          =     2                                               CA =   2 x CB x VB
                                                CBVB                   1                                                       VB

                                                CA =   2 x 0.0925 x 20/25    =  a moldm-3 (correct evaluation to 3 sig. figs)
                                                            VA      

                                    Conc. of A in gdm-3   
                                    M(HC) = 1 + 35.5      =  36.5gmol-1

                                    Conc. of A in (gdm-3)  = a x molar mass (HC1)
                                    = a x 36.5
                                    = b gdm-3
                 

 

                        (iv)       Volume of the gas evolved
                                    From reaction equation
                                    1 mole of Na2CO3  =  1 mole of CO2 at stp
                                    n(CO2)  =   n (Na2CO3)  =  0.0925 mole
                                        … Vco2    =    n x Vm
                                                   =  0.0925 x 22.4dm3
                                                   =  2.07dm3 at stp
                                    … Vco2    liberated in 25 cm3 of Na2CO3 will be 2,07 x 25 x   25
                                                                                                                                                                                  1000
                                                                           = 0.0518 dm3/51.80 cm3 at stp