This question was attempted by nearly all the candidates and the performance was poor.
In (a)(i) most candidates could not calculate the amount of the acid used. 
To calculate the amount in mole of either the acid or base, it is expected that the volume in dm3 is multiplied by the molar concentration in mol dm-3 however, majority of the candidates were using MaVa = na
MbVb     nb  which is not applicable in this case.

1. some candidates assumed values for (a)(i) to solve (a)(ii).
2. a number of candidates lost marks because of poor knowledge of how to express mole ratio mathematically.

 

In (b)(i) most candidates could not use the mole ratio obtained to determine the basicity of the acid;

1. Most candidates either gave wrong acid or gave wrong reason for the choice of the acid;
2. Many candidates could not balance the expected equation.

The expected answers were:

(a)        (i)         Let the concentration of acid in mol dm-3 = CY
                                Amount of acid, nY  =   24.80 CY
                                                               1000
                                                        =   0.0248 CY

 

 

 

1. Molar mass of Na2 CO3, C = (2 x 23) + 12 + (3 x 16) = 106 g mol-1

Conc. of Na2 CO3  Cs  5.83
                                   106
                           =  0.0550 mol dm-3
Amount of Na2 CO3, ns  = 25.0 x 0.0550
                                                  1000
                                       =   0.001375 mol
                                       =   0.00138 mol
                                                        

1. Mole ratio of acid to Na2 CO3  = ny : ns

The amount of acid, nY is either equal to, or multiples of Na2 CO3, ns

                        EITHER
                        If ny : ns  ⇒  0.0248 CY  = 0.00138
                                                         CY = 0.0556 mol dm-3
                        Hence, ny : ns = (0.0248 x 0.0556) : 0.00138
                                               = 0.00137 : 0.00138
                                                =      1      :        1

                        OR
                        If ny  = 2 x  ns ⇒  0.0248CY = 2 x 0.00138
                                                                        = 0.00276
                                                            CY  =  (0.00276)/0.0248
                                                                        = 0.111
                        Hence, ny : ns =  (0.0248 x 0.111) : 0.00138
                                                =  0.00276 : 0.00138
                                               
                                                =  0.00276 : 0.00138
                                                =  0.00138   0.00138
                                                =       2       :       1
ALTERNATIVE METHOD
1(a)      (i)         Let the concentration of acid, CY, in mol dm-3 = 0.0500 or 0.100.
                        (The only two reasonable options)
                        Amount of acid, nY     =    24.80 CY
                                                                  1000
                                                            =   0.0248  CY
                                                                 (0.00124 mol. or 0.00248 mol.)
                                                                 
            (ii)        Molar mass of Na2 CO3  = (2 x 23) + 12 + (3 x 16) = 106 g mol-1     
                       
                        Conc. of Na2 CO3, Cs  =   5.83
                                                                 106
                                                          =   0.0550 mol dm-3 correct evaluation
                        Amount of Na2 CO3, ns  = 25.0 x 0.0550
                                                                       1000
                                                            =   0.001375 mol
                                                            =   0.00138 mol
                                                           
            (iii)       Mole ratio of acid to Na2 CO3  = ny : ns
                        The amount of acid, nY is either equal to, or multiples of Na2 CO3, ns
EITHER
ny : ns      =   0.00124  :  0.00138

                   =   0.00124     0.00138
                        0.00124     0.00124
                                                  1        :        1
                        OR
                        nY  :  nS  =    0.00248 : 0.00138
                                                                                               
                                         =  0.00248 : 0.00138
                                             0.00138   0.00138
                                                1.80   :      1
                                        =       2       :      1
 (b)       EITHER
            (i)         When mole ratio of acid to Na2CO3 is 1 : 1; basicity of acid Y = 2                

                       (ii)        Y is H2SO4                                                                                                                                                                                  Since the mole ratio of acid Y to Na2CO3 is 1 : 1, one mole of acid Y provides two                         moles of hydrogen ions.
            (iii)       H2SO  + Na2CO3                Na2CO3  + H2O  + CO2
         
            OR

1. When mole ratio of acid Y to Na2CO3 is 2 : 1; basicity of acid Y = 1
2. Y is HCI/HNO3

Since the mole ratio of acid to Na2CO3 is 2 : 1, then, one mole of acid Y will provide one mole of hydrogen ions.

1. 2HCI  + Na2CO3                     2NaCl  +  H2O   +  CO2

OR
2HNO3  + Na2CO3                    2NaNO3 + H2O + CO2