The question was attempted by many candidates and the  performance was good.
                      In (a),  candidates were able to correctly carry out the titration experiment, tabulate  the burette readings and calculate the average volume of acid used.  However, some of them lost marks because they  lacked the knowledge of concordancy in precision measurement and cancelled out  their tables to retabulate in order to conform with the supervisor’s titre  values.
                      In (b)(i), most  candidates calculated correctly the concentration of E in moldm-3 as follows:
                          
                          (i)   CDVD     =     1       
        CE  VE           1
                            CE     =     CDVD     
                           VE          
                              =     0.095 xVD
                       VE
   
          =     P moldm-3  say
                                                                    OR
  1000 cm3  of D   contain 0.095 moles of HCI
 \ VD  of D will contain 0.095 x VD   moles of HCI
                                              1000
                      From the equation, 
                      1mole of HCI  reacts with 1  mole of NaHCO3
                       ∴   0.095 x VD   mole   HCI  will react with    0.095 x VD   moles of    NaHCO3
             1000                                                             1000
             
                        Hence,
                               20/25 cm3  of E contain  0.095 x VD  moles of NaHCO3
                                                      1000
       
     ∴     1000 cm3 of E will contain  0. 095 x VD  X   1000
                                                              1000               20/25
                                                =      0. 095 x VD  =   P  moldm-3  say
                                                                20/25
In (b)(ii), only  few candidates correctly calculated the solubility of  NaHCO3 thus:
                                              The dilution factor of  solution E is 10:1000 = 1:100
            Solubility of   NaHCO3   =    100 x  P moldm-3   =    Q moldm-3  
                                                              OR
CfVf  =   C1V1 
                        
C1     =    CfVf    
                   V1
         =    P x 1000
                     10
=   100 x P  = Q Say
                        
                        In (c)(i),  candidates correctly stated that damp blue litmus paper would turn  red/claret/pink or there is no effect on damp red litmus paper.
                      In (c)(ii)  candidates stated that the gas when passed into lime water turns it milky.
                      In (d),  candidates were able to correctly name one chemical industry that uses the gas and listed two other processes that release the gas into the atmosphere to  include:
                      (i)   manufacture  of Na2CO3/solvay process/food/drinks industry (aerated  drinks, soda water/beer/dry ice)
                      (ii)  -    respiration
       -    burning of fossil fuel
-    decay of organic  matter   
-    volcanic eruption
-    fermentation of sugars  etc.