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98                  Basic physical chemistry

                              [HA]initial = [HA(aq)] + [ A - (aq)]   (5.35)
            where [HA] initial  is the amount of acid in the water before any dissocia­
            tion occurs, and [HA(aq)]  and [A - (aq)] are the equilibrium concentra­
            tions of HA(aq) and A - ( aq)  in  Reaction (5. 3 1 ) .  Note  that  Eq.  (5 . 3 5) is
            written as an approximate relation because it assumes that the mass of
                                                         -
            hydrogen atoms that  have  dissociated from the A  ( aq)  is  small  com­
            pared  to  the  masses  of the  species  in  Eq.  (5.35).  Equation  (5.35)  is
            called the mass balance relation.
              The fourth relationship between the concentrations of the  species in
            Reactions  (5 . 3   I )   and  (5.32)  expresses  the  fact  that  the  solution  is
            electrically neutral ;  therefore,  the total concentration of positive ions
            must equal the total concentration of negative ions

                            [Hp + (aq)] =  [ A - (aq)] + [OH - ( aq)]   (5 .36)

            This  is  called the charge balance relation.  In practice,  the solution of
               s
            Eq .   (5 . 3 3)  through  (5 .36)  may  involve  some  laborious  algebra,  but
            judicious approximations can often simplify the solution, as illustrated
            in the next section.



                                5 . 1 2   The pH  of rainwater
                     5
            Exercise  . 2   suggested that the absorption of C02 in rainwater to form
            a weak solution of carbonic acid would give natural rainwater a pH of
            about  . 6 .   We  can  now  consider this exercise quantitatively  by  using
                  5
                                   i
            some of the concepts just  n troduced.
               As  we have seen, carbonic acid is  polyprotic
                        H2C0 (aq) + H20(l ) µ  H  C0 3 ( aq) + H 3 0 + ( aq)   (5 .37)
                             3
                        HC0 3 (aq) + H20(l) µ C05 - ( aq) + H3Q + (aq)   (5.38)

            with successive acid-dissociation constants at 25°C of  Kai = 4 . 2   x  1 0 - 7
            and Ka = 2  5 .0 x 1 0 - 1 1 •   Therefore,

                         [H 3 0 + (aq)][HC03 (aq)]
                              [H 2COiaq) ]    =  K  ai = 4 . 2 x 1 0 - 1   (5 . 3 9)

             and,


                                                                      (5 .40)
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