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94                   Basic physical chemistry

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            acid-base con u gate pair provided that neither the  acid nor  the  base is
             very  strong  [e.g. ,  acetic  acid,  HC2H30 ,  and  the  acetate  ion,
                                                    2
            C2H30 2 ( aq)] .  If a  small  amount  of a  strong  acid  is  added  to  s u ch  a
             solution,  most  of  the  added  H30 + (aq)  will  combine  with  the  weak
             base  of the buffer (to form the con u gate  acid of that weak base),  so
                                            j
             that  the  H3Q + ( aq)  concentration  and  pH  of  the  solution  will  not
                                     l
            change  very  much.  Similar y ,   if a  small  amount  of a  strong  base  is
            added  to the  solution,  most of the  oH - ( aq)  will  combine  with  the
            weak acid of the buffer.  From the equilibrium equation for the ioniza­
             tion of acetic acid, the reverse reaction of (5. 2 3), we have

                                        [C 2 H30i (aq)][H30 + (aq)]
                        Ka  (acetic acid)
                                                 H 0
                                             [HC 2 3 2 (aq)]
            Therefore,  to  achieve  a  specific  pH,  or  [H3Q + ( aq)] ,  the  ratio  of the
            concentration of acetic acid to acetate ion must be
                                                   +
                              [HC2H30 (aq)]   [H3Q (aq)]
                                      2
                              [C 2 H30i (aq)]   Ka(acetic acid)
             We  can  generalize  the  above  ideas  as  follows.  Let  us  represent  the
             weak acid in a buffer by HA and  the corresponding  salt  by MA. The
             equilibrium reaction for the acid dissociation is

                                HA(aq) � H + (aq) + A  - (aq)
             and the acid-dissociation constant is given by
                                      [H  + (aq)][A  - ( aq)]
                                  K                                   (5 .26)
                                          [HA(aq)]
                                   a
             Or, since A - ( aq) is the base and HA(aq)  the acid in  the  buffer,
                                             [acid]
                                   [H + (aq)]  = [base]  K a          (5 .27)


             From Eq. (5. 2 6)
                                                   [A - (aq)]
                             logKa - log[H  (aq)] + log
                                         +
                                  _
                                                   [HA(aq)]
             or,
                                                [A  - ( aq)]
                                - logK = pH - log  [HA(aq)]
                                      a
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