Page 136 - Thermodynamics of Biochemical Reactions
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7.5 Dissociation of  a Diprotic Acid   133


         The plot of  NH versus pH is the titration curve for H,A.  Note that In P can be
         calculated by integrating (mH/[Hf])d[Hf].  Applying equation 7.5-4 to equation
         7.5-3 yields

                                                                          (7.5-5)

         At  very  high  pH,  the  binding  of  Hi  approaches  zero,  and  at  very  low  pH  it
          approaches  2.
             There is another way  to look at this binding, and that is to assume that the
          two groups are independent. In this case the dissociation  reactions are written
                                HA-  = H+ + A2-       Kl                  (7.5-6)

                                HAH = Hf + HA-         K2                 (7.5-7)

          In other words,  K~  is the dissociation  constant for the left hydrogen atom and K,
          is the dissociation  constant for the right hydrogen atom. In this case the binding
          is simply the sum of the bindings at the two sites:


                                                                          (7.5-8)

          This equation can be rearranged to





          Thus
                                                                         (7.5- 10)


                                       K,  = rcl  + K,                   (7.5-1 1)
          If  IC~ <<  K,,  K, is essentially equal to  K~  and K,  is essentially  equal to K,.  When
          the groups are independent, the binding polynomial is given by





             K,  and K, can  be evaluated by curve fitting  the plot  of  P  (equation  7.5-3)
          versus [H'],  or better log P  versus log[H+].  If  these two dissociation  constants
          have nearly the same magnitude, a quadratic has to be solved to evaluate K, and
          Kz:
                                     K,  5  JK:   - 4K,K2
                                 K2 =                                    (7.5- 13)
                                              2
          Note  that  K, cannot  be  equal  to  K,.  If  the  two  groups  are  identical  with
          dissociation constants K  (in equations  7.5-6 and 7.5-7), K,  = 42 and K2 = 2~.
             If  K1  = K2,

                                     P=(l+F)2                            (7.5- 14)


                                                                         (7.5- 15)

          Equation 7.5-4 yields


                                                                         (7.5-  1 6)
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