-- 116 -


                 C2              C  K
          d4      e e , d, de             d32     (6.26)
               K02 K03          Kol(Ko2)2

Substituting these expressions into (6.24) reduces this equation to


               cd32 + 8d3 + y = 0                 (6.27)

where

                 HC 2 + (H-1)C K04
                   e   1 ( 2 e                    (6.28)

                   HC  (       (H+l) Ce2          (6.29)
                   + (H-1
                 Ko 2         K02 K03


             y = H(Ce - 1) + Ce                   (6.30)


        The solution of equation (6.27) is


             d3            V     ay1± 2           (6.31)

The coefficients a, 3, and y can be either positive or negative. The application of Descartes' rule of signs shows that there is only one possible combination of

   S  8, ± Y, ± Vi - 4ay/s2) which yields a realpositive value of the hydrogen ion density, d3. Once a, 3, and y are numerically evaluated, the proper choice of the radical sign determines the one acceptable solution of equation (6.27). The value of d3 is obtained from (6.31) and is