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Mansoura National University
            Pharm D-Clinical Pharmacy Program                 Level 1             Pharm. Anal. Chem. 1 (PC 101)


               Applications of Acid-Base Titrations

                          Titration curves for polyprotic (polybasic) acids
          •  A polyprotic (polybasic) acid can release more than one
                +
              H  ion in successive steps.
          •  A polybasic acid may be treated as a mixture of acids;
              each stage of dissociation is treated as a separate
              monobasic acid.



          ✓  A polyprotic acid can be titrated stepwise with good
              end-point breaks (inflections) for each proton if:
                   the difference in pK  values is at least 4  [i.e. (pK  ̶  pK ) ≥ 4 ]
                                          a                                 a2      a1
                                                           4                              4
               or the ratio of K  values is at least 10   [i.e. "        " /"     "   ≥ 10 ]
                                   a                                               
                  of each two successive protons.

                                       -                   more than   Less than
                              +
                  H A   ⇌ H  + H A  ………………….. K      ,     pK
                    3               2                           a1           a1
                       -      +       2-
                  H A  ⇌ H  + HA  ………………….. K      ,     pK
                    2                                           a2           a2
                      2-      +     3-
                  HA  ⇌ H  + A  ……………………. K      ,     pK
                                                                a3          a3



                                       Example#1: Sulfuric acid (H SO ):
                                                                              2    4

           H SO  is a strong diprotic acid. Its ionization can be expressed as follows:
             2   4
                               +         -
                 H SO  → H  + HSO  [complete ionization, pK  is too small (supposed 0)]
                   2   4                4                            a1
                       -      +        2-
                 HSO  ⇌ H  + SO  …………………..….. pK  = 2.1
                       4              4                               a2
            ✓  The difference between the two pK  values is < 4 so stepwise
                                                      a

               titration of each proton is not possible and only one inflection

               (one equivalence point) is obtained.


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