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1400-CH09  9/9/99  2:13 PM  Page 329






                                                                          Chapter 9 Titrimetric Methods of Analysis  329

                 The principle of the conservation of electron pairs is easily extended to other com-
                 plexation reactions, as shown in the following example.



                     EXAMPLE  .9
                            9
                                         –
                     The concentration of Cl in a 100.0-mL sample of water drawn from a fresh
                     water acquifer suffering from encroachment of sea water, was determined by
                     titrating with 0.0516 M Hg(NO 3 ) 2 . The sample was acidified and titrated to the
                     diphenylcarbazone end point, requiring 6.18 mL of the titrant. Report the
                                    –
                     concentration of Cl in parts per million.
                     SOLUTION
                     Conservation of electron pairs requires that
                                                 –
                                         Moles Cl =2 ´moles Hg 2+
                                                               –
                     Making appropriate substitutions for the moles of Cl and Hg 2+
                                           g Cl -
                                                 = 2  ´M Hg ´V Hg
                                              AW Cl
                     and rearranging leaves us with
                                           –
                                        g Cl =2 ´M Hg ´V Hg ´AW Cl
                     Substituting known values and solving gives

                              2 ´0.0516 M ´0.00618 L ´35.453 g/mol = 0.0226 g Cl –
                                        –
                     The concentration of Cl in parts per million, therefore, is
                                       mg Cl -   22 6  mg
                                                   .
                                               =         =226  ppm
                                         liter   0.1000 L




                 Finally, quantitative problems involving multiple analytes and back titrations also
                 can be solved by applying the principle of conservation of electron pairs.



                            9
                     EXAMPLE  .10
                     An alloy of chromel containing Ni, Fe, and Cr was analyzed by a
                     complexation titration using EDTA as the titrant. A 0.7176-g sample of the
                     alloy was dissolved in HNO 3 and diluted to 250 mL in a volumetric flask. A
                     50.00-mL aliquot of the sample, treated with pyrophosphate to mask the Fe
                     and Cr, required 26.14 mL of 0.05831 M EDTA to reach the murexide end
                     point. A second 50.00-mL aliquot was treated with hexamethylenetetramine
                     to mask the Cr. Titrating with 0.05831 M EDTA required 35.43 mL to reach
                     the murexide end point. Finally, a third 50.00-mL aliquot was treated with
                     50.00 mL of 0.05831 M EDTA, and back titrated to the murexide end point
                                               2+
                     with 6.21 mL of 0.06316 M Cu . Report the weight percents of Ni, Fe, and
                     Cr in the alloy.
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