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              112    Modern Analytical Chemistry


                                                  EXAMPLE  .4
                                                         5
                                                  A third spectrophotometric method for the quantitative determination of the
                                                  concentration of Pb 2+  in blood yields an S samp of 0.193 for a 1.00-mL sample of
                                                  blood that has been diluted to 5.00 mL. A second 1.00-mL sample is spiked
                                                  with 1.00 mL of a 1560-ppb Pb 2+  standard and diluted to 5.00 mL, yielding an
                                                  S spike of 0.419. Determine the concentration of Pb 2+  in the original sample of
                                                  blood.
                                                  SOLUTION
                                                  The concentration of Pb 2+  in the original sample of blood can be determined
                                                  by making appropriate substitutions into equation 5.7 and solving for C A . Note
                                                  that all volumes must be in the same units, thus V s is converted from 1.00 mL to
                                                         –3
                                                  1.00 ´10 mL.
                                                                                           .
                                                                0 193                     0 419
                                                                 .
                                                                        =
                                                                                                         3 –
                                                               æ 100 mL   ö  æ 100 mL   ö      æ 100 ´ 10  mL ö
                                                                               .
                                                                                                 .
                                                                 .
                                                            C A ç      ÷  C A ç      ÷ + 1560 ppb   ç        ÷
                                                                 .
                                                                               .
                                                               è 500 mL   ø  è 500 mL   ø      è    5.00 mL  ø
                                                                   .
                                                                                  .
                                                                  0 193          0 419
                                                                        =
                                                                  .       0 200C A + 0 312 ppb
                                                                                    .
                                                                           .

                                                                 0 200C A

                                                                           .
                                                                .
                                                      .
                                                     0 0386C A + 0 0602 ppb = 0 0838C A

                                                                           .
                                                                 .
                                                                0 0452C A = 0 0602 ppb
                                                                     C A =133.  ppb
                                                                       A
                                                                          2+
                                                  Thus, the concentration of Pb in the original sample of blood is 1.33 ppb.
                                                  It also is possible to make a standard addition directly to the sample after mea-
                                              suring S samp (Figure 5.6). In this case, the final volume after the standard addition is
                                              V o + V s and equations 5.5–5.7 become
                                                                           S samp = kC A
                                                                        æ     V o         V s  ö
                                                                S spike =  k C A   +  C S     ÷                5.8
                                                                        ç
                                                                        è   V o + V s   V o + V s ø
                                                                                Add V  of C S
                                                                                    S
                                                     V o                   V o
              Figure 5.6
              Illustration showing an alternative form of
              the method of standard additions. In this  Total concentration  Total concentration
              case a sample containing the analyte is  of analyte       of analyte
              spiked with a known volume of a standard                 V         V
              solution of analyte without further diluting  C A   C A   o  + C S  S
                                                                      o
                                                                                o
              either the sample or the spiked sample.                V + V S   V + V S
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