Analytical Chem istry - DePauw University

170 Analytical Chemistry 2.0CA⎛S⎞A+ 0.006⎝⎜SIS⎠⎟samp=-12.11 ppb2. 80 + 0.006== 133 . ppb-1211 . ppbThe concentration of Pb 2+ in the sample of blood is 1.33 ppb.In some circumstances it is not possible to prepare the standards so thateach contains the same concentration of internal standard. This is the case,for example, when preparing samples by mass instead of volume. We canstill prepare a calibration curve, however, by plotting (S A /S IS ) std versus C A /C IS , giving a linear calibration curve with a slope of K.5DLinear Regression and Calibration CurvesIn a single-point external standardization we determine the value of k A bymeasuring the signal for a single standard containing a known concentrationof analyte. Using this value of k A and the signal for our sample, wethen calculate the concentration of analyte in our sample (see Example5.1). With only a single determination of k A , a quantitative analysis usinga single-point external standardization is straightforward.A multiple-point standardization presents a more difficult problem.Consider the data in Table 5.1 for a multiple-point external standardization.What is our best estimate of the relationship between S std and C std ?It is tempting to treat this data as five separate single-point standardizations,determining k A for each standard, and reporting the mean value.Despite it simplicity, this is not an appropriate way to treat a multiple-pointstandardization.So why is it inappropriate to calculate an average value for k A as donein Table 5.1? In a single-point standardization we assume that our reagentblank (the first row in Table 5.1) corrects for all constant sources of determinateerror. If this is not the case, then the value of k A from a single-pointstandardization has a determinate error. Table 5.2 demonstrates how anTable 5.1 Data for a Hypothetical Multiple-Point ExternalStandardizationC std (arbitrary units) S std (arbitrary units) k A = S std / C std0.000 0.00 —0.100 12.36 123.60.200 24.83 124.20.300 35.91 119.70.400 48.79 122.00.500 60.42 122.8mean value for k A = 122.5