4. (15 points) The following calibration data were obtained for the analysis of

Question

4. (15 points) The following calibration data were obtained for the analysis of Zn using atomic absorption spectrometer Zn Concentration No. of replicates Std. deviation (s) 0.000 5.000 10.00 Mean Value of Signal (S) 0.0654 0. 607 1.26 0.011 0.063 0.103 5 A least-square analysis of these data yielded the following equation S = 1.26 CZn + 0.013 Calculate a) S/N ratio at 5.00 ppm b) Analytical sensitivity at 10.0 ppm c) Detection limit of signal (Sm) d) Lowest concentration that can be analyzed using this method? Limit of quantification e)

Explanation / Answer

Answer:

a) For any set of data the S/N ratio is,

S/N = (smean of signal)/(standard deviationof the signal)

So, for 5.00 ppm sample S/N ratio is = 0.607/0.063  = 9.635

b) The analytical sensitivity is = (Signal of the corresponding sample)/(Concentration of the sample)

   So for 10 ppm sample, analytical sensitivity = 1.26/10    ppm-1 = 0.126 ppm-1             

c) Detection Limit (Sm) = (Sample concentration ×3 Standard deviation)/(Mean signal)

        So, for   0 ppm sample, Sm = (0 3 0.011)/0.0654 = 0 ppm

               For 5 ppm sample set, Sm = (5 3 0.063)/0.607 = 1.5568 ppm

                For 10 ppm sample set, Sm = (10 3 0.103)/1.26 = 2.4523 ppm

Generally the detection limit has taken as average of two or more set of measurements, so average detection limit = (0+1.5568+2.4523) ppm = 1.3364 ppm

d) So lowest concentration can be analyzed is 1.3364 ppm.

e) Limit of quantification = Signal at zero conc. + 10 * standard deviation for zero conc.

                                         = 0.0654 + (10 * 0.011)

                                         = 0.1754

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