Question

5) A method for the determination ofcadmium in drinking water was developed. A calibration (Absorbance vs. ppm concentration) was performed and the best fit line was determined to have the equation y 0.256x-0.0019. The limit of detection was estimated to be 0.05 ppm. A blank sample was prepared and analyzed seven times giving the following responses: 0.003, 0.008, 0.002, 0.010, 0.004, 0.005, 0.012 AU (average 0.006 0.0038 AU). A 100 ppb solution was also prepared and measured seven times, giving the following responses: 0.025, 0.033, 0.018, 0.024, 0.027, 0.031, and 0.019 AU (average = 0.0253 0.0056 AU). What is the signal detection limit, the limit of detection, and the limit of quantification for the method? (15 points)

Answer 1

According to the best fit line obtained here, i.e. y = 0.256x - 0.119, sensitivity, b = slope of the equation = 0.256
The limit of detection = 0.05 = 3 * S/b ; where, S: standard deviation of the samples and b: sensitivity.

Now, for the readings of the blank sample, Mean value (Mean_blank) = 0.0063 and SD_blank = 0.0038
And for the low concentrated sample, Mean value = 0.253 and SD_{low concentration sample} = 0.0056

Hence, Signal detection limit = Mean_blank + 1.645 * SD_blank
  = 0.0063 + 1.645 * 0.0038 = 0.0125
Limit of detection = Signal detection limit + 1.645 * SD_{low concentration sample}
  = 0.0125 + 1.645 * 0.0056 = 0.0217

Limit of quantification = Mean_blank + 10 * SD_blank = 0.0063 + 10 * 0.0038 = 0.0443.