Question

A sample of pure water was spiked with 0.565 ng/mL silver ion. Ten replicate determinations of the spiked water sample gave 0.575, 0.555, 0.532, 0.518, 0.559, 0.539, 0.543, 0.547, 0.525, and 0.510 ng/mL silver ion. Determine the mean percent recovery of the spike and the detection limit (in ng/mL) of the analytical method used for silver ion determination

Answer 1

mean recovery of spike = 0.575 + 0.555 + 0.532 + 0.518 + 0.559 + 0.539 + 0.543 + 0.547 + 0.525 + 0.510/10 = 0.5403

mean % recovery of spike = 0.5403 x 100 = 54.03%

detection limit = 3 x standard deviation

standard deviation = sq.rt.[sum(mean-value)^2]

= 1.20409 x 10^-3 + 2.1609 x 10^-4 + 6.889 x 10^-5 + 4.9729 x 10-4 + 3.4969 x 10^-4 + 1.69 x 10^-6 + 7.29 x 10^-6 + 4.489 x 10^-5 + 2.3409 x 10^-4 + 9.1809 x 10^-4/10

= 3.5421 x 10^-4

standard deviation = 0.01882

detection limit = 3 x 0.01882 = 0.05646 ng/ml