Question

A sample of pure water was spiked with 0.670 ng/mL silver ion. Ten replicate determinations of the spiked water sample gave 0.680, 0.660, 0.637, 0.623, 0.664, 0.644, 0.648, 0.652, 0.630, and 0.615 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

Pure Watersample spiked with 0.670 ng/mL of Silver ion.

% recovery = (spiked sample reading / standard sample concentration) x 100

mean % recovery = sum of % recovery / 10

Readings for silver ion obtained (ng/mL)         % recovery

0.680                                                                101.49254

0.660                                                                 98.50746

0.637                                                                 95.07462

0.623                                                                 92.98507

0.664                                                                 99.10447

0.644                                                                 96.1194

0.648                                                                 96.71641

0.652                                                                 97.31343

0.630                                                                 94.02985

0.615                                                                 91.79104

mean % recovery of the spike = sum of % recovery/10 = 96.313429

To calculate Detection limit

standard deviation is needed

mean average of readings = 0.6453

calculate variance

(0.680-0.6453)^2 = 1.20409 x 10^-3

(0.660-0.6453)^2 = 2.1609 x 10^-4

(0.637-0.6453)^2 = 6.889 x 10^-5

(0.623-0.6453)^2 = 4.9729 x 10^-4

(0.664-0.6453)^2 = 3.4969 x 10^-4

(0.644-0.6453)^2 = 1.69 x 10^-6

(0.648-0.6453)^2 = 7.29 x 10^-6

(0.652-0.6453)^2 = 4.489 x 10^-5

(0.630-0.6453)^2 = 2.3409 x 10^-4

(0.615-0.6453)^2 = 9.1809 x 10^-4

variance = 3.5421 x 10^-4

standard deviation = sq.rt(variance) = 0.0188204

Thus, Detection limit = 3 x standard deviation = 3 x 0.0188204 = 0.0564612 ng/ml