A sample of pure water was spiked with 0.535 ng/mL silver ion. Ten replicate det

Question

A sample of pure water was spiked with 0.535 ng/mL silver ion. Ten replicate determinations of the spiked water sample gave 0.545, 0.525, 0.502, 0.488, 0.529, 0.509, 0.513, 0.517, 0.495, and 0.480 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. Mean percent recovery: Number 95.38 Detection limit Number 0.1785 % recovery ng/ mL There is additional feedback available! ew this feedback by clicking on the Incorrect. ttom divider bar. Click on the divider ar again to hide the additional feedback. Close

Explanation / Answer

Average Silver Found is

0.545+0.525+0.502+0.488+0.529+0.509+0.513+0.517+0.495+0.480 / 10 = 0.5103 ng/mL

Cspiked = 0.5103 ng/mL, Cunspiked =0,Cadded = 0.535 , Yblank =0,

Standard Deviation(s) of the above data is 0.019838 ,

% Recovery = [(Cspiked–Cunspiked) / Cadded] * 100 = [(0.5103-0)/ 0.535 ]*100 = 95.38 %

Detection Limit =>

Ydl = Yblank + 3s = 0 + 3*standard Deviation(s) = 0+3(0.019838) = 0.0595 ng/mL

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