Question

A suspect in a shooting incident had his hands swabbed with 5% nitric acid. The swabs...

A suspect in a shooting incident had his hands swabbed with 5% nitric acid. The swabs were then extracted into 20 cm3 of nitric acid. A portion of the sample solution was analysed by atomic emission spectroscopy and a response of 120 mV obtained. Next a 0.010 cm3 of a standard solution containing 40 µg of lead per cm3 was added to 5 cm 3 of the sample solution and the analysis repeated. A response of 177.0 mV was obtained from the "spiked" sample. Calculate the amount of lead recovered by the swabs in µg.

For the analysis of lead (Pb) in sample

Response for sample solution (Rx) = 120 mV

Concentration of standard in solution (Cs) = (40 ug x 0.010 cm^3/cm^3) x 5 cm^3/5.010 cm^3 = 0.4 ug Pb

response of sample + standard (Rx + Rs) = 177 mV

let Cx be the concentration of lead in unknown sample

Cx/(Cx + Cs) = Rx/(Rx + Rs)

Cx/(Cx + 0.4) = 120/177

177Cx = 120Cx + 48

concentration of lead in dilute sample (Cx) = 0.842 ug

So the amount of lead recovered by the swabs = (0.842 x 5.01/5)(20/5) = 3.375 ug

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