Question

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.

Answer 1

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