Help

[chloeukc]

Help I clearly shouldnt be doing a degree in forensics if I cant do this!!

 

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 125 mV obtained. Next a 0.100 cm3 of a standard solution containing 40 µg of lead per cm3 was added to 10 cm 3 of the sample solution and the analysis repeated. A response of 157 mV was obtained from the "spiked" sample. Calculate the amount of lead recovered by the swabs in µg.

[Fuzzwood]

The additional 4000 µg of lead added to half of the original sample caused a raise of 32 mV. Considering you used only half of the sample, that means you started with 62.5 mV. This means; 4000 µg of lead is the equivalent of 94.5 mV (157 - 125/2). Dividing those gives the response of the detector: 4000/94.5 = 42.4 µg of lead/mV

 

As the sample gave a signal of 125 mV, this means there was 42.4 * 125 = 5291 µg of lead in the swabs.