I would proceed this way.
A = abc and since these measurements were made with the same length cell (I assume) we an ignore that. We need to
1. graph the data for a working curve (A vs C) or
2. determine a from the data and that is easily done.
A = a*c
0.120 = a*0.1 and
0.360 = a*0.3. Solv for a and in both cases a = 1.2. That allows to you determine c for the 100 mL sample.
A = a*c; 0.215 = 1.2*c and calculate c in ppm. I'll estimate that at about 0.18 ppm or so. That's c in the 100 mL. What was it in the 1.00 L? That's approx
0.18 x 100/0.05 = about 360 ppm
You know 1 ppm is 1 g Pb in 1o^6 mL solution.
Convert 360 ppm to g in the 1000 mL, then
%Pb = (g Pb in sample/g sample)*100 = ?
Post your work if you get stuck.
Any help/advice on answering this question would be a big help! :)
The analysis of lead (Pb) in a sample of solder was carried out using atomic absorption spectroscopy as follows.
1.013 g of solder was completely dissolved in a small volume of nitric acid and the solution made up to 1.000 L with distilled water. A 50.0 μL aliquot of this was then diluted to 100.0 mL with distilled water, and when analysed by atomic absorption in an air/acetylene flame it gave an absorbance of 0.215 at 283.3 nm. The instrument was calibrated, after zeroing with distilled water, with standard solutions of lead containing 0.100 and 0.300 ppm Pb which gave absorbances of 0.120 and 0.360 respectively.
Calculate the wt % of Pb in the solder.
2 answers
DrBob222, thank you for the help!
1 answer