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So I have to find the concentration of lead in seawater. I prepared a 50 mL sample of seawater containing 0.1 M KCl, 50 mM HNO3 and 30 ppm HG2+. I took 15 ml of the above sample and spiked the sample with 1 ppm Pb2+ and Cd2+ to get a final concentration of 0.01 ppm Pb2+ and Cd2+. Then i spiked the sample again with the 1ppm Pb2+ and Cd2+ to make 0.025 ppm Pb2+ and Cd2+. I continued to spike the sample until I go 0.05, 0.075, and 0.1 ppm Pb2+ and Cd2+. For lead my results were:

0 ppm = -719 mV
0.05 ppm = -708 mV
0.075 ppm = -702 mV
0.1 ppm = -698 mv

The above is for the seawater sample. My standards are below:

0.05 ppm = -719 mV
0.1 ppm = -718 mV
0.5 ppm = -708 mV

Can someone please show me how to get this concentration?

2006-11-03 01:30:41 · 1 answers · asked by Anonymous in Science & Mathematics Chemistry

1 answers

Your question makes no sense.
If I understand correctly you are making artificial sea water by making a solution of KCl, HNO3,Hg(+2), then you are adding Pb(+2) and Cd(+2) to known concentrations... measuring potential (what method?) and you want to determine the concentration of...what? You know the concentrations of all species (unless a reaction is taking place)

2006-11-03 22:14:00 · answer #1 · answered by bellerophon 6 · 0 0

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