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AAS pollution help (1 Viewer)

BabzBrah

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Can someone please explain to me how to do this question I've tried it many times and failed, I currently can't do any of these type of questions but if i find out the method I should learn how to do them.

In order to measure the concentration of mercury in sea water near an industrial plant, a team of chemsits used AAS. They first calibrated their instrument by admitting some standard solutions to it and determining the absorbance for each. They then measured the absorbance level of several water samples collected from different locations near the industrial plant. For their standards they made a stock solution by dissolving 0.106g mercury [as mercury (II) nitrate] in water and making the volume to 250mL in a volumetric flask, then they quantitatively diluted (by pitette) 1mL, 2mL and 4mL of this solution to 100mL (volumetric flask). Absorbances for the standards and the samples are shown below.

Standards (ml) Absorbance
1 0.083
2 0.164
4 0.331

Samples Absorbance
A 0.32
B 0.46
C 0.21
D 0.055

Draw a calibration curve of absorbance versus concentration in parts per million (ppm) for the standards. Take 1.00mL of aqueous solution as 1.00g. Determine the concentration of mercury (in ppm) in each of the samples.


What I did which turned out wrong was to put the weight of mercury of that of Hg(NO3)2 and find the amount of mercury in the 0.106g sample. So 200.6/312.61 x 0.106 = 0.0680g mercury. Then divided that by 250 to get 2.72mg/L

so thats 2.72ppm in 1ml
5.44ppm in 2ml
10.88ppm in 4ml I drew the calibration curve and what not and still got it wrong. Please help me and let me know what mistakes im making? or what im not doing. Thanks :))
 

golgo13

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I'll prolly need another member to confirm, cause AAS calculation is really rusty but, the first bit up until you give me divide by 250 looks ok, it should be 2.72mg/ml not 2.72mg/L. Other than that i'm gonna need some answers to reverse engineer how you got it wrong XD. Only other thing i could see is that the 0.106g sample is already taken into account the mercury available, in which case just do 0.106g/200.6g
 

BabzBrah

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Thanks for the reply, Answers are A: 16.4ppm, B 23.6ppm C 10.8ppm D 2.9ppm
 

golgo13

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ok so i did a quick plot on excel and the equation is y=45.45x+2.117 where y is the PPM and x is the ABS

Now going off the standard ABS
1ml 0.083 = 5.88935 ppm
2ml 0.164 = 9.5708 ppm
4ml 0.331 = 17.16095 ppm

ppm from memory is mg/L so you can back track from there where it went wrong should be ok i think when you think about it this way :) hope that kind of helps
 

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