Hello,
I am using an ICP-OES for elemental determination for waste water samples. The procedure is to digest 100 mL of the waste water with 2 mL HNO3 + 1 mL HCl, until the volume is reduced to 20 mL. Then the solution will be diluted with ultra pure water to 50 mL and this solution will be measured for elemental determination on an ICP-OES.
How can I calculate the dilution factor in that case? And as an example, Cu = 300 ppb (measured), what will be the initial concentration in the waste water sample.
In advance, thank you very much for your help.
Is a simple, general "rule" or calculating method.
1. You give the sample weight in [g] (In this case 0.5 g)
2. You give the final volume (after digestion) in [ml] (In this case 100 ml)
3. The calculating factor = volume / weight = 100 / 0.5 = 200 (ml/g)
4. Your final result in [mg/kg = ppm] in original solid sample will be
AAS result in [ppm = mg/l] * calculating factor = 0.192 * 200 = 38.4 ppm = 38.4 mg/kg
Note:
This calculating method is true and works if your AAS result is given in ppb = microg/l. In your case 0.192 mg/l = your AAS result = 192 microg/l.
Your final result will be AAS result * calculating factor = 192 * 200 = 38400 microg/kg = 38400 ppb.
If AAS (or ICP or ICP-MS) result is given in mg/l your final result will be mg/kg = ppm.
If AAS (or ICP or ICP-MS) result is given in microg/l your final result will be microg/kg = ppb.
mg/l --> mg/kg, microg/l --> microg/kg, because the calculating factor volume/weight (ml/g) works as a dilution factor.
Dear Lady@Amandine Tisserand,
This is an simple question for analytical chemistry operator. Actually, you only need to know your sampling volume (100 mL) and final volume (50 mL), The dilution factor is the value obtained by dividing the final volume (50 mL) and the sampling volume (100 mL) that is generally ＞ 1, but your operation is obviously to concentrate the sample, and then digest it, so it is less than 1.
Your measured value (eg. 300 ppb for Cu) should be multiplied by the dilution factor (50/100=0.5), and hence, the real value of the waste water is 300*0.5=150 ppb. I hope my answers will help you.
Good luck!
G'day Amandine,
Junhang's explanation of concentration and dilution factors is as good as I could give.
I hope it helps.
Cheers.,
Adam
C real = C measured x Vf/ Vi x dilution factors
C real = 300 x 50/100 = 300x 0.5 = 150
dilution factors = 0.5
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