if 0.25 gram powder were digested in 6.5 ml tri-acids and raised the final volume to 50 ml with distilled water. ICP-OES quantified Arsenic concentration as 0.07 microgram/ml. What will be its final concentration in mg/kg?
Carsten Bahr
Dear Asghar,
I hope I understand your question correctly.
I would like to suggest the following calculation:
You would like to determine the arsenic content (mg/kg) of a solid sample.
1) 0.25 g sample were digested in a final volume of 50 mL
2) Analysis of the solution: 0.07 µg/mL As
3) Total arsenic: 0.07 µg/mL x 50 mL = 3.5 µg As
4) Arsenic content of the sample: 3.5 µg / 0.25 g = 14 µg/g (= 14 mg/kg)
Best regards
Carsten
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