In the ICP-MASS laboratory, 0.1082 g human hairwas digested using 10 mL HNO 3 an

Question

In the ICP-MASS laboratory, 0.1082 g human hairwas digested using 10 mL HNO3 and 1 mL 30% H2O2, which solution was evaporated to ~3 mL and then diluted to 250 mL using 2% HNO3. An internal standard method was utilized to make the calibration curve: Tm is used as an internal standard for Pb. After the ICP-MASS analysis, the following data were obtained.

Concentration of Pb (ppb)

Concentration of Pb (ppb)

Peak Area for Pb Peak Area for Tm 0.5 28366.5 2076442.5 2.5 120383.1 2095643.2 5 238702.2 2130486.6 10 458487.0 2208478.3 unknown solution in the 250 mL flask 332968.8 2377603.0 Blank solution 1000.0 2193060.0

Explanation / Answer

1) For callibration first a known concentration of pb solution is mixed with tm internal standard and the value of IRF(internal response factor ) calculated,

Then the IRF value is used in calculting the concentration of the unknown sample.

IRF=Ais/Cis * Cx/Ax

Ax=area of pb

Cx=conc of pb

Ais =area of Tm

Cis=conc of pb

note =1 ppb=10^-9g/ml

IRF(from 1st reading)=2076442.5/20 * 10^-9 g/ml * 0.5*10^-9g /ml/28366.5=2928.02

Cx for unknown sample=IRF *Cis *Ax/Ais

          Cx=2928.02 *20** 10^-9 g/ml*332968.8/2377603.0=8.02*10^-7 g/ml

Cx (blank)=2928.02*20** 10^-9 g/ml*1000/2193060.0=26.70 *10^-9 g/ml=0.267 *10^-7 g/ml

corrected Cx with blank =8.02*10^-7 g/ml -0.267 *10^-7 g/ml=7.753 *10^-7 g/ml

now we know 7.753 *10^-7 g/ml was present in 250 ml flask of the analyte

So amount present =7.753 *10^-7 g/ml*250 ml=1938.25 *10^-7 g=193.28 *10^-6 g=193.28 ug

Density of hair solution=1g/ml

So mass of hair in 250 ml solution=1g/ml*250 ml=250 g

mass of pb in hair sample=193.8ug/250*10^-3 kg=775.2 ug/Kg

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