A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate (

Question

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate (i.e. remove) proteins. After ccntrifugation, the solution was brought to a pH of 2.7 and was extracted with two 5-mL portions of methyl isobutyl ketone containing the organic lead complexing agent APCD. The extract was aspirated into an air-acetylene flame yielding an absorbance of 0.428 and 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.225 and 0.475 ppm Pb were treated in the same way and yielded absorbances of 0.387 and 0.583. Calculate the concentration of lead (in ppm) in the sample assuming Beer's law is followed.

Explanation / Answer

Absorbances of the standard=0.387,0.583

Concentration of the standard=0.225 ppm,0.475ppm

Absorbance=A=elc

e=molar absorbtivity with units

l=path length in cm-1 ppm-1

c=conc in ppm

1)0.387=0.225 ppm*e*l gives e*l=0.387/0.225=1.72

2)0.583=0.475 ppm*e*l gives el=0.583/0.475=1.23

taking average 1.72+1.23/2=1.47

So for absorbance of 0.428 and el=1.47(*volume =5ml after extraction)

c=0.428/1.47=0.291 (answer)

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