Question

Atomic absorption spectrometry was used with the method of standard additions to determine the concentration of...

Atomic absorption spectrometry was used with the method of standard additions to determine the concentration of cadmium in a sample of an industrial waste stream. For the addition, 10.0 μL of a 1000.0 μg/mL Cd standard was added to 10.0 mL of solution. The following data were obtained: Absorbance of reagent blank = 0.042 Absorbance of sample = 0.390 Absorbance of sample plus addition = 0.733.

What was the concentration of the cadmium in the waste stream sample?

Later, the analyst learned that the blank was not truly a reagent blank, but water. The absorbance of the actual reagent blank, was 0.084. Calculate the cadmium concentration using the new information of the blank.

Calculate the percent error caused by using water instead of the reagent blank.

Homework Answers

Answer #1

Absorbance of reagent blank = 0.042

Absorbance of sample = 0.390

Actual sample of absorbance = 0.390 - 0.042

= 0.348

Absorbance of sample plus addition = 0.733.

Actual sample of sample plus addition = 0.733 - 0.042

= 0.691

Absorbance of addition = 0.691 - 0.348

= 0.343

Mass of Cd = 1000 g

Molar mass of Cd = 112.4 g/mol

Moles of Cd = 1000 x 10-6 / 112.4

= 8.897 x 10-6 moles

Concentration of Cd = 8.897 x 10-6 moles / mL

Moles of CD in 10 L = (8.897 x 10-6 moles / mL) * (10L) * (1 mL / 103L)

= 8.897 x 10-8 moles

Volume of Cd = 10 L = 10 x 10-3 mL

Volume of water = 10 mL

Total volume = 10 + 10-2

= 10.01 mL = 10.01 x 10-3 L

Concentration of Cd added = 8.897 x 10-8 moles / 10.01 x 10-3 L

= 8.888 x 10-6 moles / L

Using Beer's law,

A = ebc

0.343 = eb * 8.888 x 10-6

eb = 3.859 x 104 L /mol

Again for sample absorbance -

As = eb * Cs

0.348 = 3.859 x 104 L /mol * Cs

Cs = 9.017 x 10-6 moles / L

Absorbance of actual reagent black = 0.084

Absorbance of sample = 0.390

Actual sample of absorbance = 0.390 - 0.084

= 0.306

Substituting in beer's law -

A = eb * C

0.306 = 3.859 x 104 L /mol * C

C = 7.929 x 10-6 moles / L

Percent error = Cs - C / Cs * 100

= (9.017 x 10-6 - 7.929 x 10-6 ) / (9.017 x 10-6 ) * 100

= 12.06 %

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