Question

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 = 0.041 Absorbance of sample = 0.476 Absorbance of sample plus addition = 0.821 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 0.10. Calculate the cadmium concentration using the new information of the blank.Calculate the percent error caused by using water instead of the reagent blank.

Answer 1

Blank abs = 0.041

Sample abs = 0.476

Actual Abs (corrected ) = Sample - Blank = 0.476 -0.041 = 0.435

Sample + addition = 0.821

Actual Abs for Addition = 0.821 -0.435 = 0.386

Mass of Cd:

mass = C*V = (10*10^-3 mL)(1000 *10^-6 / mL)

mass of Cd = 0.00001 grams

mol = mass/MW = 0.00001 / 112.4110 = 8.8959*10^-8 mol

Find [Cd+2]

[Cd+2] = mol/V = (8.8959*10^-8)/(10*10^-3) = 0.0000088959 M

[Cd+2] = 0.0000088959 M

Apply Beer Law

A = e*l*C

A = absorbance of sample

e is the molar absorptivity , the typical units are 1/M-cm

l size of cuvette, typically reported in cm

c is the molar concentration, in mol per ltier or M

A = m*c

m = A/c = 0.386 / 0.0000088959 = 43390.775 Now..

A = 43390.775*C

C = A/43390.775

C = 0.435/43390.775 = 0.0000100 M

if water is used:

A will be approx = 0.476 instead of 0.435

Error = (0.435-0.476 )/0.476 *100 = 8.6%