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 = 0.034 Absorbance of sample = 0.318 Absorbance of sample plus addition = 0.753 What was the concentration of the cadmium in the waste stream sample? Calculate percent error caused by using water instead of the reagent blank?

Answer 1

Reagent Blank:

Given the absorbance of reagent blank, Ao = 0.034

Unknown sample:

Corrected absorbance, A1 = 0.318 - 0.034 = 0.284

Suppose the concentration of the unknown sample be, "C1"

Volume of unknown sample, V1 = 10.0 mL

Now A1 = E*b*C1

=> 0.284 = E*b*C1 ----- (1)

Unknown sample + standard sample or spiked sample(10.0 μL of a 1000.0 μg/mL Cd):

Corrected absorbance, A(u+s) = 0.753 - 0.034 = 0.719

concentration of standard, C2 = (1000 μg / mL)

Volume of standard, V2 = 10.0 μL * (10^-3 mL / 1μL) = 0.01 mL

amount of unknown sample = C1*V1 = C1 * 10.0 mL

amount of standard = C2*V2 = (1000 μg / mL) * 0.01 mL = 10 μg

total volume of the unknown+standard, Vt = V1+V2 = 10.0 mL + 10.0 μL * (10^-3 mL / 1μL) = 10.01 mL

=> Concentration of unknown + standard, C(u+s) = (C1*V1 + C2*V2) / Vt

=> C(u+s) = (C1 * 10.0 mL + 10 μg) / 10.01 mL

Hence A(u+s) = 0.719 = E*b*C(u+s)

=> 0.719 = E*b* [ (C1 * 10.0 mL + 10 μg) / 10.01 mL] ------ (2)

Dividing equaiton-(2) by (1) gives:

0.719/0.284 = E*b* [(C1 * 10.0 mL + 10 μg) / 10.01 mL] / E*b*C1

=> 0.719/0.284 = (C1 * 10.0 mL + 10 μg) / (C1*10.01)

=> 2.5317 = (C1 * 10.0 mL + 10 μg) / (C1*10.01)

=> (C1 * 10.0 mL + 10 μg) = 2.5317 * (C1*10.01)  

=> (C1 * 10.0 mL + 10 μg) = 25.3422*C1

=> 25.3422*C1 - C1 * 10.0 mL = 10 μg

=> C1 = 0.6518  μg/mL

Hence concentration of the cadmium in the waste stream sample = 0.6518  μg/mL (Answer)

When we use water instead of reagent blank, there is no need to subtract 0.034 from absorbance. Hence

0.753/0.318 = (C1 * 10.0 mL + 10 μg) / (C1*10.01)

Solving the above equation gives

C1 = 0.7298

=> Percent error = [(0.7298 - 0.6518) / 0.6518] * 100 = 11.96 % (Answer)