An environmental chemist working for the Environmental Protection Agency (EPA) w
ID: 563139 • Letter: A
Question
An environmental chemist working for the Environmental Protection Agency (EPA) was directed to collect razor clams from a heavily-contaminated river superfund site and analyze them for their Cd2 content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist dried the clams at 95° C overnight and ground them in a scientific blender, resulting in approximately 50 g of homogenized dry weight. A representative 94.63 mg sample was taken from the approximately 50 g of dry material and dissolved in 100.0 mL of 0.1 M HCl to create a sample solution. Using the method of standard additions, the chemist prepared five standard solutions in 100.0-mL volumetric flasks, each containing 5.00-mL aliquots of the sample solution. Varying amounts of a 96.0 ppb (g/L) Cd2 standard were added to each of the flasks, which were then brought to volume with 0.1 M HCl. The Cd2 content of the solutions was then analyzed using GFAAS, resulting in the following absorbance data.
Determine the amount of Cd2 per gram of dry clam. Express your final result as milligrams of Cd2 per gram of dry clam.
This is all information given by the book for this problem.
2+ 5.00 5.00 5.00 5.00 5.00 0.00 2.50 5.00 7.50 10.00 0.080 0.119 0.163 0.200 0.241Explanation / Answer
Method of standard addition
Given,
Concentration of Cd2 standard = 96 ppb = 0.096 mg/L
concentration of standard when 5 ml of Cd2 added to solution = 0.096 mg/L x 5 ml/100 ml
= 0.0048 mg/L [Cs]
Absorbance of the solution of (5 ml sample + 5 ml Cd2 standard) diluted to 100 ml = 0.163 [A2]
Absorbance of the solution of (5 ml sample + 0 ml Cd2 standard) diluted to 100 ml = 0.080 [A2]
Concentration of unknown sample in 100 ml diluted solution = Cx
So using the formula of standard addition,
Cx/(Cx + Cs) = A1/A2
feed values from above,
Cx/(Cx + 0.0048) = 0.080/0.163
0.163Cx - 0.080Cx = 0.000384
Cx = 0.000384/0.083 = 0.00463 mg
Concentration of Cd2 in original 100 ml solution = 0.00463 mg x 100/5 = 0.0926 mg
Concentration of Cd2 in 50 g sample = 0.0936 mg x 50 g/0.09463 g = 49.456 mg
So,
the amount of Cd2 in dry clam = 49.456 mg/50 g = 0.99 mg Cd2/g dry clam