environmental chemist working for the Environmental Protection Agency (EPA) was
ID: 900011 • Letter: E
Question
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 cadmium (Cd2+) content using graphite furnace atomic absorption spectrometry (GFAAS). The chemist dried the clams at 95 degree C overnight and ground them in a scientific blender, resulting in ~ 50 g of homogenized dry weight. A representative 61. 00 mg sample was taken from the ~ 50 g of dry material and dissolved in 100. 00 mL of 0. 1 M HCI to create a sample solution. Using the method of standard additions, the chemist prepared five standard solutions in 100. 00-mL volumetric flasks, each containing 5. 00-mL aliquots of the sample solution. Varying amounts of a 87. 0 ppb (mug. L^-1) cadmium standard were added to each of the flasks, which were then brought to volume with 0. 1 M HCI. The solutions were then examined for their Cd2+ content using GFAAS, resulting in the following absorbance data. Sample Vol. (mL) cd2+ Standard Vol. (mL) Absorbance Determine the amount of Cd2+ per gram of dry clam. Express your final result as milligrams of Cd2+ per gram of dry clam.Explanation / Answer
Cd2+ conc., originated from std.sol 87 ppb, in 100 mL flasks
Absorbance
(corrected for errors)
Measured Absorbance – Absorbance in the first flask (without std. additions)
0 ppb
0.080
0.000
2.17 ppb
0.120
0.040
4.35 ppb
0.160
0.080
6.52 ppb
0.200
0.120
8.70 ppb
0.240
0.160
First solve the concentration measurement in the solutions prepared in 100 mL flasks.
The calibration is linear and the slope is
0.160 A.units / 8.70 ppb = 0.0184 a.u/ppb
The absorbance for the unknown itself is 0.080. Then the concentration of the unknown is
0.080 a.u. / 0.0184 a.u/ppb = 4.35 ppb in 100 mL flask (for AAS measurement).
In the 5 mL sample aliquot
Cd2+ conc. = 4.35 ppb x 100 mL/5 mL= 87 ppb = 87 x 10-9 g/mL
In 61 mg sample, the content of Cd2+ is
87 x 10-9 g/mL x 100 mL = 8.70x10-3 mg Cd
61 mg sample (dry clam) contains …….8.70 x10-3 mg Cd
1000 mg (1g)……………………………….x
X = 142.6 x10-3 mg Cd /g dry clam
Rounded result:
0.143 mg Cd /g dry clam
Cd2+ conc., originated from std.sol 87 ppb, in 100 mL flasks
Absorbance
(corrected for errors)
Measured Absorbance – Absorbance in the first flask (without std. additions)
0 ppb
0.080
0.000
2.17 ppb
0.120
0.040
4.35 ppb
0.160
0.080
6.52 ppb
0.200
0.120
8.70 ppb
0.240
0.160
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