Question

In: Chemistry

An environmental chemist working for the Environmental Protection Agency (EPA) was directed to collect razor clams...

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 69.58 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 85.0 ppb (μg/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.

S

Sample Vol. (mL) Cd2 Standard Vol. (mL) Absorbance

5.00

0.00 0.080
5.00 2.50 0.119
5.00 5.00 0.163
5.00 7.50 0.200
5.00 10.00 0.241

Solutions

Expert Solution

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


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