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 73.11 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 93.0 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.

Sample 5	cd2+ standard volume 0	absorbance 0.080
5	2.5	0.119
5	5	0.163
5	7.50	0.200
5	10.0	0.241

Determine the amount of Cd2 per gram of dry clam. Express your final result as milligrams of Cd2 per gram of dry clam.

Answer 1

Ans. Part A: Determining [Cd²⁺] in 100.0 mL aliquot using standard addition method:

Calculate the final [Cd²⁺] due to standard addition (addition of different volumes of the Cd²⁺ addition) as shown in table. Plot [Cd²⁺] vs respective absorbance. Generate the trend line equation.

The trendline equation for the Abs vs concertation graph is “y = 0.0173x + 0.08” in the form of y = m x + c. where, y-axis = Absorbance, and x-axis = [Cd²⁺] in ppb; Y-Intercept = c ; Slope = m

Now, Concertation of un-spiked aliquot = (Intercept / Slope) concertation units

= (0.08 / 0.0173) ppb

= 4.624 ppb

Hence, [Cd²⁺] in 100.0 mL un-spiked aliquot = 4.624 ppb

Part B: The sample preparation (from fresh clams to final aliquot analyzed GFAAS) includes following steps-

Step 1: Fresh clams dried to a mass of 50.0 g (approx..)

Step 2: 73.11 mg of above dried sample is diluted upto 100.0 mL. Label it as solution 1.

Step 3: 5.0 mL of solution 1 is mixed with standard Cd²⁺ according and final volume made upto 100.0 mL. Label it as solution 2. Solution 2 is directly analyzed in GFAAS.

# Calculating total Cd²⁺ content in solution 2

Total amount of Cd in un-spiked solution 2 = [Cd²⁺] x Volume in liters

                                                            = (4.642 ug/ L) x 0.100 L     ; [1 ppb = 1 ug/L]

                                                            = 0.4642 ug

# Calculating [Cd²⁺] and Total Cd²⁺ content in in solution 1: 100 mL of solution 2 is prepared from 5 mL of solution 1. So, total Cd content of un-spiked solution 2 is equal to the amount of Cd in 5 mL of solution 1.

That is, Cd content of solution 1 = 0.4642 ug / 5 mL = 0.09284 ug/ mL

Now,

Total amount of Cd in solution 1 = [Cd²⁺] of solution 1 x volume of solution 1

                                                            = (0.09284 ug / mL) x 100 mL

                                                            = 9.284 ug

#Step 3: Calculating [Cd²⁺] content in dry sample: Sample 1 is prepared from 73.11 mg dried clam. So, total Cd content of solution 1 is equal to total Cd content in 73.11 mg clam sample.

That is, Cd content of dried clam = 9.284 ug Cd²⁺ / 73.11 mg clam               

                                                            = (9.284 x 10^-3) mg Cd²⁺ / (73.11 x 10^-3) g clam

                                                            = 0.1269 mg Cd²⁺ / g clam

Therefore, Cd²⁺ content in clam = 0.127 mg Cd²⁺ / g clam    (by dry mass)