Question

A 400.0 mg sample of hair from a young child is digested in acid and diluted to 25.00 mL. A 10.00 mL aliquot is diluted to 25.00 mL and analyzed for cadmium (112.41 g/mol) using AAS. The absorbance reading was 0.130. A second 10.00 mL aliquot was spiked with 1.00 mL of a 0.400 ppm standard cadmium solution followed by dilution to 25.00 mL. The absorbance reading was 0.180. Calculate the cadmium content in the hair sample in ppm.

Answer 1

Ans. Part 1: Make working solution of hair sample:

Given, 400.0 mg hair sample is digested and diluted to 25.00 mL to make the stock solution. 10.00 mL of stock solution is diluted to 25.00 mL to give solution 1.

Let the mass of Cd in 400.0 mg sample = X

Now,

            I. [Cd] in stock solution = Mass of Cd / Volume of solution

                                                = X / 25.0 mL

                                                = 0.04X / mL

            II. Using C1V1 (Stock soln.)= C2V2 (solution 1)

            Or, C2 = (C1 V1) / V2 = (0.04X mL^-1 X 10.0 mL) / 25.0 mL = 0.016X/ mL

Hence, [Cd] in solution 1 = 0.016X/ mL

Part 2: Calculate increase in [Cd] due to standard addition (spiking)

1.0 mL 0.400 ppm standard Cd solution is mixed with solution 1 and finally diluted to 25.0 mL.

Using C1V1 (Stock Cd soln.)= C2V2 (standard addition or spiked solution)

            Or, C2 = (C1 V1) / V2 = (0.400 ppm X 1.0 mL) / 25.0 mL = 0.016 ppm

So, spiking increases the net [Cd] in the working solution by 0.016 ppm.

Part 3: Calculate [Cd] in un-spiked solution 1

Given, Abs of un-spiked solution 1 = 0.130

            Abs of spiked solution 1 = 0.180

            Increase in abs = 0.180 – 0.130 = 0.050

Since, the increase in intensity by 0.050 units is solely due to addition of Cd-standard, an abs of 0.050 units is equivalent to 0.016 ppm [Cd] - the increase in [Cd] due to spiking.

So,

            0.050 units of abs is equivalent to 0.016 ppm Cd

     Or, 0.130 units     -           -           -           -     (0.016 / 0.050) x 0.130 ppm Cd

                                                                             = 0.0416 ppm

That is,

the intensity of 0.130 unit for un-spiked solution 1 corresponds to [Cd] = 0.0416 ppm

Thus, [Cd] in solution 1 = 0.0416 ppm

Part 4: Calculate [Cd] in hair

From Part 1, [Cd] in solution 1 = 0.016X/ mL               - statement 1

From part 3, [Cd] in solution 1 = 0.0416 ppm              - statement 2

Comparing statements 1 and 2-

            0.016X/ mL = 0.0416 ppm

            Or, X = (0.0416 ppm x mL) / 0.016 = 2.6 ppm mL

            Hence, X = 2.6 ppm mL

# Now,

            Cd content in stock solution = 0.04X/ mL

= (0.04 x 2.6 ppm mL)/ mL

= 0.104 ppm

Hence, Cd content in stock solution = 0.104 ppm.

Since stock solution is made from 0.400 mg of hair, the total Cd content of stock solution must be equal to that of the 0.400 hair sample.

            Hence, Cd content in hair = 0.104 ppm