Question

Standard addition in atomic absorption spectrophotometry: As part of an investigation of Pb in biological ecosystems, some weeds are gathered along a highway where they have been exposed to emissions from gasoline engines. A representative sample of the weeds weighing 6.250 g is ashed to destroy organic matter, and the inorganic residue is dissolved and diluted to 100mL. A portion of this solution is aspirated into an acetylene-air flame, and an absorbance value of 0.125 is found using the 283.31-nm Pb line from a hollow cathode source. Next, 0.100 mL of a standard solution containing 55.0 μg of Pb per milliliter is added to 50 mL of the sample solution. When this "spiked" sample is aspirated into the flame, the absorbance is found to be 0.180. Calculate the Pb content of the plant material in micrograms per gram, i.e., parts per million (ppm).

Answer 1

Ans. Part 1: Calculate increase in [Pb] due to standard addition (spiking)

0.100 mL of 55.0 ug/ mL standard Pb solution is mixed with 50.0 mL of stock solution.

Final volume of the spiked solution = 0.100 mL (standard Pb soln.) + 50.0 mL stock

= 50.100 mL

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

            Or, C2 = (C1 V1) / V2 = (55.0 ug mL^-1 x 0.100 mL) / 50.100 mL = 0.1098 ug/ mL

So, spiking increases the net [Pb] in the stock solution 0.1098 ug/ mL.

Part 2: Calculate [Pb] in un-spiked stock solution

Given, Abs of un-spiked stock solution = 0.125

            Abs of spiked stock solution = 0.180

            Increase in Abs = 0.180 – 0.125 = 0.055

Since, the increase in Abs by 0.0.055 units is solely due to addition of Pb standard, an Abs of 0.055 units is equivalent to 0.1098 ug/mL [Pb] - the increase in [Pbl] due to spiking.

So,

            0.055 units Abs is equivalent to 0.1098 ug/mL Pb

     Or, 0.125 units     -           -           -      (0.1098 / 0.055) x 0.125 ug/mL Pb

                                                                             = 0.2495 ug/mL

That is, the Abs of 0.125 unit for un-spiked stock solution corresponds to [Pb] = 0.2495 ug/mL

Thus, [Pb] in stock solution = 0.2495 ug/mL

Part 3: Calculate Pb content in Plant sample.

Given, 6.250 g plant sample is digested and diluted to 100.00 mL to make the stock solution. Upon aspiration, this solution gives absorbance of 0.125.

We have, [Pb] in stock solution = 0.2495 ug/mL

Now,

Total Pb content in stock solution = [Pb] in stock solution x volume of stock solution

                                                            = (0.2495 ug/ mL) x 100.0 mL

                                                            = 24.95 ug   

Since 100.0 mL stock solution is prepared from 6.250 g plant sample, the total Pb content in 100.0 mL stock solution must be equal to the total Pb content in 6.250 g plant sample.

Hence, total Pb content in 6.250 g plant sample = 24.95 ug

Now,

            [Pb], ppm in plant sample = Mass of Pb in ug / Mass of plant sample in grams

                                                            = 24.95 ug / 6.250 g

                                                            = 3.992 ug/ g

                                                            = 3.99 ppm