Why is atomic emission more sensitive to flame instability than atomic absorptio

Question

Why is atomic emission more sensitive to flame instability than atomic absorption? A 5.00 ml sample of blood was treated with trichloroacetic acid to precipitate the proteins. After centrifugation, the resulting solution was brought to a pH of 3 and was extracted with two 5.0 ml portions of methyl isobutyl ketone containing the organic lead complexing agent APCD. The extract was aspirated directly into an air-acetylene flame yielding an absorbance of 0.44 at 283.3 nm. Five milliliter aliquots of standard solutions containing 0.25 and 0.45 ppm Pb were treated in the same way and yielded absorbances of 0.396 and 0.599. Calculate the concentration of Pb (ppm) in the sample assuming that Beer's law is observed

Explanation / Answer

6. Atomic emission spectroscopy deals with excitation of atoms from the excited state while atomic absorption spectroscopy deals with atoms in the ground state. For any element, the ground state is much highly populated while the excited state has a significantly low population of atoms. The population of atoms in the excited state is given by the Boltzmann distribution and Boltzmann distribution function is temperature-dependent. Hence, atoms in the excited state are much highly affected by temperature variations and thus, atomic emission spectroscopy is highly sensitive to flame instability.

7. Prepare a calibration curve of absorbance vs concentration for the standard samples.

Use the linear equation to find out the concentration of Pb in the diluted unknown sample.

Given y = 0.44 for the diluted unknown, plug in y = 0.44 and obtain x.

0.44 = 1.0205x + 0.1949

===> 1.0205x = 0.2451

===> x = 0.24

The concentration of Pb in the diluted unknown sample is 0.24 ppm.

Now, remember that you took 5 mL of the blood serum and diluted to 10.0 mL by using two 5 mL portions of methyl isobutyl ketone. Hence, the volume of the diluted unknown solution is 10 mL while we had 5 mL of the concentrated solution.

Therefore, concentration of Pb in the original solution (5 mL sample prepared initially) is (0.24 ppm)*(10.0 mL/5.0 mL) = 0.48 ppm (ans).

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