INSTRUCTOR MANUAL
HEAVY METALS IN LAKE NAKURU
ATOMIC SPECTROSCOPY
Q1. Give the electron configuration for sodium (Na, element number 11). In what atomic orbital
does the electron with the most energy reside?
The electron configuration for sodium is 1s2 2s2 2p6 3s1. The highest energy electron is located
in the 3s orbital.
Q2. For the sodium atom, if an appropriate amount of energy were absorbed, to what atomic
orbital would an electron be promoted if the lowest energy excited state were the result of the
absorption?
The next highest energy atomic orbitals after the 3s are the 3p orbitals, to which the transition
representing promotion of an electron to the lowest energy excited state is given by
3s1  3p1
Q3. Given that the energy difference between the ground state and the first excited electronic
state (E) for the sodium atom is 3.373 × 10-19 J, calculate the frequency, , corresponding to a
photon possessing this energy. Next, calculate the wavelength (in nm) for this photon.
E = h= (3.373 x 10-19 J)
= E / h = (3.373 x 10-19 J) / 6.626 x 10-34 J s)
=
5.091 x 10-14 s-1
E = (hc) /  = (3.373 x 10-19 J)
 = (hc) / E = [(6.626 x 10-34 J s)(3.00 x 108 m/s)] / (3.373 x 10-19 J) = 5.89 x 10-7 m = 589 nm
Q4. A common unit for metal analysis is parts-per-million, or ppm. For solids ppm is expressed
as mg/kg while for aqueous solutions mg/L is used. The current maximum contaminant level
(MCL) set by the EPA for lead in drinking water is 0.015 mg/L.2 At this concentration, how many
atoms of lead are present in 1 L?
(0.015 mg/L)(1g/1000mg)(1 mol Pb/207.2g)(6.022 x 1023molecules Pb/1 mol) =
4.4 x 1016 molecules Pb/L
Q5. A common wavelength used for measuring lead emission is 220.353 nm.
a. Calculate E for this transition.
b. Calculate the ratio of atoms in the excited state to atoms in the ground state for this
value of E at room temperature (298K).
c. Repeat the calculation at 6000K, a routine operating temperature for an ICP torch.
d. Can you hypothesize as to why atomic emission measurements are generally made at
high temperatures?
a)  = (220.353 nm)(1 m/ 109 nm) = 2.20353 x 10-7m
E = (hc) /  = [(6.626 x 10-34 J s)(3.00 x 108 m/s)] / (2.20353 x 10-7m) = 9.02 x 10-19 J
b) (Nexcited /Nground ) = e -E/kT = e -(9.02E-19 J) / (1.381E-23 J/K)(298K) = e -219 = 7.75 x 10-96
c) (Nexcited /Nground ) = e –(9.02E-19 J )/ (1.381E-23 J/K)(6000K) = e -10.9 = 1.85 x 10-5
d) Emission measurements are made as excited state electrons relax back to ground state,
releasing excess energy as photons. Unless a very high temperature is employed, the
signal will be too small to measure accurately.
Q6. What are molecules?
Molecules are compounds that result when bonds are formed between atoms.
Q7. Consider the relative temperatures of the three atomization sources described in the
previous section. Are there cases in which higher temperatures might cause something other
than atomization? What group(s) of elements would you expect to be most susceptible?
The next level of energy above that required to promote electrons to higher atomic orbitals leads
to the loss of an electron from the atom, or ionization.
The elements with the lowest ionization energies are those found in Group I and Group II of the
periodic table (alkali and alkaline earth metals).
Q8. Speculate as to possible sources of background interference when using FLAA. How might
they be eliminated or reduced?
Interference of two types may be present: spectral and non-spectral.
Most obvious spectral interference is the continuum output of the flame itself, which may reduce
the analyte signal at the source wavelength. A second interference of this type can arise from
un-atomized molecular forms of the analyte, whose absorption spectra are broad and may
include the wavelength of the atom. Both can be solved by separating the continuum spectrum
from the atomic absorption of the analyte. This is most commonly achieved by alternating
between two sources, the hollow-cathode single wavelength source and a source lamp whose
emission is a continuum across a wide range of wavelength. Subtraction of one signal from the
other at the detector allows a correction.
Non-spectral interference sources include the formation of thermally stable compounds from
analyte and interferent. Matrix modifiers are available that preferentially bind the interferent
while allowing atomization of the analyte. Increased temperature of the flame can also allow
atomization from the refractive compound. This commonly means substituting a nitrous
oxide/air mixture for acetylene/air.
Ionization of analyte, more common for alkali and alkaline earth metals, can depopulate the
ground state, leading to a reduced number of atoms that can absorb radiation from the source.
Ionization suppression can be accomplished by the addition of an easily ionized element, which
maintains a large number of free electrons within the flame, and reduces the ionization of
analyte.
Matrix differences between the sample and standard solutions can often lead to differences in
the observed signals for each. Generally, the matrix of the standard is as close a match as
possible to the sample. In some cases, it will be necessary to use the method of standard
addition to correct for serious matrix effects.
A complete discussion of background correction in FLAA can be found in Chapter 3 of reference
6.
Q9. How does the design of the graphite furnace allow for such improved sensitivity for metals
over flame AA?
The entire sample is atomized instantaneously in the confined space of the graphite tube,
leading to a much larger population of analyte atoms and hence a larger measured signal.
Q10. For the metals in Table 4, use the average final dry concentration to calculate the
dissolved concentration (mg/L, ppm) of each metal in a liquid sample prepared from 5.00 g of
dried sediment or suspended solids diluted to a final volume of 50 mL for analysis by atomic
spectroscopy.
Chromium in dry sediment:
[(67 g/g) x (5.00 g) x (1g/1000 g)] / ( 0.050 L) = 6.7 mg/L = 6.7 ppm
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Concentration
5.0 g dry sediment in
50 mL final volume (mg/L)
Concentration
5.0 g suspended solids in
50 mL final volume (mg/L)
Trace metal
Average
Average
Chromium
6.7
0.83
Copper
2.4
1.9
Lead
2.2
1.17
Zinc
14.7
7.4
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Q11. If no further information were given, which technique would be your first choice in
measuring the four metals in both of the matrices if you were to perform a follow-up study to that
done by Nelson? Explain your answer.
The concentrations of each metal following the preparation method in Q10 are easily measured
by all three techniques. If all three are available, the best choice in most every instance is the
ICP-OES, which combines a wide dynamic range, with effective elimination of most chemical
interferences, and in many cases simultaneous multi-element analysis.
Q12. What are matrix effects? What about the chemistry of Lake Nakuru might be the cause of
the severe matrix effects observed for measurements of Cr, Cu, and Pb by atomic
spectroscopy?
Harris defines a matrix effect as “a change in the analytical signal caused by anything other than
analyte”.14 In atomic spectroscopy, examples of matrix effects include the suppression of
analyte signal as atomization is limited by the formation of a refractive compound (a phosphate
salt, for example) or a contribution to the observed signal from the continuum spectrum of an
un-atomized species. Lake Nakuru is known to be highly alkaline, with extremely high
concentrations of dissolved ions10, likely causes of the observed matrix effects.
Q13. Which of the three methods you have studied is best able to overcome matrix effects?
Why is this so?
ICP-OES has the best capabilities for overcoming matrix effects. Most chemical interferences
are eliminated by the high temperature of the plasma. Background emission from the plasma
itself can be easily removed by subtraction of the emission intensity at a position away from an
emission line of interest from the intensity at the emission line. This is especially straightforward
for one of the solid state detection techniques that measure a continuous background across a
band of wavelengths. Spectral interferences can most often be minimized or eliminated by
selection of a different wavelength band for the observation of emission intensity for an element
of interest. Having multiple bands to choose from for each element also lends itself well to
improving linearity of response, where a weaker emission band can be chosen for high
concentration species in the sample.
Q14. If the original 50 mL digestion solution, for which you calculated metal concentrations
previously, were further diluted by a fact of 50x to reduce matrix effects. Would this alter your
answer(s) from Q11 as to choice of method for Cr, Cu, and Pb?
The concentrations that result from further dilution of the sample are given in the table below.
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Concentration
5.0 g dry sediment in
50 mL final volume (mg/L)
following 50x dilution
Concentration
5.0 g suspended solids in
50 mL final volume (mg/L)
following 50x dilution
Trace metal
Average
Average
Chromium
0.13
0.017
Copper
0.048
0.038
Lead
0.044
0.0234
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Looking at the concentration levels of the solutions analyzed (ppb level) ICP-OES would be the
method of choice as it can accurately measure metals in the ppb range. This is also the best
method for measuring multiple elements in a complex matrix. The choice of method has not
changed from Q11.
Q15. Suppose you were to construct calibration plots for each of the four metals of interest in
this study between the LOD and LOL values suggested in Method 200.7. Would 1 gram
samples of Lake Nakuru sediment with the metal concentrations found in the 1998 report (Table
4 of this module), diluted to a final volume of 100 mL, have analyzed values that fall between
the limits on these calibration plots?
Table 1 of EPA Method 200.7 gives the following detection and calibration limits for the four
metals of interest:
Limit of Detection (g/L)
Limit of Calibration (mg/L)
Chromium
6.1
5
Copper
5.4
2
Lead
42
10
Zinc
1.8
5
Calculation for Cr in dry sediments:
(1 g x 67 g/g) / (0.100 L) = 670 g/L which lies between the LOD and the LOL for this metal
by this method.
Similar calculations for each of the other metals in both matrices give the same result.
Q16. Use the graphical method for multiple standard additions to constant final volume to
determine the concentrations of each of the four metals in Lake Nakuru sediments and
suspended solids. This graphical method is discussed in Harvey, Chapter 5.12 Prepare a table
of concentrations similar to Table 4 in this module for the most recent concentrations. How
have the concentrations of each changed since 1996?
All samples were analyzed sequentially using the ICP-OES conditions specified in Sections 10.1
and 10.2 of Method 200.7.11 A spreadsheet file was supplied to you that contains tables of
emission intensity, corrected for background, as a function of added standard concentration of
each element. Standard addition plots and calculations of current-day concentrations are found
in a second supplied spreadsheet file.
Table 4b. Metal Concentrations in Lake Nakuru Sediments and Suspended Solids found in
Nelson Study and in Current-day Study
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Concentration in
dry sediments (g/g)
Concentration in
suspended solids (g/g)
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Trace
metal
Nelson
1996
Current-day
Nelson
1996
Current-day
Chromium
67
81.6
8.3
21.1
Copper
24
92.4
19
24.5
Lead
22
55.1
11.7
20.1
Zinc
147
197
74
71.6
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