Analytical Cumulative Examination
April 5, 2002
KEY
H. Hara, H. Kobayashi, M. Maeda, A. Ueno, Y. Kobayashi, Anal Chem. 2001, 73, 5590-5595.
General Suggestions: It is to your advantage to attempt to answer questions. You may receive
partial credit for an incorrect answer if you demonstrate that this answer is the result of some
logical thought. It is also possible to hurt your exam score by providing incorrect information
that is not backed by evidence of reasoning. A logical explanation of each answer is as important
as a “correct” answer. It is not sufficient for you to just understand the answer to a question –
your written answer must effectively communicate your understanding to the graders of the
exam. Look at what you have written for each question, and make certain that your answer
accomplishes this.
1. (6)
Define the term “speciation” and briefly explain why it is important for
environmental analysis.
There is a journal article dedicated to IUPAC Recommendations for definitions of
terms related to chemical speciation: D. M. Templeton et al., Pure Appl. Chem.
2000, 72, 1453-1470. I only expected you to give a reasonable definition of the
term, but here is the “official” definition: “Distribution of an element amongst
defined chemical species in a system”.
The introduction to the article discusses why speciation is important. The different
chemical forms of the same element may exhibit different toxicity and different
transport rates and persistence in the environment.
2. (16)
The equation below describes the response of an ion-selective electrode. Briefly
define each term in the equation. Draw a sketch of an ISE, focusing on the sensor
membrane. Briefly explain how the ion-selective electrode uses a difference in
chemical potential to produce a voltage.
0.0592


E = const. ± β
log  AA + ∑ ( k A, X AX )
n
X


This question is virtually identical to a cumulative exam question from Spring 2001.
The key from this earlier exam is still posted on the department web. The following
is adapted from that key.
E is the potential difference measured across the ion-selective membrane, the basis
of the analytical measurement.
The constant takes into account small, non-ideal potential differences across all the
interfaces in the complete electrochemical cell. It is evaluated by standardization.
The β term should be close to 1.00 and is called the electromotive efficiency.
Basically it takes into account non-ideal behavior of the electrode.
n is the charge of the ions that are being measured. The response of the ISE will be
opposite for cations and anions, and a smaller potential change will be observed for
ISE's designed to detect ions with higher charge (e.g. +2, -2, +3, -3, etc.)
The 0.0592 constant is the value for 2.303×RT/F at 25°C.
AA is the activity of the target analyte. The response of the ISE will be linear with
respect to the log of the activity of the target analyte.
The final term is the selectivity term. It takes into account the relative response of
the ISE to the activity (AX) of any ion other than the target ion. Ideally the value of k
would be 0 for all ions other than the ion of interest.
Diagrams of ISE's can be found in any undergraduate analytical chemistry textbook
and in Bard and Faulkner. Please look at any of these sources for a reasonable
example.
I like the general explanation of an ISE in D. C. Harris, "Quantitative Chemical
Analysis, 5th Edn." pp 382-384. When the concentration of an ion is different on
either side of the selective membrane of an ISE, a chemical potential exists: ∆G = RTln(A1/A2). Only the ion of interest can migrate through the membrane (ideally) to
equalize the difference in chemical potential. As this occurs, excess charge builds
up on one side of the membrane relative to the other, generating an electrical
potential and ∆G = -nFE. At equilibrium: -nFE = -RTln(A1/A2). This can be
rearranged to a simpler form of the equation in the question. Recall that very little
current actually passes across the membrane, and the concentration difference of
target analyte across the membrane does not change significantly as a result of the
measurement.
3. (10)
Briefly describe the separation mechanism for ion-exchange chromatography.
Suggest a functional group, which could be attached to a stationary phase and would
be useful for a separation of free aluminum and aluminum fluoride complexes.
Explain why you selected this functional group.
In ion-exchange chromatography retention is based on ionic interaction between
charged analytes and charged functional groups covalently bound to a stationary
phase.
In order to separate positively charged aluminum and aluminum fluoride complexes,
the functional group on the ion-exchange stationary phase must have a negative
charge. It is also sensible to suggest a functional group with a pKa such that it will
have a negative charge in aqueous solutions at moderate pH values. For example, a
simple alcohol probably would not be a good functional group for this application.
Appropriate functional groups include: RSO3- RCOO-.
4. (16)
The authors use lumogallion as a postcolumn derivatization reagent for fluorescence
detection of aluminum although they don’t describe the reagent or its use in any
detail. The chemical structure of the reagent is shown below (from Reference 8).
Draw a plausible reaction of lumogallion with Al3+ including a structure for the
reaction product. Very few reactions are suitable for postcolumn derivatization with
fluorescence detection. Discuss the most important characteristics of a reaction used
for postcolumn derivatization. Be sure to consider reaction rates and the optical
properties of the reactants and reaction products.
Cl
N
N
OH HO
OH
SO3H
My best guess is that aluminum ions bind to the two hydroxyl groups in the center of
the structure as drawn above. I was unable to find an easily accessible literature
reference containing data about the structure of the complex. Complexation to two
hydroxyl groups would lead to a reasonably stable complex and a seven-member
ring. The resulting structure would be more rigid than the reagent and would contain
three conjugated rings. Therefore, it would be more likely to fluoresce than the
uncomplexed reagent.
For postcolumn derivatization, it must be possible to detect the reaction products
(with analytes) selectively in the presence of high concentrations of the
derivatization reagents. In this case, the reagents must be fluorogenic. In other
words, the reagents must be nonfluorescent, and the reaction products must be highly
fluorescent. It is also important that the reaction not produce side products, which
would interfere with analyte detection. The reaction also has to be very fast. The
reaction has to be completed in the time between postcolumn reagent addition and
elution past the detector. If the reaction is too slow, analytes will elute past the
detector, unreacted and undetected.
5. (16)
Figure 2 presents data for the equilibrium distribution of aluminum species as the
ratio of F- to Al3+ is changed. These data were generated from chromatograms like
those shown in Figure 4. The experimental data closely match predicted equilibrium
distributions as shown in Figure 2. What does all of this tell you about the kinetics of
the reactions of F- and Al3+ in water? Considering the kinetics of the reactions of Fand Al3+ in water, can you think of any aspects of the sample preparation procedure,
which could lead to data, which do not match the predicted equilibrium
distributions?
The model used in Figure 2 assumes that the various Al-F complexes are at
equilibrium. The separation method resolves the different complexes, which were at
equilibrium. If the reactions of F- and Al3+ were fast relative to the separation
technique, it would not be possible to separate the different species as is shown in
Figure 4. If the reactions were very fast, one peak would be observed representing
the average retention for the distribution of chemical species at equilibrium. If the
reaction rates were “similar” to the time scale of the separation, very broad peaks
would result which would be difficult to use for quantitative analysis. The data in
Figure 4 and in Figure 2 suggest that the reaction rates are slow compared to the
time scale of the separation (~16 min).
This information tells you that it is important to allow enough time for samples to
reach equilibrium after preparing them. If the sample were analyzed too quickly after
preparation, it is possible that the system would not be at equilibrium when it was
analyzed. Then the experimental data would not fit the model as well, since the
model assumes the species are at equilibrium.
6. (16)
The data in Figure 2 indicate that F- and Al3+ exist in several forms in aqueous
solutions. Quantitative measurements of all Al3+ species (total Al3+) were based on
calibration using Al3+ standards, which did not contain F-. What does this assume
about detection of the different Al3+ species? If this assumption were false, how
would this affect the data in Table 1?
The data in Figure 2 are measurements of solutions containing aluminum and
fluoride, but the calibration was carried out using solutions containing only
aluminum. This calibration approach assumes that the response of the detector
(postcolumn derivatization with fluorescence detection) is the same for free
aluminum and aluminum fluoride complexes. This is a big assumption. It is
reasonable to think that the final fluorescent complexes between the reagent and
aluminum or aluminum-fluoride complexes are the same. It is also reasonable to
expect, however, that the reaction rate between the free aluminum and aluminum
fluoride complexes with the derivatization reagent are different. This difference
could lead to a difference in detector response to the different aluminum complexes.
This would lead to inaccurate determination of the relative concentrations of the
different species listed in Table 1.
7. (10)
Total and free F- concentrations were measured using an ion-selective electrode for
F-. In one measurement CDTA was added to the sample solution. This EDTA
analog presumably forms a complex with Al3+. The article does not specify which
experiment (with or without CDTA) represents a measurement of free F- and which
represents total F-. Suggest which experiment you think measures free F- and which
measures total F-. Explain your answer.
In the presence of Al3+, much of the fluoride ion will be bound to the metal ion and
would not be measured by the ion-selective electrode. This measurement represents
free fluoride ion. I expect that CDTA, like EDTA, forms very strong complexes
with Al3+. It is likely that CDTA would displace all F- bound to Al3+. A measurement
of F- in the presence of CDTA represents total F-.
8. (10)
The authors note several precautions taken to avoid contaminating the samples and
instruments with Al3+ such as avoiding the use of sapphire plunger in the pump,
using highly purified water to make solutions and adding a column to trap Al3+ from
solvents exiting the pump in the HPLC system. The authors still point out that it is
difficult to reduce Al3+ backgrounds to below about 10-8 M. Suggest other potential
sources of trace Al3+, which might contribute to this background.
One student answer pointed out that this background might represent the limit of
how much Al3+ the trapping column can remove from the solvent. Even if the
trapping column performed perfectly, any solvent or reagent added downstream
(including the sample solvent) could be contaminated with Al3+. Many of you also
pointed out that the stainless steel columns and fittings might contain trace levels of
aluminum. The stationary phase for the separation might contain trace levels of
aluminum. The reagents used for derivatization might contain trace levels of
aluminum. There are a host of likely and unlikely possible contamination sources.