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This module provides a real world context for introducing fundamental quantitative techniques (pH, Ion Selective Electrodes, Ion chromatography, and Titrimery) used in chemical analysis of water samples. Using active learning pedagogy, module explores sampling and analyzing the base inorganic ions and acidity parameters in freshwater samples. The module frames the study using the environmental impacts of acid rain on the habitat of the endangered Atlantic Salmon as a case study.
Wild Atlantic Salmon (Salmo Salar) in Maine, eastern Canada, and northern Europe are on the brink of extinction. The salmon populations segments in the northeast United States have either been extirpated or are now listed on the federal endangered species list. Rivers in Canada and Norway are heavily remediated to maintain minimal populations. Both freshwater and marine survival are low, and most populations are maintained only with a vigorous hatchery and stocking program. In order to make salmon populations self-sustaining, both freshwater and marine survival must be improved. During the past decade over 100,000 young salmon are stocked annually in a number of rivers with fewer than 100 returning as adults. In freshwater, water quality problems from a combination of related factors result in poor fish condition and low survival as they undergo the smoltification process while moving into salt water. Solving this complex problem requires both the approach and the tools used by analytical chemists. In the process of thinking about the problem, you will have the chance to apply knowledge you have gained in your chemistry coursework and to learn in greater detail important concepts related to modern chemical analysis, by developing a sampling plan, choosing an appropriate analytical method, and collecting and analyzing data.
Effects of Acid Rain on Atlantic Salmon Populations
Facts, Observations and Evidence
The Atlantic salmon is an anadromous fish species, spending most of its adult life in the ocean coming back to spawn in freshwater rivers. The adult creates a redd, where it lays the eggs, in the fall. After hatching the salmon undergo several stages of their lifecycle in the freshwater. The fry leave the spawning grounds in search of food and develop into parr. After two years, the fry undergo the process of smoltification, where their physiology changes so that they can live in salt water. At this stage the smolt leaves the river and heads to the ocean, where most ultimately head to feeding grounds off of Greenland. The salmon usually live in the ocean for 2-3 years before they return to the river they were hatched in.
In the United States, the historical grounds of the Salmon ranged on the northeast coast south to the rivers of Long Island Sound. Currently, remnants survive only in 8 rivers in Maine, the largest of which is the Penobscot River. Historically in Maine, salmon landings were as high as 90 million tons in the late 1800s. (Baum 1997) By the 1940s the commercial fisheries in Maine were closed, and by the 1990s, less than 2000 adult salmon were returning annually to all of the rivers in Maine, despite over 13 million juvenile salmon stocked annually.
To identify the cause of the minimal returns, the mortality in the river and the sea run mortality were determined. To minimize the potential mortality as parr and fry, a number of smolts are stocked shortly before they are set to leave for the sea. Some of the smolts were pit-tagged with radio labels to identify near shore mortality. Data collected demonstrated a significant mortality from the time when the fish reached the estuary and made it to the open ocean. This mortality would correspond to heavy predation in the bay, or failure of the smolt to survive the transition from fresh to salt water.
During the smoltification process, the salmon has significant biological changes in the gill structure for osmoregulatory processes. In fresh water the salmon needs to maintain a higher salt content in its blood as compared to the water it is in. When in seawater, the salmon needs to maintain a lower salt content in the blood compared to the water it is in. During this change in the gill structure, the salmon is most sensitive to stressors or damage to the gill. Studies have shown that gill deformities can be caused by low pH and by high aluminum. In both of these cases the deformities occur as the cation binding to the gill structure.
The National Resource Council in 2004 specifically stated that the effects of acid rain on smolts is one of the most significant factors impeding the recovery of the Atlantic salmon; “The problem of early mortality as smolts transition from freshwater to the ocean and take up residence as post-smolts needs to be solved. If, as seems likely, that the difficulty of the transition is due in part to water chemistry, particularly acidification, the only methods of solving the problem are changing the water chemistry and finding a way for the smolts to bypass the dangerous water.
The related Brook Trout has had similar extirpation from streams and ponds in many northeast locations due to low pH.
The question to be answered is: What is causing the increased mortality of Atlantic salmon?
Some hypotheses as to possible (and perhaps intertwined) causes include:
Low pH. Salmonids (Atlantic Salmon, Brook Trout, and related species) are known to be sensitive to pH, with stress occurring below pH 5, and mortality below pH 4.5. The pH of the aquatic ecosystem can be lowered due to natural acid inputs, such as dissolved organic carbon (DOC), or through anthropogenic inputs, predominantly acid rain. The acid rain is created from nitrogen oxides and sulfur oxides emissions. These may be episodic or chronic conditions.
High Aluminum. Aluminum will bind to the gills of the salmonids and irreversibly damage the gills, preventing the fish from uptaking oxygen. Aluminum will get into the water systems predominately from leaching from the ground when acid rain impacts a system.
Elevated Temperature Salmonids are temperature sensitive. When stream temperatures reach over 22ºC, salmon are severely compromised. When temperatures reach over 26ºC, mortality often ensues. The temperature of the stream can change due to increases in air temperature, reduction in boreal cover, or reduction in underground cold-water stream inputs.
Increased predation. There are two major predators. As the fish swim to the ocean as smolts, they are met by a host of predators such as cormorants and seals. The populations of both of these predators have increased in recent years. Another predator is humans, and overfishing is of concern. Starting in the 1950s, salmon have been caught in their ocean feeding grounds off of Greenland and Newfoundland. The catch has steadily declined with the decreasing populations.
Identifying Possible Analysis Methods
There are three hypotheses that have been put forward that could be responsible for the decline of the Atlantic Salmon. The purpose of this exercise is to identify analysis methods that could be used to test the chemical species that contribute to each of these hypotheses. Among the various analytical methods you find that may be applicable to this measurement, identify their strengths and weaknesses (e.g., sensitivity, expense, ease of use, reproducibility).
Hypothesis 1. Episodic acid rain events overcome the buffering capacity of the water system and decrease the pH leading to stress and mortality in Atlantic salmon parr and smolts.
Q1. Using available literature and other sources of information, identify possible analytical methods that could be used to monitor acidity.
Hypothesis 2. Acid rain has increased the leaching of calcium from freshwater, reducing the nutrients available for salmonids. This loss of calcium prevents salmon from properly undergoing the smoltification process resulting in mortality when the fish head to the ocean.
Q2. Using available literature and other sources of information, identify possible analytical methods that could be used to detect calcium and measure their levels in components of the water (e.g., free, organic bound).
Hypothesis 3. Acid rain has increased the leaching of aluminum, resulting in a high concentration of free aluminum ion resulting in gill damage, leading to the main cause of mortality of Atlantic Salmon during the smoltification process.
Q3. Using available literature and other sources of information, identify possible analytical methods that could be used to detect aluminum and measure their levels in components of the water (e.g., free, organic bound).
The rest of the module primarily focuses on the first two hypotheses. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/01_Identifying_the_Problem.txt |
Designing a Sampling Plan for Water Analysis
Purpose: The purpose of this module is to design a sampling plan for surface waters [i.e. ponds, lakes, rivers and streams]. Different approaches may be necessary depending on whether water will be collected from a stream, lake, or other sources of water, but the underlying steps should find broad application in the design of any sampling plan. This module does specifically use the scenario of the Atlantic Salmon survival in acid impaired streams, but is easily applied to other scenarios where surface water analysis is important.
Learning Outcomes:
At the end of this assignment you will be able to
1. Define various sampling strategies
2. Assess the benefits and limitations of different sampling strategies
3. Determine an appropriate sampling plan for the analysis of surface waters
Assignment:
An important aspect of designing an experiment is collecting a representative sample, but defining a representative sample can be a difficult problem to answer. In the context of an analytical chemistry measurement, representative means that the concentration of the analyte in the sample analyzed is a good representation of its levels in the material or system being tested. If sampling is not done correctly it could be the “weak link” in an analysis, leading to inaccurate results. Often, sampling can be a source of error that is overlooked.
Below is a map of the confluence of the East Machias and Machias rivers, two rivers systems with genetically distinct Atlantic Salmon populations. There are multiple lakes, rivers, and even ponds in this area that salmon migrate through. The design of the sampling plan will focus on the breadth of surface water sources, and is easily adaptable to almost any body of surface water.
Q1. Identify the key questions that must be considered when designing a sampling plan to obtain a representative sample.
Where to Sample?
How do you obtain a representative sample? One approach is to take a grab sample. Consider a solid sample divided out as a grid, below. How do you choose where to grab from? One approach is to choose randomly.
Q2. Pick eight random samples from the grid laid out below. How do you ensure your sampling is random?
Now take a look at the following samples divided into grids with the analyte of interest identified (colored squares).
Q3. Would you consider the samples above to be heterogeneous or homogeneous?
Q4. Did your random sampling affect the potential accuracy or precision of your measurement of the analyte for the samples in grid A or grid B? If so how?
Although we would like to assume that a sample is homogenous it is often not. When trying to sample a heterogeneous sample, you need to consider both distributional and constitutional heterogeneity.
Distribution heterogeneity is caused by segregation of the sample (i.e. settling).
Constitutional heterogeneity is a fundamental property of a material and is caused by differences in particle size and/or composition.
Q5. Each of the previous grids is an example of one of these cases. Can you identify which sample is which?
Q6. How does distribution heterogeneity affect accuracy and precision?
Q7. How does constitutional heterogeneity affect accuracy and precision?
Q8. Do you see a scenario where distribution heterogeneity could be magnified by mixing and/or sampling?
If samples are mixed well, distributional heterogeneity is insignificant and only constitutional heterogeneity is present. It is important to note, however, that the spatial or temporal distribution of an analyte might also be important in the analysis and mixing may not always be the best choice.
Types of Sampling Plans:
What we have highlighted above is an example of random sampling, which is often applied to a grid design as shown above. Random sampling strategies can be applied to any target population (i.e. evaluating a solid sample or measuring water quality parameters within a creek). However, random sampling can at times be expensive and not necessarily cost effective, as you often need a greater number of samples to ensure that your samples represent the target population. If you know something about the target population, other sampling methods may be possible or appropriate. Here are a few examples.
Selective (Judgmental) Sampling - this is at the opposite extreme of random sampling, and is done if you have prior information about the target. For example, if you wanted to evaluate the metal content in pennies you may not select coins that are corroded or choose coins from a specific mint date.
Ease of Access Sampling - Many times on ponds, rivers, or streams the ultimate sampling site is based on being able to access the point, as long as this sampling site allows the initial question to be answered. When considering the location, one needs to consider safety during access and whether the land is public or private.
Systematic Sampling - Sampling the target population at regular intervals in space or time. This is often considered to fall between the extremes of random and selective sampling.
Stratified Sampling – The population may be divided into sub populations (groups) that are distinctly different (this might be size of sample, type of sample, depth of sample). Then, the overall sampling within the groups is randomly conducted and the samples are pooled.
Cluster Sampling- is a sampling technique where the population is divided into groups or clusters and random samples are selected from the cluster for analysis. The main objective of cluster sampling is to reduce costs by increasing sampling efficiency.
Q9. What is the advantage of implementing a judgmental sampling scheme over random sampling if one knows the point source for the discharge an analyte into a system?
Q10. Assume you have a chosen a selective sampling plan to evaluate pollution from a point source into a pond. Use the diagram below and words to describe your sampling plan.
Q11. Use a grid design (as we have previously done) to show how you would conduct systematic sampling of the pollutant. Is there an advantage to what you might learn using this sampling method? What are the disadvantage(s)?
Q12. Describe how stratified sampling might be applied to evaluate the pollutant in the lake? In general, what is the advantage of stratified sampling over cluster sampling?
What Type of Sample to Collect?
When implementing a sample plan often grab samples are used. In some cases composite sampling (combining a set of grab samples into one sample) is more useful. This strategy may be used if there is interest in the target population’s average composition over time in space or a single sample does not supply sufficient material for analysis.
Q13. What is the main disadvantage of grab and composite samples?
Q14. Can you think of any control studies you might want to do when compositing samples?
Q15. If you are beginning to monitor a site or want to compare your data to a similar source, have there been any standard sampling protocols used for prior monitoring?
Another factor that can affect your sampling plan is whether the analyte of interest varies temporally. So you should think about whether there will there be any chemical changes over time that would impact your sampling plan.
Consider the fluctuation of dissolved oxygen over time in a lake as shown in the graph below.
Q16. Based on the graph above, describe a sampling procedure that would allow you to obtain a representative sample?
Another variable that can fluctuate in natural water is the pH. The graph below shows pH data continuously collected for a Maine river site over an entire year.
Q17. Is there any systematic pattern to the data in the graph above?
Q18. Can you think of some event(s) that may account for the acidic spikes in the pH?
Q19. How would this data impact when you might choose to sample a site to assess if acid rain is having an impact on the ecosystem?
Q20. Describe what else you would need to know to determine when to perform your sample collection if you are addressing whether acid rain is impacting a site. What other data might you need to look up or consider that would contribute to the changes in pH? Would this data affect when you choose to sample?
Data in the previous graph covered an entire year. The following data was collected every minutes over a 3 day period.
pH measured every 15 minutes over a period of three days starting at midnight.
Q21. What physical or chemical processes might contribute to the pH fluctuations? How might this affect your sampling plan?
Q22. Another factor to consider in a sampling plan is the sample holding time. Can you think of ways in which sample holding time may impact the concentration of different species such as nitrate, pH, and Ca2+ in the sample?
The guidelines on preservation methods and holding time for some common analytes are listed below.
Preservation Methods and Maximum Holding Times for Selected Analytes in Natural Waters and Wastewaters
Analyte
Preservation Method
Maximum Holding Time
Ammonia
cool to 4 oC; add H2SO4 to pH < 2
28 days
Chloride
none required
28 days
metals—Cr(VI)
cool to 4 oC
24 hours
metals—Hg
HNO3 to pH < 2
28 days
metals—all others
HNO3 to pH < 2
6 months
Phosphate
Cool to 4 oC
48 hours
Nitrate
Cool to 4 oC
48 hours
pH
none required
analyze immediately
Q23. Why would adjusting the solution pH to below pH 2.0 be needed for preserving samples for metal analysis? How can the solution be adjusted to pH 2 without significantly diluting the sample?
Minimizing Variance - What is the Minimum Amount of Sample? How Many Samples are Needed?
When sampling, you need to make sure the sample is not too small, so that the composition is not substantially different from the target population. You also want to ensure that you collect an appropriate number of samples for analysis.
The variance of the (1) analysis method and (2) sampling approach both contribute to the variance in a given analysis. Sampling variance can be improved by collecting more samples of the proper size. Increasing the number of times a sample is analyzed improves the method’s variance.
Q24. It is known from analyses conducted in 2008 that the % relative sampling error for water hardness by EDTA titration is 0.8%. How many samples should you collect to limit the relative standard deviation for sampling to 1.0% within the 95% confidence level? Is this a feasible task? (May help by referring to the Analytical Chemistry 2.0 by David Harvey section 7.2: How Many Samples to Collect, http://collection.asdlib.org/?p=452)
Q25. Based on your results from the prior question, if the cost of collecting a sample is \$25 and the cost of analyzing a sample is \$50 what budget should you allocate for the project and what sampling strategy would be most effective for the given number of samples?
Design Your Sampling Plan
Of course we would like to collect as many samples as possible to minimize our variance. However, in every lab there are limitations such as those above related to cost per-sample collected and cost-per analysis.
Q26. Using the map from the beginning of this module that has the area around Machias, where the confluence of the Machias, Middle, and East Machias Rivers empty into the Machias Bay. Design your sampling plan. Think about random, systematic, clustering, etc. sample strategies. Will you take grab samples or pool samples together?
Protocols associated with actually collecting samples
Equipment and sampling procedures are typically specific to the specific water quality parameter one wishes to determine. However, there are two fundamental tasks that are common to any water analysis: the preparation of the sampling containers and the sample collection.
Preparation of sampling containers.
For some analyses the EPA recommends (see the EPA Monitoring and Assessing Water Quality website (http://water.epa.gov/type/rsl/monitoring/vms50.cfm) to:
1. Wear latex gloves.
2. Wash each sample bottle or piece of glassware with a brush and phosphate-free detergent.
3. Rinse three times with cold tap water.
4. Rinse three times with distilled or deionized water.
For other analyses the EPA recommends to:
1. Wear latex gloves.
2. Wash each sample bottle or piece of glassware with a brush and phosphate-free detergent.
3. Rinse three times with cold tap water.
4. Rinse with 10 percent hydrochloric acid.
5. Rinse three times with deionized water.
Q27. Why are different procedures recommended? Why do you think acid washing of the glassware is recommended for some analyses? What is the purpose of using a phosphate-free detergent?
Q28. For some analytes such as sodium, plastic containers made of either high-density polyethylene or polypropylene might be preferable to glass. Why would this be the case? In addition, the EPA states that all containers and glassware must be “dedicated” to a specific analysis. What would be the drawback of reusing glassware for a different analysis? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/02_Sampling_and_Sample_Preparation.txt |
Analysis of cations and anions by Ion-Selective Electrodes (ISEs)
Purpose: The purpose of this assignment is to introduce potentiometric measurements of ionic species by ion selective electrodes (ISEs)
Learning Outcomes: Upon completion of this module, students will be able to:
1. Identify electrodes and measurement devices used in potentiometry.
2. Predict how analyte concentration (or activity) controls the potential of a potentiometric measurement.
3. Choose appropriate measurement conditions to minimize interferences.
4. Correct for differences in ionic strength among calibration standards and samples.
5. Construct an appropriate calibration curve for potentiometric determinations and account for changes in analyte concentration due to sample dilution.
References: Modules on the theory and operation of ISEs may be found in the Analytical Sciences Digital Library (ASDL) collection. The following hyperlinks will direct the reader to some ASDL resources on potentiometry.
1. Analytical Sciences Digital Library. Potentiometry: e-learning module. http://community.asdlib.org/activelearningmaterials/analytical-electrochemistry-potentiometry/ (accessed April 3, 2014).
2. Harvey, D. Analytical Chemistry 2.0, Chapter 11. http://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Text_Files.html (accessed April 3, 2014)
The EPA field manual includes practical aspects of field and lab measurements of pH.
1. US EPA Field Manual: http://water.usgs.gov/owq/FieldManual/Chapter6/6.4_ver2.0.pdf
Membrane-based ISEs are widely used in the determination of ionic species. Such determinations fall under the category of a direct potentiometric measurement, which you have experienced if you have ever made a pH measurement. A typical direct potentiometric measurement requires the use of an indicator electrode, a reference electrode, and a high-impedance voltmeter. An example of equipment needed to perform a potentiometric determination is shown in Figure 1.
Figure 1. An experimental setup for the direct potentiometric measurement of sodium ion in aqueous solution.
The two electrodes pictured in Figure 1 represent an electrochemical cell. The electrode on the right is the sodium indicator electrode (sodium ISE). The electrode on the left is a reference electrode. The sodium ISE has a glass membrane that responds specifically to sodium ions. This glass membrane physically separates two solutions: one inside the electrode with a constant sodium ion concentration; one outside the membrane that is the solution you are analyzing. The electrical potential difference generated by the indicator electrode depends on the sodium concentration of the outer solution. Therefore, the function of the indicator electrode is to respond to changes in the analyte concentration in a predictable manner.
The reference electrode provides a known and stable potential to compare against the indicator electrode potential. An assumption when using an ISE system is that the potential of the reference electrode is independent of the concentration of the analyte and matrix of the sample being analyzed. The electrical potential is displayed on the high-impedance voltmeter. The voltmeter has a high impedance to minimize current flow, which prevents changes to the chemical composition of the reference electrode and to the sample (so, for example, if measuring the concentration of H+ with a pH meter, there will not be any reduction of the H+ to hydrogen gas). High impedance, which typically is greater than 1012 ohms, is also needed to minimize errors in the measured potential.
One of the most common examples of ion selective electrodes is a pH electrode – pH electrodes are selective toward the H+ ion. We will start our exploration of ISEs by understanding how a pH electrode works. pH electrodes also use a glass membrane (this membrane is about 0.1 mm thick, so is quite fragile), but in this case it is a type of glass more sensitive toward the H+ ion instead of the Na+ ion. The internal solution has a fixed and known concentration of H+. The external solution is the sample whose pH you wish to measure. The concentration of H+ in the external solution varies depending on the sample being analyzed. An important point is that the glass used to manufacture the membrane has some sodium ions (Na+) in it. A key factor in the functioning of the membrane is that the inner and outer surfaces of the glass form a very thin hydrated gel layer when in contact with water (Figure 2). The internal and external hydrated gel layers are only about 10 nm thick, which is much less than the 0.1 mm thickness of the glass membrane, so a layer of dry glass always separates the two hydrated gel layers. Cations from the solution have the ability to migrate into the hydrated gel layer. For a pH electrode, that means that H+ ions from the solution will displace some of the Na+ ions in the glass that makes up the hydrated gel layer.
Figure 2. Representation of the glass membrane in a pH electrode. Note: the width of the hydrated gel layers in the representation is too large relative to the width of the dry glass portion of the membrane.
Many of you may have used a pH electrode before. If you did, it likely appeared to consist of only a single electrode. In actuality, it is a two-electrode system, but it is designed in such a way that the reference electrode is incorporated into the glass membrane electrode. The design of the combination-electrode system incorporates an electrical contact between the reference and indicator electrode that is necessary to complete the electrical circuitry and produce a potential reading.
Q1: Will a more acidic sample displace more, the same or less Na+ from the hydrated gel layer?
Different concentrations of H+ and Na+ in the hydrated gel layer cause different junction potentials. Note that the junction potential at the interface of the membrane with the internal solution never changes because the concentration of H+ is constant in the internal solution. The junction potential at the interface of the membrane with the outer solution changes for different samples with different pH. The specific reason why varying concentrations of H+ and Na+ can be found in the outer gel layer has to do with something called the mobility of the ions.
Q2: What do you think is meant by mobility of ions?
Q3: Which ion do you think has a higher mobility, H+ or Na+?
As mentioned earlier, a reference electrode is used because its junction potential stays fixed no matter what the external solution. Therefore, the only junction potential in the entire circuit of an ion selective electrode that changes is the one at the interface of the ion selective membrane and the external (sample) solution. In the case of a Na+ ion selective electrode a glass membrane of a different composition than the glass membrane in a pH electrode is used– namely one that is more responsive to sodium ions migrating into the glass. Other ion selective electrodes can be fabricated provided a membrane exists that is selective toward the ion one wishes to measure.
Q4: Do you think other cations (e.g., Li+. K+) may have some ability to migrate into the hydrated gel layer of a pH electrode? If so, is this a problem?
The potential measured by the voltmeter is described by Equation $\ref{1}$.
$\mathrm{E_{cell} = E_{ind} - E_{ref} + E_{lj}} \label{1}$
In eq $\ref{1}$, Ecell represents the potential of the electrochemical cell, Eind represents the half-cell potential of the indicator electrode, Eref represents the half-cell potential of the reference electrode, and Elj represents the liquid-junction potential between the sample solution and the outside membrane of the indicator electrode.
Of particular interest is the relationship between Ecell and the concentration of the analyte. Remember, Ecell is measured but there is only one junction potential (Elj) in the entire system that changes (at the frit), so a measurement of Ecell is essentially a measure of the one varying junction potential.
For the Na+ ion selective electrode, the varying junction potential only depends on the concentration of Na+. However, the situation is more complicated than just using the concentration of Na+, because ion-selective electrode measurements are most commonly performed in solutions with ionic strengths that are greater than zero. In these solutions, there is a difference between the formal concentration (i.e. how the solution was prepared in the lab) and the effective concentration or activity of the analyte.
Q5: Consider a solution that has some Na+ and very high concentrations of K+Cl-. What effect do you think this might have on the activity of Na+ in the solution?
The relationship between activity and concentration for sodium is illustrated in Equation $\ref{2}$.
$\mathrm{a_{Na}= γ_{Na} [Na^+]} \label{2}$
In eq $\ref{2}$, aNa is the sodium ion activity (mol L-1) , [Na+] is the sodium ion concentration (mol L-1), and γNa is the activity coefficient for the sodium ion. As we just discussed, as the sample ionic strength increases, there is a greater probability that analyte ions will interact with oppositely charged ions from the supporting electrolyte(s) dissolved in the sample. This effectively decreases the concentration of the “free ion”, which is represented by a decrease in the activity coefficient. As the ionic strength of a solution approaches zero, the activity coefficient approaches one, and under infinitely dilute conditions, the analyte activity and analyte concentration are equal. The relationship between the oxidized and reduced forms of sodium written as a reduction reaction can be described in Equation $\ref{3}$.
$\ce{Na+ (aq) + e- → Na (s)} \label{3}$
The half-cell potential of the indicator electrode responds to changes in the activity of the analyte as described by the generalized form of the Nernst equation in Equation $\ref{4}$:
$\mathrm{E_{ind}= E_{ind}^o- \dfrac{RT}{nF} \ln\left(\dfrac{1}{a_{Na}} \right)} \label{4}$
In eq $\ref{4}$, E° is the indicator electrode potential under standard conditions (298 K, 1.00 M Na+), R is the molar gas constant (8.314 J K-1 mol-1) , T is the absolute temperature (K) , n is the number of moles of electrons in the half-reaction, and F is Faraday’s constant (96485 C mol-1).
Q6: If the indicator electrode potential under standard conditions is -0.100 V, what is the indicator electrode potential at 298 K if the activity of the sodium ion is 0.10 M?
Q7: How does the indicator electrode potential change in the previous question if the temperature is increased by 10 degrees?
A sodium ion selective electrode must be calibrated before it can be used to measure the concentration of Na+ in an unknown sample.
Q8: How would you go about calibrating a sodium ion selective electrode?
We just discussed how the ionic strength of a solution impacts the activity of Na+. Suppose you wanted to analyze the sodium concentration of a low ionic strength sample such as natural pond water.
Q9: Can you think of a way to mitigate possible effects of ionic strength to insure that your calibration procedure and sample analysis provide an accurate measurement of the concentration of Na+ in the unknown?
If one keeps the ionic strength high and constant, then the Nernst equation can be expressed in terms of analyte concentrations (and not activities) because the activity coefficient of all samples and standards are equivalent and knowledge of activity is no longer critical.
Q10: Would this proposed way to mitigate possible effects of ionic strength be utilized in pH measurements?
Q11: In the potentiometric determination of sodium ion of a mineral water sample, indicate if either of the following supporting electrolytes can be used for ionic strength adjustment: a 4.0M NH3 – NH4Cl buffer (pH 10) or 4.0M NaCl.
Q12: What would be the general criteria you would need to use in selecting a suitable supporting electrolyte for an analysis using an ion selective electrode?
The concentration of each calibration standard can be expressed as a formal concentration. The concentration term in the Nernst equation is often converted from base e to base 10, which can also be expressed as a p-function, shown in Equation $\ref{5}$.
$\mathrm{pNa= - \log[Na^+]} \label{5}$
Assuming a temperature of 298 K, and the constants R, T and F combined into a single value, under those conditions, the Nernst equation takes on the following form for cationic analytes (shown for sodium in Equation $\ref{6}$):
$\mathrm{E_{cell}= E_{cell}^o- \dfrac{0.05915}{n} pNa} \label{6}$
Q13: Based on the relationship in eq $\ref{6}$, how would you construct a calibration that links the changes in electrode potential to changes in the concentration of the sodium ion?
Q14: What is the expected slope of a potentiometric calibration curve for sodium at 35°C? What effect does temperature have on the slope of a potentiometric calibration curve?
As discussed earlier, indicator electrodes do not have a specific response to a given analyte, but have a wide range of responses to a group of analytes that are similar in charge and size. The electrode is designed to exhibit the greatest response for the target analyte, but the presence of chemically similar analytes in a sample may interfere with the determination of the target analyte and bias the potentiometric response. The selectivity of an ion-selective electrode is expressed by Equation $\ref{7}$
$\mathrm{E_{ind}=E_{ind}^o- \dfrac{0.05915}{n_{Analyte}} \log\{a_{Analyte}+ K_{Analyte,Interferent} (a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} } \} } \label{7}$
Expressed in Equation $\ref{8}$, the selectivity coefficient $\mathrm{(K_{Analyte,Interferent})}$ is a ratio of analyte to interferent activities where each species influences the indicator electrode potential to the same degree.
$\mathrm{K_{Analyte,Interferent}=\dfrac{a_{Analyte}}{(a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} }}} \label{8}$
Q15: If a sample has a sodium concentration of 1.0 x 10-3 M, and the sodium ISE has a selectivity coefficient of KNa,H = 30, what sample pH would cause a 1% error in the sodium ISE response?
Q16: Evaluate whether it is best to use alkaline or acidic conditions to determine the sodium ion concentration by ISE?
Whenever a method calibration is performed using linear regression (i.e. a best-fit line or trendline, as it is called in Microsoft Excel), it is understood that extrapolating beyond the concentration range used in the regression analysis can lead to biased results. Typically when the analyte concentration in a sample is greater than the analyte concentration for the most concentrated standard, the sample is diluted so that the analyte concentration is between the lowest and highest standard on the calibration curve. The original sample concentration is calculated using the dilution equation, shown in eq $\ref{9}$.
$C_1 V_1= C_2 V_2 \label{9}$
In eq $\ref{9}$, C1 is the analyte concentration of the original (undiluted) sample, V1 is the volume of the original sample, V2 is the volume of the diluted sample, and C2 is the analyte concentration of the diluted sample. If the sample is diluted prior to analysis, the response of the diluted sample is used as the y-value in the calibration equation and the analyte concentration of the diluted sample is the x-value determined using the calibration equation. The analyte concentration of the original sample is calculated using the dilution equation.
Q17: The table below contains sodium ISE calibration data. If the cell potential measured in a sample is ‑0.115 V, determine the sodium concentration (mol L-1) in this sample.
[Na+] (M)
Ecell (V vs SCE)
1.0 x 10-4
-0.221
1.0 x 10-3
-0.164
1.0 x 10-2
-0.107
1.0 x 10-1
-0.048
Q18: In the previous question, the sample was prepared by pipetting 5.00 mL of the original water sample and 2.00 mL of an ionic strength adjustment buffer into a 100 mL volumetric flask and diluting to the mark with distilled water. Determine the sodium concentration (mol L-1) in the original water sample. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/03_pH___Ion_Selective_Electrodes.txt |
Titrimetry
Purpose: This module provides an introduction to Titrimetry. In this module the basic theory and applications of titrimetry in water quality analysis will be presented at a level that assumes a basic general chemistry background.
Learning Objectives: At the end of this content, students will be able to:
1. Identify the key components of general titrimetric methods.
2. Evaluate the feasibility for using a titrimetric method for a specific sample.
Titrimetry is often used in determining the concentration of acids or bases in solutions, but can also be used in determining the concentration of other analytes. In any titration, you add a titrant to a titrand. You need to know the chemical reaction that is occurring so that you know the stoichiometry between the titrant and the titrand.
Q1. In an acid-base titration, what is the chemical reaction that is occurring and how would you determine the endpoint, when all of the unknown had reacted with the titrant?
Q2. What might be the difference between using an acid-base indicator and a pH electrode in measuring the endpoint for an acid-base titration?
If you think about this scenario, you have basically identified that there is some way to discriminate when a stoichiometrically equivalent amount of two different solutions are mixed.
In a broad sense, titrimetric methods require two solutions to be combined while accurately measuring at least one of the volumes, and often both volumes. The method also requires some way to determine an endpoint to be able to use stoichiometry to calculate the amount of the unknown analyte in the sample. In order for the titrimetric method to be viable, the chemical reaction must be complete and fast.
Since you are using a chemical reaction to compare the amounts of two chemicals, the minimum concentration that can be detected usually is rather high (normally over 100 ppm). For example, it is difficult to use any acid or base solution with a concentration below 0.001 M, primarily due to impacts of water ionization or the presence of dissolved CO2, which produces some carbonic acid in the solution.
Alkalinity or Acid Neutralizing Capacity (ANC)
Purpose: The purpose of this assignment is to introduce concepts of titrimetry as they pertain to the determination of ANC and water hardness of water samples.
Learning Objectives: At the end of this assignment, students will be able to:
1. Understand the chemistry occurring in the analysis of ANC.
2. Be able to determine ANC in a water sample.
Acid Neutralizing Capacity is the capacity of a water sample to neutralize added acid (the standard method can be found in the EPA National Field Manual for the Collection of Water Quality Data, chapter 6.6, at http://water.usgs.gov/owq/FieldManual/). ANC is measured on a water sample as collected, while alkalinity is measured on filtered (0.45 μm) samples. Alkalinity is often measured to check the charge balance of a solution or to identify how susceptible a water system is to acidification from inputs such as acid rain.
Q1. How is ANC different than the pH of a solution?
Q2. What are some likely chemical species present in natural water that would neutralize an added acid?
Q3. How could particles filtered out of a water sample neutralize an acid?
Q4. If pKa1 of carbonic acid is 6.352 and pKa2 is 10.329, what carbonate species would primarily be present at pH 7.0? As you adjust the pH to around 4.5, how would the dominant species change?
The primary method to determine ANC is by titrating the water sample with sulfuric acid to neutralize the bicarbonate species.
Q5. For most surface water samples, the method calls for a sample volume of either 50 mL or 100 mL—How would you accurately measure this volume?
Q6. The analysis requires a known concentration of H2SO4, usually 0.1600N or lower. How would you prepare and standardize this titrant?
During the titration, the solution is stirred and the pH is measured after each addition of titrant. The endpoint of the titration is the point at which all of the bicarbonate species has been neutralized. Therefore, as you titrate the solution, care must be taken to titrate slowly to a pH of 4.0 or lower, to obtain sufficient data to define the endpoint.
Q7. Sketch an approximate plot of how the pH of a solution of calcium carbonate (a representative weak base) would change as it is titrated with a strong acid. Where is the endpoint of the titration on this plot?
To analyze the ANC data, you would create an analogous plot of the pH versus of the volume of titrant, and calculate the change in pH per change in volume for each data point. The greatest change in pH per change in volume will yield the equivalence points for the titration. (If you start above pH 9, you should have two places of really high change in pH per change in volume)
A plot of representative data can be seen below, where the greatest change in pH per volume can be seen at 6 mL and 12 mL additions.
A
Water Hardness
Analytes other than acids and bases can also be measured via titrimetry. One such use of titrimetry is the determination of water hardness (EPA method 130.2-- http://www.epa.gov/region6/6lab/methods/130_2.pdf). Hard water (water containing a high concentration of multivalent ions such as Ca2+ and Mg2+) has a tendency to form an insoluble scale on the inside of pipes. Hard water can also form a sparingly soluble precipitate with soap (“soap scum”). This precipitate makes it hard to get suds to form and good detergent action in water high in Ca2+ and Mg2+, hence the name “hard water”. A complexing molecule such as ethylenediaminetetraacetic acid (EDTA) is commonly used to determine the concentration of calcium and magnesium in natural water samples.
Theory
The chemical structure for the fully deprotonated form of EDTA is shown below:
There are six binding sites on the EDTA molecule. Four of these sites are the carboxylic acid groups and two of these sites are the nonbonding electron pairs on the nitrogen atoms. All six of these sites are used to interact electrostatically with multivalent cations to form a coordination compound. In this case, EDTA serves as the electron donor group or ligand.
Q1. How can you ensure that the four carboxylic acid sites are deprotonated to ensure the ability to bind with Ca2+ and Mg2+?
Ligands with two or more electron donor groups belong to a class of compounds called chelates. The word chelate is derived from the Greek word for claw. Because EDTA has six electron donor groups, it is referred to as a hexadentate chelating agent. The EDTA molecule can wrap around a multivalent cation such that the electron donor groups are positioned around the central metal ion to form an octahedral structure. Below is a three-dimensional representation of an EDTA molecule bound to a ferric ion.
The presence of so many electron donor groups on the EDTA molecule provides it with some beneficial properties with respect to complexometric titrations.
1. EDTA forms a 1:1 complex with metal ions, so the titration endpoint is well defined.
2. Metal-EDTA complexes are highly stable so the reaction lies almost completely on the product side (Kf >>1), a necessary requirement for titration work.
Q2. What metals would EDTA likely form stable complexes with in most water samples? How selective is this process?
Q3. The endpoint of a titration is determined using an indicator. What would be the general features of an indicator that could be used to determine the endpoint of a water hardness titration?
There is a class of indicators that change color as a metal binds to the indicator. For EPA method 130.2 Eriochrome Black T (EBT) is used to determine total water hardness. In the case of EBT, the indicator is wine-red in the presence of metal ions and blue when the indicator does not have a metal bound to it.
Q4. When the EBT is added to your sample, what color would you expect for the indicator?
This method is used to determine total water hardness, which is the combined concentration of Ca2+ and Mg2+. With a slight modification it can also be used to determine the individual concentrations of Ca2+ and Mg2+ in solution. The modification involves adjusting the pH to a higher or more basic value. Since the pH is a higher value, a different indicator, hydroxynapthol blue (HB) is used. In the case of HB, the indicator is raspberry-red in the presence of metal ions and lavender when no metal ions are complexed. Let us consider how this modification works to allow the analysis of an individual analyte.
Q5. How might raising the pH potentially cause metals to precipitate?
The chart below lists the solubility product constants of the hydroxide complexes of several common cations in water.
Q6. Which metal would precipitate last as its hydroxide complex? This is the metal then available for titration using the HB indicator after you adjusted the solution pH to 12.0.
Cation
Equilibrium
Ksp
Al3+
Al(OH)3 (s) ↔ Al3+ (aq) + 3 OH- (aq)
4.6 x 10-33
Ca2+
Ca(OH)2 (s) ↔ Ca2+ (aq) + 2 OH- (aq)
6.5 x 10-6
Fe2+
Fe(OH)2 (s) ↔ Fe2+ (aq) + 2 OH- (aq)
8 x 10-16
Fe3+
Fe(OH)3 (s) ↔ Fe3+ (aq) + 3 OH- (aq)
1.6 x 10-39
Mg2+
Mg(OH)2 (s) ↔ Mg2+ (aq) + 2 OH- (aq)
7.1 x 10-12
Q7. If you have measured the total water hardness using EBT as the endpoint and the concentration of the Ca2+ using HB as the endpoint, how could you determine the concentration of Mg2+?
Q8. Have you made any assumptions in performing this calculation?
Q9. The concentration of the titrant needs to be known when performing a titration. This concentration is determined by standardizing the titrant. How could you standardize the EDTA solution?
An arbitrary scale has been set up to define the “hardness” of water (the ppm concentrations represent the combined concentration of Ca2+ and Mg2+).
“Soft” = 0-60 ppm
“Moderately hard” = 61-120 ppm
“Hard” = 121-180 ppm
“Very hard” = over 180 ppm | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/04_Titrimetry_Alkalinity_and_Water_Hardness.txt |
Purpose: This module provides an introduction to Ion Chromatography (IC). In this module the basic theory and applications of IC will be presented at a level that assumes a basic general chemistry background.
Learning Objectives: At the end of this content, students will be able to:
1. Explain the basic principles, operation and application of IC.
2. Explain the chemical basis for stationary phase effects and mobile phase effects.
3. Differentiate between stationary phases used in anion exchange and cation exchange.
4. Explain the basis for the common IC detection methods.
5. Describe the general process of analyzing a sample by IC.
Ion Chromatography (IC) is a useful tool for determining the presence and concentration of ions in samples and is utilized in numerous settings including environmental analyses such as the determination of anions (PO43-, Cl-, NO3-, etc) in surface waters. Current IC methods are often used to quantify concentrations in the low ppm level, depending upon the specific instrumentation used.
IC is a subset of liquid chromatography methods: ion exchange, ion exclusion, ion pair chromatography. IC methods were first reported around 1850 when H.Thomson and J.T. Way used various clays as an ion exchange and extracted labile calcium, magnesium, and ammonium ions. In 1927, the first zeolite column was used to remove Mg2+ and Ca2+ from water. Cation exchange using a sulfonated polystyrene/divinylbenzene column was developed in the 1940s as part of the Manhattan project. Very large columns were used to concentrate and purify the radioactive nucleotides required for the atom bomb. In the late 1940s anion exchange was performed with the attachment of a quaternary ammonia on the polystyrene/divinylbenzene support. (For additional description of the history see Small, H J Chem Ed, 2004, 81(9), 1277-1284.)
Basic Principles:
Ion-exchange is the basic principle behind the removal of cations and anions from drinking water using most commercial, such as Brita®, water filters. Ion-exchange is also a natural process that occurs with clay substrates, resulting in the mobility of cations in soils.
Ion exchange columns used in chromatographic applications have a fixed ion that is covalently bound to a solid support. This solid support is packed inside the column, producing the stationary phase. The mobile phase that is pumped through the column in ion exchange chromatography uses water as the solvent. The fixed ion must have an oppositely charged counterion to balance the change. The figure below represents an ion exchange resin that would be used for the analysis of anions. The two ions are held in place by electrostatic forces. Central to the functioning of ion-exchange resins is that the counterion is exchangeable. In the figure below, the anion currently associating with the resin can be displaced by another anion that is in the aqueous mobile phase. The ion in the mobile phase that is used to displace anions from the resin is referred to as the eluent ion.
The basic process of chromatography using ion exchange can be represented in 5 steps: eluent loading, sample injection, separation of sample, elution of analyte A-, and elution of analyte B-, shown and explained below. Elution is the process where the ion of interest is moved through the column. Elution happens because the eluent ion is constantly pumped through the column. The chemical reactions below are for an anion exchange process.
Step 1: Run the eluent anion through the column to displace any other anions bonded to the resin. This loading step saturates the resin surface with only the eluent anion. Note, for reproducible chromatographic results, it is essential that all the ionic sites on the column are occupied by eluent ions before starting the analysis.
(key: Eluent ion = , Ion A = , Ion B = )
The following reaction can be used to represent the eluent ion (E-) displacing any bound anion (represented by X-) that were initially associated with the resin:
$\ce{Resin^+-X^- + E^- \rightleftharpoons Resin^+-E^- + X^-}\nonumber$
Step 2: A small sample (often less than 100 µL) containing A- and B- is injected onto the column. This sample could contain many different ions, but for simplicity this example uses just two different ions.
Step 3: After the sample has been injected, the continued addition of eluent anion in the mobile phase causes A- and B- to move through the column by an ion-exchange process.
Q1. Write the chemical reaction for the association of A- and B- with the ion exchange resin.
Q2. Write the chemical reaction for the elution of A- and B- from the ion exchange resin.
Although not rigorously correct, a helpful way to examine the separation of A- and B- is to think of the reactions you wrote to answer Q2 as equilibrium processes.
Q3. Write the equilibrium expression constant for A- and B-. In chromatographic separations, this term (Kc) is referred to as the distribution coefficient.
Q4. Do you think the magnitude of the distribution coefficients are the same for A- and B-? Why or why not?
Q5. If the distribution coefficient of A-() is smaller than the distribution coefficient of B-(), draw a sketch of the elution process that is similar to the figure in Step 1 but at a point where A- and B- are partway through the column.
Step 4: As the eluent continues to be added, A- eventually moves through the column in a band and ultimately elutes from the column.
Q6. Suppose B- had a very strong affinity for the resin, what would happen to its elution time?
Q7. Is this a desirable or undesirable situation if you were trying to analyze A- and B- in a mixture?
Q8. Is there a situation you can think of when it might be desirable for B- to have a very strong affinity for the resin?
Step 5: The eluent eventually causes the elution of B- off the column.
$\ce{Resin+-B- + E- \rightleftharpoons Resin+-E- + B-}\nonumber$
The overall 5 step process can be represented pictorially:
Stationary phase (or resin) composition
There are a number of different resins or stationary phases that have been developed for use in IC.
Q9. If you want to separate cations, what would be different about the stationary and mobile phases?
All of the resins used in ion exchange columns consist of very fine particles.
Q10 Would water flow easily through a column containing very fine particles?
Q11. If not, how could you get the water through the column?
Q12. Considering that the column is packed with very fine particles, what must be done to surface water samples before injecting them onto the column?
Detection Methods
In order for an ion exchange separation to be useful information, you need detect the ions as they elute from the column.
Q13. Can you think of a way to detect the presence of ionic substances in water?
Q14. Would this detection method distinguish between Na+ and Ca2+?
Q15. If you utilize this detection method with the chromatographic separation, how important is the selective response of the detector?
It is important to remember that the Na+ and Ca2+ only move through the column because of the continuous presence of the eluent ion in the mobile phase.
Q16. Will the eluent ion respond to the detection method you thought of above to measure the presence of Na+ and Ca2+.
We will not go into the details of how the process works, but modern ion chromatographic systems are designed in such a way that the eluent ion either undergoes a chemical reaction or is removed from the system after the column such that the only ionic substance left in solution is the analyte ion. This suppression of the signal from the eluent ion allows for the selective detection of only the analyte ions as they elute from the column.
The results from the detection give rise to a chromatogram, which shows the detectors response as a function of elution time. If all of the analyte came out at the same time the chromatogram would look like:
In reality, the peaks are actually broadened due to factors we will not discuss. Thus the peaks will look more like:
Q17. How will the chromatogram change as you increase the concentration of A- and B- injected into the column? Make sure to label the axes of your chromatogram.
Q18. Since neither axis in the chromatogram you drew above is concentration, how can we calibrate the detector response to determine the concentration of A- and B-?
You now have an outline of the basic ion chromatography process. The exact way the sample is loaded onto the column varies with the instrument. This process is often either a direct injection or using an inline automated sampling system. The sample is eluted off of the column, through the detector. The signal from the detector is then converted into the chromatograph. The peak areas then can be converted to a concentration using a calibration curve. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/05_Ion_Chromatography.txt |
Environmental Analysis: Effects of Acid Rain on Atlantic Salmon Populations
Sections of this overall module:
• Sample Gran Plot (Excel)
How this module has been used:
There are several different experimental approaches that can be pursued as identified in the different hypotheses.
This project has been used as an analytical problem based project in Environmental Chemistry, and in a Quantitative Analysis leaning Analytical Chemistry course where the students cover the basic analytical methods and then
• Collect water samples,
• Obtain field pH measurements over time during spring snow melt,
• Measure Acid Neutralizing Capacity (ANC) to identify sites where the water system will be sensitive to acid rain impacts,
• Measure nutrient levels (phosphorus and nitrogen containing compounds),
• Measure water hardness (titration) and concentrations of Ca2+ and Mg2+ via IC.
The surface waters in the Machias and East Machias watershed are very low conductivity (often about 20 µS/cm), resulting in needing the trace analysis via ion chromatography (IC) rather than titration. The Ca2+ concentrations in these watersheds often run about 2 ppm. IC is used as it was the instrument readily available for undergraduate use. Atomic absorption spectroscopy could readily be used, and this process is described in other modules.
These low ionic strength waters also make measuring pH a challenge and require very careful calibration, validation, and measurements.
There is also a remediation project where clam shells are used as a source of calcium carbonate to buffer smaller streams from episodic acid rain events. This project has been used to obtain upstream and downstream analysis to identify the changes in water chemistry from the dissolving shells.
There is sample data provided so this could also be used as a dry lab for students to walk through the data analysis process.
This module in primarily focused on the first two hypotheses spelled out in the Identifying the Problem section.
06 Instructors Guide
Facts, Observations and Evidence
The Atlantic salmon is an anadromous fish species, spending most of its adult life in the ocean coming back to spawn in freshwater rivers. The adult creates a redd, where it lays the eggs, in the fall. After hatching the salmon undergo several stages of their lifecycle in the freshwater. The fry leave the spawning grounds in search of food and develop into parr. After two years, the fry undergo the process of smoltification, where their physiology changes so that they can live in salt water. At this stage the smolt leaves the river and heads to the ocean, where most ultimately head to feeding grounds off of Greenland. The salmon usually live in the ocean for 2-3 years before they return to the river they were hatched in.
In the United States, the historical grounds of the Salmon ranged on the northeast coast south to the rivers of Long Island Sound. Currently, remnants survive only in 8 rivers in Maine, the largest of which is the Penobscot River. Historically in Maine, salmon landings were as high as 90 million tons in the late 1800s. (Baum 1997) By the 1940s the commercial fisheries in Maine were closed, and by the 1990s, less than 2000 adult salmon were returning annually to all of the rivers in Maine, despite over 13 million juvenile salmon stocked annually.
To identify the cause of the minimal returns, the mortality in the river and the sea run mortality were determined. To minimize the potential mortality as parr and fry, a number of smolts are stocked shortly before they are set to leave for the sea. Some of the smolts were pit-tagged with radio labels to identify near shore mortality. Data collected demonstrated a significant mortality from the time when the fish reached the estuary and made it to the open ocean. This mortality would correspond to heavy predation in the bay, or failure of the smolt to survive the transition from fresh to salt water.
During the smoltification process, the salmon has significant biological changes in the gill structure for osmoregulatory processes. In fresh water the salmon needs to maintain a higher salt content in its blood as compared to the water it is in. When in seawater, the salmon needs to maintain a lower salt content in the blood compared to the water it is in. During this change in the gill structure, the salmon is most sensitive to stressors or damage to the gill. Studies have shown that gill deformities can be caused by low pH and by high aluminum. In both of these cases the deformities occur as the cation binding to the gill structure.
The National Resource Council in 2004 specifically stated that the effects of acid rain on smolts is one of the most significant factors impeding the recovery of the Atlantic salmon; “The problem of early mortality as smolts transition from freshwater to the ocean and take up residence as post-smolts needs to be solved. If, as seems likely, that the difficulty of the transition is due in part to water chemistry, particularly acidification, the only methods of solving the problem are changing the water chemistry and finding a way for the smolts to bypass the dangerous water.
The related Brook Trout has had similar extirpation from streams and ponds in many northeast locations due to low pH.
The question to be answered is: What is causing the increased mortality of Atlantic salmon?
Some hypotheses as to possible (and perhaps intertwined) causes include:
Low pH. Salmonids (Atlantic Salmon, Brook Trout, and related species) are known to be sensitive to pH, with stress occurring below pH 5, and mortality below pH 4.5. The pH of the aquatic ecosystem can be lowered due to natural acid inputs, such as dissolved organic carbon (DOC), or through anthropogenic inputs, predominantly acid rain. The acid rain is created from nitrogen oxides and sulfur oxides emissions. These may be episodic or chronic conditions.
High Aluminum. Aluminum will bind to the gills of the salmonids and irreversibly damage the gills, preventing the fish from uptaking oxygen. Aluminum will get into the water systems predominately from leaching from the ground when acid rain impacts a system.
Elevated Temperature Salmonids are temperature sensitive. When stream temperatures reach over 22ºC, salmon are severely compromised. When temperatures reach over 26ºC, mortality often ensues. The temperature of the stream can change due to increases in air temperature, reduction in boreal cover, or reduction in underground cold-water stream inputs.
Increased predation. There are two major predators. As the fish swim to the ocean as smolts, they are met by a host of predators such as cormorants and seals. The populations of both of these predators have increased in recent years. Another predator is humans, and overfishing is of concern. Starting in the 1950s, salmon have been caught in their ocean feeding grounds off of Greenland and Newfoundland. The catch has steadily declined with the decreasing populations.
Identifying Possible Analysis Methods
There are three hypotheses that have been put forward that could be responsible for the decline of the Atlantic Salmon. The purpose of this exercise is to identify analysis methods that could be used to test the chemical species that contribute to each of these hypotheses. Among the various analytical methods you find that may be applicable to this measurement, identify their strengths and weaknesses (e.g., sensitivity, expense, ease of use, reproducibility).
Hypothesis 1. Episodic acid rain events overcome the buffering capacity of the water system and decrease the pH leading to stress and mortality in Atlantic salmon parr and smolts.
Q1. Using available literature and other sources of information, identify possible analytical methods that could be used to monitor acidity.
Hypothesis 2. Acid rain has increased the leaching of calcium from freshwater, reducing the nutrients available for salmonids. This loss of calcium prevents salmon from properly undergoing the smoltification process resulting in mortality when the fish head to the ocean.
Q2. Using available literature and other sources of information, identify possible analytical methods that could be used to detect calcium and measure their levels in components of the water (e.g., free, organic bound).
Hypothesis 3. Acid rain has increased the leaching of aluminum, resulting in a high concentration of free aluminum ion resulting in gill damage, leading to the main cause of mortality of Atlantic Salmon during the smoltification process.
Q3. Using available literature and other sources of information, identify possible analytical methods that could be used to detect aluminum and measure their levels in components of the water (e.g., free, organic bound).
The rest of the module primarily focuses on the first two hypotheses. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/06_Instructors_Guide/01_Identifying_the_Problem.txt |
This section of material is designed to be approached after students read through the Identifying the Problem module (or an analogous scenario), so they have some idea of what analytes and samples sources the sampling program would focus on addressing. This is meant to have students can work in groups on the questions that are provided. The answers to the questions are provided below.
I have used this by giving the question sets to the students with about 10 minutes left in a class period where I had them discussing Q1. Then they did Q2-5 between classes- so they began thinking about the questions on their own. The next class period they worked in small groups of 3-4 to discuss those answers and proceeded through much of the rest of the assignment. This ended up taking about 90 minutes of class time.
Q1. What key questions must be considered when designing a sampling plan?
Students should be given time to brainstorm on key issues to consider when designing a sampling plan. The instructor may want to provide the answers listed below only after the students may have had a chance to come up with their own answers. Eventually students should be guided to consider issues such as:
1. Where in the watershed should we collect water samples?
2. What type of samples should we collect?
3. When should we collect the sample?
4. What is the minimum amount of sample for each analysis?
5. How many samples should we analyze?
6. How can we minimize the overall variance for the analysis?
Q2. Pick eight random samples from the grid laid out above. How do you ensure you sampling is random?
One way to get random samples is to use Excel or other random number generator. To get 8 random grids label the boxes 1-8, row 1, 9-16, row 2, etc. for 64 boxes. Then use Excel to generate 8 random numbers between 1 and 64; for example: 63, 35, 25, 46, 7, 53, 43, 5.
You might ask the students to discuss whether they think a random approach represents the best way to sample. They may realize that the answer depends in part on what you may already know about the system you are sampling. If there is a specific site of concern or point source of the chemical, then random sampling might not be the best option.
Now take a look at the following grids with the analyte of interest identified (colored squares).
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
Grid A
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
Grid B
Q3. Would you consider the samples above to be heterogeneous or homogeneous?
The analyte is heterogeneous because it is not evenly distributed throughout the entire grid.
Q4. Did your random sampling affect the potential accuracy or precision of your measurement of the analyte for the samples in grid A or grid B? If so how?
The random sampling would have been better for grid B because the analyte is more spread out than in grid A where the analyte is confined or stratified.
Q5. Each of the previous grids is an example of one of these cases. Can you identify which sample is which?
Sample B exhibits constitutional heterogeneity while Sample A exhibits distributional heterogeneity.
Q6. How does distribution heterogeneity affect accuracy and precision?
The answer to Q6 is included with the answer to Q7 below.
Q7. How does constitutional heterogeneity affect accuracy and precision?
When you overlay the sampling scheme with Sample A and Sample B neither sampling scheme is effective, as shown below. The analyte in sample “A” was sampled once (35) as it was in Sample “B” (53). With either sample, unless the sampling scheme could take this heterogeneity into account, the accuracy and precision of the measurement would be compromised.
A. Distributional Heterogeneity
B. Constitutional Heterogeneity
Q8. Do you see a scenario where distribution heterogeneity could be magnified by mixing and/or sampling?
Sampling is often by weight or by “grab”. In this case settling may alter the sample composition.
Q9. What is the advantage of implementing judgmental sampling over random sampling if one knows the point source for the discharge an analyte into a system?
The advantage is that you can get larger number of relevant samples which should decrease the standard deviation of the average value measured for that sample. The cost should also decrease.
Q10. Assume you have chosen a selective sampling plan to evaluate pollution from a point source into a pond. Use the diagram below and words to describe your sampling plan.
If we use a purely random grid over the pond we will be unable to tell what the effect of the point source is because we would have sampled only once at the source.
A selective method might be the following:
A sample is taken at the point source and for comparison a sample is taken at a distance from the point source.
Q11. Use a grid design (as we have previously done) to show how you would conduct systematic sampling (regular intervals in space and time) of the pollutant. Is there an advantage to what you might learn using this sampling method? What are the disadvantage(s)?
Here we have set up a grid along regular intervals. Because we collect only 8 samples, we may or may not collect a representative sample from within each grid point as we are only collecting one sample.
Q12. Describe how stratified sampling (random sampling within sub populations) might be applied to evaluate the pollutant in the lake? In general, what is the advantage of stratified sampling over cluster sampling?
In this example the trajectory along the longest distance from the point source is sampled a total of four times, with random grabs to be co-joined into a single sample.
Q13. What is the main disadvantage of grab and composite samples?
You cannot use them continuously for real time in situ monitoring. If you have an analyte that could vary over time, such as from episodic acid rain, your grab sample may not identify the problem.
Q14. Can you think of any control studies you might want to include when compositing samples?
You might want to retain portions of the grabs and analyze them separately.
Q15. If you are beginning to monitor a site or want to compare your data to a similar source, have there been any standard sampling protocols used for prior monitoring?
This will depend on the project you are initiating. For the water quality monitoring for the Atlantic Salmon project there are some guidelines in EPA standard methods, or regional EPA management protocols to ensure data can be compared. However, most sites such as the EPA Volunteer Stream Monitoring web site does not describe a particular sampling method because each sampling plan must be designed for the specific analytical question being addressed.
Q16. Based on the graph above, describe a sampling procedure that would allow you to obtain a representative sample.
A strategy would be to collect multiple samples at regular intervals of time during a 24-hour span of time.
Q17. Is there any systematic pattern to the data in the graph above?
There is not a systematic pattern although spikes in pH appear more prominent in late fall and winter and early summer months.
Q18. Can you think of some event(s) that may account for the acidic spikes in the pH?
Spikes may be due to acidic snow or rain fall, snowmelt, or other acidic water inputs (flushing of a bog system) that are episodic in nature.
Q19. How would this pH data affect when you might choose to sample a site to assess if acid rain is impacting the ecosystem?
Sampling would have to be conducted at regular intervals to account for known fluctuations. If considering the impact of acid rain, the sampling protocol may focus on sampling around and after rainfall events. If historical data are available, they may aid in better understanding the impact of any changes in pH.
Q20. Describe what else you would need to know to determine when to perform your sample collection if you are addressing whether acid rain is impacting a site. What other data might you need to look up or consider that would contribute to the changes in pH? Would this data affect when you choose to sample?
One consideration is the natural pH functions due to changes in CO2. In the early morning CO2 levels tends to be higher due to respiration that occurs during night time. As sun rises, plants and algae begin photosynthesis thereby consuming CO2 and causing the pH to rise (more basic) as the day progresses. Algae blooms can significantly increase this effect.
You also would need baseline conditions, as some bodies of water are naturally acidic due to dissolved organic matter (e.g. fulvic and humic acids). Thus you would either need historical data to demonstrate a chronic acidification.
Q21. What physical or chemical processes might contribute to the pH fluctuations? How might this affect your sampling plan?
As stated above, CO2 functions may affect pH. Water temperature affects solubility of gasses, thus changing the amount of CO2 dissolved in water.
Q22 Another factor to consider is the sample handling time. Can you think of ways in which sample holding time may impact the concentration of different species such as nitrate, pH, and Ca2+ in the sample?
Some analytes may precipitate or degrade over time if the sample is not properly handled. For example, calcium and magnesium tend to precipitate as hydroxides. Therefore, sample pH will have to be adjusted below 2 if the analysis is not conducted right away. Nitrates are quickly degraded by bacteria, so concentration of nitrates may change over time if the sample is not refrigerated. pH can change due to processes that change the CO2 concentration in water system (degasing or increasing solubility of the gas as temperature changes, respiration of organisms in the sample), or some redox processes.
Q23. Why would adjusting the solution pH to below pH 2.0 be needed for preserving samples for metal analysis? How can the solution be adjusted to pH 2 without significantly diluting the sample?
Adjusting the pH to below pH 2.0 would prevent the formation of metal hydroxide precipitates. This concentration of H+ can also help limit adsorption to glass surfaces.
Q24. It is known from analyses conducted in 2008 that the % relative sampling error for water hardness by EDTA titration is 0.8%. How many samples should you collect to limit the relative standard deviation for sampling to 1.0% within the 95% confidence level? Is this a feasible task?
Answering this question requires reading Harvey, specifically section 7.2: How Many Samples to Collect.
The minimum number of samples can be calculated using the equation:
$n_\textrm{samp} = \dfrac{t^2s^2_\textrm{samp}}{e^2}\nonumber$
where t is the value for the t test which depends on the confidence level, ssamp is the relative standard deviation for sampling, and e is the percent relative sampling error.
Because the value of t depends on nsamp, the solution is found iteratively. We start for a value of n = ∞ and t (0.05, ∞) = 1.960.
$n_\textrm{samp} = \dfrac{(1.960)^2(1.0)^2 }{(0.8)^2} = 6.0\nonumber$
Letting nsamp = 6, t (0.05, 6) = 2.447
$n_\textrm{samp} = \dfrac{(2.447)^2(1.0)^2}{(0.8)^2}= 9\nonumber$
Letting nsamp = 9, t (0.05, 9) = 2.262
$n_\textrm{samp} = \dfrac{(2.262)^2(1.0)^2} {(0.8)^2}= 8\nonumber$
Letting nsamp = 8, t (0.05, 8) = 2.306
$n_\textrm{samp} = \dfrac{(2.306)^2(1.0)^2}{(0.8)^2}= 8\nonumber$
Because two successive calculations give the same value for ssamp, we have an iterative solution to the problem. We need at least 8 samples to achieve a percent relative sampling error of ±0.80% at the 95% confidence level.
Q25. If the cost of collecting a sample is $20 and the cost of analyzing a sample is$50 what budget should you allocate for the project and what sampling strategy would be most effective for the given number of samples?
If we have 8 samples, the cost of collecting will be $160 and cost of analysis$400, for a total of \$560. Most effective sampling plan would be judgmental.
Q26. Below is an image of the area around Machias, where the confluence of the Machias, Middle, and East Machias Rivers empty into the Machias Bay. Design your sampling plan. Think about random, systematic, clustering, etc. sample strategies. Will you take grab samples or pool samples together?
One way to use this question is to have the students discuss these questions in groups and put together a final plan. The plans could either be presented to the class or turned in as a graded written assignment.
The last part of this module discusses proper preparation of sampling containers. The procedures included within this guide adhere to EPA guidelines for water monitoring.
Q27. Why are different procedures recommended? Why do you think acid washing of the glassware is recommended for some analyses? What is the purpose of using a phosphate-free detergent?
To answer this question, students may be directed to the EPA Monitoring and Assessing Water Quality website (http://water.epa.gov/type/rsl/monitoring/vms50.cfm). Acid washing is required for nitrate and phosphate analyses. Obviously, phosphate-free detergent is necessary to eliminate any possible contributions of phosphate that may have adhered to the container during washing. Acid washing will remove any traces of nitrates or phosphates from containers.
Q28. For some analytes such as phosphorous, plastic containers made of either high-density polyethylene or polypropylene might be preferable to glass. Why would this be the case? In addition, the EPA states that all containers and glassware must be “dedicated” to a specific analysis. What would be the drawback of reusing glassware for a different analysis?
Glass and some types of plastic containers have positive ion-exchange sites that can interact with negative ions in solution. Phosphate can therefore adhere to the surface of glass and be lost. Phosphorous is also known to leach out of glass containers. Rinsing containers with dilute HCl helps saturate the sites and minimize losses due to adsorption.
Since most of the analyses are performed at trace level, dedicated containers should be used to avoid cross contamination.
Additional Resources: For further discussion on sampling amounts see section 7.2 in Harvey, D. Chapter 7, Collecting and Preparing Samples. (http://collection.asdlib.org/?p=452) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/06_Instructors_Guide/02_Sampling_and_Sample_Prepar.txt |
Because H+ binds more strongly than Na+ to the –SiO- function in glass, the H+ will displace more Na+. As the sample becomes more acidic, there are more H+ present. This greater H+ concentration will displace more Na+ from the hydrated gel.
Q2: What do you think is meant by mobility of ions?
Students will usually consider Mobility to mean how well something moves around. The mobility of ions can be defined as the rate at which they move around.
The smaller an object is the more mobile it will tend to be. For ions the mobility will be based on its effective ionic radius.
This is a challenging question – based on effective ionic radii in water, Na+ is smaller, so it should have a higher mobility in solution, but not necessarily in glass membrane. H+ is smaller than Na+, so it would have a higher mobility in glass.
Q4: Do you think other cations (e.g., Li+. K+) may have some ability to migrate into the hydrated gel layer of a pH electrode? If so, is this a problem?
Yes they will be able to migrate into the gel layer. This can cause a problem in that those ions will also displace Na+ giving rise to a higher membrane potential. This is a cause of the sodium error in pH measurements.
For the Na+ ion selective electrode, the varying junction potential only depends on the concentration of Na+.
It should be noted that liquid junction potentials will happen whenever two different electrolytes come in contact. Thus in the combination pH electrode, a potential can arise at the frit between the electrode and the solution. This junction potential arises due to the difference in mobility as the electrolyte moves across the boundary. This potential can be difficult to reproduce or known with any accuracy.
In solutions with a very low ionic strength, e.g. the surface waters of interest in the salmon declines, there is greater driving force for diffusion across the frit. Thus the propensity for junction potential errors becomes important to consider. This junction potential can be unstable, leading to drift; therefore periodic checks of the calibration (often hourly) may be needed. Calibration needs to be with low ionic strength buffers to better reflect the samples of interest, and make it so that the potential across the glass membrane is the only potential that is changing. It can also be important to note that the electrode design is important, therefore there are recommended electrodes for some types of environmental analyses when measuring surface or rain waters of low ionic strength.
This issue generally leads to a validation of the pH measurement by measuring the pH of distilled water that is in equilibrium with air. The pH should be around 5.7, thus pH readings should stabilize rather quickly between 5.6 and 5.8 to demonstrate the pH electrode is responding appropriately.
Q5: Consider a solution that has some Na+ and very high concentrations of K+Cl-. What effect do you think this might have on the activity of Na+ in the solution?
The high concentration of K+Cl- will affect the ionic strength of the solution, this will affect the activity coefficient of Na+. This can be calculated more formally using the extended Debye-Huckel equation (Equation 6.63 in the Harvey Analytical 2.0 text)
The relationship between activity and concentration for sodium is illustrated in Equation 2.
$\mathrm{a_{Na}= γ_{Na} [Na^+]} \tag{2}$
In eq 2, aNa is the sodium ion activity (mol L-1) , [Na+] is the sodium ion concentration (mol L-1), and γNa is the activity coefficient for the sodium ion. As we just discussed, as the sample ionic strength increases, there is a greater probability that analyte ions will interact with oppositely charged ions from the supporting electrolyte(s) dissolved in the sample. This effectively decreases the concentration of the “free ion”, which is represented by a decrease in the activity coefficient. As the ionic strength of a solution approaches zero, the activity coefficient approaches one, and under infinitely dilute conditions, the analyte activity and analyte concentration are equal. The relationship between the oxidized and reduced forms of sodium written as a reduction reaction can be described in Equation 3.
$\ce{Na+ (aq) + e- → Na (s)} \tag{3}$
The half-cell potential of the indicator electrode responds to changes in the activity of the analyte as described by the generalized form of the Nernst equation in Equation 4:
$\mathrm{E_{ind}= E_{ind}^o- \dfrac{RT}{nF} \ln\left(\dfrac{1}{a_{Na}} \right)} \tag{4}$
In eq 4, E° is the indicator electrode potential under standard conditions (298 K, 1.00 M Na+), R is the molar gas constant (8.314 J K-1 mol-1) , T is the absolute temperature (K) , n is the number of moles of electrons in the half-reaction, and F is Faraday’s constant (96485 C mol-1).
Q6: If the indicator electrode potential under standard conditions is -0.100 V, what is the indicator electrode potential at 298 K if the activity of the sodium ion is 0.10 M?
$\mathrm{E_{ind}= -0.100\:V- \dfrac{ \dfrac{8.314\:J}{K\:mol}\times 298\:K}{1\:mol\: e^- \times \dfrac{96485\:C}{mol}} \ln \left(\dfrac{1}{0.1\:M}\right)}\nonumber$
Therefore Eind = -0.1591 V
Q7: How does the indicator electrode potential change in the previous question if the temperature is increased by 10 degrees?
$\mathrm{E_{ind}= -0.100\:V- \dfrac{\dfrac{8.314\:J}{K\:mol}\times 308\:K}{1\:mol\: e^- \times\dfrac{96485\:C}{mol}} \ln\left(\dfrac{1}{0.1\:M}\right)}\nonumber$
Therefore Eind = -0.1611 V
Thus the measured potential is dependent on the temperature. This temperature effect has led to most meters have an automatic temperature correction.
Q8: How would you go about calibrating a sodium ion selective electrode?
You basically have to measure the response as a function of concentration. There are several practical ways that this is done. Most ISE are standardized by doing standard addition, where you start with distilled water and ionic strength adjuster. You then add increments of the analyte to the solution and measure the potential for each step.
The electrode could be also be calibrated using different solutions of known concentration (similar to how the pH electrode is calibrated using at least two solutions of known pH).
Q9: Can you think of a way to mitigate possible effects of ionic strength to insure that your calibration procedure and sample analysis provide an accurate measurement of the concentration of Na+ in the unknown?
The ionic strength needs to be rather constant, so the ionic strength is adjusted using the addition of an ionic strength adjuster solution. This solution contains a high concentration of ions that do not interfere or mask the indicator electrode response.
Q10: Would this proposed way to mitigate possible effects of ionic strength be utilized in pH measurements?
The ionic strength adjustor solution would have to have species that would not affect pH, thus greatly limiting the possible solution. In, practice this does not work well for measuring pH, as most ionic strength adjustments will have trace contamination that will affect pH, or the cations will introduce additional errors when the cation interacts with the glass membrane (the source of the Na+ error in pH measurements).
Q11: In the potentiometric determination of sodium ion of a mineral water sample, indicate if either of the following supporting electrolytes can be used for ionic strength adjustment: a 4.0M NH3 – NH4Cl buffer (pH 10) or 4.0M NaCl.
The 4.0 M NaCl would introduce sodium ions into the solution, which affects the concentration of what you are trying to measure. Therefore the ammonia buffer solution would be able to be used for the ionic strength adjuster.
Q12: What would be the general criteria you would need to use in selecting a suitable supporting electrolyte for an analysis using an ion selective electrode?
You want a supporting electrolyte that:
1. Is different than analyte.
2. Will not interfere with the concentration of the analyte (it does not react or bind with the analyte)
3. It will not mask the presence of the analyte
4. The indicator electrode does not respond to it.
5. It will not foul the electrode or any electrode junction.
Q13: Based on the relationship in eq 6, how would you construct a calibration that links the changes in electrode potential to changes in the concentration of the sodium ion?
In the Nernst equation (eq 6) the cell potential varies with the log of the activities. The measured electrode potential is graphed versus the log of the concentration. The line of best fit is linear. The equation of the line of best fit is used to determine the concentration of the unknown.
Q14: What is the expected slope of a potentiometric calibration curve for sodium at 35°C? What effect does temperature have on the slope of a potentiometric calibration curve?
$\mathrm{E_{ind}= E^0-\dfrac{\dfrac{8.314\:J}{K\:mol}\times 308\:K}{1\:mol\: e^-\times\dfrac{96485\:C}{mol}} 2.303 \log(Na^+)}\nonumber$
If you plot the potential as a function of log concentration then the slope would be 0.0611 since ln x = 2.303 log x
Q15: If a sample has a sodium concentration of 1.0 x 10-3 M, and the sodium ISE has a selectivity coefficient of KNa,H = 30, what sample pH would cause a 1% error in the sodium ISE response?
First calculate the Eind for the solution of 1.0 x 10-3 M
$\mathrm{E_{ind}=-.100\:V- \dfrac{0.05915}{1\:mol} \log\{1\times10^{-3} \}}\nonumber$
$\mathrm{E_{ind}=0.07745\:V}\nonumber$
A 1% error in the response would give a
$\mathrm{E_{ind}=0.07822\:V}\nonumber$
Putting this back into the original equation
$\mathrm{0.07822\:V=-.100\:V- \dfrac{0.05915}{1\:mol} \log\{x\}}\nonumber$
Thus x = 0.0476
Putting this into
$\mathrm{0.0476=\{a_{Analyte}+ K_{Analyte,Interferent} (a_{Interferent} )^{n_{Analyte}⁄n_{Interferent}} \}}\nonumber$
$\mathrm{0.0476=\{1\times10^{-3}+ 30(a_{Interferent} )^1 \}}\nonumber$
Leads to an H+ concentration of 0.00155, or pH 2.8
Q16: Evaluate whether it is best to use alkaline or acidic conditions to determine the sodium ion concentration by ISE?
You want the alkaline conditions, as in acidic conditions you have a higher H+ activity, leading to a higher % error.
Q17: The table below contains sodium ISE calibration data. If the cell potential measured in a sample is ‑0.115 V, determine the sodium concentration (mol L-1) in this sample.
[Na+] (M)
Ecell (V vs SCE)
1.0 x 10-4
-0.221
1.0 x 10-3
-0.164
1.0 x 10-2
-0.107
1.0 x 10-1
-0.048
In Excel the graph would look like:
The line of best fit is -0.115V = 0.0576x +0.009
The log of Na+ is -2.152
Thus the concentration of Na+ is 7.0 x 10 -3 M
Q18: In the previous question, the sample was prepared by pipetting 5.00 mL of the original water sample and 2.00 mL of an ionic strength adjustment buffer into a 100 mL volumetric flask and diluting to the mark with distilled water. Determine the sodium concentration (mol L-1) in the original water sample.
$C_1\times \mathrm{5.00\:mL= 7.0\times 10^{-3}\: M \times 100\:mL}\nonumber$
Thus the original concentration is 0.14M | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/06_Instructors_Guide/03_pH___Ion_Selective_Electro.txt |
Titrimetry
Q1. In an acid-base titration, what is the chemical reaction that is occurring and how would you determine the endpoint, when all of the unknown had reacted with the titrant?
You need some indicator to identify when the endpoint occurs. For an acid-base titration, you can have change in solution color due to acid/base indicator, or you can determine the endpoint by measuring pH as a function of added titrant. The inflection point of this graph will be at the endpoint.
Q2. What might be the difference between using an acid-base indicator and a pH electrode in measuring the endpoint for an acid-base titration?
The indicator changes in the pH range of the indicator, therefore you have to select the correct acid-base indicator. The pH electrode consistently allows pH to be measured, and requires a titration curve be prepared in order to determine the endpoint. An acid/base indicator is much quicker-but you need to know the pH of the endpoint. Following the titration with a pH electrode can give you additional information, such as something about how the solution is buffered, and the pH of the endpoint.
Alkalinity or Acid Neutralizing Capacity (ANC)
Resources:
Rounds, S.A., 2006, Alkalinity and acid neutralizing capacity (ver. 3.0): U.S. Geological Survey Techniques of Water-Resources Investigations, book 9, chap A6, sec 6.6, July 2006, accessed [June 25, 2012], from http://water.usgs.gov/owq/FieldManual/Chapter6/section6.6/ .
Q1. How is ANC different than the pH of a solution?
pH is simply a measure related to the amount of H+ in solution. ANC deals with how much acid can be neutralized, therefore it relates to the buffer capacity of the system. The greater the ANC, the less likely that body of water will be affected by acidic inputs.
Q2. What are some likely chemical species present in natural water that would neutralize an added acid?
Carbonate species (HCO3-, CO32- ) and partially deprotonated natural organic acids (humic acid, fulvic acids) can both act to buffer natural water systems.
Q3. How could particles filtered out of a water sample neutralize an acid?
Filtering a water sample will remove suspended particles, and natural organic acids can form micelle-like aggregates that are large enough to be filtered, particularly in the presence of metal ions. Therefore filtering can remove some of the deprotonated organic acids.
Q4. If pKa1 of carbonic acid is 6.352 and pKa2 is 10.329, what carbonate species would primarily be present at pH 7.0? As you adjust the pH to around 4.5, how would the dominant species change?
At pH 7.0, since this is a little more basic than the pKa1 of carbonic acid, the primary species would be the HCO3-. As you adjust the pH to 4.5 you will continue to protonate this species, causing the dominant carbonate species to be the carbonic acid.
Q5. For most surface water samples, the method calls for a sample volume of either 50 mL or 100 mL?
Will either use volumetric pipet or some graduated cylinders that are rated TD (to deliver).
Q6. The analysis requires a known concentration of H2SO4, usually 0.1600N or lower. How would you prepare and standardize this titrant?
This concentration can either be purchased directly, or prepared by diluting from a more concentrated solution. Since H2SO4 has two H+, the normality of the acid solution is twice the molarity of the acid solution. The USGS Field Manual has the directions for preparing and standardizing this solution in section 6.6.2
Q7. Sketch an approximate plot of how the pH of a solution of calcium carbonate (a representative weak base) would change as it is titrated with a strong acid. Where is the endpoint of the titration on this plot?
This would be a diprotic acid titration curve with two equivalence points.
The endpoints can be determined by the inflection point method. This works if the sample has a high enough ANC to allow for an accurate determination of the inflection point.
In samples with low alkalinity, such as the salmon rivers in Maine, the ANC can be calculated using a Gran Method rather than an inflection point.
ANC data analysis:
1. In Excel enter sample name, volume and pH values between 4.5 and 3.5 with their corresponding titrant volumes added (Vi).
2. Calculate Gran F for each pH.
Gran F = (Sample Volume + Vi) x 10(4-pH)
3. Perform a data regression to calculate the ANC (milliEQ/L) and R2 using Gran F as the X axis and Vi as the Y axis.
The R2 value should be greater than or equal to 0.990.
ANC (milliEQ/L) = (Acid Normality X Regression Constant X 1000)/ Sample Volume. A sample plot is presented in the Excel sheet (ANC_samplegranplots) with this learning module)
Alkalinity data analysis:
Alkalinity is a related parameter calculated in mg/L CaCO3 using the equation:
$T_{alk} = \left(\dfrac{V_B * N_{H_2SO_4}}{V_S}\right)\left(\dfrac{1\: mole\: CaCO_3}{2\: eq\: H^+}\right)(MW_{CaCO_3})\nonumber$
$T_{alk}$ : Total alkalinity (mg/L CaCO3)
$V_B$ : Total titrant volume required to reach bicarbonate equivalence point (L)
$N_{H_2SO_4}$ : Normality of sulfuric acid (eq/L)
$V_S$ : Sample volume titrated (L)
$MW_{CaCO_3}$ : Molecular weight of calcium carbonate (1.00087 x 105 mg/mol)
This analysis makes the assumption that there are low concentrations of other titratable species such as ammonia, silicic acid, or borate.
Additional possible exercises:
If you have measured the concentration of other common cations and anions, you can usually calculate a mass balance once you have calculated alkalinity.
Water Hardness
Q1. How can you ensure that the four carboxylic acid sites are deprotonated to ensure the ability to bind with Ca2+ and Mg2+?
You need to have the solution pH above the pKa4 for EDTA.
Q2. What metals would EDTA likely form stable complexes with in most water samples? How selective is this process?
In most water samples the common cations EDTA would complex with would be Ca2+, Mg2+, and in some samples Al3+, or iron species (Fe2+, Fe3+)
Q3. The endpoint of a titration is determined using an indicator. What would be the general features of an indicator that could be used to determine the endpoint of a water hardness titration?
The indicator would need to have some aspect where it changes color when a cation is bonded to it. There is a class of indicators that change color as a metal binds to the indicator. For EPA method 130.2 Eriochrome Black T (EBT) is used to determine total water hardness. In the case of EBT, the indicator is wine-red in the presence of metal ions and blue when the indicator does not have a metal bound to it.
Q4. When the EBT is added to your sample, what color would you expect the indicator?
You would expect the indicator to be a wine-red color, as the EBT would have the cations in the water sample to bind to it. (Do note that EBT does degrade when in solution, and thus can only be stored for a couple of weeks before a new indicator solution needs to be prepared.)
As you titrate the solution with EDTA, the EDTA binds with the metal and once it has complexed all of the cations in solution, the EBT will change color as it is not complexed.
Q5. How might raising the pH potentially cause metals to precipitate?
Metals that are not Group 1 metals will precipitate as the hydroxide if the pH is raised high enough.
Q6. The chart below lists the solubility product constants of the hydroxide complexes of several common cations in water. Which metal would precipitate last as its hydroxide complex? This is the metal then available for titration using the HB indicator after you adjusted the solution pH to 12.0.
Cation
Equilibrium
Ksp
Al3+
$\ce{Al(OH)3 (s) ↔ Al^3+ (aq) + 3 OH- (aq)}$
4.6 x 10-33
Ca2+
$\ce{Ca(OH)2 (s) ↔ Ca^2+ (aq) + 2 OH- (aq)}$
6.5 x 10-6
Fe2+
$\ce{Fe(OH)2 (s) ↔ Fe^2+ (aq) + 2 OH- (aq)}$
8 x 10-16
Fe3+
$\ce{Fe(OH)3 (s) ↔ Fe^3+ (aq) + 3 OH- (aq)}$
1.6 x 10-39
Mg2+
$\ce{Mg(OH)2 (s) ↔ Mg^2+ (aq) + 2 OH- (aq)}$
7.1 x 10-12
Raising the pH above pH 12 will precipitate all of the cations except for Ca2+. This process masks the other ions by precipitating them out. After filtering the sample, titration with EDTA is performed. Thus Ca2+ is the only species dissolved in solution, and when this solution is titrated with EDTA, the EDTA will only complex with the Ca2+. Thus the titration at pH 12 with HB as the indicator will determine just the Ca2+.
Q7. If you have measured the total water hardness using EBT as the endpoint and the concentration of the Ca2+ using HB as the endpoint, how could you determine the concentration of Mg2+?
Since the concentration of the Al3+ and the iron species are usually very low to negligible, the Mg2+ can be determined by taking the difference in results between the titration with the EBT and the HB analyses.
Q8. Have you made any assumptions in performing up this calculation?
The assumption that the concentrations of Al3+ and the iron species are usually very low to negligible compared to either the calcium or magnesium.
Q9. The concentration of the titrant needs to be known when performing a titration. This concentration is determined by standardizing the titrant. How could you standardize the EDTA solution?
The EDTA is standardized by titrating it was a solution of known Ca2+ concentration. EBT is usually used as the indicator.
Experimental analysis:
The common procedure for this analysis is:
Solutions needed:
• 0.00250 M EDTA solution
• Eriochrome Black T indicator solution (0.1 g in 25 mL of methanol)
• Ammonium Buffer (pH 10): dissolve 67.6 g NH4Cl in 572 mL of NH4+ and dilute to 1 liter with distilled water.
• Hydroxynaphthol blue indicator (solid)
• 50% NaOH
Waste information:
• Neutralize the solutions to between pH 5 and 8, then flush down the drain.
The volume of your sample will depend upon the hardness of the water. Generally a 50.00 mL sample of freshwater is reasonable. Add 3.0 mL of ammonium buffer and 6 drops Eriochrome black T indicator. Titrate with EDTA solution and note when the color changes from wine red to blue. You may want to practice finding the end-point several times using tap water. If the water is rather hard, you may have to increase the EDTA concentration, or you can decrease the sample volume. Save a solution at the endpoint to compare for the titration of your sample.
The water hardness results are usually reported with units of ppm CaCO3, which you calculate using the following equation:
Calculation of Water Hardness
$\dfrac{(V_{EDTA})(C_{EDTA})\left(\dfrac{1\: mmole\: M^{n+}}{1\: mmole\: EDTA}\right)\left(\dfrac{1\: mmole\: CaCO_3}{1\: mmole\: M^{n+}}\right)\left(\dfrac{100.09\: mg\: CaCO_3}{1\: mmole\: CaCO_3}\right)}{(V_{sample})} = ppm\: CaCO_3\nonumber$
Where:
(VEDTA) is the corrected titration volume of EDTA standard titrated (in liters)
(CEDTA) is the concentration of the EDTA standard (in mM)
(VSample) is the volume of the water sample (in liters) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/06_Instructors_Guide/04_Titrimetry_Alkalinity_and_.txt |
This instructor’s guide includes sample answers to the questions in the IC module, and additional information at times.
I have given this module as a take home assignment for the students to work on and read through, then had them present their answers to each other in small groups. This resulted in the students having discussion on where their answers were different and did well in getting to the underlying principles.
Q1. Write the chemical reaction for the association of A- and B- with the ion exchange resin.
$\ce{Resin^+-E- + A- <=> Resin^+-A- + E- }\nonumber$
$\ce{Resin^+-E- + B- <=> Resin^+-B- + E- }\nonumber$
Q2. Write the chemical reaction for the elution of A- and B- from the ion exchange resin.
$\ce{Resin^+-A- + E- <=> Resin^+-E- + A- }\nonumber$
$\ce{Resin^+-B- + E- <=> Resin^+-E- + B- }\nonumber$
Q3. Write the equilibrium expression constant for A- and B-. In chromatographic separations, this term (Kc) is referred to as the distribution coefficient.
Q4. Do you think the magnitude of the distribution coefficients are the same for A- and B-? Why or why not?
If they are different ions, then they would have different distribution constants, because they would have different affinities for the resins. These differences will arise due to differences in size, charge, and other characteristics.
Q5. If the distribution coefficient of A-() is smaller than the distribution coefficient of B-(), draw a sketch of the elution process that is similar to the figure in Step 1 but at a point where A- and B- are partway through the column.
Q6. Suppose B- had a very strong affinity for the resin, what would happen to its elution time?
As the affinity for the resin increases the elution time increases. Thus you could get to a point where the elution time is so long it does not appear to come off the column.
As the sample is injected onto the column, the two different analytes briefly displace the eluent as the counter-ion to the charged resin. The analyte is briefly retained at the fixed charge on the resin surface. The analytes are subsequently displaced by the eluent ions as the eluent is added to the column. The different affinities (see the chemical reactions in the basic process section) are the basis for the separation. The Kf value of each reaction is also known as the selectivity coefficient. The greater the difference between the Kf values for the two analytes, the more the two analytes will be separated during the ion chromatography process. In reality, the interaction between the solvent and the analyte can also have an impact on the order each analyte is eluted. For a more in-depth analysis of predicting the retention order see the material by Dr. Thomas Wenzel. (http://www.bates.edu/x65385.xml)
Q7. Is this a desirable or undesirable situation if you were trying to analyze A- and B- in a mixture?
You want have a difference in elution times, but you do not want the affinity to be so strong that the ion does not readily come off the column. The longer the elution time the more peak broadening will happen (this links to later brief information on peak broadening.)
Q8. Is there a situation you can think of when it might be desirable for B- to have a very strong affinity for the resin?
Water filtration systems, such as water softeners, usually have an ion exchange resin where the affinities are strong enough to effectively remove ions from the water. The Mg2+ or Ca2+ displaces the Na+ on the resin. The affinity for the Mg2+ or Ca2+ ions needs to be much stronger than the affinity for Na+ in order to effectively soften the water. In sample pretreatment you may want very strong affinities to bind all of the ions and then extract with a different solvent.
Q9. If you want to separate cations, what would be different about the stationary and mobile phases?
The resin would need to be negatively charged, and the mobile phase would be positively charged. The common cation exchange resins are based on either polystyrene-divinylbenzene (PS-DVB) or methacrylate polymers. The surface of these polymers (Figure 1) is functionalized with a negatively charged sulfonated group (-SO3- ). The cation in the eluent or the analyte of interest is the counter-ion in the vicinity of the charged functional group.
The surface of the polymer is functionalized with a quaternary amine (-N+R3) for anion exchange (see Figure 2). The quaternary amine provides a positive charge to the surface, attracting negatively charged anions in the liquid phase. Just like the cation exchange resin, the anion of the eluent or the analyte of interest exists as the counter-ion in the vicinity of the positive charge residing on the amine.
Q10. Would water flow easily through a column containing very fine particles?
As the resin particle become finer, they pack together closer, leaving less room for the solution to flow through. This, in conjunction with the more resin-solvent interactions, will increase the resistance to flow. You then have to apply pressure to force the solutions through the column.
Q11. If not, how could you get the water through the column?
As the water becomes less likely to flow, you need to apply a force to push the water through the column. This is done with a pump that can handle higher pressures (often 200-1000 psi) such as a double piston high pressure pump to force the mobile phase through the column.
Q12. Considering that the column is packed with very fine particles, what must be done to surface water samples before injecting them onto the column?
If the column particles are very fine, the surface water sample will likely need filtered to remove any detritus that would foul up the column. Surface water samples are filtered through at least 0.45 µm filters but may be filtered through as small as a 0.20 µm, similar to how other solutions are filtered for use in IC. The filter type is one that must not introduce an error.
Q13. Can you think of a way to detect the presence of ionic substances in water?
The most common factor to all ionic substances is that they have a charge, thus a detection method that is based on measuring the charge is frequently used. Ions in solution conduct electricity, so the conductivity of a solution will change as the concentration of ions change. This is the common basic detection. There are other possibilities that are more selective.
Conductivity would not distinguish between the two ions. Therefore you need the two ions to be separated on the column.
The selectivity in the overall method comes from a good separation of the ions, so the detection method does not need to be very selective. The detection does need to be quantifiable and applicable to a broad range of ions.
Q16. Will the eluent ion respond to the detection method you thought of above to measure the presence of Na+ and Ca2+.
Yes it will. Thus eluent suppression will usually be used.
Q17. How will the chromatogram change as you increase the concentration of A- and B- injected into the column? Make sure to label the axes of your chromatogram.
x-axis is elution time, y-axis is conductivity. As you increase the concentration of an ion, the height and area of the peak corresponding to that ion will increase.
Q18. Since neither axis in the chromatogram you drew above is concentration, how can we calibrate the detector response to determine the concentration of A- and B-?
You have to calibrate the detector by analyzing standard samples with known concentrations of both A- and B-. The conductivity is plotted as a function of concentration to determine the response of the detector. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Effects_of_Acid_Rain_on_Atlantic_Salmon_Populations/06_Instructors_Guide/05_Ion_Chromatography.txt |
This module provides a context for introducing fundamental techniques used in chemical analysis (spectrophotometry, atomic absorbance spectroscopy and ion selective electrodes) along with considerations about sampling and sample preservation. Using an active learning approach, the module explores some fundamental water quality parameters such as the concentration of inorganic cations and anions that may aid in understanding why certain ponds provide a more suitable habitat for the Columbia Spotted frogs and aquatic snails.
The End Creek Wetland Restoration Project is located 15 km north of La Grande, in Union County, Oregon, USA. The approximately 550 acre restored wetland is the only natural grassland/floodplain area of any size at the north end of the Grande Ronde Valley which is mostly agricultural lands. Consequently, it protects a rare island of habitat for mammals, fish, amphibians, and reptiles and is an important harbor for threatened, endangered, or sensitive plant and animal species. The wet meadows and ponds provide stopover sites for migratory waterfowl traveling north and south through the intermountain west, and nesting and feeding habitat for those that stay. Drained and converted to agricultural land by the end of the last century and operated as a dairy farm until the mid-50s, the property became part of the Wetland Reserve Program in 2005. In 2006 plans began to restore the creek to its original sinuosity with the goals of improving fish habitat and restoring original vegetation. Creek restoration helped rebuild the water table and feed natural ponds. In 2008 Columbia Spotted Frogs were discovered in one of the ponds and a breeding population study began. Over the next few years the spotted frog population expanded, preferentially breeding in certain ponds, but has lately dwindled. In addition, three families of aquatic snails were identified but their distribution differs between ponds and seasonally.
Using analytical tools, you will learn how to approach a problem of analytical significance, from designing a sampling plan to choosing appropriate analytical methods and analyzing the data to determine whether any significant difference exists in the water composition and quality that may justify differences in frog and snail populations.
• Data Sets for Analysis (Excel)
• Assessment Questions: For Assessment questions that accompany this module, please contact Anna Cavinato ([email protected])
End Creek: Spotted Frogs and Aquatic Snails in Wetlands A Water Quality Investigation
Columbia Spotted frogs (Rana luteiventris) are found from Alaska and most of British Columbia to Washington east of the Cascades, Idaho, and portions of Wyoming, Nevada, and Utah. The Great Basin population range includes eastern Oregon, southwestern Idaho, and the northern drainages of Nevada. They live in spring seeps, meadows, marshes, ponds and streams, and other areas where there is abundant vegetation. They often migrate along riparian corridors between habitats used for spring breeding, summer foraging and winter hibernation. Columbia spotted frogs typically have a light-colored stripe along the jaw and are light to dark brown or olive on their backs with varying numbers of irregular black spots. The coloration of their underside ranges from white to yellow, and mottling is present to varying degrees. The species is currently a ‘candidate species’ for listing under the Endangered Species Act. The largest known threat to spotted frogs is habitat alteration and loss, specifically loss of wetlands used for feeding, breeding, hibernating, and migrating. Reduction or loss of habitat can be attributed at least in part to recent drought conditions, wetland degradation, and poor water quality. Inorganic nitrogen, particularly in the form of nitrate (NO3-), poses a threat because of its known toxicity. Its main toxic action on aquatic animals is due to the conversion of oxygen-carrying pigments (e.g., hemoglobin, hemocyanin) to forms that are incapable of carrying oxygen (e.g., methemoglobin) (1). Other threats include predation by non-native species such as bull frogs and diseases such as Chytrid fungus (2).
Columbia Spotted Frog Breeding Survey:
Adult Columbia Spotted frogs are occasionally found at End Creek and egg masses were discovered in one pond for the first time in April 2008. In 2009, 2010, and 2011 all of the ponds at End Creek were surveyed for breeding activity of Columbia spotted frogs. In 2009, 56 egg clusters were identified in the East Pond, and no breeding activity in any of the other ponds. In 2010, a total of 18 egg clusters were found, with 13 clusters in the East Pond, one in the South Pond, and four in the North Pond. In 2011 14 egg clusters were found in the East Pond and 2 in the South Pond. In 2012 only 3 egg clusters were located, suggesting that the spotted frog population is quickly declining (Fig. 1).
Figure 1. End Creek wetland area.
Aquatic Snails at End Creek:
Three families of pulmonate snails occur in the ponds at End Creek, Physidae, Lymnaeidae, and Planorbidae (Figs. 2-4). Pulmonate snails contain a lung-like organ that allows them to store oxygenated air and remain under water for long periods of time. They also use this air reserve to adjust their buoyancy in the water. Because of their capacity to store air, they are better adapted to survive in water with low oxygen content than other types of snails.
Fig. 2. Physidae
Fig. 3. Lymnaeidae
Fig. 4. Planorbidae
These snails are annual species. Adults lay eggs in late winter; eggs hatch in early spring, and the snails grow throughout the summer months. They reach adult size by fall, lay eggs in late winter, and then die. Because of this seasonal life cycle, the time of sampling appears to influence how many snails are found in the samples. In spring, snails presumably are still so small that they are easily missed in the sampling process. By fall, they are larger and are more easily located.
Snails require calcium in order to build their shells; consequently, their growth rates may be influenced by calcium availability (3). Some snails may be more tolerant of low-calcium environments than others. Other aspects of water quality that may influence snail distribution include pH and water buffering capacity. Acidic water can cause loss of calcium from snail shells. Some species may be more tolerant of low pH than others as long as sufficient environmental calcium is available. Sources of acidification may include atmospheric deposition of sulfur and nitrogen oxides from automobiles and industrial emissions. Atmospheric CO2 also can diffuse into water, creating carbonic acid and lowering pH. At pH levels below 5.7, many organisms cannot survive (3). Nitrate toxicity needs to be considered as well. Much like noted for spotted frogs, nitrate can be toxic to aquatic invertebrates and its toxicity increases with concentration and exposure time (1).
Samples of aquatic invertebrates, including snails were collected from two ponds in May 2008 and October 2008 (Tables 1 and 2). In May, a large number of Planorbid snails were found in the North Pond, but in October, only Physid snails were identified in that pond. In May, few snails were found in the South Pond, and Lymnaeids were the most abundant. In October, all three snail families were present in higher numbers. In spring, on average, snails represented about 11% of the total number of invertebrates collected from both ponds. In fall, in both ponds, snails represented about 16% of the total aquatic invertebrates collected in our samples.
Table 1. Snails collected in two ponds in May 2008.
North Pond
South Pond
Family
Number
% of total sample
Number
% of total sample
Physidae
2
0.39
1
1.72
Lymnaeidae
1
0.20
4
6.90
Planorbidae
67
13.11
0
0.0
Total
70
13.7%
5
8.62%
Table 2. Snails collected in two ponds in October 2008.
North Pond
South Pond
Family
Number
% of total sample
Number
% of total sample
Physidae
46
16.4
16
4.73
Lymnaeidea
0
0.0
25
7.4
Planorbidae
0
0.0
15
4.44
Total
46
16.4%
56
16.57%
In summary, these data suggest that Lymnaeid snails are most abundant in the South Pond and Physid snails are most abundant in the North Pond. It is difficult to understand the dynamics of the Planorbid snails. They were abundant in the North Pond in spring, but did not appear in any samples from that pond in fall. In the South Pond, they did not appear in any samples in spring, but were well-represented in fall sampling. Obviously, further studies will be necessary to better assess snail population distribution in different ponds and over time.
Q1. What are some possible water quality parameters that could affect invertebrate and amphibian populations in a fresh water environment? You may want to research if any information is available on recommended levels of specific ions that may positively or negatively impact these populations.
We could then ask the following questions:
1. Is water quality at the End Creek ponds potentially responsible for the decrease observed in spotted frog population?
2. Are there differences in water chemistry that influence snail family distribution in the ponds at End Creek?
Following are some water quality parameters that you may want to consider as you undertake this investigation.
Nutrients. Nutrients such as nitrogen and phosphorous are essential to plants and aquatic organisms. However, in larger quantities, they can vastly reduce water quality by triggering accelerated plant growth and algae blooms. These, in turns, may lead to large amounts of decaying plant material causing low dissolved oxygen and, ultimately, death of fish and other aquatic species. Nitrogen species can be particularly toxic to freshwater invertebrates and amphibians (1). Due to the agricultural practices around End Creek, one can speculate that nitrogen and phosphorous sources may come from residual fertilizers. Possible forms of nitrogen can include ammonia (NH3), nitrates (NO3-) and nitrites (NO2-). Phosphorous is typically found as phosphate (PO43-) and can be in the form of organic phosphorous (associated with carbon-based molecules) or inorganic phosphorous (4).
Water hardness. Calcium and magnesium compounds along with other metals contribute to water hardness. Typical guidelines classify water with 0-60 mg/L calcium carbonate as soft; 61-120 mg/L as moderately hard; 121-180 mg/L as hard; and more than 180 mg/L as very hard (5). Snails require calcium to grow their shell and differences in calcium concentrations may play a role as to why snails may be found preferentially in certain ponds (3).
Dissolved oxygen. The amount of oxygen dissolved in water is measured as Dissolved Oxygen (DO). It is typically reported in mg/L or % saturation. The level of dissolved oxygen varies with temperature and altitude and fluctuates seasonally and over a 24-hour period. In ponds dissolved oxygen can fluctuate greatly due to photosynthetic oxygen production by algae during the day and the continuous consumption of oxygen due to respiration. As a result of these processes, dissolved oxygen typically reaches a maximum during the late afternoon and a minimum around sunrise (6). The solubility of oxygen also increases in colder water and decreases at higher altitudes. Decreased availability of dissolved oxygen negatively impacts aquatic organisms and in extreme conditions can cause death.
pH. pH affects many chemical and biological processes in the water. For example, different organisms flourish within different ranges of pH. The largest variety of aquatic animals prefers a range of 6.5-8.0. pH conditions outside this range reduce the diversity in the stream because it stresses the physiological systems of most organisms and can reduce reproduction. Low pH can also allow toxic elements and compounds to become mobile and “available” for uptake by aquatic plants and animals. This can produce conditions that are toxic to aquatic life, particularly to invertebrates such as snails. Changes in acidity can be caused by atmospheric deposition (acid rain), the surrounding rock, and certain wastewater discharges.
Total solids. Total solids include dissolved solids (typically cations and anions of salts that will pass through a filter with 2 micron pores) and suspended solids (silt and clay particles, algae, small debris and other particulate matter with size larger than 2 micron). Because the particles absorb heat, greater amounts of suspended solids contribute to increased water temperature thus decreasing availability of dissolved oxygen. Suspended particles can be carriers of toxins, particularly pesticides and herbicides, in water bodies in proximity to agricultural land. They can clog fish gills and affect egg development. Dissolved solids affect the water balance in cells. If the amount of dissolved solids is too low, aquatic organisms will tend to swell as water moves inside the cells. Levels of dissolved solids that are too high will cause organisms to shrink due to water moving out of cells (4).
Identifying possible analysis methods
Water quality tests can be performed on site using inexpensive field kits that provide fast response with minimal sample preparation. Alternatively, water can be collected and brought back to a laboratory where analyses can be conducted using more sophisticated instrumentation.
The purpose of this module is to lead you to identify appropriate methods of analysis best suited to test possible differences in the water quality of the ponds located at End Creek. As you research these methods, assess them in terms of their sensitivity, ease of implementation, cost, speed, etc.
Experimental Design for this Project
Once you decide on the most appropriate methods of analysis, the next step will be to design an experiment that will provide meaningful data so that sound conclusions can ultimately be drawn. It may be helpful as you embark in this investigation to refer to guidelines and protocols such as those provided by the Environmental Protection Agency (EPA) (4) or the American Public Health Association (APHA) (7). You may want to specifically refer to the EPA Volunteer Stream Monitoring: A Methods Manual (http://water.epa.gov/type/rsl/monitoring/stream_index.cfm). Often one may have to adapt a method based on available equipment or other limitations. Therefore, it is important to understand all aspects of a given method and be able to show that your method is valid.
Analysis of Samples from End Creek ponds
In our case study we will develop a sampling plan for water samples to be collected at three ponds where egg clusters and snails have been identified. Some tests could be conducted on site and compared to more sophisticated analyses conducted in the laboratory. For this purpose water will be collected, brought back to the lab and stored for further analyses. Due to time and cost constraints, there are limitations on the number of samples that can be processed (limited to 30).
The subsequent parts of this module will examine the following two questions:
How do you decide where and when to collect samples to ensure they are representative?
How will you process the water for storage to ensure that analytes are not degraded or lost during storage and how long can you store your water sample?
A series of assignments are presented to help you understand how to design a water quality monitoring experiment and address questions such as those posed above. Through this process, you will also be introduced to important analytical chemistry concepts that underlie the analytical methods discussed. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/01_Identifyin.txt |
Adapted from “A Study of Onondaga Lake Water” – Chemistry in Context: Applying Chemistry to Society Lab Manual
The End Creek Wetland Restoration Project is located 15 km north of La Grande, in Union County, Oregon. The approximately 550 acre restored wetland embodies many qualities of high value to the community. The wetland is the only natural grassland/floodplain area of any size at the north end of the Grande Ronde Valley. It forms a natural corridor between the Blue Mountains and the Grande Ronde River as it transects through extensive, uninterrupted agricultural lands. Consequently, it protects a rare island of habitat for mammals, fish, amphibians, and reptiles and is an important harbor for threatened, endangered, or sensitive plant and animal species. The wet meadows and ponds provide stopover sites for migratory waterfowl traveling north and south through the intermountain west, and nesting and feeding habitat for those that stay. Drained and converted to agricultural land by the end of the last century and operated as a dairy farm until the mid-50s, the property became part of the Wetland Reserve Program in 2005. In 2006 plans began to restore the creek to its original sinuosity with the goals of improving fish habitat and restoring original vegetation. Creek restoration helped rebuild the water table and feed natural ponds. In 2008 Columbia Spotted Frogs were discovered in one of the ponds and a breeding population study began. However, over the following years this population has dwindled down and research shows no correlation to water quality parameters. We will therefore focus our research on aquatic snails. Three families of aquatic snails were identified but their distribution differs between ponds and seasonally. In this investigation, you will learn how to identify some metal ions and anions that are connected to water quality. The presence of these ions may play a role as to why certain ponds provide a more suitable habitat for certain types of aquatic snails. The ultimate goal will be for you to determine whether any significant difference exists in the water composition across three different ponds that may justify differences in snail populations.
The question to be answered is:
Are there differences in water chemistry that influence snail family distributions in the ponds at End Creek?
Background
You will be investigating the presence of positive and negative ions in water samples using a variety of chemical and physical tests. Presence of ionic species in solution will influence the conductivity. Higher concentrations of ions will result in a higher conductivity. Flame tests can be used to qualitatively identify the presence of specific ions. Much like in fireworks, when metal ions are heated in a flame, electrons from lower energy levels absorb energy and are promoted to higher levels. When these electrons relax back to the ground state, they emit light of characteristic wavelength. By observing the color emitted by known solutions of metal ions, the presence of metals can be determined in an unknown sample.
You will also make use of chemical reactions to identify ions based on the formation of insoluble salts or highly colored compounds. You will first investigate different reagents and determine which reagent is best suited for the identification of a specific ion. Once you have developed your own analytical scheme on a set of known positive and negative ions, you will use the same chemical reactions to analyze water sampled at End Creek and determine whether there are differences in the chemical composition of the three ponds.
Overview
1. Test the electrical conductivity of water from the three End Creek ponds.
2. Use the flame test to test for presence of Na+, Ca2+ and Fe3+ in standard samples.
3. Use the flame test to test for presence of Na+, Ca2+ and Fe3+ in the End Creek ponds.
4. Conduct chemical tests on samples of known composition to identify best reagents to analyze for presence of Ca2+ and Fe3+.
5. Once you have identified the best reagents, test the End Creek pond water for presence of Ca2+ and Fe3+.
6. Conduct chemical tests on samples of known composition to identify best reagents to analyze for presence of Cl-, SO42- and PO43-.
7. Once you have identified the best reagents, test the End Creek pond water for presence of Cl-, SO42- and PO43-.
Materials Needed
Chemicals
Equipment
Dropper bottles with dilute solutions of:
• Sodium chloride
• Calcium chloride
• Iron (III) chloride
• Sodium oxalate
• Potassium thiocyanate
• Sodium sulfate
• Sodium hydrogen phosphate
• Silver nitrate
• Barium nitrate
• Plastic well plates
• Small stirring rods
• Disposable pipets
• Ni-Cr wire for flame test
• Conductivity tester
• Bunsen burner
Experimental Procedure
I. Test for conductivity
1. Place a small amount of water from each pond in a well plate well. Immerse the two leads of the conductivity tester in each water sample making sure the two wires do not touch. Record your observations.
II. Flame test for metal ions
1. Using separate, clean wells, place a few drops each of sodium chloride, calcium chloride, and iron (III) chloride.
2. Light the Bunsen burner and adjust the flame so to obtain a bright blue inner cone (instructor will demonstrate). Wash the Ni-Cr wire with deionized water and bring wire into the flame. If the wire is clean, no color should be emitted.
3. Dip the wire in the calcium chloride solution and position it right above the inner blue cone where the hottest part of the flame is located. Observe the color emitted from calcium and record your observations.
4. Clean the wire and then repeat flame test with sodium chloride and iron (III) chloride. Compare the colors and record your observations.
5. Using clean wells, place a few drops each of the pond water samples. Make sure your Ni-Cr wire is clean! Dip the wire in each water sample and bring the wire into the flame. Do you observe any color? Record your results.
III. Chemical test for positive ions
1. In this investigation you will discover which reagent is best suited to analyze a water sample for presence of sodium, calcium and iron ions. Available reagents are sodium oxalate (Na2C2O4) and potassium thiocyanate (KSCN). A suitable reagent is one that will uniquely react with the analyte of interest. Signs of chemical reactions include change of color, precipitation, evolution of gasses, etc.
2. Using the well plate, standard solutions of sodium chloride, calcium chloride and iron (III) chloride, devise a strategy to identify the most suitable reagent that can be used to identify presence of sodium, calcium and iron ions. Jot down an experiment design and check with the instructor before proceeding.
3. Once you have identified the best reagents for the identification of metal ions in water, test the three pond water samples. Record your observations. Based on these results, decide whether any metal ions are present in any of the three ponds.
4. Clean up: empty any contents of the well plates into proper disposal bottle located under the hood. Rinse well plate and shake to remove any residual water.
IV. Chemical test for negative ions
1. In this investigation you will discover which reagent is best suited to analyze a water sample for presence of anions such as sulfate, chloride and phosphate. Available reagents are silver nitrate (AgNO3), barium nitrate (Ba(NO3)2) and calcium chloride (CaCl2). Again, look for signs of chemical reactions which may include change of color, precipitation, evolution of gasses, etc.
2. Using the well plate and standard solutions of sodium sulfate, sodium chloride and sodium hydrogen phosphate, devise a strategy to identify the most suitable reagent that can be used to identify presence of sulfate, chloride and phosphate ions in water. Jot down an experiment design and check with the instructor before proceeding.
3. Once you have identified the best reagents for the identification of anions in water, test the three pond water samples. Record your observations. Based on these results, decide whether any anions are present in any of the three ponds.
4. Clean up: empty any contents of the well plates into proper disposal bottle located under the hood. Rinse well plate.
V. Questions to be answered after completing experiment
1. What do the results of the conductivity test indicate about what is present in the End Creek water samples? Is there a difference in the conductivity of the three samples?
2. What color do sodium, calcium and iron ions produce in the flame? Based on the results from the flame test, which of these ions are present in the pond water samples?
3. Describe which reagent you chose, respectively, for the analysis of calcium and iron ions and write the balanced chemical reaction for each test. Based on your results, which metal ions are present in the pond water?
4. Research the solubility rules for anions and cations. Was any reagent suitable for testing for sodium ion? Based on solubility rules, why would it be hard to devise a test for sodium based on the formation of a precipitate?
5. Describe which reagent you chose respectively for the analysis of sulfate, chloride and phosphate and write the balanced chemical reaction for each test. Based on your results, which anions are present in the pond water?
6. Review all the results gathered through this exploration. Based on these results, are there any differences in water composition among the ponds at End Creek? Summarize your conclusions. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/02_Qualitativ.txt |
Designing a sampling plan for End Creek ponds
Purpose: The purpose of this assignment is to address questions related to designing a sampling plan for surface waters [i.e. ponds, lakes, rivers and streams] where several water quality parameters will be tested. The following example describes the specific case of sampling ponds such as those found on the End Creek Wetland restoration site. While different approaches may be necessary depending on whether water will be collected from a stream, lake, or natural spring, the fundamental steps described in this document should find broad application in the design of any sampling plan.
Learning Outcomes:
At the end of this assignment you will be able to
1. Define various sampling strategies
2. Assess the benefits and limitations of different sampling strategies
3. Determine an appropriate sampling plan for analysis of surface waters
Assignment:
An important aspect of designing an experiment is collecting a representative sample. What is a representative sample? That is often a difficult question to answer. In the context of an analytical chemistry measurement, representative means that the concentration of the analyte in the sample analyzed is a good representation of its levels in the material or system being tested. If sampling is not done correctly it could be the “weak link” in an analysis, leading to inaccurate results. Often, sampling can be a source of error that is overlooked. Below is an actual map of the End Creek wetland. We will focus our sampling efforts specifically on the North, South and East ponds.
DESIGNING A SAMPLING PLAN
Q1. What key questions must be considered when designing a sampling plan?
Where to Sample?
How do you obtain a representative sample? One approach is to take a grab sample. Consider a solid sample divided out as a grid, below. How do you choose where to grab from? One approach is to choose randomly.
Q2. Pick eight random samples from the grid laid out below. How do you ensure your sampling is random?
Suppose that our analyte of interest (colored squares) was distributed across the sample in the manner shown below.
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
Grid A
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
Grid B
Q3. Would you consider the samples above to be heterogeneous or homogeneous?
Q4. Did your random sampling affect the potential accuracy or precision of your measurement of the analyte concentration for the samples in grid A or grid B? If so how?
Although we would like to assume that a sample is homogenous it is often not the case. When trying to obtain a representative sample from a heterogeneous material or system, you need to consider both distributional and constitutional heterogeneity.
Distribution heterogeneity is caused by segregation of the sample (i.e. settling).
Constitutional heterogeneity is a fundamental property of a material and is caused by differences in particle size and/or composition.
Q5. Each of the previous grids is an example of one of these cases. Can you identify which sample shows distributional heterogeneity and which shows constitutional heterogeneity?
Q6. How does distribution heterogeneity affect accuracy and precision?
Q7. How does constitutional heterogeneity affect accuracy and precision?
Q8. Do you see a scenario where distribution heterogeneity could be magnified by mixing and/or sampling?
If samples are mixed well, distributional heterogeneity is insignificant and only constitutional heterogeneity is present. It is important to note, however, that the analyte’s spatial or temporal distribution might also be of importance in the analysis and mixing may not always be the best choice.
Types of Sampling Plans:
What we have highlighted above is an example of random sampling, which is often applied to a grid design. Random sampling strategies can be applied to any target population (i.e. evaluating a solid sample or measuring water quality parameters at End Creek). However, random sampling can at times be expensive and is not necessarily cost effective, as you often need a greater number of samples to ensure that your samples represent the body of water. If you know something about the body of water under study, other sampling methods may be possible or appropriate. Here are a few examples.
Selective (Judgmental) Sampling – this is at the opposite extreme of random sampling, and is done if you have prior information about the target. For example, if you wanted to evaluate the metal content in pennies you might exclude from your study coins that are corroded or select coins from a specific mint date.
Ease of Access Sampling – Many times on ponds, rivers, or streams the ultimate sampling site selected is a location where access is available. This can be acceptable as long as the sampling site allows the initial question to be answered. When considering the location, one needs to consider safety during access and whether the land is public or private.
Systematic Sampling – Sampling the target population at regular intervals in space or time. This approach is often considered to fall between the extremes of random and selective sampling.
Stratified Sampling – The population may be divided into sub populations (groups) that are distinctly different (this might be size of sample, type of sample, depth of sample). Then, the overall sampling within the groups is randomly conducted and the samples are pooled.
Cluster Sampling – is a sampling technique where the population is divided into groups or clusters and random samples are selected from the cluster for analysis. The main objective of cluster sampling is to reduce costs by increasing sampling efficiency.
Q9. What is the advantage of implementing a judgmental sampling scheme over random sampling if one knows the point source for the discharge an analyte into a system?
Q10. Assume you have chosen a selective sampling plan to evaluate pollution from a point source into a pond. Use the diagram below and words to describe your sampling plan.
Q11. Use a grid design (as we have previously done) to show how you would conduct systematic sampling of the pond to measure levels of the pollutant. Is there an advantage to what you might learn using this sampling method? What are the disadvantage(s)?
Q12. Describe how stratified sampling might be applied to evaluate the pollutant in the lake? In general, what is the advantage of stratified sampling over cluster sampling?
What Type of Sample to Collect?
When implementing a sample plan often grab samples are used. In some cases composite sampling (combining a set of grab samples into one sample) is more useful. This strategy may be used if:
There is interest in the target population’s average composition over time in space.
A single sample does not supply sufficient material for analysis.
Q13. What is the main disadvantage of grab and composite samples?
Q14. Can you think of any control studies you might want to do, when compositing samples?
Q15. Does the EPA Volunteer Stream Monitoring web site suggest a particular sampling method?
Consider the natural fluctuation of dissolved oxygen in a lake over time that is shown in the graph below.
Q16. Based on the graph above, describe a sampling procedure that would allow you to obtain a representative sample.
Another variable that can fluctuate in natural water is the pH. The graph below shows pH data continuously recorded in a river over an entire year.
Q17. Is there any systematic pattern to the data in the graph above?
Q18. Can you think of some event(s) that may account for the acidic spikes in the pH?
Q19. How would this pH data affect when you might choose to sample a site to assess if acid rain is impacting the ecosystem?
Q20. Describe what else you would need to know to determine when to perform sample collection if you are addressing whether acid rain is impacting a site. What other physical or chemical parameters might you need to consider that could contribute to changes in pH? Is it really the "data" that would affect when you would choose to sample?
Data in the previous graph covered an entire year. The following data was collected every fifteen minutes over a 3 day period.
pH measured every 15 minutes over a period of three days starting at midnight.
Q21. What physical or chemical processes might contribute to the pH fluctuations? How might this variation affect your sampling plan?
Q22. Another factor to consider is the sample handling time. Can you think of ways in which the sampling handling time may impact the concentration of species in a sample?
Minimizing Variance - What is the Minimum Amount of Sample? How Many Samples are Needed?
When sampling you want to make sure the sample is not too small, so that the composition is not substantially different from the typical composition of the body of water. You also want to ensure that you collect an appropriate number of samples for analysis.
The variance of the (1) analysis method and (2) sampling approach both contribute to the variance in a given analysis. Sampling variance can be improved by collecting more samples of the proper size. Increasing the number of times a given sample is analyzed improves the method’s variance.
Q23. It is known from analyses conducted in 2008 that the % relative sampling error for water hardness by EDTA titration is 0.8%. How many samples should you collect to limit the relative standard deviation for sampling to 1.0% within the 95% confidence level*? Is this a feasible task?
Q24. If the cost of collecting a sample is \$20 and the cost of analyzing a sample is \$50 what budget should you allocate for the project and what sampling strategy would be most effective for the given number of samples?
For further discussion on sampling amounts see section 7.2 in Harvey, D. Chapter 7, Collecting and Preparing Samples.
*For a review of confidence levels see section 4.4 in Harvey, D. Chapter 4, Evaluating Analytical Data.
Design Your Sampling Plan
Of course we would like to collect as many samples as possible to minimize our variance. However, in every lab there are limitations such as those above related to cost per-sample collected and cost-per analysis.
Q25. Below is a picture of one of the three ponds at End Creek. The darker shades indicate areas of deeper water. Design your sampling plan. Think about random, systematic, clustering, etc. sample strategies. Will you take grab samples or pool samples together?
How do we go about collecting samples?
Equipment and sampling procedures are typically specific to the specific water quality parameter one wishes to determine. However, there are two fundamental tasks that are common to any water analysis: the preparation of the sampling containers and the sample collection.
Preparation of sampling containers.
For some analyses the EPA recommends to:
1. Wear latex gloves.
2. Wash each sample bottle or piece of glassware with a brush and phosphate-free detergent.
3. Rinse three times with cold tap water.
4. Rinse three times with distilled or deionized water.
For other analyses the EPA recommends to:
1. Wear latex gloves.
2. Wash each sample bottle or piece of glassware with a brush and phosphate-free detergent.
3. Rinse three times with cold tap water.
4. Rinse with 10 percent hydrochloric acid.
5. Rinse three times with deionized water.
Q26. Why are different procedures recommended? For which analytes is acid washing required and why? What is the purpose of using a phosphate-free detergent?
Q27. For some analytes such as phosphorous, plastic containers made of either high-density polyethylene or polypropylene might be preferable to glass. Why would this be the case? In addition, the EPA states that all containers and glassware must be “dedicated” to a specific analysis. What would be the drawback of reusing glassware for a different analysis?
Sample collection.
The EPA recommends two procedures for water collection, either using disposable whirl-pak bags or screw cap bottles. The following are excerpts from the EPA Monitoring and Assessing Water Quality website (http://water.epa.gov/type/rsl/monitoring/vms50.cfm).
“For Whirl-pak® Bags
1. Label the bag with the site number, date, and time.
1. Tear off the top of the bag along the perforation above the wire tab just prior to sampling . Avoid touching the inside of the bag. If you accidentally touch the inside of the bag, use another one.
2. Wading. Try to disturb as little bottom sediment as possible. In any case, be careful not to collect water that contains bottom sediment. Stand facing upstream. Collect the water sample in front of you.
Boat. Carefully reach over the side and collect the water sample on the upstream side of the boat.
3. Hold the two white pull tabs in each hand and lower the bag into the water on your upstream side with the opening facing upstream. Open the bag midway between the surface and the bottom by pulling the white pull tabs. The bag should begin to fill with water. You may need to "scoop" water into the bag by drawing it through the water upstream and away from you. Fill the bag no more than 3/4 full!
4. Lift the bag out of the water. Pour out excess water. Pull on the wire tabs to close the bag. Continue holding the wire tabs and flip the bag over at least 4-5 times quickly to seal the bag. Don't try to squeeze the air out of the top of the bag. Fold the ends of the wire tabs together at the top of the bag, being careful not to puncture the bag. Twist them together, forming a loop.
5. Fill in the bag number and/or site number on the appropriate field data sheet. This is important! It is the only way you will know which bag goes with which site.
6. If samples are to be analyzed in a lab, place the sample in the cooler with ice or cold packs. Take all samples to the lab.
Whirl-pak® bag
For screw-cap bottles
1. Label the bottle with the site number, date, and time.
2. Remove the cap from the bottle just before sampling. Avoid touching the inside of the bottle or the cap. If you accidentally touch the inside of the bottle, use another one.
3. Wading. Try to disturb as little bottom sediment as possible. In any case, be careful not to collect water that has sediment from bottom disturbance. Stand facing upstream. Collect the water sample on your upstream side, in front of you. You may also tape your bottle to an extension pole to sample from deeper water.
Boat. Carefully reach over the side and collect the water sample on the upstream side of the boat.
4. Hold the bottle near its base and plunge it (opening downward) below the water surface. If you are using an extension pole, remove the cap, turn the bottle upside down, and plunge it into the water, facing upstream. Collect a water sample 8 to 12 inches beneath the surface or mid-way between the surface and the bottom if the stream reach is shallow.
5. Turn the bottle underwater into the current and away from you. In slow-moving stream reaches, push the bottle underneath the surface and away from you in an upstream direction.
6. Leave a 1-inch air space (Except for DO and BOD samples). Do not fill the bottle completely (so that the sample can be shaken just before analysis). Recap the bottle carefully, remembering not to touch the inside.
7. Fill in the bottle number and/or site number on the appropriate field data sheet. This is important because you will know which bottle goes with which site.
8. If the samples are to be analyzed in the lab, place them in the cooler for transport to the lab.”
For further discussion on designing and implementing a sampling plan see sections 7.2 and 7.3 in Harvey, D. Chapter 7, Collecting and Preparing Samples. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/03_Sampling.txt |
Sample preservation and preparation
Purpose: The purpose of this assignment is to address questions related to sample preservation and preparation for analysis.
Learning Outcomes:
At the end of this assignment you will be able to
1. Suggest a method to preserve a sample for analysis
2. Understand the time frame within which a sample needs to be analyzed
Assignment:
Once a water sample is collected and possibly filtered if particulate matter is present, it is important to ensure that its chemical composition is not altered prior to analysis. Samples can only be stored for a limited period of time.
In many cases water samples can be stabilized for long term storage by controlling the pH, cooling, limiting light exposure or adding a chemical preservative.
Consult the literature in regards to water sample preservation and maximum holding times (refer to section 7.3 in Harvey, D. Chapter 7, Implementing the Sampling Plan).
The table below provides guidelines on preservation methods and maximum holding times for some analytes in natural waters and wastewaters.
Preservation Methods and Maximum Holding Times for Selected Analytes in Natural Waters and Wastewaters
Analyte
Preservation Method
Maximum Holding Time
ammonia
cool to 4 ºC; add H2SO4 to pH < 2
28 days
chloride
none required
28 days
metals—Cr(VI)
cool to 4 ºC
24 hours
metals—Hg
HNO3 to pH < 2
28 days
metals—all others
HNO3 to pH < 2
6 months
nitrate
none required
48 hours
organochlorine pesticides
1 1mL of 10 mg/mL HgCl2 or immediate extraction with a suitable non-aqueous solvent
7 days without extraction
40 days with extraction
pH
none required
analyze immediately
Q1. Under what conditions should the water samples be stored and within what time frame should the samples be analyzed?
Q2. What changes could take place that would cause sample degradation?
Q3. No preservation method is required for nitrates. However, a water sample can only be stored for up to 48 hours. What possible degradation could occur over an extended period of time? What chemical preservative could be added to prolong the life of the sample? How could you devise an experiment to test whether compositional changes for nitrate occur over time?
Q4. Cr(VI) must be analyzed within 24 hours. What might happen to compromise the analysis of Cr(VI) if the samples were held only longer before completing the analysis?
Q5. Preservation of water samples for metals analysis requires acidification with nitric acid below pH 2. Why is this procedure required for analysis of calcium and magnesium?
Consider the Ksp for Ca(OH)2 and Mg(OH)2.
Ksp Ca(OH)2 = 6.5 x 10-6
Ksp Mg(OH)2 = 2 x 10-13
Q6. Compare the solubility of Ca(OH)2 and Mg(OH)2 at pH 2 and pH 8. What could happen to the Ca2+ and Mg2+ ions stored over time if the pH was not adjusted to an acidic value?
05 Choosing t
Selection of an analysis method requires a consideration of the strengths and weaknesses of each potential method.
Q1: What general criteria would you consider when selecting an appropriate analysis method?
Metal cations are one important set of species in water samples. There are a number of ways that metals can be analyzed. The most common are Atomic Absorption (AA), Ion chromatography (IC), Ion-selective electrodes (ISE), and Inductively Coupled Plasma Spectroscopy (ICP).
The following links provide general background on each of the methods:
The limit of detection is an important figure of merit that can be used to compare the sensitivity of methods for the analyte of interest. For metals, Table 1 gives a general rule of thumb as to the methods that may be pertinent for different concentrations.
Table 1. Suggested analyte classification table for common cation analysis. See Analytical Chemistry 2.0 (Chapter 3 – Section 3.4) on the Analytical Sciences Digital Library website for additional information.
Analyte Classification
Concentration Range (%(w/w) & (ppm))
Appropriate Analytical Methods
Analyte Amount (mg)
Based on 25 mL sample size
Major
1 - 100
(10000 - 1000000)
Titrimetry, Gravimetry
25 - 2500
Minor
0.01 – 1.00
(100 - 10000)
Titrimetry, Potentiometry, Spectroscopy,
0.25 – 25
Trace
1 x 10-7 – 0.01
(10 ppb – 100 ppm)
Potentiometry (to low ppm) Spectroscopy (entire range)
2.5 x 10-6 – 0.25
Homework Question:
Your group will be assigned a particular analyte from a list that we wish to analyze. Your group will explore appropriate analysis methods found in the literature and make a decision about which would be preferable for the analysis we want to perform. As you read literature articles and compare methods, consider the following questions:
Q2: What analytical methods are best suited to provide the information about your analyte?
Q3: Of these, which techniques/instruments are available for the analysis?
Q4: Are there cost or timing issues that will influence the choice of method?
You may want to think about how long each analysis takes and whether it is amenable to the number of samples you need to analyze. Another consideration may be whether the technique is capable of analyzing several ions simultaneously.
Q5: Do the available methods have sufficient selectivity for the type of sample that will be analyzed?
A good example in thinking about selectivity is the difference in determining water hardness (generally the concentration of Ca2+ and Mg2+ combined) versus just the concentration of Ca2+. When one wants to determine just the concentration of Ca2+, a method that can differentiate between Ca2+ and Mg2+ is needed; whereas if one wants to determine water hardness, there is no need to differentiate between the two cations.
Q6: What sample pretreatment will be required?
Q7: What sample size (mass or volume) is needed, and can that be feasibly collected?
Q8: What is the anticipated range of analyte concentrations?
Q9: What is the limit of detection of the method and is its dynamic range appropriate for the range of concentrations of the analyte?
Q10: What is the precision of the method?
Q11: Is there a target concentration that is important for regulatory purposes?
Q12: Is the analyte in a form (solid, liquid, gas) suitable for the analytical method that you have selected?
Q13: How will the method be validated?
Q14: What type of calibration will be used?
Q15: Do any of the reagents used need to be standardized?
Q16: Can a reference standard be used to ensure accuracy?
As you identify the answer to each of these questions, you will need to balance each key attribute of a method to decide, or prioritize which method you ultimately will choose to pursue the analysis. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/04_Sample_Pre.txt |
Purpose: The purpose of this module is to introduce fundamental concepts of atomic spectroscopy, particularly flame atomic absorption spectroscopy (FAAS) as it pertains to the measurement of analytes in water samples. Calcium and magnesium are used as specific examples.
Learning Outcomes:
At the end of this assignment, students will be able to:
1. Understand the basic operation of a flame absorption spectrophotometer.
2. Select the appropriate lamp and wavelength to use for an analysis.
3. Evaluate the selectivity of the technique.
4. Investigate the quantitative aspects of flame atomic absorption spectrophotometry (e.g., use Beer’s law to calibrate the spectrophotometer and determine the concentration of an unknown amount of calcium and magnesium in a water sample).
For a discussion of fundamental concepts of atomic spectroscopy check the following links:
One common way to analyze for the presence of metal cations in aqueous samples at very low concentrations (parts per million or lower) is by flame atomic absorption spectrophotometry. An atomic absorption (AA) spectrophotometer is similar in many respects to a UV-VIS spectrophotometer except that the lamp needs to be specific to the analyte and the analyte needs to be converted to free atoms (atomization). This last requirement is accomplished by aspirating the sample into a flame that converts metal species to their elemental form. Once in the gas phase, atoms absorb radiation emitted by a special lamp (hollow cathode lamp) chosen to emit the same wavelength that the analyte will absorb. By constructing a standard calibration curve using known concentrations of the analyte under investigation, a relationship can be developed between the absorbance of the analyte at the specific wavelength and its concentration. By measuring the absorbance of an unknown sample, the concentration of the analyte in the sample can be determined.
Q1: Based on the description above, draw a block diagram and label the parts of a flame atomic absorption spectrophotometer.
Each element is characterized by the unique way its electrons are arranged in orbitals around the nucleus - its electronic configuration. These atomic orbitals are designated by type and energy as 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.
Q2: Write the electron configuration for calcium (Ca, element number 20). In what atomic orbital do the electrons with the most energy reside?
The electron configuration you wrote for the calcium atom describes the element in its lowest energy or ground state.
Q3: Calcium atoms undergo atomic absorption in the flame of the AA. Which electrons absorb the energy from the source?
Q4: Identify some possible absorption transitions for calcium.
Each atomic orbital of an element has a single discrete or quantized amount of energy. Therefore, the energies separating any two atomic orbitals is also discrete or quantized value. Only when the energy of radiation interacting with an atom exactly matches the energy gap between two levels of the atom will an absorption event occur. Because elements have different numbers of electrons, these energy differences are unique to each and can be used to identify that element. Because the energy differences between atomic levels are so specific, meaning that the absorption occurs at a very specific wavelength, the spectra are said to consist of a series of lines.
The energy necessary for a transition can be calculated using the equation:
$\mathrm{\Delta E = h\nu}\nonumber$
where ΔE represents the difference in energy between the ground and excited state, h is Planck's constant (h = 6.626 × 10-34 J s), and ν is the frequency of the absorbed radiation.
Q5: Given that the energy difference between the ground state and the first excited electronic state (ΔE) for the calcium atom is 4.687 × 10-19 J, calculate the frequency, ν, corresponding to a photon possessing this energy. Next, calculate the wavelength (in nm) for this photon.
As previously mentioned, for a sample to be analyzed by atomic spectroscopy, it must be in the gas phase. This is accomplished using an atomizer, the most common of which involves a flame, plasma or graphite furnace. Plasmas and graphite furnaces are typically used in analyses requiring the determination of metals at the parts per billion (ppb) level. However, in most water samples, the concentrations of cations such as Ca2+ and Mg2+ are in the parts per million (ppm) range, which are easily attainable by flame AA.
A variety of fuels and oxidants can be used to produce the flame, with the choice dependent upon the desired temperature, which in turn is dependent upon the atom under consideration. Routine analysis of calcium and magnesium can be accomplished with an acetylene/air flame which has a temperature of approximately 2500 K. A “slot” burner, as shown below, is common.
Q6: Why do you think a “slot” burner is used instead of the configuration of a traditional Bunsen burner?
The most common radiation source for flame atomic absorption is the hollow cathode lamp (HCL), which is illustrated below. The hollow cathode is coated with the element of interest, which in our example is calcium. The lamp is filled with a low pressure of argon gas. Argon atoms are ionized to Ar+ at the anode and these ions flow to the cathode. When they strike the cathode, they sputter off calcium atoms into the gas phase. Some of the gas phase calcium atoms are promoted to excited states in the sputtering process. These excited state atoms can emit radiation at specific lines of the element.
Q7: Do calcium and magnesium absorb at the same wavelength?
Q8: How will this affect the ability to determine both metals simultaneously?
Exploring fundamental relationships in spectrophotometric methods
Much like in UV-VIS spectrophotometry, the amount of radiation absorbed by the atoms of a given analyte in the gas phase is measured by the absorbance, which is defined as
$\mathrm{A = \log P_0/P}\nonumber$
where P0 represents the radiant power of the beam before the sample and P after the sample. The absorbance is proportional to the concentration of the analyte according to Beer’s law:
$\mathrm{A = \varepsilon bc}\nonumber$
where
A is the absorbance
ε is the molar absorptivity – include that this is a measure of the probability of the transition – the molar absorptivity is largest at λmax, meaning that this is the most probable transition
b is the path length (in this case the length of the burner)
c is the concentration
Q9: How would this type of relationship help you in determining the concentration of Ca2+ and Mg2+ in an unknown water sample?
Analyzing Ca2+ and Mg2+ in water samples
Now that we have developed general background information on flame atomic absorption spectrophotometry, we can explore specific features of its use in the analysis of Ca2+ and Mg2+ in water. You may want to research EPA methods 7140 for analysis of calcium and 7450 for analysis of magnesium.
Q10: What wavelength would you use to measure Ca2+? What wavelength would you use to measure Mg2+?
Q11: If you wanted to measure Mg2+ concentrations, what instrumental parameters would you need to know?
Q12: If you know that the concentration of calcium and magnesium in your water sample is approximately 5 to 10 ppm, suggest a strategy to appropriately calibrate the spectrophotometer for such an analysis.
Q13: If the absorbance of an unknown water sample is found to be greater than the absorbance of the highest calcium standard used to calibrate the spectrophotometer, what steps would you take to ensure that the analysis is providing accurate results?
Q14: Considering that the sample is introduced into the flame through a very thin capillary, what step would you have to take before analyzing a surface water sample that may have small amounts of detritus? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/06_Analysis_o.txt |
Purpose: The purpose of this assignment is to introduce fundamental concepts of visible (VIS) spectrophotometry as it pertains to the measurement of analytes in water samples. Phosphate will be the specific example used.
Learning Outcomes:
At the end of this assignment, students will be able to:
1. Select the appropriate wavelength to collect spectra.
2. Evaluate the selectivity and detection limits of the method.
3. Use UV-VIS spectrophotometry as a quantitative tool (use Beer’s law to calibrate the spectrophotometer and determine the concentration of an unknown amount of phosphate in a water sample).
For a discussion of fundamental concepts of UV-VIS spectrophotometry check the following links:
Often, to assess water quality, anions such as nitrates and phosphates need to be quantitatively determined at concentrations in the parts per million (ppm) range or lower. One way to measure low concentrations of many species is to use a spectrophotometric analysis. A spectrophotometric analysis is performed by illuminating the sample with radiation in the visible or ultraviolet range and detecting the amount of light that reaches the detector. Depending on the chemical properties of the species in the sample, some light will be absorbed while the remainder will be transmitted to the detector where it is converted to an electrical signal that is proportional to the amount of light absorbed, called the absorbance. By constructing a standard calibration curve using known concentrations of the analyte under investigation, a relationship can be developed between the absorbance of the analyte at a specific wavelength and its concentration. By measuring the absorbance of an unknown sample and comparing it to the calibration plot, the concentration of the analyte in the sample can be determined. This module will address the spectrophotometric analysis of phosphate.
Q1: Draw a block diagram of the components of a spectrophotometer.
When a beam of electromagnetic radiation passes through a sample, most of the radiation is transmitted but, at specific wavelengths, the radiation may be absorbed by chemical constituents within the sample. For an atom or molecular species, the absorption of light causes valence electrons to be excited from lower to higher energy states. For this transition to occur, the energy provided by the radiation has to match the difference in energy (ΔE) between the lower and higher energy states.
Q2: What molecular properties must a compound have in order to absorb UV-VIS radiation?
Q3: Do you think phosphate ions have the ability to absorb UV or visible radiation? Justify your answer.
Many species, including phosphates, can be analyzed by visible spectrophotometry by reacting them with other species, thus forming a new compound that absorbs light at a specific wavelength. The general approach to any spectrophotometric analysis is the same with differences in reagents and chemical reactions used to generate the light absorbing species.
Exploring fundamental relationships in spectrophotometric methods
When light traveling through a sample is absorbed, the radiant power of the beam decreases from P0 to P, where P0 represents the radiant power of the beam before the sample and P the power after the sample. The fraction of incident radiation transmitted through the solution is called transmittance and is represented by the equation:
$\mathrm{T = P/P_0}\nonumber$
The absorbance of a solution is related to the transmittance through a logarithmic relationship:
$\mathrm{A = - \log T = - \log P/P_0 = \log P_0/P}\nonumber$
Before moving on to the analysis of phosphate, we will examine some dye solutions to explore fundamental relationships within spectrophotometric methods.
For the following activity you can either use a real spectrophotometer or the simulator at the following website http://web.mst.edu/~gbert/ColorScan/Spectrophotometry.HTML
Prepare solutions of red dye and blue dye and record their spectra between 400 and 700 nm.
1. At which wavelength does the blue dye absorb the most? This wavelength is referred to as λmax
2. At which wavelength does the red dye absorb the most?
3. What is the value of the transmittance at λmax for each dye? What is the value of the absorbance?
4. What wavelength would you choose to quantitatively determine the concentration of the red dye? Why did you choose this wavelength?
5. What wavelength would you choose to quantitatively determine the concentration of the blue dye? Why did you choose this wavelength?
Determining method’s selectivity
A method’s selectivity refers to the extent to which it can determine particular analyte(s) in a complex mixture without interference from other components in the mixture (4,5).
Q4: Is it possible to determine the concentration of the blue dye if it is contaminated with some of the red dye?
Q5: When does it become possible to selectively measure the blue dye in presence of the red dye?
Relationship between absorbance and concentration
At the heart of a quantitative estimation of phosphates using UV-VIS spectrophotometry is Beer’s law which states the linear relationship between absorbance and concentration according to the equation:
$\mathrm{A = εbc}\nonumber$
where
A is the absorbance
ε is the molar absorptivity - a measure of the probability of the transition – the molar absorptivity is largest at λ max, meaning that this is the most probable transition
b is the path length of the cuvette
c is the concentration
Q6: Draw a representative plot of A versus c if the spectrophotometer is set to λmax.
Q7: How could you use this plot to determine the concentration of an unknown?
Analyzing phosphate by visible spectrophotometry
Now that we have developed general background information on spectrophotometric measurements, we can explore specific features of its use in the analysis of phosphate ion.
In question 3 you were asked if phosphate has the ability to absorb UV or VIS light. Most likely the anion does absorb light but such absorbance is not sensitive and specific enough to base a quantitative method of analysis. One strategy to improve analysis sensitivity is to react the species, in this case phosphate, with a reagent that forms a new molecule capable of absorbing light at a specific wavelength. The amount of phosphorous as orthophosphate (PO43-) present in a water sample may be best determined by the ascorbic acid/molybdenum blue method (1-3). In this method, ammonium paramolybdate, (NH4)6Mo7O24∙4H2O, reacts with orthophosphate in acidic solution and in the presence of potassium antimonyl tartrate, C8H4K2O12Sb2, to produce phosphomolybdic acid, H3[P(Mo3O10)4]. This acid, in turn, is reduced by ascorbic acid, C6H8O6, to produce intensely colored molybdenum blue. “Molybdenum blue” is actually not a single compound, but a complex mixture of polyoxomolybdate (POM) ions. The role of potassium antimonyl tartrate is as a catalyst.
Below is a spectrum recorded on phosphate solutions of increasing concentrations after reaction with the molybdenum blue reagent.
Q8: What wavelength would you choose to quantitatively determine orthophosphate by the molybdenum blue method?
Assessment of possible interferences
Q9: What potential interferences would limit the measurement of orthophosphate using the molybdenum blue method?
Q10: What steps could you take in the sample preparation to decrease the chances of interfering species in the orthophosphate analysis?
Q11: Looking at the interferences listed in EPA method 365.1 will any of these affect the way you will analyze your sample for orthophosphate?
Preparing calibration standards for phosphate analysis
You are now going to prepare a set of phosphate calibration standards and use them to calibrate the spectrophotometer.
Q12: Given that a stock solution of 3 mg/L phosphate (PO43-) is equivalent to a concentration of 1 mg/L of elemental phosphorous (P), how many milliliters will you have to pipette to prepare 25.00 mL of the following six standards?
Standard concentration
(mg P/L)
Milliliters of phosphate standard to pipette
0.00
0.04
0.08
0.12
0.15
0.20
Q13: What is the purpose of preparing the standard at 0.00 mg/L P? How is it used in the analysis?
Q14: The second standard is 0.04 mg P/L. How many parts per million (ppm) of P does this correspond to?
The following absorbance values were recorded when analyzing the six standards prepared above:
Standard concentration
(mg/LP)
Absorbance
0.00
0.000
0.04
0.021
0.08
0.039
0.12
0.052
0.15
0.078
0.20
0.102
Use Excel or equivalent software to plot the standard calibration curve.
Q15: What relationship do you observe between absorbance and concentration?
Q16: What parameter allows us to determine whether there is a good fit between absorbance and phosphate concentration?
Once the calibration is completed, the next step is to analyze unknown water samples for phosphate concentration. Suppose that three replicate water samples from one of the ponds at End Creek are analyzed and their absorbance is 0.025, 0.027 and 0.023, respectively.
Q17: What is the average concentration of phosphate (expressed in mg P/L) in the pond water?
Q18: What parameter allows you to assess the reproducibility of the measurement?
Evaluating the detection limit of the method
The detection limit is the smallest concentration or absolute amount of analyte which, with a stated probability, can be distinguished from a suitable blank value (5). If we define Smb as the average signal for a method blank, and σmb as the method blank’s standard deviation, the detection limit (S)DL is calculated as follow:
$\mathrm{(S)_{DL} = S_{mb} + 3σ_{mb}}\nonumber$
Q19: How would you determine the average signal and the associated standard deviation of the method blank using the UV-VIS spectrophotometer?
Q20: If you determined that the average signal from the method’s blank is 0.0081 and the standard deviation is 0.0006, what is the detection limit of the method?
Q21: Would you be able to detect phosphate in the pond water mentioned above if the absorbance of an unknown sample was 0.003?
Automated methods
Most water labs currently use automated methods of analysis for phosphate, nitrates and other routine analytes. A typical approach is based on flow-injection analysis.
Q22: What are the likely advantages of automating a method? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/07_Analysis_o.txt |
Purpose: The purpose of this assignment is to introduce potentiometric measurements of ionic species by ion selective electrodes (ISEs)
Learning Outcomes:
Upon completion of this module, students will be able to:
1. Identify electrodes and measurement devices used in potentiometry.
2. Predict how analyte concentration (or activity) controls the potential of a potentiometric measurement.
3. Choose appropriate measurement conditions to minimize interferences.
4. Correct for differences in ionic strength among calibration standards and samples.
5. Construct an appropriate calibration curve for potentiometric determinations and account for changes in analyte concentration due to sample dilution.
References: Modules on the theory and operation of ISEs may be found in the Analytical Sciences Digital Library (ASDL) collection. The following hyperlinks will direct the reader to some ASDL resources on potentiometry.
1. Analytical Sciences Digital Library. Potentiometry: e-learning module. http://community.asdlib.org/activelearningmaterials/analytical-electrochemistry-potentiometry/ (accessed April 3, 2014).
2. Harvey, D. Analytical Chemistry 2.0, Chapter 11. http://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Text_Files.html (accessed April 3, 2014)
Membrane-based ISEs are widely used in the determination of ionic species. Such determinations fall under the category of a direct potentiometric measurement, which you have experienced if you have ever made a pH measurement. A typical direct potentiometric measurement requires the use of an indicator electrode, a reference electrode, and a high-impedance voltmeter. An example of equipment needed to perform a potentiometric determination is shown in Figure 1.
Figure 1. An experimental setup for the direct potentiometric measurement of sodium ion in aqueous solution.
The two electrodes pictured in Figure 1 represent an electrochemical cell. The electrode on the right is the sodium indicator electrode (sodium ISE). The electrode on the left is a reference electrode. The sodium ISE has a glass membrane that responds specifically to sodium ions. This glass membrane physically separates two solutions: one inside the electrode with a constant sodium ion concentration; one outside the membrane that is the solution you are analyzing. The electrical potential difference generated by the indicator electrode depends on the sodium concentration of the outer solution. Therefore, the function of the indicator electrode is to respond to changes in the analyte concentration in a predictable manner.
The reference electrode provides a known and stable potential to compare against the indicator electrode potential. An assumption when using an ISE system is that the potential of the reference electrode is independent of the concentration of the analyte and matrix of the sample being analyzed. The electrical potential is displayed on the high-impedance voltmeter. The voltmeter has a high impedance to minimize current flow, which prevents changes to the chemical composition of the reference electrode and to the sample (so, for example, if measuring the concentration of H+ with a pH meter, there will not be any reduction of the H+ to hydrogen gas). High impedance, which typically is greater than 1012 ohms, is also needed to minimize errors in the measured potential.
One of the most common examples of ion selective electrodes is a pH electrode – pH electrodes are selective toward the H+ ion. We will start our exploration of ISEs by understanding how a pH electrode works. pH electrodes also use a glass membrane (this membrane is about 0.1 mm thick, so is quite fragile), but in this case it is a type of glass more sensitive toward the H+ ion instead of the Na+ ion. The internal solution has a fixed and known concentration of H+. The external solution is the sample whose pH you wish to measure. The concentration of H+ in the external solution varies depending on the sample being analyzed. An important point is that the glass used to manufacture the membrane has some sodium ions (Na+) in it. A key factor in the functioning of the membrane is that the inner and outer surfaces of the glass form a very thin hydrated gel layer when in contact with water (Figure 2). The internal and external hydrated gel layers are only about 10 nm thick, which is much less than the 0.1 mm thickness of the glass membrane, so a layer of dry glass always separates the two hydrated gel layers. Cations from the solution have the ability to migrate into the hydrated gel layer. For a pH electrode, that means that H+ ions from the solution will displace some of the Na+ ions in the glass that makes up the hydrated gel layer.
Figure 2. Representation of the glass membrane in a pH electrode. Note: the width of the hydrated gel layers in the representation is too large relative to the width of the dry glass portion of the membrane.
Many of you may have used a pH electrode before. If you did, it likely appeared to consist of only a single electrode. In actuality, it is a two-electrode system, but it is designed in such a way that the reference electrode is incorporated into the glass membrane electrode. The design of the single-electrode system incorporates an electrical contact between the reference and indicator electrode that is necessary to complete the electrical circuitry and produce a potential reading.
Q1: Will a more acidic sample displace more, the same or less Na+ from the hydrated gel layer?
Different concentrations of H+ and Na+ in the hydrated gel layer cause different junction potentials. Note that the junction potential at the interface of the membrane with the internal solution never changes because the concentration of H+ is constant in the internal solution. The junction potential at the interface of the membrane with the outer solution changes for different samples with different pH. The specific reason why varying concentrations of H+ and Na+ can be found in the outer gel layer has to do with something called the mobility of the ions.
Q2: What do you think is meant by mobility of ions?
Q3: Which ion do you think has a higher mobility, H+ or Na+?
As mentioned earlier, a reference electrode is used because its junction potential stays fixed no matter what the external solution. Therefore, the only junction potential in the entire circuit of an ion selective electrode that changes is the one at the interface of the ion selective membrane and the external (sample) solution. In the case of a Na+ ion selective electrode a glass membrane of a different composition than the glass membrane in a pH electrode is used– namely one that is more responsive to sodium ions migrating into the glass. Other ion selective electrodes can be fabricated provided a membrane exists that is selective toward the ion one wishes to measure.
Q4: Do you think other cations (e.g., Li+. K+) may have some ability to migrate into the hydrated gel layer of a pH electrode? If so, is this a problem?
The potential measured by the voltmeter is described by Equation $\ref{1}$.
$\mathrm{E_{cell} = E_{ind} - E_{ref} + E_{lj}} \label{1}$
In eq $\ref{1}$, Ecell represents the potential of the electrochemical cell, Eind represents the half-cell potential of the indicator electrode, Eref represents the half-cell potential of the reference electrode, and Elj represents the liquid-junction potential between the sample solution and the outside membrane of the indicator electrode.
Of particular interest is the relationship between Ecell and the concentration of the analyte. Remember, Ecell is measured but there is only one junction potential (Elj) in the entire system that changes, so a measurement of Ecell is essentially a measure of the one varying junction potential. For the Na+ ion selective electrode, the varying junction potential only depends on the concentration of Na+.
However, the situation is more complicated than just using the concentration of Na+, because ion-selective electrode measurements are most commonly performed in solutions with ionic strengths that are greater than zero. In these solutions, there is a difference between the formal concentration (i.e. how the solution was prepared in the lab) and the effective concentration or activity of the analyte.
Q5: Consider a solution that has some Na+ and very high concentrations of K+Cl-. What effect do you think this might have on the activity of Na+ in the solution?
The relationship between activity and concentration for sodium is illustrated in Equation $\ref{2}$.
$\mathrm{a_{Na}= γ_{Na} [Na^+] \label{2}}$
In eq $\ref{2}$, aNa is the sodium ion activity (mol L-1) , [Na+] is the sodium ion concentration (mol L-1), and γNa is the activity coefficient for the sodium ion. As we just discussed, as the sample ionic strength increases, there is a greater probability that analyte ions will interact with oppositely charged ions from the supporting electrolyte(s) dissolved in the sample. This effectively decreases the concentration of the “free ion”, which is represented by a decrease in the activity coefficient. As the ionic strength of a solution approaches zero, the activity coefficient approaches one, and under infinitely dilute conditions, the analyte activity and analyte concentration are equal. The relationship between the oxidized and reduced forms of sodium written as a reduction reaction can be described in Equation $\ref{3}$.
$\ce{Na+ (aq) + e- → Na (s)} \label{3}$
The half-cell potential of the indicator electrode responds to changes in the activity of the analyte as described by the generalized form of the Nernst equation in Equation $\ref{4}$:
$\mathrm{E_{ind}= E_{ind}^o- \dfrac{RT}{nF} \ln\left(\dfrac{1}{a_{Na}} \right)} \label{4}$
In eq $\ref{4}$, E° is the indicator electrode potential under standard conditions (298 K, 1.00 M Na+), R is the molar gas constant (8.314 J K-1 mol-1), T is the absolute temperature (K) , n is the number of moles of electrons in the half-reaction, and F is Faraday’s constant (96485 C mol-1).
Q6: If the indicator electrode potential under standard conditions is -0.100 V, what is the indicator electrode potential at 298 K if the activity of the sodium ion is 0.10 M?
Q7: How does the indicator electrode potential change in the previous question if the temperature is increased by 10 degrees?
A sodium ion selective electrode must be calibrated before it can be used to measure the concentration of Na+ in an unknown sample.
Q8: How would you go about calibrating a sodium ion selective electrode?
We just discussed how the ionic strength of a solution impacts the activity of Na+. Suppose you wanted to analyze the sodium concentration of a low ionic strength sample such as natural pond water.
Q9: Can you think of a way to mitigate possible effects of ionic strength to insure that your calibration procedure and sample analysis provide an accurate measurement of the concentration of Na+ in the unknown?
If one keeps the ionic strength high and constant, then the Nernst equation can be expressed in terms of analyte concentrations (and not activities) because the activity coefficient of all samples and standards are equivalent and knowledge of activity is no longer critical.
Q10: In the potentiometric determination of sodium ion of a mineral water sample, indicate if either of the following supporting electrolytes can be used for ionic strength adjustment: a 4.0M NH3 – NH4Cl buffer (pH 10) or 4.0M NaCl.
Q11: What would be the general criteria you would need to use in selecting a suitable supporting electrolyte for an analysis using an ion selective electrode?
The concentration of each calibration standard can be expressed as a formal concentration. The concentration term in the Nernst equation is often converted from base e to base 10, which can also be expressed as a p-function, shown in Equation $\ref{5}$.
$\mathrm{pNa= - \log[Na^+]} \label{5}$
Assuming a temperature of 298 K, and the constants R, T and F combined into a single value, under those conditions, the Nernst equation takes on the following form for cationic analytes (shown for sodium in Equation $\ref{6}$):
$\mathrm{E_{cell}= E_{cell}^o- \dfrac{0.05915}{n} pNa} \label{6}$
Q12: Based on the relationship in eq $\ref{6}$, how would you construct a calibration that links the changes in electrode potential to changes in the concentration of the sodium ion?
Q13: What is the expected slope of a potentiometric calibration curve for sodium at 35°C? What effect does temperature have on the slope of a potentiometric calibration curve?
As discussed earlier, indicator electrodes do not have a specific response to a given analyte, but have a wide range of responses to a group of analytes that are similar in charge and size. The electrode is designed to exhibit the greatest response for the target analyte, but the presence of chemically similar analytes in a sample may interfere with the determination of the target analyte and bias the potentiometric response. The selectivity of an ion-selective electrode is expressed by Equation $\ref{7}$
$\mathrm{E_{ind}=E_{ind}^o- \dfrac{0.05915}{n_{Analyte}} \log\{a_{Analyte}+ K_{Analyte,Interferent} (a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} } \} } \label{7}$
Expressed in Equation $\ref{8}$, the selectivity coefficient (KAnalyte,Interferent) is a ratio of analyte to interferent activities where each species influences the indicator electrode potential to the same degree.
$\mathrm{K_{Analyte,Interferent}=\dfrac{a_{Analyte}}{(a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} }}} \label{8}$
Q14: If a sample has a sodium concentration of 1.0 x 10-3 M, and the sodium ISE has a selectivity coefficient of KNa,H = 30, what sample pH would cause a 1% error in the sodium ISE response?
Q15: Evaluate whether it is best to use alkaline or acidic conditions to determine the sodium ion concentration by ISE?
Whenever a method calibration is performed using linear regression (i.e. a best-fit line or trendline, as it is called in Microsoft Excel), it is understood that extrapolating beyond the concentration range used in the regression analysis can lead to biased results. Typically when the analyte concentration in a sample is greater than the analyte concentration for the most concentrated standard, the sample is diluted so that the analyte concentration is between the lowest and highest standard on the calibration curve. The original sample concentration is calculated using the dilution equation, shown in eq $\ref{9}$.
$C_1 V_1= C_2 V_2 \label{9}$
In eq $\ref{9}$, C1 is the analyte concentration of the original (undiluted) sample, V1 is the volume of the original sample, V2 is the volume of the diluted sample, and C2 is the analyte concentration of the diluted sample. If the sample is diluted prior to analysis, the response of the diluted sample is used as the y-value in the calibration equation and the analyte concentration of the diluted sample is the x-value determined using the calibration equation. The analyte concentration of the original sample is calculated using the dilution equation.
Q16: The table below contains sodium ISE calibration data. If the cell potential measured in a sample is ‑0.115 V, determine the sodium concentration (mol L-1) in this sample.
[Na+] (M)
Ecell (V vs SCE)
1.0 x 10-4
-0.221
1.0 x 10-3
-0.164
1.0 x 10-2
-0.107
1.0 x 10-1
-0.048
Q17: In the previous question, the sample was prepared by pipetting 5.00 mL of the original water sample and 2.00 mL of an ionic strength adjustment buffer into a 100 mL volumetric flask and diluting to the mark with distilled water. Determine the sodium concentration (mol L-1) in the original water sample.
Using a nitrate ion selective electrode to measure nitrate in water samples
Figure 3. Schematic of a nitrate ion-selective electrode.
Because the concentration of nitrate inside the electrode chamber and in the ion-exchange membrane are different, a potential difference (voltage) is established, which can be measured with a potentiometer. The voltage depends on the concentration of nitrate in the membrane, which in turn depends on the concentration of nitrate in the water sample. Empirically, it is found that the potential difference (E) is linearly proportional to the logarithm of the concentration of nitrate in the water sample (cnitrate), according to Equation ($\ref{10}$).
$E = m × \log(c_{nitrate}) + b \label{10}$
In this equation, m is the slope of the line, and b is the y-intercept.
Q18: How could you devise a means to calibrate this type of electrode and determine the concentration of nitrate in an unknown water sample?
09 Instructor
Overview on the Use of This Module
This module is designed to guide students to investigate the role that water quality may have on the survival of aquatic species such as frogs and snails. It is developed based on water monitoring activities conducted in wetlands areas in eastern Oregon that were recently reclaimed and restored after many years of agricultural usage. Through this module, students are introduced to fundamental concepts related to sampling as well as analytical methods for the determination of cations and anions in water such as spectrophotometry, atomic absorption spectroscopy, and ion selective electrodes. These materials are designed to be modular in their format and used relatively independently. They can also be implemented at different levels of guidance to students. For example, instructors in a general chemistry course could use the module to introduce the research question but provide students with an already developed sampling plan and detailed instructions to conduct experiments. Alternatively, in a more advanced analytical course, students could be asked to develop and implement the sampling plan as well as research the most appropriate methodologies of analysis. A complete dataset is available that can be used for learning as simple tasks as constructing a calibration curve and predicting the concentration of an analyte in an unknown sample or more advanced statistical data analysis.
These materials could also be used in the classroom as active learning exercises or used as supplement to laboratory experiences, as a prelab for existing experiments or a drylab where the relevant instrumentation is not available. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/End_Creek%3A_Spotted_Frogs_and_Aquatic_Snails_in_Wetlands__A_Water_Quality_Investigation/08_Analysis_o.txt |
Lake Nakuru is located in the Rift Valley Province of Kenya, situated inside the Lake Nakuru National Park near the city of Nakuru. It may be best known for its large population of lesser flamingos (Phoeniconiais minor), at times reaching one million birds, who feed on the blue-green algae (Spirulina platenis) that thrives in the alkaline lake. Lake Nakuru has a fragile ecosystem, susceptible to climatic and human activities. During only 3 months of 1993, tens of thousands of flamingos died during a drought period and in the year 2000, 40,000 flamingos were reported dead at Nakuru National Park. Peculiar behavior, such as disorientation and a lack of mobility, were also observed. Flamingos have not only been dying at Lake Nakuru, but changing their migratory habits. Many have left Lake Nakuru, never to return. In this module, you will have the opportunity to explore reasons why the flamingos might be dying in Lake Nakuru Park. As is the case for many complex problems, addressing this question requires both the approach and the tools used by analytical chemists. In the process of thinking about the problem of the Lake Nakuru flamingos, you will have the chance to apply knowledge you have gained in your chemistry coursework and to learn in greater detail important concepts related to modern chemical analysis.
Photo of Lake Nakuru courtesy of Dr. Terry Master, Professor of Biology at East Stroudsburg University
Environmental Analysis Lake Nakuru Flamingos
To many people the image of a flamingo may bring to mind thoughts of paradise. A flamingo spends much of its time wading and dipping its bill in an inverted motion into the algae-rich water. Its bill serves as a filter to capture its primary food source, a blue-green algae. A flamingo prefers to spend its time in this manner and can be stressed by other activities such as flying and swimming (Vick, 2000). About 80 percent of the world’s flamingos reside in the Great Rift Valley of eastern Africa (Vick, 2000). Roughly 1 million of these rift valley flamingos reside at Lake Nakuru located in Nakuru National Park in Kenya.
Figure 1. A lone flamingo at Lake Nakuru (left)1; a large group of flamingos at Lake Nakuru (middle)1; and a map of Lake Nakuru National Park and surrounding area (right)2.
Lake Nakuru can appear ensconced in pink when occupied by lesser flamingos (phoeniconiais minor) stopping to feed during their annual migrations through Kenya. Hundreds of thousands of tourists come every year from throughout the world to see the spectacular views as shown above.
Lake Nakuru is an alkaline (pH = 10.5) salt lake that provides an ideal environment for the growth of arthospira fusiformis and spirulina platensis, the types of cyanobacteria consumed by the flamingos. Volcanic activity in the area supports Nakuru and other “soda lakes”, named for their high levels of carbonate and bicarbonate. The Lake Nakuru catchment basin is a “closed system”. Five seasonal rivers flow into it, but there is no exit stream and the water loss is only due to evaporation. These factors, along with the shallow depth (averaging around 2.5 m, Odada, 2006) of the lake create a fragile ecosystem, susceptible to climatic and human activities. The area around the catchment basin includes not only the national park, but also agricultural land and the growing city of Nakuru (pop. 400,000; Odada, 2006).
Figure 2. The Town of Nakuru, Kenya.
Facts, Observations and Evidence
The typical lifespan of a flamingo is 50 years (Mukhebe, 2005). Flamingos do not breed at Lake Nakuru, but at Lake Natron in Tanzania. In recent years the flamingo has been classified as “near threatened” due to its susceptibility to small changes in its fragile habitat (Bloomfield, 2008).
During only 3 months of 1993, tens of thousands of flamingos died during a drought period. In 1997, El Nino diluted the salinity of the lake, adversely affecting food sources and driving flamingos to nearby Lake Bogoria (Odada 2006). During the year 2000 40,000 flamingos were reported dead at Nakuru National Park (Vick, 2000; Astill, 2001). Peculiar behavior, such as disorientation and a lack of mobility, were also observed (Astill, 2001). Flamingos have been observed not only to be dying at Lake Nakuru, but to be changing their migratory habits. Many have left Lake Nakuru, never to return (Mukhebe, 2005). Some of these birds were found dead along ponds around the city’s sewage treatment plant.
Heavy metal pollutants, such as lead, mercury and arsenic have been found in the tissue of flamingos from both Lake Nakuru and Bogoria which are 60 km apart (Astill, 2001; Wanjiru, 2001). These were attributed to effluents from local industries. Other pollutants, such as organochlorine pesticides were also found in flamingo body tissues (Kairu, 1994; Wanjiru, 2001). Heavy metals and pesticide residues have been found to be present in both the waters and sediments of Lake Nakuru, particularly during the rainy season (Mavura, 2003).
The waste treatment and disposal in the city cannot keep up with the population growth and development in the area. Some of the excess waste that accumulates is deposited in the lake via stormwater and wind (Odada, 2006).
The question to be answered is: What is causing the changes in behavior and death of the Lake Nakuru flamingos?
Some hypotheses as to possible (and perhaps intertwined) causes include:
Pollution. Toxins get into the lake via runoff from various sources, including fertilizers, organochlorine pesticides, and effluents from industrial plants. Some scientists suspect that organochlorine pesticides first accumulate in the cyanobacteria that the flamingos feed on (Mukhebe, 2005) and then are transferred to the tissue of the flamingos (Kairu, 1994; Wanjiru, 2001). Fertilizers could be enhancing the growth of cyanobacteria, which are hypothesized to accumulate organochlorine pesticides (Mukhebe, 2005). When the flamingos eat the cyanobacteria, the organochlorine compounds bioaccumulate into their tissues (Kairu, 1994).
Drought. The Kenyan Wildlife Service cited dramatic decreases in lake water volume since the last El Nino cycle. Corresponding changes in the lake salinity in the mid to late 1990’s were noted to have adversely affected the flamingo’s food supply (Odada, 2006). Specifically, the type of phytoplankton residing in the lake changed to a smaller size that could not be captured by the flamingo, thus reducing the amount of algae that the flamingos are able to capture. This phenomenon could be happening again as a result of increased settlement and farming in the area which consumes water from the rivers that run into Lake Nakuru.
Multiple factors may be contributing to the increased death rate. For example, in addition to decreasing flamingo food supplies, as the lake dries up pollutants and toxins that are already present become more concentrated. It is possible that the upswing in flamingo deaths is due to a factor that is as yet unrecognized and altogether different from those hypothesized above.
Identifying Possible Analysis Methods
There are three hypotheses that have been put forward that could be responsible for the death of flamingos at Lake Nakuru. The purpose of this exercise is to identify analysis methods that could be used to test the chemical species that contribute to each of the three hypotheses. Among the various analytical methods you find that may be applicable to this measurement, identify their strengths and weaknesses (e.g., sensitivity, expense, ease of use, reproducibility).
Hypothesis 1. High levels of pesticides found in flamingo tissues interfere with their endocrine systems leading to the high levels of disease and death that have recently been observed.
Q1. Using available literature and other sources of information, identify possible analytical methods that could be used to detect pesticides and measure their levels in components of the lake (e.g., water, sediment).
Q2. Using available literature and other sources of information, identify possible analytical methods that could be used to detect heavy metals and measure their levels in components of the lake (e.g., water, sediment).
Hypothesis 3. Increasing salinity due to decreases in lake water levels has decreased the food supply of flamingos and hunger and starvation are key reasons for the recently observed increase in the flamingo death rate.
Q3. Using available literature and other sources of information, identify possible analytical methods that could be used to measure the salinity of the water and the chemical constituents that account for the salinity.
Experimental Design for this Project
The other sections of this website are designed to support the investigation of the hypothesis that organochlorine pesticides in Lake Nakuru are responsible for the deaths of the flamingos. The next step is to design how the experiment will be conducted.
All analyses face certain real world limitations, such as cost, availability and compound stability. It is also important that the experiments be performed in such a way that the results are of sufficient quality to allow a scientifically sound conclusion to be drawn from the data. Standard methods such as those published by the Environmental Protection Agency (EPA) or Association of Analytical Communities (AOAC) can be useful in designing the experimental plan. You often may have to adapt a method based on available equipment or other limitations. Therefore, it is important to understand all aspects of a given method and be able to show that your method is valid.
Analysis of Samples from Lake Nakuru
In our case study we will develop a sampling plan for water samples to be collected by local scientists in Kenya and shipped back to the United States for analysis. Therefore, there are limitations on the cost of sample preparation and shipping, as well as sample stability. As a result, the analysis will be limited to 200 samples (the maximum capacity of the GC-MS autosampler). Taking this into consideration:
• How do you decide where to collect samples to ensure they are representative?
• What sample preparation method will you utilize (considering cost, effectiveness, and availability)?
A series of assignments are presented within this website to help you understand how to design a GC-MS experiment utilizing EPA method 525.2, and address questions such as those posed above. Through this process, you will also be introduced to important analytical chemistry concepts that underlie the analytical methods discussed. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/01_Identifying_the_Problem.txt |
Toxic trace metals are possible culprits for the ongoing deaths of large numbers of lesser flamingos at Lake Nakuru, Kenya. In this section we explore the possible role of heavy metals in these deaths and examine instrumental methods utilized to evaluate levels of copper, zinc, lead and chromium present in Lake Nakuru sediment and suspended solid samples. These methods include anodic stripping voltammetry (ASV), atomic spectroscopy, and x-ray fluorescence spectroscopy (XRF). Data sets are provided for each technique so that current levels can be calculated and compared to those contained in a report published in 1998.
• Introduction – Heavy Metals in Lake Nakuru
The purpose of this module is to explore the hypothesis that heavy metal contamination is responsible for the flamingo deaths at Lake Nakuru. Water chemistry of the lake will be considered as it pertains to the respective chemical state, concentration, sampling and sample preparation of toxic metals found there. In the next section, relevant methods of analysis for these metals will be investigated.
• Metals Analysis by X-ray Fluorescence
The purpose of this unit is to introduce the fundamental principles of x-ray fluorescence spectroscopy (XRF), apply this method to the analysis of simulated data for heavy metals in suspended particulate and sediment samples from Lake Nakuru and critically contrast XRF with alternative methods for determination of heavy metal concentrations.
• Metals Analysis by Anodic Stripping Voltammetry
The purpose of this learning module is to introduce students to the electrochemical technique of anodic stripping voltammetry (ASV), which is particularly good at detecting metals at the ppb level.
• Metals Analysis by Atomic Spectroscopy
This section will provide an introduction to atomic spectroscopy. The basic concepts of flame atomic absorption spectroscopy (FLAA), graphite furnace atomic absorption spectroscopy (GFAA), and inductively coupled plasma optical emission spectroscopy (ICP-OES) will be presented.
• Instructor’s Guides: For instructor’s guide in a PDF or Word document, please contact Cindy Larive. ([email protected])
• Introduction
• X-ray Fluorescence
• Anodic Stripping Voltammetry
• Key for Atomic Spectroscopy
• Assessment Questions: For Assessment questions that accompany this module, please contact Cindy Larive ([email protected])
Heavy Metals
Learning Outcomes
At the end of this assignment students will be able to:
• Understand the rudiments of stabilization of water samples.
• Know the definition of water alkalinity and be able to calculate total alkalinity.
• Distinguish between soluble and particulate bound metals and understand Langmuir adsorption behavior.
Purpose
The purpose of this module is to explore the hypothesis that heavy metal contamination is responsible for the flamingo deaths at Lake Nakuru. Water chemistry of the lake will be considered as it pertains to the respective chemical state, concentration, sampling and sample preparation of toxic metals found there. In the next section, relevant methods of analysis for these metals will be investigated.
Introduction
You have already read that toxic trace metals are a possible culprit for the deaths of 40,000 lesser flamingos at Lake Nakuru, Kenya during the time periods of August – November 1993 and August – September 1995. (Nelson 1998) Local industries such as tanneries, electroplating facilities, textile mills, battery factories, and paint packaging are the most likely sources of the heavy metals. In this section we further explore this possibility and examine in greater detail the sampling and sample preparation steps that are necessary to evaluate the levels of several candidate metals present in Lake Nakuru water and sediment samples.
The focus of our investigation will be on the metals copper (Cu), zinc (Zn), lead (Pb) and chromium (Cr). Often such metals are referred to as “heavy” or “toxic” metals but, unfortunately, there is not a clear definition of these terms in the scientific literature. (Duffus, 2002) The idea that the density of metals may be correlated to their toxicity is not true, and some metals may be required for life at trace levels but toxic at higher levels. To further confuse matters the oxidation state of the metal can be important in determining both its chemical properties and biological effects. For example, for chromium, Cr(III) is much less harmful than Cr(VI).
Three items must be shown to support the hypothesis that heavy metals are responsible for flamingo deaths:
1. Demonstration of the identities of heavy metals present in the lake (qualitative analysis) and determination of the concentration levels (quantitative analysis) of each toxic metal present in the lake.
2. What is the potential for exposure of the flamingos to each of these metals?
1. Do the toxic trace metals exist in the lake in forms that will be consumed by the flamingos?
3. If exposed, is the dose toxic to the flamingos?
The paper by Nelson et al. (1998) reports the results of a thorough study of heavy metals in water samples taken from Lake Nakuru during 1996. Both water and sediment samples were collected and analyzed by the research team. In addition to evaluating the levels of metal samples in the water samples, adsorption equilibria were studied for the binding of Cu, Zn, Pb and Cr to suspended solids collected from Lake Nakuru and to the algae S. platensis grown in pure culture.
Read through the article “Model for Trace Metal Exposure in Filter-Feeding Flamingos at Alkaline Rift Valley Lake, Kenya” (Nelson 1998) and consider the following:
Q1. What is the likely route of exposure (i.e. eating, drinking, or some other means)?
Q2. Why do you think the authors emphasize the importance of particulate bound metals rather than metal ions dissolved in the water? Is part of the rationale specific to flamingos? If so, why? Are there other reasons?
Lake Nakuru is referred to in this paper as a highly alkaline lake. Although the lake is basic, with a pH of 10, water alkalinity is not directly related to the water pH, but instead to its buffering capacity. Alkalinity is determined by assessing the ability of the water to neutralize added acid. It is typically calculated as the total alkalinity (AT). Because carbonate (CO32-) and bicarbonate (HCO3-) are usually the most significant components in natural waters, alkalinity is measured by titrating a water sample with a strong acid until pH 4.5 where bicarbonate or carbonate has been consumed. Although the most important determinant of alkalinity is the amount of carbonate, other weak bases can also contribute to total alkalinity, as indicated below. Because of its importance in many surface waters, AT is normally expressed as mg/L CaCO3, even though this unit is not particularly meaningful for Lake Nakuru, as Nelson et al. report a [Ca2+] of 0 in Table 1.
$\mathrm{ A_T= [HCO_3^- ] + 2 \times [CO_3^{2-} ] + [SO_4^{2-} ] + 2 \times [PO_4^{3-} ] + [F^- ] + [NH_3] + [OH^- ] - [H^+]}\nonumber$
You will notice that in the equation for AT some ions are multiplied by a factor of two, and that this factor is not related to their charge. Alkalinity can be measured by titrating a sample with a strong acid until the buffering capacity of the ions is consumed. Normally, environmental water samples are buffered by CO32- and HCO3- so the endpoint of the alkalinity titration is functionally set to pH 4.5. The [CO32-] is multiplied by 2 as two moles of H+ would be required to fully neutralize the carbonate ion. The [SO42-] is multiplied by a factor of 1 because it would not be converted to H2SO4 under the conditions of the titration. Similarly [PO43-] would not be converted only to H2PO4- at pH 4.5 therefore it is multiplied only by a factor of 2.
Q3. Why do you think that [F-] is included in this calculation, while [Cl-] is not?
Q4. Using the data in Table 1 (Nelson, 1998), calculate the alkalinity of the Lake Nakuru, in mg/L CaCO3.
The Methods section of the Nelson paper suggests that the water samples collected may not have been acidified, although, acidification with 5% nitric acid is often used to stabilize water samples especially those collected for metal ion analysis (EPA-1669, 1996). If samples are to be measured separately for dissolved metals they should be filtered before acidification.
Q5. Why could it be important to filter suspended particles or sediment from the water sample before acidification?
Section 8.4.3 of EPA method 1669 says “Preservation of aliquots for metals other than trivalent and hexavalent chromium—Using a disposable, precleaned, plastic pipet, add 5 mL of a 10% solution of ultrapure nitric acid in reagent water per liter of sample. This will be sufficient to preserve a neutral sample to pH <2.”
Q6. Assume that you collect and filter a 1 liter water sample from Lake Nakuru. Using the total alkalinity you calculated in Q4, would addition of 5 mL of 10% nitric acid reduce the pH of your water sample to < 2? If not, how do you suggest that this protocol should be altered?
The paper by Nelson et al. examines adsorption to both particulate matter and the algae. The particulate matter was derived from sediment samples collected from Lake Nakuru at the locations where the water samples were collected. The algae Spirulina platensis was grown in the laboratory under conditions similar to those in Lake Nakuru. For the results for binding to the suspended sediments, the authors write:
“Adsorption isotherms for metal adsorption to Lake Nakuru sediments in reconstituted lake water at pH 10.0 were linear for all four metals with respect to dissolved metal concentration over the range of concentrations considered, and distribution coefficients (Kd) were calculated for each metal (Fig. 1).”
Q7. Examine the results presented in Figure 1 of the paper by Nelson et al. Why do you think that the experimental results deviate most from the line at low concentrations of the dissolved metal?
Q8. For adsorption to algae (Figure 2 of Nelson et al.) the adsorption behavior is not linear. The authors suggest that the results for Cr, Cu and Zn are consistent with that of a Langmuir isotherm (click here for a tutorial on the Langmuir isotherm).
Summary
The paper by Nelson et al. provides an excellent entry point for our consideration of the possible effect of toxic metals on the flamingos at Lake Nakuru. It should now be clear to you that flamingos are exposed to metals primarily by filtering suspended sediments along with their food supply. The next unit will allow you to develop a plan for the analysis of heavy metals that are adsorbed on the suspended sediment found in Lake Nakuru. Several analytical approaches will be considered: x-ray fluorescence analysis (XRF), atomic spectroscopy (FLAAS, GFAA, and ICP-AES), and anodic stripping voltammetry (ASV). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Heavy_Metals/01_Introduction__Heavy_Metals_in_Lake_Nakur.txt |
X-ray Fluorescence Spectroscopy
Learning Outcomes
At the end of this assignment students will be able to:
• Describe the process of XRF and explain how element specific information can be obtained using this technique.
• Differentiate between qualitative and quantitative analysis using XRF.
• Compare and contrast the benefits and challenges of quantitative analysis using XRF.
• Apply the knowledge gained to determine the role of XRF in the analysis of heavy metals in Lake Nakuru sediment and suspended solids samples.
Purpose
The purpose of this unit is to introduce the fundamental principles of x-ray fluorescence spectroscopy (XRF), apply this method to the analysis of simulated data for heavy metals in suspended particulate and sediment samples from Lake Nakuru and critically contrast XRF with alternative methods for determination of heavy metal concentrations. The content is appropriate for students in a course in analytical chemistry or instrumental analysis. This introduction will be achieved via an inquiry-based learning style.
Recommended reading prior to completion of this exercise:
Introduction
What analytical chemist hasn't longed for a Star Trek tricorder? In addition to the medical and engineering tricorders, a variety of other tricorders were "used" on the original TV series for geological (GEO) and metallurgical (MET) and Biological (BIO) measurements. Landing on a new planet, the tricorder gave the Star Trek crew the ability to rapidly scan the elemental composition of surrounding rocks, evaluate alloys of structures and survey animal and plant life. To those of us burdened by collection of representative samples that must be transported back to the laboratory for laborious processing and sample preparation steps, and finally an instrumental analysis – the ability to get instantaneous elemental composition was as fantastic as transporters, phasers and Klingon cloaking devices. With recent developments of handheld XRF devices this fantasy has become a scientific reality. In this module we will examine the scientific principles that underlie the method of XRF, examine its advantages and limitations and apply this method to the analysis of sediment samples from Lake Nakuru.
X-ray fluorescence is a luminescence-based field-portable method that can provide rapid elemental analyses, relatively inexpensively. The battery operated devices shown below operate essentially in a point and acquire mode that allows measurements to be made outside the laboratory, for example determining whether heavy metals are present in toys or other consumer products, or sorting scrap metal alloys for recycling.4 The non-destructive nature of this method makes it ideal for the compositional analysis of priceless art and antiquities.
XRF instruments are also available for use in a more standard laboratory spectrometer format in which the sample is inserted into the instrument. These benchtop instruments can be more sensitive and accurate, but generally are also more expensive.
How can x-rays generate fluorescence emission? In molecular fluorescence measurements, UV-visible light is used to excite valence electrons into an excited state.1 Fluorescence occurs when these excited state electrons relax back to the ground state emitting photons. Compared with the light used for excitation, the wavelength corresponding to the maximum fluorescence intensity is red shifted (i.e. at a longer wavelength and lower energy) because the excited electrons relax to the ground vibrational and rotational states of the excited electronic state before emitting a photon.1 Atomic fluorescence is a similar phenomenon, except that the electrons are excited thermally in a flame or plasma and fluorescence results when these excited electrons relax.
X-ray fluorescence occurs by an analogous but subtly different process illustrated in Figure 2. X-ray radiation is much higher energy than UV-visible light. X-rays are referred to as ionizing radiation because the energy of an x-ray photon can provide enough energy to eject an electron from an inner shell atomic orbital creating a positive ion. As shown in Figure 2, higher energy outer shell electrons fill the vacancies in the lower energy orbitals created by electron ejection. The excess energy of the electron that fills the vacancy is emitted as a secondary x-ray photon, generating the fluorescence signal. As in molecular fluorescence, the photons emitted by fluorescence are lower in energy than the source radiation. Also, because the energy (and hence wavelength) of the fluorescence depends on differences in energy of the inner shell atomic orbitals that do not participate in molecular bonds, it is characteristic of the elemental composition of the sample independent of its chemical form.
The material provided in the tutorial "Introduction to Energy-Dispersive X-ray Fluorescence (XRF) – an Analytical Chemistry Perspective" (reference 2) will likely be helpful in answering the questions below.
Q1. Figure 3 shows a periodic table with the relevant energies (in electron Volts; eV) for the X-ray emission of various elements. What do the letters K, L and M represent?
The characteristic x-rays listed for each element in Figure 3 are labeled as K, L, M or N to denote the shells they originated from, with x-rays originating from the K-shell having the highest energy. Another designation: $\alpha$, $\beta$ or $\gamma$, is used to indicate x-rays that originated from the original shell of the transition electrons that fill the vacancy created by the ejected electron. For example, a K$\alpha$ x-ray is produced from a transition of an electron from the L to the K shell, while a K$\beta$ x-ray results from the transition of an electron from the M to a K shell, etc. Since within the shells there are multiple orbits of higher and lower energy electrons, a further designation is made as $\alpha$1, $\alpha$2 or $\beta$1, $\beta$2, etc. to denote transitions of electrons from these orbits into the same lower shell.
Q2. For lead, atomic absorption measurements are typically made at a wavelength of 283.3 nm. Calculate the energy in joules of a 283.3 nm photon and the L$\alpha$ and M$\alpha$ photons listed for lead in the Periodic Table in Figure 3. Note that the values in Figure 4 are given in keV. Offer an explanation for the relative order of the energies you calculated.
Q3. Examine the values in the periodic table in Figure 3. What are the trends in energy as you move along a row of the periodic table? What about moving down a column? Can you explain these trends in terms of differences in atomic structure? How might these differences in energy be useful in the analysis of the amounts of different elements in a complex sample?
XRF measures the energy and intensity of secondary x-rays produced, as illustrated in Figure 2. The emitted x-rays are called "secondary" because they are produced as a result of irradiation from a higher energy "primary" source. Backscattering of the x-ray source radiation also occurs and is a source of interference that spectrometers try to reduce or remove, for example using filters or polarization methods. Backscattered radiation does not interact with the sample and has the same wavelength as the source radiation but is scattered from the sample in all directions. In some cases, backscattered x-rays can be used to help normalize the data and compensate for self-absorption and differences in sample density.
The power of XRF for elemental analysis is in its ability to report on the presence of a wide range of elements, its ease of use (little sample preparation is required) and the non-destructive nature of the measurement. The results of XRF can be quantitative, however to match the accuracy and precision of alternative methods like atomic absorption spectroscopy, special care is required in the preparation and analysis of the sample.2 An example spectrum for an energy-dispersive (explained below) x-ray fluorescence (EDS-XRF) measurement is shown in Figure 4.
Q4. Locate absorption bands in Figure 4 that might be attributed to the Cr and Zn K$\alpha$ transitions listed in Figure 3.
The EDS spectrum in Figure 4 was produced using a solid-state silicon-based detector with a digital processor that can rapidly analyze the electronic pulses caused by the incident x-rays. The detector monitors the energy and number of the x-ray photons over a defined integration time window. The signal is processed using a multichannel analyzer that accumulates the electrical signals to produce a digital spectrum. As illustrated in Figure 4, the energy in eV of the fluorescent x-rays reflects the elemental composition of the sample. The photon emission rate in terms of the number of counts per second (cps) at a given voltage is proportional to the concentration of the element producing a particular signal.
XRF measurements made with field-portable devices are most often used for qualitative analysis. The presence of a metal of concern (e.g., lead in children's toys or house paint) can be easily identified. The non-invasive nature of XRF analyses makes it an excellent screening tool for field measurement as illustrated by this short video Quantitative analyses can be performed using XRF provided that care is taken in calibration and efforts are made to overcome matrix effects.2 As demonstrated in several examples in reference 2, calibration plots for XRF can compare favorably with those produced by atomic spectroscopy, however such careful quantitative experiments are typically performed in the laboratory using standard additions and comparison with a standard reference material (SRM) to validate the measurement. Although the method of standard additions typically provides more accurate quantitative results, when this method is used we can no longer consider XRF to be a nondestructive analysis method because the sample is irreparably altered by addition of the standard.
EPA method 6200 describes the use of field-portable XRF (FPXRF) for screening for 26 analytes including Cr, Cu, Pb and Zn, in soil and sediment samples. To quote section 1.2 of the method:
"This method is a screening method to be used with confirmatory analysis using other techniques (e.g., flame atomic absorption spectrometry (FLAA), graphite furnace atomic absorption spectrometry (GFAA), inductively coupled plasma-atomic emission spectrometry, (ICP-AES), or inductively coupled plasma-mass spectrometry, (ICP-MS)). This method’s main strength is that it is a rapid field screening procedure. The method's lower limits of detection are typically above the toxicity characteristic regulatory level for most RCRA analytes. However, when the obtainable values for precision, accuracy, and laboratory-established sensitivity of this method meet project-specific data quality objectives (DQOs), FPXRF is a fast, powerful, cost effective technology for site characterization."
Table 1 of EPA Method 6200 lists "Interference Free Lower Limits of Detection" in silica sand for a variety of metals; the values relevant to this study are listed below.
Table 1. Interference Free Lower Limits of Detection (LLOD) in Silica Sand Taken from EPA Method 6200, pg. 25.
Analyte
LLOD (mg/kg)
Chromium (Cr)
150
Copper (Cu)
50
Lead (Pb)
20
Zinc (Zn)
50
Table 2 lists the levels for chromium, copper, lead, and zinc reported in Lake Nakuru sediments and suspended solids during a study by Nelson, et al. in 1998.8
Table 2. Metal Concentrations in Lake Nakuru Sediments and Suspended Solids found in Nelson Study8
Concentration in dry sediments ($\mu$g/g) Concentration in suspended solids ($\mu$g/g)
Trace metal Average (11 sites) Range Average Standard deviation
Chromium 67 10 – 280 8.3 0.82
Copper 24 5 – 95 19 2.9
Lead 22 4 – 100 11.7 0.9
Zinc 147 44 – 630 74 31
Q5. Compare the levels of chromium, copper, lead and zinc for the sediment samples in Table 2 to the limits of detection taken in Table 1 from EPA Method 6200. Do you think that XRF is a suitable method for detecting the presence each of the metals listed in Lake Nakuru sediments? Justify your response. Which is more relevant to consider in this analysis: the average values or ranges of values across the 11 samples tested?
Q6. Similarly examine the values for suspended solids reported in Table 2 and compare these to the LLOD values in Table 1. Do you think that XRF would be a suitable method for detecting the presence of each of the metals in suspended solid samples from Lake Nakuru? Do you need to consider the standard deviation in answering this question?
Q7. Suppose you wanted to use XRF to quantify the levels of these metals in Lake Nakuru samples. Estimate the expected limit of quantitation for each of the metals in Table 1?
Q8. The values in Table 1 are based on silica sand which has a fairly uniform size and composition. Would you expect that the samples from Lake Nakuru sediments and suspended solids would have a more complex matrix than silica sand? If so, what effect might this sample matrix have on your ability to quantify the levels of metals in these samples?
Q9. A common format involving the use of XRF in analyses is to use it to screen samples in the field to select those above a certain value for collection and subsequent analysis in the laboratory using a method like atomic spectroscopy. Describe how you might incorporate XRF into a plan for sampling the sediment samples at Lake Nakuru. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Heavy_Metals/02_Metals_Analysis_by_X-ray_Fluorescence.txt |
Learning Outcomes
After reading about ASV and working through this learning module, you should be able to:
• Describe some advantages of anodic stripping voltammetry.
• Outline the general steps in an ASV experiment.
• List the important parameters that affect sensitivity and detection limits in an ASV experiment.
• Perform calculations for determination of metal ion content in unknowns analyzed by ASV. These calculations will utilize the single standard addition quantitation method.
Let’s say that you wanted to determine the heavy metal identities and concentrations in sediment and suspended solid samples obtained from present day Lake Nakuru. What technique would you use for this determination? You could use Atomic Absorption Spectroscopy (AAS) as was done in the article (Nelson 1998), but another technique, which is particularly good at detecting metals at the ppb level is anodic stripping voltammetry (ASV). This learning module will focus on ASV.
Purpose
The purpose of this learning module is to introduce students to the electrochemical technique of anodic stripping voltammetry (ASV). The content is appropriate for students in a course in analytical chemistry or instrumental analysis. This introduction will be achieved via an inquiry-based learning style.
Content specific to voltammetric methods and ASV can be found in the citations listed in the reference section of this document and are available free of charge at the following websites:
• in the Analytical Electrochemistry module (Kelly 2009) at the ASDL website (click here to link directly to the ASV section)
• A chapter on voltammetric techniques can be found in the eText Analytical Chemistry 2.0 found in the ASDL collection (Chapter 11; specifically section 11.4 discusses stripping voltammetry).
• Additionally, you can perform a search with the key words anodic stripping voltammetry at the ASDL website (click here for the voltammetry section of the ASDL collection accessed on June 26, 2012).
Theory
Read the Anodic Stripping Voltammetry section of the Analytical Electrochemistry eLearning module and answer the following questions. If you have no background in voltammetric measurements and instrumentation, read the Basics of Voltammetry section of the Analytical Electrochemistry module. Once you have read the background material, the questions below will help you ascertain and apply important aspects of ASV theory and practice.
Q1. How would you prepare the solid samples obtained from Lake Nakuru for an ASV analysis? In what physical form should they be for ASV analysis? What are the expected concentration levels in your samples from Lake Nakuru?
All voltammetric techniques require a working electrode – this is where the electrochemical reactions occur and where the analytical signal is generated.
Q2. What is the most common type (i.e. material used) of working electrode used in ASV? You should be able to find this information in the material referenced in the Theory section of this document.
An advantage of mercury as an electrode material is that mercury forms amalgams (solid solutions) with many metals allowing preconcentration of more of the analyte metal at the electrode surface. The metal in solution is cationic, but the mercury amalgam forms with the neutral metal analyte.
Q3. What must happen to the metal at the electrode surface to form the neutral species?
ASV is useful for detecting metal ions at the ppb level. The preconcentration step is key for achieving these low detection limits. When this preconcentration step utilizes electrochemical accumulation, the metal must be reduced and form an amalgam in the mercury (drop or film). In a mercury electrode, the concentration of metals is 100-1000 times that in solution.
Q4. Look up the reduction potentials of the following metals of interest (Cu2+, Pb2+ and Zn2+). You can find a table of reduction potentials in the back of most analytical textbooks. These are generally reported with respect to the standard hydrogen electrode (SHE). We will continue this convention throughout the module. You can also find a table of values here (from the online textbook Analytical Chemistry 2.0 Appendix 13, p. 1107). Would you be able to discriminate between these three metals based on their reduction potentials?
Judging from the separation between potentials it appears that these metals could be resolved in an ASV measurement. However, in practice it can be difficult to discriminate between Cu and Zn. Intermetallic compounds between Cu and Zn can form in the Hg amalgam, particularly at thin Hg film electrodes (Copeland 1974). To remove one of the interfering ions during a determination, a complexing agent can be added. For example, during the determination of zinc, sulfide can be added to selectively complex with copper (Lau 1998).
Q5. What potential (versus SHE) would be necessary to simultaneously pre-concentrate zinc and copper?
Q6. Using the symbol M2+ to represent a divalent metal cation, write the appropriate redox reaction for the reduction of the cation at a mercury electrode. In addition to the applied potential, what other experimental variable is used to “drive” the metal cation to the electrode?
The following figure depicts the potential-time waveform for an ASV experiment.
The potential labeled 1a in Figure 1 is the deposition/accumulation/preconcentration potential. A DC potential is applied at 300-500 mV more negative than the E0 of the metal with the most negative reduction potential (if analyzing a mixture). This would be ~-1.1 V since the Zn2+ (E0 = -0.7618 V) has the most negative reduction potential of the three metals listed in Table 1. The reduction reaction, indicating the amalgamation with mercury is shown below.
$\ce{M^2+(aq) + 2e- \xrightarrow{Hg} M(Hg)} \nonumber$
The duration of the deposition step is determined by the analyte concentration – it should be longer for lower concentration analytes. Generally, the deposition time is between 1 and 30 minutes. The solution containing the analyte is stirred during this step (or the electrode can be rotated) to ensure faster and more reproducible transport to the electrode.
The interval indicated by 1b is the quiescent time (quiet time). During this period the potential is still applied but the stirring is stopped for 30-60 sec prior to analysis.
In the region labeled 2 in Figure 1, the potential is scanned to oxidize the metal back to its original state. The linear ramp as shown in Figure 1 is the simplest method for re-oxidizing the amalgamated metal:
$M(Hg) \rightarrow M^{2+}(aq) + 2e^- + Hg \nonumber$
The resulting plot of current versus potential for this oxidation process is called a voltammogram. The observed peak current is proportional to the analyte concentration.
Q7. In the space below, sketch out the voltammogram (plot of current versus potential) expected from the experiment outlined in Figure 1.
Q8. What values would you choose as the start and end of the potential ramp for an analysis of zinc?
Q9. Summarize the quantitative and qualitative nature of the ASV experiment.
Q10. The Lake Nakuru samples will most likely contain measurable concentrations of copper, lead and zinc. Describe how multiple metals can be analyzed in a single ASV experiment. Draw what the voltammogram would look like. What starting and ending potentials would you use?
ASV is very sensitive to experimental conditions. Important variables include electrode size (described by Hg film area or Hg drop size), deposition time, rest time, stirring rate, and scan rate during stripping. An example set of experimental conditions is given below:
• Working electrode = Hg or Hg film
• Deposition time = 5 min (depends on concentration)
• Deposition potential = -1.1 V (This chosen potential is with respect to the standard hydrogen electrode. The Potentiometry e-learning module found in the ASDL collection provides a review of Reference Electrodes. The Electrochemical Methods Chapter in Analytical Chemistry 2.0 also discusses various reference electrodes.)
• Rest time = 30 s
• Stirring rate = 600 rpm
• Scan rate during stripping = 200 mV/s.
• Final potential (after anodic scan) = 600 mV
Q11. How could the experimental conditions be modified to increase the analyte signal during an analysis? Hint: think about what variables will increase the amount of metal deposited and look at the equation for ip. (This equation can be found in the ASV section of the Analytical Electrochemistry module).
The use of potential waveforms other than the linear scan discussed here can improve detection limits. These are called “step” methods. For example, in the paper Locatelli 1999, differential pulse anodic stripping voltammetry (DPASV) is used for the determination of Cu, Pb, Cd, and Zn. Differential pulse voltammetry improves detection limits by minimizing the background current portion of the current and therefore increasing signal to noise. DPASV is discussed in more detail in the ASV section of the Analytical Electrochemistry module.
Advantages of ASV
Anodic stripping voltammetry, along with certain other electrochemical techniques, is capable of determining the oxidation state of a metal (this is called speciation). For example, as seen in a table of Standard Redox Potentials, copper(I) reduces at 0.520 V and copper(II) reduces at +0.159 V. An ASV experiment could be designed to discriminate between these two oxidation states during the potential stripping scan. As we have observed, ASV is also capable of simultaneous determination of multiple metals. An atomic spectroscopic (AS) technique, such as atomic absorption spectroscopy, could not discriminate between the two oxidation states as all metal in the sample is reduced to neutral gas phase atoms during the atomization step. Additionally, electrochemical techniques generally cost less than their spectroscopic counterparts, and are amenable to miniaturization and to in situ monitoring. ASV requires minimal sample pre-treatment and is generally not affected by high salt content. However, ASV is limited in the number of metals it is able to determine when compared to AS techniques. For example, when Hg-based working electrodes are used, the metal must form an amalgam with mercury. Table 11.11 in Analytical Chemistry 2.0 supplies a list of metals that can typically be determined by ASV.
Special Instructions for trace analysis.
Because of the low concentrations of the metals in samples from many natural systems, special care must be taken during sampling and sample handling. Analytes can be lost via adsorption to sample containers and glassware. Additionally, these containers and glassware can be sources of contamination. Glassware and containers must be thoroughly washed and even soaked in deionized water to remove any possible contaminants. Any solution containing trace metal amounts should be stored in a special Teflon bottle. As previously discussed, electrochemical techniques can provide information about the oxidation state of a metal. However, physical and chemical changes made to the sample can change the original oxidation state. Do not add substances known to be oxidizing or reducing agents to your samples. Check for trace metal impurities in the reagents used.
Quantitation methods. Either the standard addition or external calibration methods can be used for quantitation in ASV. In this module the standard addition method has been used to decrease matrix effects from the lake water. For a review of the standard addition method, see Chapter 5: Standardizing Analytical Methods in Analytical Chemistry 2.0.
Data and Analysis
In the following section, you will review the data reported in Nelson 1998 for the analysis by atomic spectroscopy and XRF of Cr, Cu, Pb, and Zn in suspended solids and dried sediments from Lake Nakuru. Next, you will be supplied with simulated ASV data from Lake Nakuru sediments so that you can compare current levels of the four target metals to those obtained in the 1996 study. As a starting point, let us consider Figures of Merit common to the ASV experiment, as well as some practical considerations for application of the technique.
Figures of Merit, Anodic Stripping Voltammetry
Precision and Accuracy. You can typically expect a precision of ~2-4% relative standard deviation and similar accuracy in the range of 2-5% relative error.
Detection Limits and Linearity. You have seen that ASV is an extremely sensitive technique, with detection limits commonly in the ppb range. Part per trillion detection limits can be reached for some analytes by using an increased deposition time. ASV is typically performed on dilute solutions and the linearity is generally over two orders of magnitude.
Interferences. The metals being analyzed must have redox potentials sufficiently different so that the peaks are resolved and proper quantitation can be performed. Typically up to five metals can be simultaneously determined by ASV. Only certain metals can be determined by ASV. Common metals are Cd, Pb, Zn, Tl, In, and Cu.
Practical Considerations for Anodic Stripping Voltammetry
Cost. ASV is a relatively inexpensive technique. The electrochemical instrumentation and electrodes are a fraction of the cost of other metal analysis methods (e.g. ICP and AA systems).
Speed. The per sample run time depends on the detection limits required. The lower the sample concentrations, the longer the analysis time. The primary factor is the deposition time. For ppm to ppb concentrations this can be a few seconds to a minute. For ppb-ppt detection limits, the deposition time can be a few to many minutes.
Data from Nelson Study, 1998
Table 2 gives the levels for chromium, copper, lead, and zinc found in Lake Nakuru sediments and suspended solids collected by Nelson, et al. in 1996. (Nelson 1998)
Table 2. Metal Concentrations in Lake Nakuru Sediments and Suspended Solids (Nelson 1998)
Concentration in dry sediments (µg/g) Concentration in suspended solids (µg/g)
Trace metal Average (11 sites) Range Average Standard deviation
Chromium 67 10 – 280 8.3 0.82
Copper 24 5 – 95 19 2.9
Lead 22 4 – 100 11.7 0.9
Zinc 147 44 – 630 74 31
Q12. Are the levels observed for each of these four metals in sediments and suspended solids amenable to analysis by ASV? Note that the Nelson paper does not prepare samples for ASV analysis. In your answer, consider the sample preparation steps described by Locatelli 1999, which is described in the next section.
Sample Preparation and Data Analysis.
In this section, you will be given the chance to obtain practical experience evaluating ASV data representative of the current levels of the metals of interest in simulated Lake Nakuru sediment and suspended solid samples. These samples were collected and prepared in a manner similar to that described in the literature (Locatelli 1999). Briefly,
“Two samplings . . . were made with a plexiglass device. Single carrots of superficial sediment (height 5 cm) were drawn out and stored in polyethylene bottles. The samples, dried at 60 C for 48 hr, were sieved through a 40 mesh sieve to eliminate coarse material, then through a second 150 mesh sieve, and lastly powdered by means of a corundum ball mill. The sediments were then mineralized with the acidic mixture HNO3-HCl. . . .
Approximately 1.0 g, weighed accurately in a pyrex digestion tube, was dissolved in 10 mL of 1:1 diluted 69% (m/m) nitric acid. The tube was inserted into the cold home-made block digester, raising gradually the temperature up to 95 C and keeping it for all the time of mineralization. After 15 min 10 mL, and after 30 min further addition of 5 mL of concentrated nitric acid were made. At the end, 5 mL of 37% (m/m) hydrochloric acid was added, maintaining the temperature at 95 C for a further 15 min. After cooling the digest was diluted to 100 mL. The solutions so obtained were diluted 1:20 before voltammetric and spectroscopic measurements.
Voltammetric curves were recorded with a Metrohm Model VA 646 polarographic processor equipped with a Metrohm Model VA 647 Stand. An Ag/AgCl (KCl sat.) electrode and a platinum wire were used as reference and auxiliary electrode, respectively. The voltammetric cell was rinsed every day with suprapure concentrated HNO3 to avoid metal contamination of the sample. The solutions were thermostated at 20±0.5 C and deaerated with pure nitrogen prior to analysis, while a nitrogen blanket was maintained above the solutions during analyses. The solutions were stirred with a Teflon-coated magnetic stirring bar in the electrolysis step.”
Differential Pulse ASV measurements. The instrumental experimental conditions used are shown in the table below.
Table 3: Experimental conditions for the determination of copper, lead, and zinc by DPASV in sediments*
Ed Electrolysis potential (mV/Ag, AgCl, KCl satd.) –1150
Ef Final potential (mV/Ag, AgCl, KCl satd.) +50
dE/dt Potential scan rate (mV/s) 10
f Pulse repetition (s) 0.1
ΔE Pulse amplitude (mV) 50
v Pulse duration (ms) 40
ts Sampling time (ms) 8
td Electrolysis time (s) 240
tr Rest time (s) 15
t Purging time prior of the electrolysis (s) 300
u Stirrer speed (r.p.m.) 600
*Note: conditions modified from Locatelli 1999 who were determining Cu, Pb, Cd, and Zn in sediment samples.
Below are simulated sample data acquired for samples taken from Lake Nakuru. Sample preparation was similar as described above by Locatelli 1999. Briefly, 1 g sediment samples were acid digested and diluted to 100 mL. This solution was diluted by a factor of 20 with appropriate electrolyte. The analysis was performed on this second solution. The single standard addition method of quantitation was used (section 5.3.3 in Analytical Chemistry 2.0). A 10.00 mL aliquot of sample solution was pipetted into the sample container and purged with nitrogen as described above. The sample underwent the anodic stripping procedure outlined above. Upon collection of this voltammogram, 100.0 µL of the standard solution (5.000 ppm) was added to the 10.00 mL sample and the ASV procedure was performed on this “spiked” sample. The data for the peak current was recorded in tabular format and is shown below. You will use this data to calculate the concentration of each heavy metal in the sediment samples in µg/g (as was reported in Nelson 1998). Don’t forget to account for any dilutions performed during the sample preparation. You may hand-calculate the concentration for sample 1 then set up a spreadsheet to calculate the concentrations of the metals for all three samples.
Q13. Recently, sediment samples were collected from Lake Nakuru and analyzed by ASV in the manner described above. Given the sample data in Table 4, calculate the current concentration levels of the heavy metals (Cu, Pb and Zn) in the sediment samples obtained from Lake Nakuru. What do the data tell you about the heavy metal concentrations in the sediments at Lake Nakuru? Have the concentrations increased or decreased since the 1996 sampling?
Table 4: Sample Data for Copper, Lead and Zinc Determination in Sediment Samples in Lake Nakuru
Sample number
ip (µA)
Sample
Sample + 100 µL spike
Cu
Pb
Zn
Cu
Pb
Zn
1
0.0985
0.1011
0.2134
0.2000
0.3045
0.3152
2
0.0996
0.1008
0.2159
0.2015
0.3026
0.3165
3
0.1000
0.1205
0.2200
0.2099
0.3225
0.3251
Note: All samples analyzed were a 10.00 mL aliquot. The 100.0 µL standard spike contained a target concentration of 5.000 ppm of each metal of interest. The exact concentration (depending on standard mass weighed out) can be determined via calculation.
Data Analysis – Suspended Solids
Q14. Recently, suspended solid samples were collected from Lake Nakuru and analyzed by ASV in the manner described above. Given the sample data in Table 5, calculate the current concentration levels of the heavy metals (Cu, Pb and Zn) in suspended solids in Lake Nakuru. What do the data tell you about the heavy metal concentrations in the suspended solids at Lake Nakuru? Have the concentrations increased or decreased since the 1996 sampling?
Table 5: Sample Data for Copper, Lead and Zinc Determination in Suspended Solid Samples in Lake Nakuru
Sample number
ip (µA)
Sample
Sample + 100 µL spike
Cu
Pb
Zn
Cu
Pb
Zn
1
0.0999
0.1511
0.1985
0.4697
0.9513
0.4545
2
0.1001
0.1623
0.1752
0.4898
0.9253
0.4326
3
0.0959
0.1506
0.1863
0.4800
0.9222
0.4863
Note: All samples analyzed were a 10.00 mL aliquot. The 100.0 µL standard spike contained a target concentration of 5.000 ppm of each metal of interest. The exact concentration (depending on standard mass weighed out) can be calculated.
Chromium measurement
You may have noticed that the data analyzed so far was for lead, copper and zinc, while the Nelson 1998 paper has reported data for chromium. If we wish to use voltammetry for chromium measurement, the technique of choice is Adsorptive Stripping Voltammetry. The key difference between Adsorptive and Anodic stripping voltammetry is in the deposition (accumulation) procedure. In this alternative technique, the accumulation step is purely adsorptive (i.e. the analyte “sticks” onto the electrode) rather than electrolytic (electrochemical deposition). The stripping step for the adsorptive method can involve either anodic or cathodic removal of the analyte from the electrode.
Q15. Let’s see if you are paying attention . . . During the accumulation step in adsorptive stripping voltammetry, does a redox reaction occur?
Note also that the Nelson 1998 paper measured chromium by graphite furnace atomic absorption spectroscopy. In natural samples, chromium can exist in the Cr(III) oxidation state or in various compounds in the Cr(VI) oxidation state. Graphite furnace AAS is an excellent choice for a total chromium measurement. As previously mentioned, one advantage of electrochemical over spectroscopic methods is the speciation capabilities. While adsorptive stripping methods have been developed to differentiate between Cr(III) and Cr(VI) species (Dominguez 2002), the data below will reflect a total chromium measurement. This analysis will keep this introductory material simpler and allow comparison with the graphite furnace AAS data in Nelson 1998.
In a typical experiment for chromium measurement a HMDE (Golimowski 1985) or mercury-based electrode (Bas 2006 and Wang 1997) is employed as the working electrode. A complexing agent is added to complex with chromium(III), forming a surface-active complex that adsorbs to the electrode surface. During this deposition step, the electrode is held at a potential sufficiently negative to reduce any Cr(VI) to Cr(III). The Cr(III) complex adsorbs to the electrode surface. Subsequently, during the stripping step, the electrode is cathodically scanned and the complex is reduced, giving rise to the analytical signal. The reduction is a one electron process:
$\ce{Cr(III)} + e^- \rightarrow \ce{Cr(II)}\nonumber$
Differential Pulse Adsorptive Stripping Voltametry measurements of chromium. The instrumental experimental conditions used are shown in the table below.
Table 6: Experimental conditions for the determination of chromium in sediments*
Ea Accumulation potential (mV/Ag, AgCl, KCl satd.) -1000
Ei Initial scan potential mV/Ag, AgCl, KCl satd.) -1000
Ef Final potential (mV/Ag, AgCl, KCl satd.) -1600
dE/dt Potential scan rate (mV/s) 10
f Pulse repetition (s) 0.1
ΔE Pulse amplitude (mV) 50
v Pulse duration (ms) 20
ts Sampling time (ms) 8
ta accumulation time (s) 180
tr Rest time (s) 15
t Purging time (s) 300
u Stirrer speed (r.p.m.) 600
*Note: conditions modified from Wang 1997 and Golimowski 1985.
Simulated sample data acquired for Lake Nakuru sediment samples are given in Table 7. Sample preparation was similar to that described by Locatelli 1999. Briefly, 1 g sediment samples were acid digested and diluted to 100 mL. This solution was diluted by a factor of 20 with appropriate electrolyte. The analysis was performed on this second solution. The single standard addition method of quantitation was used (section 5.3.3 in Analytical Chemistry 2.0). A 10.00 mL aliquot of sample solution was pipetted into the sample container and purged with nitrogen as described above. The sample underwent the adsorptive stripping procedure outlined above. Upon collection of this voltammogram, 100.0 µL of the standard solution (5.000 ppm) was added to the 10.00 mL sample and the adsorptive stripping procedure was performed on this “spiked” sample. The data for the peak current was recorded in tabular format and is shown below. Use this data to calculate the concentration of each heavy metal in the sediment samples in µg/g (as was reported in Nelson 1998). Don’t forget to account for any dilutions performed during the sample preparation. Hand-calculate the concentration for sample 1 then set up a spreadsheet to calculate the concentrations of the metals for all three samples.
Q16. Recently, sediment and suspended solid samples were collected from Lake Nakuru and analyzed by ASV in the manner described above. Given the sample data in Tables 7 and 8, calculate the current concentration levels of chromium in the sediment samples obtained from Lake Nakuru. What do the data tell you about chromium concentration in the sediments and suspended solids in Lake Nakuru? Has the concentration increased or decreased since the 1996 sampling?
Table 7: Sample Data for Chromium Determination in Sediment Samples in Lake Nakuru
Sample number
ip (µA)
Sample
Sample + 100 µL spike
Cr
Cr
1
0.1111
0.2522
2
0.1212
0.2855
3
0.1359
0.2989
Note: All samples analyzed were a 10.00 mL aliquot. The 100.0 µL standard spike contained a target concentration of 5.000 ppm of Cr(III). The exact concentration (depending on standard mass weighed out) can be determined on the corresponding Excel spreadsheet.
Table 8: Sample Data for Chromium Determination in Suspended Solid Samples in Lake Nakuru
Sample number
ip (µA)
Sample
Sample + 100 µL spike
Cr
Cr
1
0.1015
0.8986
2
0.1985
0.9563
3
0.1574
0.9254
Note: All samples analyzed were a 10.00 mL aliquot. The 100.0 µL standard spike contained a target concentration of 5.000 ppm of Cr(III). The exact concentration (depending on standard mass weighed out) can be determined on the corresponding Excel spreadsheet.
Q17. If your sample was known to contain metal ions that cannot be measured by ASV, what other techniques would you suggest trying? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Heavy_Metals/03_Metals_Analysis_by_Anodic_Stripping_Volt.txt |
Learning Outcomes
At the end of this assignment students will be able to:
• Understand and apply the underlying theory involved in major areas of atomic spectroscopy.
• Utilize their knowledge of atomic spectroscopy to choose between methods for specific analyses.
• Make use of spreadsheets and graphical analysis to solve problems involving standard addition in elemental analysis.
Purpose
This section will provide an introduction to atomic spectroscopy. The basic concepts of flame atomic absorption spectroscopy (FLAA), graphite furnace atomic absorption spectroscopy (GFAA), and inductively coupled plasma optical emission spectroscopy (ICP-OES) will be presented. Next, the application of each of these techniques to the determination of chromium, copper, lead, and zinc in Lake Nakuru sediment and suspended solid samples will be evaluated and compared to a previous study.
Introduction
Atomic spectroscopy comprises an important set of analytical techniques. The results of atomic spectroscopy experiments are quantitative measurements of elemental concentrations. Because the determination of levels of various metals is important in many clinical assays, products of the food and beverage industry, evaluation of environmental quality and impacts, and in a wide range of manufacturing concerns (from pharmaceuticals to steel), this method is widely applied. Atomic spectroscopy instruments can be divided into three basic types, depending on whether the phenomenon measured is based on light absorption, emission or fluorescence. Although atomic fluorescence can provide lower detection limits for some metal ions, these instruments are less commonly available therefore, our discussion in this unit will focus on analysis using atomic absorption and emission techniques.
Starting small. We all learned in our first course in chemistry that the most basic forms of matter are the elements, and that the smallest unit of an element is the atom. Elements display properties unique to themselves, including the number of electrons located in what are referred to as atomic orbitals. These atomic orbitals are designated by type and energy as 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.
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 you just wrote for the sodium atom describes the element in its lowest energy or ground state. Each of the atomic orbitals for an element is separated from the next by a discrete or quantized amount of energy, with the energies separating successively higher energy orbitals being different but also discrete. Only when the energy of radiation interacting with an atom exactly matches the gap between two energy levels of the atom will an absorption (or emission) event occur. Not surprisingly, these energy differences are unique to an individual element, and thus can be used to identify that element.
Atomic spectroscopy is based on the quantized changes in atomic energy following the absorption (or emission) of light. In absorption, the energy of the atom increases as an electron is promoted from an orbital of lower energy (the ground state) to one of higher energy (the excited state). In this process, light of the appropriate frequency, $\nu$, (or wavelength, $\lambda$) is absorbed according to Equation \ref{1}, where ΔE represents the difference in energy (in Joules) between the excited and ground states, h is Planck's constant (h = 6.626 × 10-34 J s), and c is the speed of light in a vacuum (c = 2.998 × 108 m/s).
$\mathrm{\Delta E = h\nu = \dfrac{hc}{\lambda}} \label{1}$
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?
Q3. Given that the energy difference between the ground state and the first excited electronic state (ΔE) for the sodium atom is 3.373 × 10-19 J, calculate the frequency, $\nu$, corresponding to a photon possessing this energy. Next, calculate the wavelength (in nm) for this photon.
Atomic emission is, in essence, the reverse of the absorption process. Atoms in the excited state will emit (or give off) light when the electron relaxes back to the lower energy ground state. Again, because this energy is quantized, the light emitted is of a specific frequency, indicated by Eq 1.
A graphical depiction of this process is provided in Figure 1.
Astronomers were among the first to recognize the potential of atomic spectroscopy measurements for elemental analysis. We normally see the light from stars (like our sun) as white light. This light appears white because it is comprised of a continuum of all wavelengths. However, astronomers realized if they carefully analyzed the wavelengths of light emitted by a star, there were missing frequencies, or lines, at very specific wavelengths that corresponded to absorption of light by elements in the atmosphere. (For a more detailed discussion, and example calculations see Harvey, Analytical Chemistry 2.0, Chapter 10, Section 10.1.1.)1
It is important to remember that in the analysis of Lake Nakuru samples, our measurements will not be made for single atoms, as illustrated in Figure 1, but rather will result from the net absorption or emission from the large numbers of a particular atom in our sample, called a population. Within this population, the fraction of the atoms that will be in the ground or the excited state depends on the sample temperature as defined by the Boltzmann equation, Equation \ref{2}, where Nground is the population of atoms in the ground state, Nexcited is the population of atoms in the excited state, ΔE is the difference in energy between the two states (Eexcited - Eground), T is the temperature in Kelvin and k is the Boltzmann constant (1.381 × 10-23 J/K).
$\mathrm{\dfrac{N_{excited}}{N_{ground}} = e ^{- \Delta E/kT}} \label{2}$
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?
Q5. A common wavelength used for measuring lead emission is 220.353 nm.
1. Calculate ΔE (Joules) for this transition.
2. Calculate the ratio of atoms in the excited state to atoms in the ground state for this value of ΔE at room temperature (298K).
3. Repeat the calculation at 6000K, a routine operating temperature for an ICP torch.
4. Can you hypothesize as to why atomic emission measurements are generally made at high temperatures?
A Short Aside: Atomic vs. Molecular Spectroscopy
In order to better understand atomic spectroscopy, you will want to know how it differs from molecular spectroscopy, a technique you may be more familiar with. Start by considering the differences between atoms and molecules.
Q6. What are molecules?
Simply put, molecules are the combination of atoms that result when bonds are formed between the atoms. Long ago, you learned that when atoms approach each other and their valence electron densities begin to overlap (bonding), atomic orbitals can be combined to form molecular orbitals (remember, for example, s + p = sp). Without going into detail here, since our focus is on atomic spectroscopy, the formation of bonds introduces many, many more possible electron energy levels that can be addressed by the absorption or emission of radiation. In addition to the electronic ground and excited states possible for atoms, molecules possess both vibrational and rotational levels within each of the electronic states, with the energy separations between these new levels much lower than those between the electronic ones.
Since the atomic orbitals are discrete and few in number, atomic spectra consist of very narrow absorption bands at single wavelengths corresponding to the various ΔE values between the orbitals. Molecular spectra are much more complicated than atomic spectra as a consequence of the increased number of accessible energy levels for molecules. Broad absorption bands composed of many, many individual transitions centered on the most favorable absorption wavelength characterize these spectra. This difference is illustrated in Figure 2, below.
Bottom) Illustration of (left) energy levels for a molecule, with So representing a singlet ground state for the molecule and S1, S2 representing singlet excited state energy levels, and (right) shape of a typical molecular absorption spectrum.3
The narrow absorption lines for atomic spectroscopy provide its major advantage over molecular spectroscopy. There is rarely any overlap between the spectra of different elements in a complex sample, and with the appropriate detection system, as many as 80 elements can be determined in the same sample rapidly in sequence, or in many cases, simultaneously.
Techniques for Atomic Spectroscopy
A common goal for all techniques in atomic spectroscopy (AS) is the isolation and quantification of unbound, gas phase atoms from the sample matrix regardless of how the atoms were combined in the sample. The block diagram in Figure 3 lists the general components needed to perform atomic spectroscopy.
Typically, a liquid sample is introduced into the atomizer through a nebulizer (FLAA, ICP) which breaks the liquid into small particles that can easily be evaporated in either a flame or a plasma. In GFAA, microliter volumes of liquid sample are introduced directly into a small graphite tube (approximately 0.7 x 3 cm) using a micropipettor or autosampler. Atomization is accomplished in the high temperature of the flame (2300 – 3400 K) or the plasma (6000 – 8000 K), or by electrothermal heating in the graphite tube (1500 – 3000 K).
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?
1. Flame Atomic Absorption Spectroscopy (FLAA) In FLAA, atoms of interest are isolated, or atomized, in a flame. A variety of fuels and oxidants can be used to produce the flame, with the choice dependent upon the desired temperature, which in turn is dependent upon the atom under consideration. Routine analysis can be accomplished with an acetylene/air flame which has a temperature of approximately 2500 K. A “slot” burner is common, which allows for an optical path length across the flame of up to 10 cm. An acetylene/air flame is shown in Figure 4.
The burner assembly for FLAA consists of fuel and oxidant lines (at bottom left in Figure 4) which introduce the gases into the nebulizer where they are mixed and fed into the slot burner head, where they are ignited. The updraft caused by the flame pulls the liquid sample through a small “sipper” (at center of black nebulizer) into the nebulizer where large droplets are eliminated, and the smaller liquid droplets swept into the flame. The flame serves to strip away the solvent and causes atomization of the sample. On either side of the flame are quartz windows that allow the excitation source (at left) to pass through the flame and into the radiation detector (at right). The large clear tube at bottom center is the waste collection line for excess sample.
The most common radiation source for FLAA is the hollow cathode lamp (HCL) which incorporates two electrodes inside an evacuated tube containing an inert gas at 1 – 5 torr. The cathode is made from the element of interest, or coated with that element and is maintained at a potential of about 500 V negative of the anode. This applied potential is sufficient to ionize the fill gas molecules (Ar Ar+ + e-, for example), with the ions being accelerated towards the negatively charged cathode. The inert gas cations strike the cathode with enough force to “sputter” excited state atoms of the element of interest into the gas phase. These excited state atoms emit radiation of characteristic wavelengths that are used to excite gas phase atoms in the flame, with the amount of radiation absorbed by the sample being proportional to the concentration of the atom in the sample. In FLAA, a different HCL is most often required for each element you wish to analyze. A typical lamp is shown in Figure 5.
Single-element lamps limit FLAA to sequential analysis if more than one element is to be determined in a sample. The advantage of narrow, specific absorption bands, however, allows for simple detection systems to be employed. Typically, a minimal monochromator and photomultiplier tube detector suffice for routine analyses.
Q8. Speculate as to possible sources of background interference when using FLAA. How might they be eliminated or reduced?
In FLAA, the primary interference is from the continuum emission from the flame itself. It is possible to distinguish the signal originating in the flame from that of the atomic absorption by electronically pulsing the source lamp (HCL) on and off, and subtracting the signal from the flame from that produced by the combination of the flame + the lamp. A mechanical chopper can also be used for this purpose. Alternatively, background absorption in the flame can be minimized by use of a continuum source (usually a D2 lamp) alternated with the HCL.4 The element being determined effectively absorbs light from the primary source only, while background absorption occurs equally with both beams. The difference between the two signals gives the true atomic absorption signal.
You were asked in Q7 to consider what processes might occur in addition to atomization as the temperature of the atomizer increased. One possibility is that enough energy is transferred to the atoms that they are promoted to an excited energy state. This occurs in the flame primarily for the alkali and alkaline earth elements, whose excitation energies are the lowest. Consequently, these elements can be determined using flame atomic emission spectroscopy (FLAE), where the radiation emitted from the excited state atoms is measured. In a previous course you may have performed flame tests in which you visually observed the color of light emitted when solutions containing metals such as lithium, sodium, potassium or calcium were introduced into a flame using an inert wire loop. You will see in a later section that excitation and ionization occur at the very high temperatures of the ICP, making that atomizer most effective for atomic emission spectroscopy, and for ion detection by mass spectrometry. The development of ICP-based spectroscopy has allowed it to mostly replace FLAE for routine analysis, except in cases where cost is a factor or the increased sensitivity of the ICP is unnecessary. Figure 6 shows the red emission of atomic lithium in an acetylene/air flame.
2. Graphite Furnace Atomic Absorption Spectroscopy (GFAA) The graphite furnace is generally a cylindrical graphite tube placed in the optical path of the spectrophotometer, with rapid atomization accomplished by the application of high electrical potential at two contact points. Two types of tubes are in common use: a) the transversely heated graphite tube in which the electrodes are applied perpendicular to the light path, allowing for even heating along the entire length of the tube, and b) the longitudinally heated tube where electrode contacts are at the ends of the tube, frequently resulting in uneven heating. Both types of tube are shown in Figure 7.
The graphite tube is commonly fitted with what is called a L’vov (after its inventor) platform, a curved surface attached at a single point to the outside walls of the tube which allows for the sample to be heated radiatively as the temperature is increased. This results in more even heating of the sample and better reproducibility of response. The platform can be seen in Figure 8.
Aliquots of liquid sample (5 – 50 μL) are introduced through a small hole located in the center of the tube. The technique is also amenable under certain conditions to solid samples. The GFAA assembly is mounted within an enclosed, water-cooled housing so that temperature can be quickly lowered between runs. Inert gas, typically argon, is used to protect the tube from oxidation at high temperatures, and to purge oxygen from the tube to prevent the formation of metal oxides during the atomization step. One type of furnace arrangement is shown in Figure 9.
The furnace typically allows 2 – 3 orders of magnitude lower limits of detection than the flame, and requires significantly less sample.
Q9. How does the design of the graphite furnace allow for such improved sensitivity for metals over flame AA?
Normally, a GFAA measurement is made in five steps. The first of these is the drying step, in which the temperature of the furnace is raised slowly to a temperature slightly above the boiling point of the solvent, and held there for 30 – 60 seconds. Some methods specify that this step be repeated at a temperature slightly higher than the first drying temperature to assure complete solvent removal and to prevent “bumping” at higher temperatures. The second step is referred to as the charring step, in which the organic compounds and other low boiling substances in the sample are thermally decomposed. The charring, or pyrolysis temperature is typically between 600 and 1000 oC. Argon is allowed to flow through the tube during these first two steps to remove smoke and other vapors produced in drying and pyrolysis. The third step, atomization, is dependent somewhat upon the element of interest, but is routinely between 1500 and 2000 oC. The argon flow is stopped during the 5 – 10 second measurement window, trapping the atomized sample in the small volume of the graphite tube where its absorbance is integrated over time. After the sample has been atomized, the temperature is raised quickly to a value several hundred degrees higher than the atomization temperature and held there for 10 – 20 seconds with a high rate of argon flow. This clean out step assures that any sample residue remaining in the tube is removed. The last step is the cool down step in which the furnace is allowed to return to ambient temperature (with water cooling) before the next sample is introduced.
The GFAA signal observed for Pb at a concentration of 5 μg/L is shown in Figure 10. In contrast to FLAA, where the observed absorbance signal for a sample is constant as long as the sample is being aspirated, the observed signal in GFAA is transient as a result of atomization of the entire amount of analyte during the measurement step. The signal increases during atomization, and then decreases as the vaporized atoms diffuse out of the furnace. Generally, the integrated peak area (AU x s) is used in quantification of the element of interest.
Most GFAA methods include the addition of a matrix modifier to the sample volume introduced into the furnace. Matrix modifiers reduce the loss of analyte during charring by enhancing the volatility of the matrix so that it is removed at a lower temperature or by making the analyte less volatile so that a higher atomization temperature can be employed. Commonly used matrix modifiers include palladium metal and Mg(NO)2. Dilute solutions of these modifiers, on the order of 0.1% m/v, are added in small amounts to the sample being analyzed. Added masses of modifiers range from 3 – 50 μg in the analyzed sample volume, depending upon the analyte.
Generally, the same detection system can be used for GFAA as for FLAA. Similarly, most GFAA systems will employ the same D2 continuum correction. In addition, many GFAA systems include what is known as a Zeeman correction system that allows background correction at higher absorbance values than D2 correction, and for matrices that are more spectrally complex. Zeeman correction is also applicable at all wavelengths.6 Figure 11 shows the instrumental arrangement for one type of Zeeman correction.
The basis for Zeeman correction is the observation that the absorption (or emission) bands observed for atoms can be split into multiple lines in the presence of an applied magnetic field. In the simplest example, a single absorption band can be split into two components observed at slightly lower and slightly higher wavelengths than the original. By pulsing the magnetic field during the atomization step, the signal observed at the original wavelength when the magnet is on (background only) can be subtracted from the signal observed when the magnet is off (background + analyte) to give the corrected signal.
3. Inductively Coupled Plasma – Optical Emission Spectroscopy (ICP-OES) The atomization source for this technique, known as an inductively coupled plasma, operates at significantly higher temperatures than do the flame and the furnace. As a result, many of the atoms are excited into higher energy states, making this technique more amenable to the measurement of emission rather than absorption. Since the plasma can serve as both the atomizer and the excitation source, a separate source like the HCL is unnecessary in this technique.
As we will see, the sensitivity and dynamic range of this technique are far better than either of the two we have studied previously. Except for its much higher initial cost and an increased cost of routine operation, this method satisfies nearly every critical figure of merit for atomic spectroscopy.
Plasma is a distinct phase of matter, composed of highly ionized gas containing high concentrations of ions and free electrons. The plasma for ICP is initiated by ionizing a flowing stream of argon by the injection of free electrons with a Tesla coil. The high energy of the plasma is maintained by inducing the charged particles to rotate within a fluctuating magnetic field generated in an RF (radio frequency) coil. Resistance to this motion by unionized argon generates considerable resistive heating resulting in temperatures of 6000 - 10,000 K. A schematic of an ICP torch is shown in Figure 12.
It is composed of concentric tubes, generally quartz, allowing three distinct paths of argon flow. The inner gas flow, through the capillary injection tube, carries the sample aerosol from the nebulizer into the plasma. An intermediate gas flow, through the argon plasma gas inlet, serves to keep the plasma localized above the injection tube and away from the intermediate flow tube. This flow also reduces the buildup of carbon at the injector tip when organic samples are introduced.8 The outer flow, labeled Ar tangential flow in Figure 12, serves to keep the quartz walls of the torch cool. Because of the small spacing between the intermediate and outer flow tubes, argon follows a tangential path through the tube at high velocity.
As was mentioned before, the energy of the plasma is sufficient to populate many excited states of resident atoms at the same time, making ICP-OES an ideal method for multielement analysis. For many elements significant numbers of excited state ions also result. This generally allows for multiple spectral lines to be available for each element, with many elements emitting most strongly from excited ionic states. Quantitative information is obtained most commonly from comparison of emission intensity of unknown to calibration plots of emission intensity as a function of concentration. Linear calibration plots for ICP-OES are linear over 4 – 6 orders of magnitude, superior to both FLAA and GFAA.
Figure 13 shows the ICP torch assembly mounted inside an ICP-OES instrument. This arrangement has the torch installed horizontally, with the nebulizer flow entering at the left. Two of the argon flow lines are visible at bottom left, with the copper induction coil wrapped around center of the glass torch.
The instrument illustrated in Figure 13 allows two different optical paths by which emitted radiation can be measured. At right, the cylindrical window allows introduction of radiation to the detector from an axial direction, that is, head-on to the top of the ICP torch. An alternative pathway allows radiation to be measured radially to the torch, with the optical window leading to the detector located at a right angle to the torch, center bottom of figure. This design allows for a wider dynamic analytical range, with the axial arrangement giving better linearity of response at low concentrations of analyte, resulting from the increased path length through the torch from which the emission is being measured. The radial path allows for the best linearity of response at high concentrations of analyte.
The argon ICP plasma torch displays a very intense continuum emission that appears white in color. A photograph of the torch in operation is found in Figure 14.
Detection systems in ICP-OES most commonly employ a diffraction grating that allows radiation emitted from the elements in the plasma to be directed onto either multiple PMTs or high sensitivity solid state detectors for simultaneous intensity measurement at multiple wavelengths. Included in the latter type are the photodiode array detector (PDA), the charge injection device (CID), and the charge coupled device (CCD). A complete discussion of detector systems for emission spectroscopy can be found in reference 9.9
For ICP-OES, 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.
Summary
An excellent guide to the strengths and limitations of FLAA, GFAA, and ICP-OES (as well as additional techniques not covered in this module) can be found in Reference 13. A general decision matrix for choice of method reprinted from that source is given below as Table 1.13 A comparison of several important instrumental parameters for FLAA, GFAA, and ICP-OES are found in Table 2. Lastly, specific information on the achievable limits of detection (LOD) and limits of linearity (LOL) for the four metals being considered here are included for the three techniques in Table 3.
Table 2. Important Instrumental Parameters for FLAA, GFAA, and ICP-OES13
FLAA GFAA ICP-OES
Detection Limit Ranges 0.9 – 100 ppb 0.005 – 2 ppb 0.02 – 100 ppb
Typical System Costs $15 – 25K$30 – 50K \$60 – 100K
Analytical Working Range*
(orders of magnitude)
4 2.5 10
*concentration range over which quantitative results can be obtained without recalibration
Table 3. Comparison of Detection Limits (LOD) and Limit of Linearity (LOL) for FLAA, GFAA, and ICP-OES (all values are in μg/L, ppb)
FLAA GFAA ICP-OES
Metal LOD13 LOL* LOD13 LOL* LOD13 LOL15
Cr 1.5 104 0.014 4.4 0.4 105
Cu 3.0 104 0.004 1 0.2 105
Pb 15 105 0.05 16 1 105
Zn 1.5 104 0.02 6 0.2 105
* Limits of linearity for AA and GFAA are estimated from analytical working range estimates in Reference 13.
Data and Analysis
In the following section, you will evaluate the data reported in Nelson, et al.10 for the analysis by atomic spectroscopy and XRF of Cr, Cu, Pb, and Zn in suspended solids and dried sediments from Lake Nakuru. Next, you will be supplied with simulated data from sediments and suspended solids so that you can compare those levels of the four target metals to those obtained in 1996. As a starting point, let us look again at the results from that previous investigation.
The procedure used by Nelson to prepare samples of Lake Nakuru sediments and suspended solids for analysis by atomic spectroscopy can be summarized as follows:
5 g samples of filtered and dried solids were digested for 1 hour in 50 mL of high purity HNO3. The digested samples were filtered through 0.45 μm PVDF membranes and the filtrate diluted to a final volume of 50.00 mL in a volumetric flask with purified water.
Final concentrations for each of the four metals found in sediments and suspended solids by Nelson are given in Table 4 on a dry weight basis (μg/g). The results for Lake Nakuru suspended solids were obtained using XRF, a technique you will encounter in a separate section of this module.
Table 4. Metal Concentrations in Lake Nakuru Sediments and Suspended Solids found in Nelson Study10
Concentration in dry sediments (µg/g) Concentration in suspended solids (µg/g)
Trace metal Average (11 sites) Range Average Standard deviation
Chromium 67 10 – 280 8.3 0.82
Copper 24 5 – 95 19 2.9
Lead 22 4 – 100 11.7 0.9
Zinc 147 44 – 630 74 31
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.
The concentrations you calculated in Q10 for each of the metals in sediment and suspended solids may suggest to you which method you learned about earlier in this module would be most appropriate for each of the metals if further study were performed.
Q11. If no further information were given, which technique would be your first choice in measuring the concentration of the four metals in both of the matrices if you were to perform a follow-up study to that done by Nelson? Explain your choice(s).
Returning to the Nelson paper, if you dig further into the procedure for suspended sediment sample preparation beyond the initial 50 mL dilution you will find the following:
…samples were diluted with distilled-deionized water (10 to 50x) to reduce observed matrix effects, and the method of standard additions was used to account for any remaining matrix effects.
This additional dilution was required in the analysis for each of the metals except for Zn, for which no further dilution was required to reduce observed matrix effects.
Q12. In general, 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?
Q13. Which of the three methods you have studied is best able to overcome matrix effects? Why is this so?
Q14. If the original 50 mL digestion solution, for which you calculated metal concentrations previously, were further diluted by a factor of 50x to reduce matrix effects, would this alter your answer(s) from Q11 as to choice of method for Cr, Cu, and Pb?
Simulated Analysis of Present-day Lake Nakuru Metals
As a follow-up to the study conducted by Nelson, et al. on samples collected from Lake Nakuru in 1996, you will now have a chance to evaluate simulated data for Cr, Cu, Pb, and Zn levels in sediments and suspended solids. You may have come to the conclusion in the previous section that the best method available to evaluate multiple metal species in complex matrices is that of ICP-OES. EPA Method 200.7 (Revision 4.4) Determination of Metals and Trace Elements in Water and Wastes by Inductively Coupled Plasma-Atomic Emission Spectrometry11 will be the method of choice for this analysis. It can be obtained free online here.
The method is applicable to 32 metals, including the four of interest, in various matrices. For this study, it has been applied to samples of sediment and filtered suspended solids that were dried to constant weight at 60 oC. In our analysis, 1 gram samples of solid were massed exactly and digested in nitric and hydrochloric acids, then taken up into a final volume of 100 mL. The one exception was that for analysis of lead in suspended solids – in that case, a 3 gram sample was required to provide an appropriate instrument response.
Table 1 of Method 200.7 gives the optimal wavelength for each of the metals, along with an estimated method detection limit (LOD) and an upper concentration limit to which calibration plots are typically linear (LOL). Normally, the lowest concentration that is reliably quantitated (LOQ) is taken as 10x the method detection limit.
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?
Because matrix effects were an issue in the 1998 report, we have chosen to use the method of standard additions in the measurement of metal concentrations by ICP-OES. Background information on standard addition methods can be found in Harvey, Chapter 5.12
Sediment samples. Four 1.0000 g samples of dried sediment were accurately massed into separate digestion vessels, and treated according to Method 200.7. Following the extraction, the liquid extracts were quantitatively transferred to separate 100 mL volumetric flasks. Standard additions were then made for each of the four metals so that the four flasks had final added standard concentrations of 0.00, 0.10xLOL, 0.20xLOL, and 0.50xLOL (all in ppm). For example, zinc was added in sufficient quantities to give final added concentrations of 0.00, 0.50, 1.00 and 2.50 ppm following dilution to 100 mL with reagent water. 1000 ppm standards for each metal (ICP grade) can be obtained commercially or prepared according to instructions in Section 7.0 of Method 200.7.11
Suspended solid samples. Analysis of Cr, Cu, and Zn: four 1.0000 g samples of dried suspended solids were treated in the same way as for sediment (previous section). Standard solutions were added at the same levels as before in each of the four volumetric flasks prior to dilution with reagent water. Analysis of Pb: four 3.0000 g samples of dried suspended solids were digested as before, transferred to separate 100 mL volumetric flasks, appropriate amounts of Pb standard added to each flask, and dilution to final volume accomplished with reagent water.
All samples were then analyzed sequentially using the ICP-OES conditions specified in Sections 10.1 and 10.2 of Method 200.7.11 A spreadsheet file will be supplied to you that contains tables of emission intensity, corrected for background, as a function of added standard concentration of each element.
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 1998? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Heavy_Metals/04_Metals_Analysis_by_Atomic_Spectroscopy.txt |
Could toxic pesticides like DDT be responsible for the deaths of large numbers of lesser flamingos at Lake Nakuru, Kenya? While many organochlorine pesticides including DDT have been banned for decades in the US due to their adverse effects on bird populations, especially bald eagles, they are still used for mosquito control in tropical regions of Africa where malaria is epidemic. In addition, East Africa has become an international dumping ground for stockpiles of obsolete pesticides. In this section we explore the possible role of organochlorine pesticides in the flamingo deaths and examine the use of gas chromatography – mass spectrometry (GC-MS) to separate, detect and quantify pesticides in Lake Nakuru water samples.
• Instructor’s Guide:
• ** For instructor’s guide in a PDF or Word document, please contact Cindy Larive. ([email protected])
• EPA Method
Contributors and Attributions
• Dr. Heather A. Bullen (University of Northern Kentucky)
• Dr. Alanah Fitch (Loyola University – Chicago): [email protected]
• Dr. Richard S. Kelly (East Stroudsburg University): [email protected]
• Dr. Cynthia K. Larive (University of California – Riverside): [email protected]
• Content from ASDL.
Pesticides
Learning Outcomes
At the end of this assignment students will be able to:
• Define various sampling strategies.
• Assess the benefits and limitations of different sampling strategies.
• Determine an appropriate sampling plan for an analysis.
An important aspect of designing an experiment is collecting a representative sample. What is a representative sample? That is often a tough question to answer. It is important to ensure samples are representative for the analyte of interest. If sampling is not done correctly it could be the “weak link” in an analysis, leading to inaccurate results. Often, sampling can be a source of error that is overlooked.
Purpose
The purpose of this assignment is to address questions related to designing a sampling plan.
Designing a Sampling Plan
The key questions which must be considered are:
1. From where within the target population should we collect samples?
2. What type of samples should we collect?
3. What is the minimum amount of sample for each analysis?
4. How many samples should we analyze?
5. How can we minimize the overall variance for the analysis?
Map of Lake Nakuru (www.jgmccue.com)
Where to Sample?
How do you obtain a representative sample? One approach is to take a grab sample. Consider a solid sample divided out as a grid, below. How do you choose where to grab from? One approach is to choose randomly.
Q1. For any given analysis you would have multiple samples for validation of results. Assume you would like eight grab samples of the solid to be evaluated. Pick eight random grab samples from the solid sample divided as a grid below. How do you ensure your sampling of the solid is random?
Now take a look at the following solid samples divided into grids with the location of the analyte of interest (within the solid sample) identified (colored squares).
A
B
Q2. Would you consider the samples above to be heterogeneous or homogeneous?
Q3. Did your random sampling (Question 1) affect the potential accuracy or precision of your measurement of the analyte for the samples in grid A or grid B? If so how?
Although we would like to assume that a sample is homogenous it is often not. When trying to sample a heterogeneous sample, you need to consider both distributional and constitutional heterogeneity.
• Distribution heterogeneity is caused by segregation of the sample (i.g. settling).
• Constitutional heterogeneity is a fundamental property of a material and is caused by differences in particle size and/or composition.
Q4. Each of the previous grids is an example of one of these cases. Can you identify which sample is which?
Q5. How does distribution heterogeneity affect accuracy and precision?
Q6. How does constitutional heterogeneity affect accuracy and precision?
Q7. Do you see a scenario where distribution heterogeneity could be magnified by mixing and/or sampling?
If samples are mixed well, distributional heterogeneity is insignificant and only constitutional heterogeneity is present. It is important to note, however, that the analyte’s spatial or temporal distribution might also be of importance in the analysis and mixing may not always be the best choice. For example, maybe you are interested in the amount of heavy metals in sediment samples, but are also interested in how far into the sediment the metals have leached. This would be a case where mixing would not be appropriate.
Types of Sampling Plans:
What we have highlighted above is an example of random sampling, which is often applied to a grid design as shown above. Random sampling strategies can be applied to any target population (i.e. evaluating a solid sample or determining the presence of pesticides in the water of Lake Nakuru). However, random sampling can at times be expensive and not necessarily cost effective, as you often need a greater number of samples to ensure that your samples represent the target population. If you know something about the target population, other sampling methods may be possible or appropriate. Here are a few examples.
• Selective (Judgmental) Sampling - this is at the opposite extreme of random sampling, and is done if you have prior information about the target. For example, if you wanted to evaluate the metal content in pennies you may not select coins that are corroded or choose coins from a specific mint date.
• Systematic Sampling - Sampling the target population at regular intervals in space or time. This is often considered to fall between the extremes of random and selective sampling.
• Stratified Sampling – The population may be divided into sub populations (groups) that are distinctly different (this might be size of sample, type of sample, depth of sample). Then, the overall sampling within the groups is randomly conducted and the samples are pooled.
• Cluster sampling- is a sampling technique where the population is divided into groups or clusters and random samples are selected from the cluster for analysis. The main objective of cluster sampling is to reduce costs by increasing sampling efficiency.
Q8. What is the advantage of implementing a judgmental sampling scheme over random sampling if one knows the point source for the discharge an analyte into a system?
Q9. Assume you have a chosen a selective sampling plan to evaluate pollution from a point source into a lake. Use the diagram below and words to describe your sampling plan.
Q10. Use a grid design (as we have previously done) to show how you would conduct systematic sampling of the pollutant. Is there an advantage to what you might learn using this sampling method? What are the disadvantage(s)?
Q11. Describe how stratified sampling might be applied to evaluate the pollutant in the lake? In general, what is the advantage of stratified sampling over cluster sampling?
Q12. What is a general rule with regard to sampling times or locations to increase the likelihood that samples will be representative?
What Type of Sample to Collect?
When implementing a sample plan often grab samples are used. In some cases composite sampling (combining a set of grab samples into one sample) is more useful. This strategy may be used if:
• There is interest in the target population’s average composition over time and space.
• A single sample does not supply sufficient material for analysis.
Q13. What is the main disadvantage of grab and composite samples?
Q14. Can you think of any control studies you might want to do, when compositing samples?
Q15. Does the EPA Method 525.2 suggest a particular sampling method?
Minimizing Variance - What is the Minimum Amount of Sample? How Many Samples are Needed?
When sampling you want to make sure the sample is not too small, so that the composition is not substantially different from the target population. You also want to ensure you collect an appropriate number of samples for an analysis.
The variance of the (1) analysis method and (2) sampling approach both contribute to the variance in a given analysis. Sampling variance can be improved by collecting more samples of the proper size. Increasing the number of times a sample is analyzed improves the method’s variance.
Q16. What sample size does EPA Method 525.2 suggest for analysis of pesticides in water? Why?
Q17. An analytical method has a percent relative sampling variance of 0.10% and a percent relative method variance of 0.20%. The cost of collecting a sample is \$20 and the cost of analyzing a sample is \$50. Propose a sampling strategy that provides a maximum percent relative error of ±0.50% (a = 0.05) and a maximum cost of \$700.
For further discussion on sampling amounts see section 7.2 in Harvey, D. Chapter 7, Collecting and Preparing Samples.
Design Your Sampling Plan
Of course we would like to collect as many samples as possible to minimize our variance. However, in every lab there are limitations such as those above, related to cost per-sample collected and cost-per analysis.
Q18. The autosampler on the GC-MS you will be using for pesticide analysis has 200 vial locations. How will you choose your representative samples? Below is a picture of Lake Nakuru. Design your sampling plan. Think about random, systematic, clustering, etc. sample strategies. Will you take grab samples or pool samples together?
http://www.enchanted-landscapes.com/...akuru_park.gif
http://www.ilec.or.jp/database/afr/img/afr07-01.gif
Helpful Resources
Harvey, D. Chapter 7, Collecting and Preparing Samples. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Pesticides/01_Sampling.txt |
Learning Outcomes
At the end of this assignment students will be able to:
• Define components and operation of common sample preparation methods.
• Assess the benefits and limitations of different extraction methods.
• Validate extraction efficiency.
• Determine an appropriate extraction and preconcentration method for an analysis.
In the last module, you examined a process for collecting samples from Lake Nakuru for analysis. In many analytical measurements, the samples cannot be analyzed directly without some form of pretreatment. That will also be the case with the samples from Lake Nakuru. The aqueous samples you have collected cannot be injected directly into a gas chromatograph-mass spectrometer (GC-MS), and must go through a pretreatment process prior to analysis.
Purpose
The purpose of this assignment is to address questions related to the sample preparation portion of EPA method 525-2 for measurement of organochlorine pesticides by GC-MS.
Why do samples often need pre-treatment before analysis?
There are three key issues that often arise that necessitate pre-treatment of a sample prior to analysis: (1) the sample is in the wrong physical state for the analysis method (e.g., the method requires a liquid but you have a solid sample), (2) the sample has interfering matrix components that may give either a false positive or negative reading in the measurement, and (3) the sample has too low an analyte concentration to be detected by the instrument.
Consider the aqueous samples from Lake Nakuru, and discuss the following questions.
Q1. Is the sample in the wrong physical state for the analysis method?
Q2. Does the sample have interfering matrix components that may give either a false positive or negative reading in the measurement?
Q3. Does the sample have too low an analyte concentration to be detected?
With the answer to the above questions, it is apparent that the water samples from Lake Nakuru will need to be pretreated prior to the GC-MS analysis.
Q4. Can you think of a procedure to remove the organochlorine pesticides from water?
In other laboratories, for example in organic chemistry lab, you may have performed a liquid-liquid extraction to isolate a product or remove a contaminant. It is possible to use a liquid-liquid extraction with a water-immiscible organic solvent to remove organochlorine pesticides from water. A more common form of sample pretreatment used today involves a liquid-solid extraction. In this procedure, the liquid phase (in our case, the water sample) is passed through a cartridge or disk that is packed with a solid material. The goal is to have the organochlorine pesticides adsorb to the surface of the solid while the water passes through. You are probably familiar with water purification filters that can be attached to a tap to provide better drinking water. These operate under the same principle as they are solid extraction cartridges that are designed remove components of the water that are not desirable to drink.
Take a look at the following section (7.2) from the EPA procedure:
“Liquid-Solid Extraction (LSE) Cartridges -- Cartridges are inert non-leaching plastic, for example polypropylene, or glass, and must not contain plasticizers, such as phthalate esters or adipates, that leach into the ethyl acetate and methylene chloride eluant. The cartridges are packed with about 1 g of silica, or other inert inorganic support, whose surface is modified by chemically bonded octadecyl (C18) groups. The packing must have a narrow size distribution and must not leach organic compounds into the eluting solvent. One liter of water should pass through the cartridge in about two hours with the assistance of a slight vacuum... The extraction disks contain octadecyl bonded silica uniformly enmeshed in an inert matrix. The disks used to generate the data in this method were 47 mm in diameter and 0.5 mm in thickness. Other disk sizes are acceptable and larger disks may be used for special problems or when sample compositing is carried out. As with cartridges, the disks should not contain any organic compounds, either from the matrix or the bonded silica, which will leach into the ethyl acetate and methylene chloride eluant. One L of reagent water should pass through a disc in five to 20 min using a slight vacuum.”
The procedure says that “one liter of water should pass through the cartridge in about two hours with the assistance of a slight vacuum.”
Q5. Why is this procedure done over two hours instead of 10 minutes?
The procedure also specifies the use of an octadecyl cartridge.
Q6. Why is octadecyl chosen as an extraction stationary phase for the study of pesticides? What would happen if bare silica was used as an extraction phase?
You now have the pesticides adsorbed onto the solid phase cartridge, but they are still not in a form suitable for analysis by GC-MS. The elution step involves the use of a liquid solvent to desorb the organochlorine pesticides from the solid phase.
Q7. Discuss the following aspects of the solvent you would choose to elute the organochlorine pesticides.
• Polarity
• Boiling point
• Volume
The elution step you have devised in response to this last question will likely result in an organic extract that contains some water. It is not desirable to have water in the samples that will be injected into the GC-MS instrument because water can damage the column.
Q8. How would you remove any residual water in the organic phase?
For analytes present at low concentration, the volume of the organic extract obtained in the elution step is still too large and the sample too dilute for analysis.
Q9. How would you reduce the volume of the organic extract that contains your pesticide?
Q10. What concerns might you have in the solvent reduction step?
Below is the section of the EPA method about the concentration of the extract. The process of desorption of the organochlorine pesticides from the solid cartridge into an appropriate solvent at an appropriate volume is referred to as sample reconstitution. Sometimes the organic compounds are removed with a solvent that is especially suitable for the desorption step, but this solvent is then evaporated and the compounds taken up in another solvent more suitable for the analysis.
For example, see Section 11.2.9. “While gently heating the extract in a water bath or a heating block, concentrate to between 0.5 mL and 1 mL under a gentle stream of nitrogen. Do not concentrate the extract to less than 0.5 mL, since this will result in losses of analytes. Make any volume adjustments with ethyl acetate. It is recommended that an aliquot of the recovery standard be added to the concentrated extract to check the recovery of the internal standards (see Section 7.12).”
Usually, before doing such a sample pretreatment, the analyst has done calculations to ensure that the final solutions will have sufficient concentrations of the analyte to be quantified by the instrument. Let’s assume that the lake water has an average concentration of DDT of 150 ng/L (ppt). The limit of detection (LOD) for the GC-MS procedure is 0.083 μg/L (ppb).
Q11. What would be the final extraction concentration be if you plan to use EPA Method 525.2? Can you detect this level in the lake water samples?
Q12. What does it mean to determine the precision of the pesticide analysis? What does it mean to determine the percent recovery of the pesticide?
Q13. Suppose a solid-phase extraction procedure provided a low percent recovery. What could you change in an attempt to raise this value?
The following table is real data obtained from the EPA method.
Q14. Is the solid used in the cartridge effective for removing the pesticides from water?
Q15. How is it possible to get a percent recovery of over 100?
In addition to reconstitution, additional steps are often needed to preserve or prepare samples for analysis. The EPA Method 525.5 specifies that the water samples acquired should be specifically treated to preserve the pesticides:
“8.2 Sample Dechlorination and Preservation -- All samples should be iced or refrigerated at 4 °C and kept in the dark from the time of collection until extraction. Residual chlorine should be reduced at the sampling site by addition of 40-50 mg of sodium sulfite (this may be added as a solid with stirring or shaking until dissolved) to each water sample. It is very important that the sample be dechlorinated prior to adding acid to lower the pH of the sample. Adding sodium sulfite and HCl to the sample bottles prior to shipping to the sampling site is not permitted. Hydrochloric acid should be used at the sampling site to retard the microbiological degradation of some analytes in water. The sample pH is adjusted to <2 with 6 N hydrochloric acid. This is the same pH used in the extraction, and is required to support the recovery of acidic compounds like pentachlorophenol.
8.3 Holding Time -- Results of the time/storage study of all method analytes showed that all but six compounds are stable for 14 days in water samples when the samples are dechlorinated, preserved, and stored as described in Section 8.2. Therefore, samples must be extracted within 14 days. If the following analytes are to be determined, the samples cannot be held for 14 days but must be extracted immediately after collection and preservation: carboxin, diazinon, disulfoton, disulfoton sulfoxide, fenamiphos, and terbufos. Sample extracts may be stored at 4 °C for up to 30 days after sample extraction.”
Q16. From the chemistry described in sections 8.2 and 8.3 of the EPA method, what might be the cause of pesticide degradation before extraction?
Standards, Spikes and Surrogates
Note: A discussion of standards, spikes and surrogates will also be covered in the method validation section. This section will highlight their roles in sample preparation.
Take a look at the following excerpt from the EPA Method.
“7.8 Fortification Solution of Internal Standards and Surrogates -- Prepare an internal standard solution of acenaphthene-d10, phenanthrene-d10, and chrysene-d12, in methanol, ethyl acetate, or acetone at a concentration of 500 μg/mL of each. This solution is used in the preparation of the calibration solutions. Dilute a portion of this solution by 10 to a concentration of 50 μg/mL and use this solution to fortify the actual water samples (see Section 11.1.3 and Section 11.2.3). Similarly, prepare both surrogate compound solutions (500 μg/mL for calibration, 50 μg/mL for fortification). Surrogate compounds used in developing this method are 1,3-dimethyl-2-nitrobenzene, perylene-d12, and triphenylphosphate. Other surrogates, for example pyrene-d10 may be used in this solution as needed (a 100 μL aliquot of this 50 μg/mL solution added to 1 L of water gives a concentration of 5 μg/L of each internal standard or surrogate). Store these solutions in an amber vial at 4 °C or less. These two solutions may be combined or made as a single solution.”
Q17. What is a surrogate compound?
Q18. What assumptions must be made when choosing a surrogate?
Q19. When is the surrogate added during the sample preparation (i.e. before extraction or after) and why is it added then?
Take a look at the following tables taken from the EPA Method 525.2.
Q20. What could be responsible for the different percentages of true concentration for the surrogates in the two tables?
Q21. What is an internal standard? Why is it used? At what point during the sample preparation process are internal standards added?
Q22. Why spike samples? Do you spike them before or after the sample preparation?
Helpful Resources
https://www.sigmaaldrich.com/analytical-chromatography/sample-preparation/spe/reversedphase-methodology.html
Harvey, D. Chapter 7, Collecting and Preparing Samples. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Pesticides/02_Sample_Preparation.txt |
Learning Outcomes
At the end of this assignment, students will be able to:
• Identify what characteristics of a gas chromatographic stationary phase determine the retention times for and extent of separation between injected compounds.
• Discuss the parameters used to evaluate the quality of a separation in GC.
• Understand why temperature programming is used in GC and what determines the maximum operating temperature of a column.
• Explain the figures of merit important in evaluating the performance of a gas chromatographic column.
Purpose
The purpose of this assignment is to introduce concepts of gas chromatography (GC) as they pertain to the separation and identification of organochlorine pesticides found in water samples.
Organochlorine compounds can be quantitatively determined by gas chromatography (GC) at very low levels. The chromatographic column is chosen to separate a large number of the organochlorine species through differential retention by the stationary phase. Separation can be enhanced by temperature programming of the column during separation. Although this project focuses on GC-MS, detection for gas chromatography of organochlorines is also commonly accomplished by electron capture detection (ECD).
Sources of information about compounds and the physicochemical properties affecting their separation can be found at:
NIST Webbook: http://webbook.nist.gov/chemistry/
NIH Chem IDPlus chem.sis.nlm.nih.gov/chemidplus
Compound Separation in Gas Chromatography
The most common columns used today are long (e.g., 30 or 60 m) fused silica capillary tubes. The stationary phase typically consists of a polymeric liquid that is first coated onto and then chemically bonded to the underlying glass surface of the capillary tube. The stationary phase is chosen for its interaction with the compounds to be separated and for its stability to heat. As a compound in the gas phase moves through the column encountering the bonded phase, it will spend some amount of time partitioned into the bonded phase due to an interaction with that material. The magnitude of a compound’s interaction with the stationary phase is expressed as a partition coefficient. Compounds must have different partition coefficients to be separated by a chromatographic column. The smaller the partition coefficient, the less time the compound spends within the stationary phase and the faster it elutes from the column.
Once a stationary phase has been selected, it is important to be able to predict the elution order (the order in which compounds will emerge from the column) of the chemical compounds we wish to separate. In gas chromatography, one possibility is to consider the differences in the boiling points of compounds. The boiling point of a pure compound is dependent on the intermolecular interactions between the molecules. The stronger these interactions, the higher the boiling point. Boiling points of linear alkanes are primarily controlled by dispersion forces. The more atoms in the alkane, the greater the number of polarizable electrons, the greater the energy of interaction, and, consequently, the higher will be the boiling point of the alkane. Table 1 gives the boiling points of a series of linear alkanes
Table 1. Linear Alkane Data (http://webbook.nist.gov/chemistry/)
Compounds
# C atoms
Mass
Boiling Point oC
Boiling Point K
Propane
3
44.0956
-42.0oC
231.1
Butane
4
58.122
-0.5
272.5
Pentane
5
72.148
36.0
309.2
Hexane
6
86.17
68.8
341.9
Heptane
7
100.2
98.4
371.5
Octane
8
114.22
125.6
398.7
Nonane
9
128.255
150.6
423.8
Decane
10
142.28
174.0
447.2
Undecane
11
156.3083
195
468
n-dodecane
12
170.3348
216
489
Tridecane
13
184.3614
234
507
Tetradecane
14
198.3880
250
523
Pentadecane
15
212.4146
267
540
Q1. Using the data in Table 1, create a graph of the boiling points of this series of linear alkanes as a function of mass. Describe the trend you observe in the graph.
Q2. Figure 1 contains a portion of chromatogram showing the elution of a sample of n-alkanes from a column held at a constant temperature of 87oC. Estimate the approximate time at which C16 will appear?
Figure 1. Separation of n-alkanes using hydrogen as the carrier gas. http://www.restek.com/
Table 2. Structures, molecular weights, boiling points and retention times of selected organochlorine pesticides (NIST, CHEM IDplus).
Compound
Formula CAS registry
Structure
M.W. g/mol
B.P. oC
Retention time (min) tablulated using 8081B column DB-608
1) Alpha-Lindane (HCH)
C6H6Cl6
319-84-6
290.8280
288
8.14
2) Del-lindane (HCH)
C6H6Cl6
319-86-8
290.8280
11.20
3) Gamma lindane (HCH)
C6H6Cl6
55963-79-6
290.8280
323.4
9.52
4) Lindane
C6H6Cl6
58-89-9
290.8280
323.4
5) 4-4' DDE
C14H8Cl4
72-55-9
318.0241
336
16.34
6) 4-4'-DDD
C14H10Cl4
72-54-8
320.0400
350
18.43
7) Methoxychlor
C16H15Cl3O2
72-43-5
345.6472
346
22.80
8) 4-4'-DDT
C14H9Cl5
50-29-3
354.4848
260
*
19.48
9) Aldrin
C12H8Cl6
34487-55-3
364.913
145
11.84
10) Heptachlor
C10H5Cl7
76-44-8
373.3156
135-145
*
10.66
11) Endrin Aldehyde
C12H8Cl6O
7421-93-4
380.9075
19.72
12) Dieldrin
C12H8Cl6O
60-57-1
380.9075
330
16.41
13) Endrin
C12H8Cl6O
72-20-8
380.9075
**
17.80
14) Heptachlor epoxide
C10H5Cl7O
1024-57-3
389.319
135-145
13.97
15) Endosulfan I
C9H6Cl6O3S
19595-59-6
406.924
15.25
16) Endosulfan II
C9H6Cl6O3S
33213-65-9
406.924
450
18.45
17) Endosulfan sulfate
C9H6Cl6O4S
1031-07-8
422.924
481
20.21
Q3. Using the data in Table 2, create a graph of the boiling points as a function of molecular weight. Is the trend in this graph the same as you observed for the series of n-alkanes? If not, what do you think makes it different?
Q4. Examine the molecular structures of the compounds in Table 2. These compounds can be divided into three general types. Which ones would you group together into the three different types? Within a compound group, is there a relationship between structure, molecular weight and boiling point? Do you think that the compounds in Table 2 can be separated based on their boiling points alone?
Role of the Stationary Phase
Table 3. Common Polysiloxane Stationary Phases for GC Separations. http://delloyd.50megs.com/moreinfo/gcphases.html
BONDED PHASE
TEMP °C
GENERAL USE OF PHASE
EQUIVALENTS
Methyl polysiloxane
50-325
Most frequently used phase in GC. Low selectivity, separates compounds according to boiling points. Excellent thermal stability
SE-30, OV-1, OV-101, DB-1, SPB-1, BP-1, HP-1, ULTRA-1, RTx-1, AT-1, CPSil-5
Methyl 5% Phenyl Polysiloxane
50-325
Similar to methyl polysiloxane but slightly more selective due to phenyl content. Excellent thermal stability.
SE-54, OV-23, DB-5, SPB-5, BP-5, HP-5, ULTRA 2, RTx-5, CPSil-8
Methyl 50% Phenyl Polysiloxane
40-325
Added selectivity due to higher phenyl content. Usually retains similar compounds longer than methyl silicone. Provides efficient separations of PAHS and biomedical samples such as drugs, sugars and steroids. Good thermal stability.
OV-17, DB-17, SPB-7, BP-10, HP-17, RTx-17, AT-50,
50% Trifluoropropyl 50% Methyl polysiloxane
40-300
Selectivity for compounds with lone pair electrons or carbonyl groups. Retains oxygenated compounds in the order ether, hydroxy, ester and keto Widely used as a confirmatory phase for chlorinated pesticides. Also suitable for PCB’s, phenols and nitroaromatics. Good thermal stability.
DB-210, RTx-200, HP-210
6% Cyanopropylphenyl 94% Methylpolysiloxane
30-320
An additional choice for a general purpose phase with nominal selectivity for polarizable and polar compounds. More of a boiling point phase than 007-1701; and exhibits less retention of polyaromatic compounds than 007-17. Good thermal stability.
DB-1301, RTx-1301, HP-1301
Methyl 7% Cyanopropyl 7% Phenyl Polysiloxane
280
Unique selectivity of cyanopropyl and phenyl groups provide efficient separations of derivitized sugars and many environmental samples. Not truly a polar phase. Good thermal stability
DB-1701, CPSil-19, RTx-1701, AT-1701
Methyl 25% Cyanopropyl 25% Phenyl Polysiloxane
40-240
Polar phase which provides efficient separations of polar molecules such as fatty acids and alditol acetate derivatives of sugars. Fair thermal stability
DB-255, HP-255, CPSil-43, RTx-225, AT-255
As you may have guessed from the exercises above, it is difficult to design a separation of organochlorine pesticides that depends on physical properties of the compounds alone. To alter the mobilities of the compounds to achieve an adequate separation we need to also consider their interactions with the stationary phase.
Q5. Use the information in Table 3 above to find the column recommended for a boiling point-based separation of linear alkanes. What is the chemical composition of the columns suggested for the separation of alkanes?
In Section 6.9 of EPA method 525.2 a bonded stationary phase is suggested.
“6.9 Fused Silica Capillary Gas Chromatography Column -- Any capillary column that provides adequate resolution, capacity, accuracy, and precision (Section 10.0) can be used. Medium polar, low bleed columns are recommended for use with this method to provide adequate chromatography and minimize column bleed. A 30 m X 0.25 mm id fused silica capillary column coated with a 0.25 µm bonded film of polyphenylmethylsilicone (J&W DB-5.MS) was used to develop this method. Any column which provides analyte separations equivalent to or better than this column may be used.”
Q6. From Table 3 and Figure 2 what are the compositions of the DB-5 and DB-1 stationary phases? Why is the DB-5 phase a better choice for separation of the organochlorine compounds listed in Table 2?
Figure 2. Chemical structures of common GC stationary phases.
The EPA method 525.2 states that any column that provides analytical separations equivalent or better than this one may be used.
Q7. Of the column types listed in Table 3 what might you suggest as alternatives? What figures of merit would you use to determine whether a column provides a “better” separation?
The actual column recommended by the EPA is not DB-5 but DB-5.MS. The MS column designation refers to a bonded siloxane phase with a slightly different chemical composition as shown in Figure 2, which improves the stability and can allow for higher maximum operating temperatures. This reduces column bleed, which is the loss of components of the stationary phase, especially problematic at higher temperatures. Column bleed can produce a high background, reducing the sensitivity of the MS measurements.
Q8. What is column bleed? From the structures in Figure 2, why do you think that the DB-5.MS columns have reduced column bleed than those using the standard DB-5 stationary phase?
Consider the chromatogram of a standard mixture of 20 pesticides shown below.
Figure 3. Separation of a 20 pesticide mixture using chrysene-d12 as the internal standard. The separation was performed as described in EPA Method 525.2 except that the carrier gas was at 25 cm/s.
Q9. The chromatogram in Figure 3 shows a separation of a standard containing 20 organochlorine pesticides plus an internal standard. How would you determine which peak correlates to each organochlorine compound? Be specific.
Temperature Programming and Column Stability
The chromatogram in Figure 3 was acquired by increasing the temperature as the separation progresses according to a set program. This is called temperature programming and is a common feature of gas chromatography. The effect of temperature programming on separation time and peak width is illustrated in the animation at http://www.shsu.edu/~chm_tgc/sounds/flashfiles/GCtemp.swf.
Q10. Why is temperature programming commonly used with gas chromatography?
Q11. Look at the maximum temperatures at which the stationary phases in Table 3 are stable. What is the maximum operating temperature for the DB-5 column?
Evaluating the Performance of the Column for Quality Control
Now that we have a column (stationary phase) and a temperature program to achieve separation how do we
• Ensure that we do not overload the column?
• Describe the quality of the separation?
• Establish confidence in the quality of the signal intensity?
Q12. What would an ideal chromatographic peak look like? What would a chromatographic peak look like if you overloaded the column? Why is it undesirable to overload a column?
The following discussion regarding injection is from EPA method 525.2.
“6.10.1 The GC must be capable of temperature programming and be equipped for splitless/split injection. On-column capillary injection is acceptable if all the quality control specifications in Section 9.0 and Section 10.0 are met. The injection tube liner should be quartz and about 3 mm in diameter. The injection system must not allow the analytes to contact hot stainless steel or other metal surfaces that promote decomposition.”
10.2.3 Inject a 1 µL aliquot of a medium concentration calibration solution, for example 0.5-2 µg/L…”
The instructions in section 10.2.3 indicate that a typical injection should be 1 µL. This injection size coupled with the EPA designated column size sets a limit on the maximum amount of material that can be placed onto the column. The capacity of the column, i.e. the maximum amount of analyte that should be injected, depends upon the volume of the stationary phase and the interaction between the stationary phase and the solute.
Table 4. Column Capacity in ng http://www.chem.agilent.com/cag/cabu/colselect.htm
Film Thickness (um)
Column Diameter (mm)
0.18-0.20
0.25
0.32
0.53
0.10
20-35
25-50
35-75
50-100
0.25
35-75
50-100
75-125
100-250
0.50
75-150
100-200
125-250
250-500
1.00
150-250
200-300
250-500
500-1000
3.00
400-600
500-800
1000-2000
5.00
1000-1500
1200-2000
2000-3000
Because the EPA method dictates the column used (or similar column) it also dictates the total amount of analyte that can be loaded onto the column.
“6.9 Fused Silica Capillary Gas Chromatography Column -- Any capillary column that provides adequate resolution, capacity, accuracy, and precision (Section 10.0) can be used. Medium polar, low bleed columns are recommended for use with this method to provide adequate chromatography and minimize column bleed. A 30 m X 0.25 mm id fused silica capillary column coated with a 0.25 µm bonded film of polyphenylmethylsilicone (J&W DB-5.MS) was used to develop this method. Any column which provides analyte separations equivalent to or better than this column may be used.”
If the capacity of the column is exceeded the shape of the chromatographic peak is altered from its “normal” Gaussian shape to an asymmetric shape. The figure below shows an example of an asymmetric peak shape that results from fronting. Peak tailing is also possible when columns are overloaded. An older EPA method required documentation that the chromatographic peaks are normal in shape.
If the column is overloaded one solution to the problem is to perform a “split” injection. In splitting, a portion of the sample is swept away in the carrier gas leaving only a fraction behind for injection. Split injections can be used to lower the amount injected to as low as 1% of the amount in the sample. This animation shows how a split injection is performed. http://www.instrumentalchemistry.com/gasphase/pages/split.htm
Q13. A 1 L water sample containing 0.1 µg/L of DDT was concentrated to ~1 mL by the sample preparation procedure and 1 µL injected splitless onto a 30 m X 0.25 mm id fused silica capillary column coated with a 0.25 µm bonded film of DB-5.MS. Calculate the ng of material injected. Is this value greater or less than the capacity of the column?
Our method must demonstrate its quality with respect to the baseline. In GC, bleeding of the stationary phase can occur, especially at higher temperatures. Bleeding affects the baseline noise in the spectra. Baseline noise affects the ability to quantitate a given chromatographic peak. This figure shows an example taken from an instrument manufacturer which illustrates the impact of baseline noise in a chromatogram.
The upper chromatogram has a large baseline signal that registers at about 50% of the intensity of the peak at 15.70 min. The signal to noise (S/N) is calculated as:
$\dfrac{S}{N} = \dfrac{X_p - X_b}{pp_{baseline}} \nonumber$
where the signal (S) is defined by the difference in the integrals of Xp (the analyte peak at 15.70 min) and Xb is the integral of a similar area of the baseline. The noise in this formula is calculated as the peak-to-peak noise (ppbaseline) measured in a baseline region containing no peaks. In the top spectrum, the difference between the 15.70 peak intensity (100%) and the baseline intensity (~50%) divided by the peak-to-peak fluctuation in the baseline (shown as a redline) is only 4.91, whereas in the bottom spectrum with less background, a S/N of 39.50. It should be clear that reduction of the baseline bleed (and other contributors to background signals) will greatly enhance the S/N.
Another quality check for the GC data specified by the EPA method is chromatographic resolution.
“10.2.4.1 GC Performance -- Anthracene and phenanthrene should be separated by baseline. Benz[a]anthracene and chrysene should be separated by a valley whose height is less than 25% of the average peak height of these two compounds. If the valley between benz[a]anthracene and chrysene exceeds 25%, the GC column requires maintenance.”
Q14. Evaluate the GC data below for benz[a]anthracene and chrysene, and determine if the GC column requires maintenance. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Pesticides/03_Gas_Chromatography.txt |
Learning Outcomes
At the end of this assignment, students will be able to:
• Explain the processes involved in the ionization of compounds in GC-MS.
• Predict isotopic distributions and identify chlorinated compounds by their MS isotopic signature and fragmentation patterns.
• Identify and use unique ions for MS quantitation in complex samples.
• Using a sampling scheme for Lake Nakuru, determine the concentrations of DDT from GC-MS results and draw some conclusions as to whether the levels of DDT detected in Lake Nakuru water play a role in flamingos’ death.
Purpose
The purpose of this assignment is to introduce concepts of mass spectrometry (MS) as they pertain to the qualitative and quantitative analysis of organochlorine pesticides found in water samples.
In the previous modules you have been introduced to the problem of flamingo die off events in Lake Nakuru, and the analytical chemistry considerations needed to address this problem including sampling, sample preparation, and gas chromatography. In addition to this discussion of MS analysis, the following module on validation of the analytical method addresses issues related to ensuring that the results obtained are of sufficient quality that they can be used to draw appropriate scientific conclusions. In this section, aspects of ionization and mass analysis are explored and example GC-MS results are presented.
Introduction
This section is designed to provide an introduction to mass spectrometry (MS) in the context of its utilization as a gas chromatography detector. The basic concepts of ionization, mass analysis, isotope ratios, and mass spectral interpretation are introduced. Often to solve an analytical problem one must identify a single component in a complex mixture. This is one reason that chromatography is such an important tool for chemical analysis. A goal of our experiment might be to identify a specific pesticide in a water sample containing other pesticides and their degradation products.
Q1. What important information would you need to unambiguously identify a pesticide like DDT in the presence of other similar compounds?
What is unique about a molecule?
One answer to this question could be molecular weight or mass. Consider ethanol, CH3CH2OH. What makes a molecule ethanol and nothing else? You might say “2 carbons, 6 hydrogens, and 1 oxygen” make ethanol. You could further characterize the molecule by saying it had a molar mass of (2 x 12) + (6 x 1) + (1 x 16) = 46 grams/mole. But, that said, what distinguishes it from all other molecules (such as its structural isomer dimethyl ether, CH3OCH3) with a molar mass of 46? Obviously we need more than just molecular weight to unambiguously distinguish ethanol from dimethyl ether. There are important structural differences between these two compounds that distinguish them chemically. Analytical chemists devise techniques that allow determination of both the identity (qualitative analysis) and the concentration or amount (quantitative analysis). One of the most powerful analytical techniques is mass spectrometry. Mass spectrometry is widely used in analytical chemistry and in other fields such as forensics and bioanalysis. In fact, you may have encountered a mass spectrometer while going through airport security. (Jacoby 2009) Some advantages of mass spectrometry include its sensitivity and low detection limits. Both qualitative and quantitative results can be obtained using a mass spectrometer. However, mass spectrometers are in general more costly than the electrochemical and spectroscopic techniques with which you may be familiar from your laboratory coursework.
What happens to molecules in a mass spectrometer?
Mass spectrometry (MS) is a technique in which the analysis is conducted on gas phase ions. These ions are manipulated through a mass analyzer and pass through to a detector. The basic components of a GC-MS instrument are shown in Figure 1.
Figure 1. Block diagram of the basic components of a GC-MS instrument.
What kind of information is produced by a mass spectrometer?
A mass spectrometer discriminates between different ions by their mass to charge ratio (m/z). A variety of mass analyzers are available, but for GC-MS, the most commonly used are the quadrupole mass filter and the quadrupole ion trap. These mass analyzers facilitate the movement of ions using electrostatic focusing, and allow only ions of a particular mass to charge ratio (m/z) to hit the detector at a given time, generating a signal. By scanning through the desired mass range, the ions at each m/z value can be detected in turn. The mass spectrum is a plot of ion abundance versus mass-to-charge ratio (m/z). Figure 2 shows the mass spectrum of DDT; the y axis is ion count and the x axis is m/z.
Figure 2. Mass spectrum of DDT.
The molecular weight of DDT, C14H9Cl5, is 354.486 g/mole. Although there is a group of small peaks in the mass spectrum between m/z 350 and 360, the most abundant ion, called the base peak, is at m/z 235. Clearly MS analysis is more complex than it might appear at first glance. In order to interpret the mass spectrum of DDT, we need to understand the process of ionization.
Q2. Why are ions needed instead of neutral species in mass spectrometry?
Q3. Mass spectrometers operate under high vacuum conditions. Why?
MS Ionization Methods
There are many ionization methods available for use in mass spectrometry. The ionization method used is determined by the properties of the analyte and the type of sample (e.g., gas, liquid or solid) to be introduced into the mass spectrometer. In GC-MS, the GC eluent is introduced directly into the ionization source of the mass spectrometer. Because the MS instrument is designed to selectively analyze and detect ions, the molecules in the eluent must first be ionized. The modes of ionization commonly used for GC-MS are electron ionization (EI) and chemical ionization (CI). Electron ionization is used much more frequently than chemical ionization. This learning module will focus on electron ionization. EPA method 525.2 section 6.10.3 states:
“The mass spectrometer must be capable of electron ionization at a nominal electron energy of 70 eV to produce positive ions.”
What happens during electron ionization?
Molecules must be in the vapor state for EI, making it compatible with gas chromatography. In an EI source, electrons are produced by electrically heating a wire filament. A voltage applied between the filament and an anode accelerates the emitted electrons towards the anode. In the ionization space, the energized electrons interact with analyte molecules (M) and induce the loss of a valence electron. This creates a radical cation with an odd number of electrons:
$\mathrm{M(\mathit{g}) + \mathit{e}^- → M^{+●} + 2\mathit{e}^-}\nonumber$
When one electron is lost, the resulting radical cation is referred to as the molecular ion. EI is described as a hard ionization method because it produces extensive fragmentation of the molecule. A summary of the process of EI ionization with animations can be viewed at https://www.shimadzu.com/an/gcms/support/fundamentals/molecularion_by_ei.html.
Q4. What is an advantage of having a large number of fragment ions present in the mass spectrum? Are there any disadvantages?
The analysis of organochlorine pesticides in Lake Nakuru is a targeted analysis, in that prior to the experiment we have identified a list of target analytes, or compounds of interest. However, it is highly likely that we will also detect chromatographic peaks of additional compounds. One way of identifying compounds responsible for an unknown peak in a complex sample is through comparison of the experimental mass spectrum with the entries in a database, such as the NIST webbook (http://webbook.nist.gov/chemistry/name-ser.html). The figure below shows the database spectrum for DDT.
Figure 3. EI mass spectrum of DDT from the NIST webbook, www.webbook.nist.gov/chemistry. Note that in this spectrum, ion intensity is expressed as relative abundance, with the most intense peak defined as 100%.
Q5. Compare the experimentally determined mass spectrum for DDT in Figure 3 to that from the NIST database in Figure 4. Do you think this is a good match? Why?
For complex samples, like water samples from Lake Nakuru, it is likely that even with a high resolution GC separation that we would observe overlap between eluting peaks due to compounds of similar structure. These compounds frequently display many of the same fragment ions in their mass spectra. In selected ion monitoring (SIM), an ion that is ideally unique to one particular compound (a characteristic or quantitation ion) is chosen so that compound can be quantified in the presence of interfering substances.
Q6. Table 2 of the EPA method 525.2 lists the quantitation ions for DDT as 235 and 165. Why do you think these ions were selected to determine the amount of DDT in our sample?
The molecular weight of DDT is 354.486 g/mole. This value is calculated using the average atomic masses of carbon (12.010), hydrogen (1.007), and chlorine (35.453 g/mole). These average atomic weights are calculated from the masses of the naturally occurring isotopes of each element and their relative abundances. For example, there are two naturally occurring isotopes of chlorine, 35Cl (mass 34.96885) and 37Cl (mass 36.96590). These isotopes have a relative abundance of 75.78% for 35Cl and 24.22% for 37Cl.
For a compound containing one chlorine atom, for example CH3Cl, we would calculate an average molecular weight of 50.484 g/mole. Although it is appropriate to use average molecular weights when we are analyzing large numbers of molecules, the mass spectrometer detects individual ions containing either 35Cl or 37Cl. Therefore, we would expect two MS peaks for CH3Cl+ at m/z values of 50 (M+) and 52 (M++2) Da. (Dalton (Da) is a unit that is used for indicating mass on an atomic or molecular scale.) The ratio of the M+ and (M++2) peaks will reflect the relative abundance of the chlorine isotopes. In mass spectrometry it is customary to normalize peak heights to the most abundant peak. Therefore a molecular ion containing one chlorine atom would produce relative intensities for the M+ and (M++2) peaks of 100:32.5. For a simple molecule like CH3Cl, it is straightforward to predict the relative abundance expected for the M+ and (M++2) peaks, while more complicated structures require the use of an isotopic distribution calculator like the one found www.sisweb.com/mstools/isotope.htm to calculate relative ion abundances due to the isotopic distribution of the molecular ion peak.
Q7. Using this program, calculate the isotopic abundances of the molecular ion peak for CH3Cl+. Are the calculated relative intensities of the ions at 50 and 52 what you expected? In addition to the expected peaks at 50 and 52 Da, the program lists low intensity peaks at 51 and 53. What is the source of these peaks?
Q8. Now use this program to compare the isotopic distribution patterns of CH3Cl+ to CH2Cl2+, CHCl3+ and CCl4+. What trends do you observe as the number of chlorine atoms increases?
Q9. Now calculate the isotopic distribution expected for the molecular ion (M+) of DDT. How many 35Cl atoms are in the molecule that comprises the most intense peak?
Q10. In a complex environmental sample that has pesticide contaminants at low levels, the GC-MS spectrum extracted for the DDT molecular ion or base peak may have contributions from other compounds, naturally occurring or anthropogenic, that might be confused with DDT. How could the chlorine signature in the mass spectrum be used to help with compound identification?
Q11. Propose a structure for the 235 fragment ion of DDT. Use the isotopic distribution calculator to calculate the relative abundances of the ions expected for the structure you propose. How does the predicted spectrum match the intensities in Figure 2?
In addition to the database match of the fragments in the mass spectrum and analysis of the chlorine signature of each compound, there is additional information that can be used to give confidence in our assignment of an unknown peak in the chromatogram. In GC-MS we have two pieces of information for each compound: its chromatographic retention time and its mass spectrum. Figure 4 shows the total ion chromatogram (TIC) for a standard organochlorine pesticide mixture. The TIC represents the intensity of all of the ions detected over the mass range measured (shown on the y axis) as a function of time (the x axis). Chrysene-d12 was added as an internal standard for this analysis and the GC-MS chromatogram was recorded as described in EPA method 525.2. Although this chromatogram appears complex, it is much less complicated than what would be expected for a real water sample from Lake Nakuru.
Figure 4. Total Ion Chromatogram (TIC) measured for a standard mixture of 20 organochlorine pesticides, www.restek.com/restek/prod/5670.asp.
The number plotted above each peak in Figure 4 gives its exact retention time (minutes). Each time point in the chromatogram in Figure 4 contains a mass spectrum. Integration across a chromatographic peak produces an average mass spectrum for that peak. For example, Figures 5 and 6 show the mass spectra extracted for the DDT, chrysene-d12, DDE and DDD peaks. Each figure shows the mass spectrum extracted from the chromatogram at a specific time and for comparison, the spectrum from the NIST library is shown below the experimental mass spectrum. Quantitation for chrysene uses the 240 ion, while for DDT, the intensity of the 235 can be used. DDT and DDE share several common ions, because they are closely related structurally.
Figure 5. The top mass spectra were extracted from the TIC in Figure 4 at the retention times indicated. The database mass spectra plotted below for DDT and chrysene-d12 are from the NIST webbook. In each case, the GC retention time and the MS fragmentation pattern gives a positive identification.
Figure 6. The top mass spectra were extracted from the TIC in Figure 4 at the retention times indicated. The database mass spectra plotted below for DDE and DDD are from the NIST webbook. In each case, the GC retention time and the MS fragmentation pattern gives a positive identification. Note the similarity of the DDD and DDT fragmentation patterns.
Q12. Why was chrysene-d12 selected as an internal standard rather than chrysene with a natural isotopic abundance of hydrogen nuclei? Would you expect that the chrysene and chrysene-d12 to be resolved in the GC chromatogram?
For complex samples, like water samples from Lake Nakuru, it is likely that even with a high resolution GC separation, the chromatogram could be complicated with overlapping peaks. In this case, selected ion monitoring provides improved selectivity and sensitivity. For example, our pesticide test mixture contains a number of chlorinated compounds that are likely to be present at low levels in the complex samples prepared from Lake Nakuru. Figure 7 shows the results of selected ion monitoring for that sample. Figure 7B was prepared by searching the mass spectra comprising the TIC shown in Figure 7A for ions at 235.00 (a window of ± 0.30 Da was used). The only peaks that appear in the selected ion chromatogram are those of DDD and DDT, which share the fragment at m/z 235. If we chose instead to search the chromatograph for ions at 246.00 ± 0.30 we get a chromatogram that shows intense peaks for DDE and DDT, both of which produce low intensity ions at this m/z. Selected ion monitoring can allow quantitation of analytes even when they are too low of intensity to be visible in the TIC.
Figure 7. Selected ion chromatograms measured for the pesticide test mixture.
Analysis of Data
With any analysis you must first evaluate the minimum amount of sample that your method can reliably detect. In addition, to demonstrate that analyte is not lost from the sample extraction and other steps in the procedure, you must assess percent recovery for the analyte.
Tables 1 and 2 below present data for DDT collected according to EPA Method 525.2. The data in Table 1 resulted from laboratory fortified blank (LFB) solutions that were spiked at a level of 5.00 μg/L with the internal standard (chrysene-d12) and 0.50 μg/L of DDT. One liter samples of the LFB were carried through the extraction procedure outlined in the method, and the resulting solutions analyzed for DDT concentration. Data in Table 2 resulted from the analysis of calibration standards containing between 0.50 – 10.00 mg/L DDT and 5.00 mg/L chrysene-d12 prepared by the dilution of stock standard solutions in ethyl acetate.
Table 1. Laboratory fortified blank data (0.50 μg/L DDT and 5.00 μg/L chrysene-d12)
Sample #
Intensity of m/z = 240 for chrysene-d12
Intensity of m/z = 235 for 4,4'-DDT
1
2494373
143803.1
2
2800996
148889.7
3
2985785
182870.4
4
2855369
180016.7
5
2762538
154296.0
6
2602473
152374.8
7
2289537
125819.2
Table 2. DDT calibration data
Concentration (mg/L)
Intensity of m/z = 240 for chrysene-d12 (5 mg/L)
Intensity of m/z = 235 for 4,4'-DDT
0.00
0
0
0.50
2053420
132008
1.00
2874852
313059
2.00
3115040
614663
5.00
2233038
1035790
10.00
2818628
2567623
Q13. Section 13.1.2 of EPA 525.2 describes the calculation of the method detection limit (MDL) for each analyte. Use the data in Tables 1 and 2 to calculate the MDL for DDT (μg/L). How does this value compare with those found in Tables 3-6 of EPA Method 525.2?
Q14. What is the average percent recovery for DDT indicated for the given data?
A sampling plan is provided for Lake Nakuru in Figure 8. Assuming you are limited by the number of samples you can analyze, choose the five data sites for which you would like to acquire data. Your instructor will provide these data to you.
Figure 8. Lake Nakuru sampling plan
Q15. What is the concentration of DDT (μg/L) for each of the five samples you selected?
Q16. Are the levels of DDT that you observed in your samples from Lake Nakuru potentially toxic to flamingos there?
Q17. If you were not limited to water samples, what other types of samples would you collect for analysis and why?
05 Validation of the Method
Learning Outcomes
At the end of this assignment students will be able to:
• Explain the importance of method validation.
• Identify the standard solutions generally used in a validation procedure, and explain why each is necessary.
• Calculate the detection limit of a method for a specific analyte.
Purpose
The purpose of this assignment is to address questions related to the validation of EPA method 525.2. You will need a copy of this method to complete many of the questions in this assignment.
To demonstrate that the numbers obtained in our measurements are correct and can be trusted, the entire procedure used in their determination must be validated. This process assures that the concentrations of analyte that are determined are not the result of errors in the sample workup, an interferent or matrix effect that gives a false “positive” reading, a matrix effect that diminishes the signal, or an improper instrument response. Validation procedures for EPA method 525.2 include Quality Control (Section 9.0), allowing the demonstration of method accuracy and precision, and Method Performance (Section 10.0) that tests the reproducible day-to-day operation of the instrument.
Q1. Define the terms accuracy and precision. What types of error are associated with each?
For a method as complex as EPA 525.2, error can be introduced during many different steps in the procedure, from sampling to sample preparation to instrumental analysis. The process of validation is time-consuming and tedious, but many possible sources of errors must be addressed and either eliminated or compensated for before the values you obtain have any “value”.
Section 3.0 of EPA 525.2 supplies a list of definitions of terms important in the validation process. Before continuing, you should read through that section carefully so that you can separate your FRBs from your IRCs. Additionally, Section 4.0 describes the major sources of interferences for the method.
An important part of method validation is having appropriate blanks.
Q2. What is the general definition of a blank?
Q3. Explain the difference between a field reagent blank and a lab reagent blank. What is the purpose of each?
Q4. What are fortified solutions? What is the purpose of a fortified solution?
Q5. The purpose of internal standards and surrogate analytes was covered in the section on sample preparation. Review these materials and describe how internal standards and surrogate analytes differ? What is the purpose of an internal standard and a surrogate?
Q6. Why is decafluorotriphenylphosphine (DFTPP) used in the method?
Q7. Table 3.1 gives the results of DFTPP performance tests. What are the three parts to the performance test? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Environmental_Analysis__Lake_Nakuru_Flamingos/Pesticides/04_Pesticide_Analysis_by_MS.txt |
Can the most prevalent inorganic ions be determined in Lithia water using a representative cross-section of the analytical techniques (e.g. titrimetry, potentiometry, spectroscopy) covered in a typical quantitative analysis course? The analytical chemistry laboratory is well-suited to investigate this mineral-rich water source. In this module, we will examine the role of chemical equilibria, stoichiometry, and univariate statistics in the sample preparation and characterization of Lithia water.
The discovery of mineral springs in the vicinity of Ashland, Oregon sparked the pursuit of a “spa economy” during the 1910s and 1920s. The lithium concentration in this spring, which is the second highest in the U.S., was a marketing point for town leaders in the early 20th century. Even today, Lithia water plays a visible role in the culture and history of Ashland.
These materials may be used as a term-long quantitative analysis laboratory project or as a dry lab using the questions and the data supplied in this module. If this module is used in support of a laboratory activity, instructors may either use an active learning approach, where students are assigned the problems in the modules and are asked to search for appropriate analytical procedures, or students can be assigned laboratory procedures as weekly activities.
• Assessment Questions: For Assessment questions that accompany this module, please contact Steve Petrovic ([email protected])
Contributors and Attributions
• Dr. Steven Petrovic (Southern Oregon University): [email protected]. Content from ASDL.
Lithia Water Springs Project
The discovery of natural springs in the vicinity of Ashland, Oregon sparked the pursuit of a “spa economy” during the 1910s and 1920s. The lithium concentration in this spring, which is the second highest in the U.S., was a marketing point for town leaders in the early 20th century. Envisioning flocks of tourists searching for rest cures and treatments for various ailments, bond measures were passed to fund the system of pipes used to transport Lithia water several miles to downtown Ashland. Although this approach to economic development has largely been supplanted by the Oregon Shakespeare Festival, the consumption of Lithia water is still seen as a “rite of passage” for first-time tourists. There are also those who still put faith in the curative powers of Lithia water, especially since lithium salts have more recently been prescribed as a drug for chemical depression.
Figure 1. Lithia fountain pedestal in Ashland's Plaza circa Summer 2006
What are the major inorganic components dissolved in Lithia water and how do they contribute to the characteristics of this spring water? How much lithium is in Ashland’s Lithia water, and can one safely consume enough Lithia water to ingest a therapeutically significant dose? The understanding and use of stoichiometry, equilibria and statistics can bring to light the nature of this naturally effervescent spring water.
The problem for this group of analytical chemistry students is:
1. Can the most prevalent inorganic ions be determined in Lithia water using a representative cross-section of analytical techniques (e.g. titrimetry, potentiometry, UV-Visible spectroscopy)?
2. Can univariate statistics be used to support the claim that the most prevalent inorganic ions have been determined?
To answer these questions, it is necessary to review additional background information about Lithia water and understand general guidelines for the analysis of inorganic ions in water samples.
Definitions of Analyte Levels and Choice of Analytical Methods
An analyte is a component in a mixture that is of interest to the analyst, and the sample matrix is the rest of the sample. For example, when an analyst determines the concentration of lithium in Lithia water, lithium is the analyte and the remaining components comprise the sample matrix. An analyte can be classified based on the analyte concentration as shown in Table 1.
Table 1. Suggested analyte classification table for Lithia water analysis. See Analytical Chemistry 2.0 (Chapter 3.4, Scale of Operation) for additional information.
Analyte Classification
Concentration Range (%(w/w) & (ppm))
Appropriate Analytical Techniques
Analyte Amount (mg)
Based on 25 mL sample size
Major
1 - 100
(10000 - 1000000)
Titrimetry, Gravimetry
250 – 25000
Minor
0.01 – 1.00
(100 - 10000)
Titrimetry, Potentiometry, Spectroscopy
2.5 – 250
Trace
1 x 10-7 – 0.01
(10 ppb – 100 ppm)
Potentiometry (to low ppm)
Spectroscopy (entire range)
0.00025 –2.5
If the analyst has an idea of the concentration of a given analyte, then appropriate analytical methods can be suggested and analytical procedures can be developed for the determination of that analyte. Because this project is focused on the determination of inorganic constituents in Lithia water, students should complete the questions in the Major Inorganic Constituents, Origins of Lithia Water, and Sample Handling and Treatment sections in this module. The questions of interest in the Sample Handling and Treatment section of this module requires students to refer to another module entitled Chemical Equilibria & Sample Preparation.
Major Inorganic Constituents
Purpose: The purpose of this section is to understand the various ways of expressing solute concentrations in a solution.
Learning Objectives: Upon completion of this assignment, students will be able to:
1. Use the stoichiometric composition of compounds to convert between mass/volume concentrations and molar solute concentrations
2. Identify appropriate analytical techniques for the determination of inorganic analytes in Lithia water based on analyte concentration
Figure 2. Plaque representing a 1915 analysis of Lithia water from Ashland, OR
The plaque in Figure 2 contains the results of a historic analysis of inorganic compounds from one of Ashland’s Lithia water springs. Notice that the concentration of each inorganic constituent is expressed as a salt or combination of salts. If the composition of the Lithia water is stable over time, one can estimate the concentration for each cation and anion in Lithia water prior to working in the laboratory.
For the questions below, assume that the soluble form of silica solely exists as the silicate ion (SiO32-).
Q1. Which cations and anions are represented by a single salt and which are represented by multiple salts?
Q2. Of those analytes listed on the plaque, are there any cation or anion concentrations that are unknowns based on the information provided? How can you tell?
Q3. Based on the analysis engraved on the plaque above, how would you calculate the concentration of each ion represented by a single salt? Complete these calculations, and express the solute concentration in milligrams per liter (mg L-1) and in moles per liter (M).
Q4. How does your approach to calculating ion concentrations represented by multiple salts differ from expressing ion concentrations represented by a single salt? Complete these calculations, and express the solute concentration in milligrams per liter (mg L-1) and in moles per liter (M).
Q5. Determine the number of milligrams of each analyte present in 25.00 mL of Lithia water from the 1915 analysis, assuming that the composition of Lithia water has not changed since 1915. Using these analyte quantities, determine appropriate analytical techniques based on the classification scheme in Table 1.
Q6. Once the appropriate analytical technique(s) for each analyte have been identified based on the classification scheme in Table 1, consider, discuss, and identify additional criteria that would be important in the selection of an appropriate method for each analyte in Lithia water. An appropriate starting point for this discussion would be to read Chapter 3.4 in Analytical Chemistry 2.0.
Q7. Lithium carbonate is often prescribed to control manic depression. A typical dosage is 900 mg day-1 of lithium carbonate. How many liters of Lithia water would someone have to drink to ingest lithium equivalent to a prescribed dose of lithium carbonate? Based on the volume of Lithia water you calculated in the previous question, use any electronic or hard-copy resources to determine whether there are health related issues that could arise due to the ingestion of other ions present in Lithia water.
Origins of Lithia Water
The Lithia water that is pumped to downtown Ashland originates from a chain of soda water (carbonated) springs called Lithia Springs, which are three miles east of Ashland near Emigrant Creek Road. The pumphouse for the Lithia water that is pumped to Ashland is currently on a site maintained by the Ashland Gun Club. The ground surrounding the pumphouse for the soda water spring contains deposits of tufa, which is a porous and impure limestone deposit. Although limestone is not known to be present in the underlying rocks, carbon dioxide emissions are associated with volcanic activity, and the soda spring water may be moving through coal and volcanic bedrock to finally emerge at Lithia Springs.1
Figure 3. Pumphouse used to pump Lithia water to downtown Ashland
Lithia water has an acidic effervescence, and its taste is salty and bitter, much like Alka-Seltzer tablets. Freshly collected Lithia water also has a slight sulfurous odor and taste.
References:
1. Wagner, N. S. Natural Sources of Carbon Dioxide in Oregon. Oregon Geology 1959, 21, 103-113.
Questions:
Q8. If the Lithia spring water was to leach ions from limestone deposits to form soda springs, what primary cations and anions would you expect to have dissolved in the spring water due to such leaching action?
Q9. Soda springs are characterized by significant levels of CO2. If the pH of a recently obtained sample of Lithia water is 6.4, draw a ladder diagram for carbonic acid, and predict the predominant form of carbonic acid in this sample of Lithia water.
Sample Handling & Treatment
Sometimes, freshly obtained samples need to be immediately treated with a reagent prior to chemical analysis in order to minimize changes in analyte concentration during sample storage. Long-term changes in analyte concentration may be due to physical considerations such as evaporation, but short-term changes in analyte concentration are often a result of a chemical reaction. Oxidation-reduction reactions change the oxidation state of an analyte, and can lead to a decrease in the concentration of an analyte via precipitation of less soluble compounds.
Q10. Answer Q1 – Q13 in the Chemical Equilibria and Sample Preparation document of the Lithia water module. After completing these questions, go online to the U.S. Environmental Protection Agency website (www.epa.gov) and search for documentation regarding the preservation of inorganic analytes in water samples.
Q11. Based on your search results on the U.S. EPA website and your answer to question 13 in the Chemical Equilibria and Sample Preparation document, develop a sampling plan for Lithia water so that all significant analytes can be determined.
Activity 1. At this point, if you are going to have your students analyze some of the constituents of a local water source, it would be an appropriate time to have them collect the samples. If students are interested in investigating changes in sample appearance upon standing, students should obtain a small, clean Nalgene bottle (60 – 250 mL capacity) and rinse it thoroughly with the water sample. After rinsing, fill the bottle completely with water and cap it tightly. Students should record any observations regarding the appearance of the water immediately after collection and a day or two afterwards. They should measure the pH of the water sample and record that value in their laboratory notebook. If students have already developed a sampling plan, appropriate steps should be taken to properly clean sample containers and preserve samples as needed. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/01_Statement_of_the_Problem.txt |
Learning Outcomes
The purpose of this assignment is to predict how solute concentrations are controlled by chemical equilibria, understand the chemistry involved with sample preparation, and the extent to which sample preparation is necessary prior to analyte determinations.
At the end of this assignment, students will be able to:
1. Assess the effect of pH and redox chemistry on solubility equilibria in Lithia water
2. Develop appropriate strategies for sample preparation prior to analysis of Lithia water
Redox Chemistry
In the Major Inorganic Constituents module, you learned about the main ionic solutes present in Lithia water, however, dissolved gases are also present. Of these gases, diatomic oxygen (O2) can act as an oxidizing agent:
$\ce{O2 (g) + 4 H+ (aq) + 4 e- \rightleftharpoons 2 H2O (l)} \hspace{25px} \mathrm{E^o = 1.23\: V}\nonumber$
This is a very favorable reduction half-reaction. Dissolved species that can be oxidized by dissolved oxygen need to have valence electrons available. Therefore, one can initially evaluate whether a redox reaction should occur by determining whether a dissolved chemical species has any available valence electrons that can reduce oxygen to water.
Q1. Which of the cationic solutes in Lithia water are main group ions? Which of the cationic solutes are transition metal ions? Evaluate whether main group and transition metal cations have lost all of their valence electrons. Can main group and/or transition metal cations potentially be oxidized by dissolved oxygen?
In the Analytical Chemistry 2.0 text (http://www.asdlib.org/onlineArticles/ecourseware/Analytical Chemistry 2.0/Text_Files_files/AnalChem2.0.pdf), the author introduces the concept of ladder diagrams as a vehicle for evaluating whether a redox reaction can take place based on reduction potentials. A list of standard reduction potentials is available in Analytical Chemistry 2.0. The ladder diagram in Figure 1 illustrates whether dissolved oxygen could oxidize a transition metal ion, such as tin (II).
$\ce{Sn^4+ (aq) + 2e- \rightleftharpoons Sn^2+ (aq)} \hspace{25px}\mathrm{E^o = +0.13\: V}\nonumber$
Figure 1. A ladder diagram for the redox couple of dissolved oxygen and divalent tin under standard conditions.
The overlap in the ladder diagram with dissolved oxygen and tin (IV) indicates the ability of these two species to coexist in solution, highlighting tin (II) as a reducing agent.
A more complete evaluation of such redox chemistry involves the use of the Nernst equation since it is rare to encounter redox chemistry in the environment under standard conditions (25 °C, 1 M concentration for each species in solution or 1 atm pressure for gas-phase species). The Nernst equation is shown in eq 1:
$E= E^o - \frac{RT}{nF} \ln K\tag{1}$
In eq 1, Eo is the half-reaction potential (V) under standard conditions, E is the half-reaction potential (V) under non-standard conditions, K is the equilibrium constant for a given reaction, R is the gas constant (8.314 J K-1 mol-1), T is the temperature (K), n is the number of electrons transferred and F is Faraday’s constant (96,485 C mol-1). Combining constants at 298 K and converting from natural to base 10 logarithm yields eq 2:
$E= E^o - \frac{0.05915}{n} \log K \tag{2}$
For the reduction half-reaction of oxygen to water in eq 3:
$K= \dfrac{1}{[O_2 ] [H^+ ]^4 }\tag{3}$
Assuming a dissolved oxygen concentration of 10 mg/L (3.1 x 10-4 M) and a pH of 6.0:
$E=1.23V- \dfrac{0.05915}{4} \log \dfrac{1}{[3.1\times10^{-4} M] [1.0\times10^{-6} M]^4 }=0.823V\nonumber$
An updated ladder diagram under non-standard conditions is shown in Figure 2.
Figure 2. Ladder diagram for the redox couple of dissolved oxygen and divalent tin under non-standard conditions described in the text above.
Even under these conditions, dissolved oxygen can be a potent oxidizing agent.
Q2. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ under standard conditions. Predict the predominant oxidation state for iron under these conditions.
Q3. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ assuming the following non-standard conditions: assume that the dissolved oxygen concentration is 10 mg L-1, the total iron concentration is 10 mg L-1, and that the initial ratio of [Fe2+]/[Fe3+] is 1000. Predict the predominant oxidation state for iron under these conditions and compare the result to your answer in question 2.
Q4. Refer to the plaque shown in Fig. 2 of the Major Inorganic Constituents section. What are the two predominant anions in Lithia water? Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect these anions to react with dissolved oxygen at concentrations expected in Lithia water? Provide appropriate support for your answer.
Q5. Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect main group cations to react with transition metal cations present in Lithia water? Provide appropriate support for your answer.
Q6. Briefly summarize the redox chemistry that according to your analysis can take place among the major inorganic cations in Lithia water.
Solubility Equilibria – First Principles
The solubility rules for inorganic salts in water are provided below:
Rule 1. All compounds of Group IA elements (the alkali metals) are soluble.
Rule 2. All ammonium (NH4+) salts are soluble.
Rule 3. All nitrate (NO3-), chlorate (ClO3-), perchlorate (ClO4-), and acetate (CH3COO- or C2H3O2-) salts are soluble.
Rule 4. All chloride (Cl-), bromide (Br-), and iodide (I-) salts are soluble except for those of Ag+, Pb2+, and Hg22+.
Rule 5. All sulfate ( SO42-) compounds are soluble except those of Ba2+, Sr2+, Ca2+, Pb2+, Hg22+. The Hg2+, Ca2+ and Ag+ sulfates are only moderately soluble.
Rule 6. All hydroxide (OH-) compounds are insoluble except those of Group I-A (alkali metals) and Ba2+, Ca2+, and Sr2+.
Rule 7. All sulfide (S2-) compounds are insoluble except those of Groups I-A and II-A (alkali metals and alkaline earths).
Rule 8. All sulfites (SO32-), carbonates (CO32-), chromates (CrO42-), and phosphates (PO43-) are insoluble except for those of NH4+ and Group I-A (see rules 1 and 2).
Q7. Using the solubility rules provided above and the primary inorganic constituents in Lithia water, which inorganic salts would be most likely to form a precipitate in Lithia water?
The concentration of an ion is controlled in part when a heterogeneous equilibrium is established between ions in solution and a precipitate. It is the establishment of this equilibrium that provides meaning to the concept of solubility. If one starts with pure water and adds an excess of a sparingly soluble solid to the water, the solid will dissolve until the product of the relevant ion concentrations in mol L-1, raised to their stoichiometric coefficients, is equal to the equilibrium constant for the sparingly soluble salt. The equilibrium constant expression for sparingly soluble salts is known as Ksp. An example of a solubility equilibrium is the dissolution of calcium carbonate in water shown in eq 3.
$\ce{CaCO3 (s) \rightleftharpoons Ca^2+ (aq) + CO3^2- (aq)} \hspace{25px} \mathrm{K_{sp}} = \ce{[Ca^2+][CO3^2- ]} = 4.5 \times 10^{-9}\tag{3}$
Q8. Why is there no term for the solid calcium carbonate in the Ksp expression?
Precipitation and dissolution processes control the concentration of ionic species, such as those that are being determined in Lithia water.
Solubility Equilibria – Infinite Dilution Conditions
It is important to know how to calculate ion concentrations from solubility equilibria, both under infinite dilution conditions (i.e. zero ionic strength) and under conditions of significant ionic strength. The solubility of compounds in water or other solvents can be expressed in a variety of ways. For our purposes, we will define solubility in the following way:
Solubility is the number of moles of the solid species that dissolve in one liter of solution. This is known as the molar solubility.
The example below involves calculating the solubility of calcium carbonate ignoring the effects of solution pH. Assuming that the calcium carbonate is the only source of ions in solution, it is typically assumed that the ionic strength is negligible, and that the thermodynamically-based solubility product adequately represents the extent to which the dissolution of CaCO3 occurs. The solubility of CaCO3 may be calculated using the following balanced reaction and equilibrium constant expression:
$\begin{array}{llclcl} \ce{ &CaCO3 (s) &⇌ &Ca^2+ (aq) &+ &CO3^2- (aq)}\ \ce{Initial &--- & &0 & &0}\ \ce{Change &-X & &+X & &+X}\ \ce{Equilibrium &--- & &X & &X} \end{array}\nonumber$
$\mathrm{K_{sp}=4.5\times10^{-9}=[Ca^{2+}][CO_3^{2-}]=X^2}\nonumber$
$\mathrm{X=[Ca^{2+}]=[CO_3^{2-}]=\sqrt{4.5\times10^{-9}}=6.7\times10^{-5}\:M}\nonumber$
Note, in this case, for each mole of solid calcium carbonate that dissolves, one mole of calcium ion and one mole of carbonate ion is found in solution. Because the relationship is 1:1, the concentration of calcium ion and concentration of carbonate ion is each equal to the solubility of calcium carbonate. If the salt has a stoichiometry that is not 1:1 (e.g., Ca3(PO4)2), the concentration of the calcium ion and concentration of the phosphate ion are not directly equal to the solubility of the solid in water.
Q9. Write expressions that relate the concentration of calcium ion and the concentration of phosphate ion to the solubility of calcium phosphate in water.
These equilibrium calculations can also be applied to all sparingly soluble salts under different experimental conditions. For example, pH has a pronounced effect on the solubility of metal hydroxides. The solubility product constant (Ksp) is tabulated in Table 1 for a series of insoluble and slightly soluble metal hydroxides. The solubility products for the corresponding carbonate salts are also included in Table 1.
Table 1. Solubility product constants for insoluble hydroxide& carbonate salts that could control the solubility of metal ions measured in Lithia water.
Metal Cation
Solubility Equilibrium
Ksp
Al3+
$\ce{Al(OH)3 (s) ⇌ Al^3+ (aq) + 3 OH- (aq)}$
4.6 x 10-33
No pure aluminum carbonate exists
Ca2+
$\ce{Ca(OH)2 (s) ⇌ Ca^2+ (aq) + 2 OH- (aq)}$
6.5 x 10-6
$\ce{CaCO3 (s) ⇌ Ca^2+ (aq) + CO3^2- (aq)}$
4.5 x 10-9
Fe2+
$\ce{Fe(OH)2 (s) ⇌ Fe^2+ (aq) + 2 OH- (aq)}$
8 x 10-16
$\ce{FeCO3 (s) ⇌ Fe^2+ (aq) + CO3^2- (aq)}$
2.1 x 10-11
Fe3+
$\ce{Fe(OH)3 (s) ⇌ Fe^3+ (aq) + 3 OH- (aq)}$
1.6 x 10-39
No stable ferric carbonate exists
Mg2+
$\ce{Mg(OH)2 (s) ⇌ Mg^2+ (aq) + 2 OH- (aq)}$
7.1 x 10-12
$\ce{MgCO3 (s) ⇌ Mg^2+ (aq) + CO3^2- (aq)}$
3.5 x 10-8
Q10. Calculate the solubility of calcium hydroxide in a solution that has a pH of 12. Note: The pH has been adjusted to 12 by some other means than the dissolution of calcium hydroxide. If calcium hydroxide is added to water at pH of 12, some of the calcium hydroxide will dissolve to satisfy the Ksp equilibrium expression.
Q11. What is the molar solubility and mass solubility (i.e. solubility in grams of solid per liter of solution) of each of the metal hydroxides listed in Table 1 at pH values of 10.0, 6.0, and 2.0? How is the solubility of a metal hydroxide affected by pH? Offer an explanation for this behavior using LeChatelier’s principle.
Q12. Integrating the ideas of redox chemistry and solubility expressed in this module, summarize the chemical and physical changes you would expect over time in a sample of Lithia water? What sample preparation step could be used to prevent such changes from occurring?
Q13. Collection of samples for trace metal analysis in drinking water is described in the Sample Collection, Preservation and Storage section of EPA method 200.5, Determination of Trace Elements in Drinking Water by Axially Viewed Inductively Coupled Plasma - Atomic Emission Spectrometry (http://www.epa.gov/nerlcwww/ordmeth.htm). After reading this document, suggest a general approach to preserving Lithia water prior to analysis. Would this work for all analytes of interest in Lithia water? Explain.
Solubility Equilibria – Advanced Topics
Based on the simple solubility equilibrium calculation for calcium carbonate in the Solubility Equilibria – First Principles section, the concentration of calcium ion in Lithia water would be 67 μM, or 2.7 mg L-1. However, there are three problems with this approach:
1. According to the carbonic acid ladder diagram, the predominant species is not the carbonate ion but the bicarbonate ion at the pH that is normally measured for Lithia water.
2. The ionic strength of Lithia water is not zero.
3. Lithia water contains dissolved CO2 (i.e. Lithia water is naturally carbonated), which reacts with water to form carbonic acid. According to LeChatelier’s principle, the dissociation of protons from carbonic acid shifts the carbonate equilbrium in Lithia water favoring the protonation of carbonate ion to form bicarbonate ion.
Problem 1 - Alpha Values
n titratable protons, there are (n+1) alpha values as shown in eq 4:
$1= α_{H_n A}+ α_{H_{(n-1) A^-}} +α_{H_{(n-2) A^{-2}}} +...+α_{A^{-n}}\tag{4}$
In eq 4, $α_{H_n A}$ is the fraction of the analytical concentration for the undissociated acid (no protons lost), $α_{H_{(n-1) A^-}}$ is the fraction of the analytical concentration for the weak acid that has lost one proton, $α_{H_{(n-2) A^{-2}}}$ is the fraction of the analytical concentration for the weak acid that has lost two protons, and $α_{A^{-n}}$ is the conjugate weak base with no remaining protons.
Equations to calculate alpha values as a function of [H+] and Ka are provided without proof below:
Monoprotic Acid
$\mathrm{\alpha_{HA} = \dfrac{[H^+]}{[H^+] + K_a }}\nonumber$
$\mathrm{\alpha_{A^-} = \dfrac{K_a }{[H^+] + K_a }}\nonumber$
Diprotic Acid
$\mathrm{\alpha_{H_2A} = \dfrac{[H^+]^2}{[H^+]^2 + [H^+] K_{a1} + K_{a1}K_{a2}}}\nonumber$
$\mathrm{\alpha_{HA^-} = \dfrac{[H^+]K_{a1} }{[H^+]^2 + [H^+] K_{a1} + K_{a1}K_{a2}}}\nonumber$
$\mathrm{\alpha_{A^{2-}} = \dfrac{K_{a1}K_{a2}}{[H^+]^2 + [H^+] K_{a1} + K_{a1}K_{a2}}}\nonumber$
Triprotic Acid
$\mathrm{\alpha_{H_3A} = \dfrac{[H^+]^3}{[H^+]^3 + [H^+]^2 K_{a1} + [H^+]K_{a1}K_{a2} + K_{a1}K_{a2}K_{a3}}}\nonumber$
$\mathrm{\alpha_{H_2A^-} = \dfrac{[H^+]^2K_{a1}}{[H^+]^3 + [H^+]^2 K_{a1} + [H^+]K_{a1}K_{a2} + K_{a1}K_{a2}K_{a3}}}\nonumber$
$\mathrm{\alpha_{HA^{2-}} = \dfrac{[H^+]K_{a1}K_{a2}}{[H^+]^3 + [H^+]^2 K_{a1} + [H^+]K_{a1}K_{a2} + K_{a1}K_{a2}K_{a3}}}\nonumber$
$\mathrm{\alpha_{A^{3-}} = \dfrac{K_{a1}K_{a2}K_{a3}}{[H^+]^3 + [H^+]^2 K_{a1} + [H^+]K_{a1}K_{a2} + K_{a1}K_{a2}K_{a3}}}\nonumber$
The value of this approach is that if one knows the pH and the weak acid dissociation constants for a particular weak acid, the alpha values for all forms of the weak acid can be calculated.
Q14. Calculate the alpha values for the three carbonic acid species (H2CO3, HCO3-, CO32-) at pH 6.4. Do these values agree with your predictions based on the carbonic acid ladder diagram?
The alpha values you calculated in Q14 provide the relative concentration of the various species in the carbonic acid equilibrium. A useful application of these alpha values is in the calculation of the calcium ion concentration using the calcium carbonate solubility equilibrium expression. In the previous section, the bicarbonate concentration was determined to be 0.06736 M, based on the analysis engraved on the plaque in the Major Inorganic Constituents section.
Q15. Given that the pH of the Lithia water is 6.4, what is the total concentration of all species in the carbonic acid equilibrium?
Q16. What is the concentration of carbonate ion?
Q17. Using this concentration of carbonate, calculate the concentration of calcium that you would expect in Lithia water from the dissolution of calcium carbonate.
Q18. Compare this value to the value of calcium on the plaque.
This concentration is approximately fifteen times lower than the calcium ion content determined, which indicates that the acid-base chemistry of the carbonic acid system is not the only factor that controls the calcium ion concentration in Lithia water.
Problem 2 - Effect of Ionic Strength on Solubility
The extent to which a precipitate dissociates into solution phase ions is also affected by the ionic strength of the solution. The equilibrium constants compiled in the back of Analytical Chemistry 2.0 are based on an ionic strength of zero (i.e. infinite dilution conditions). If the ionic strength of a solution is greater than zero, there is a measurable concentration of supporting electrolyte in the solution. Ionic strength is defined in the equation below:
$\mathrm{\mu=\sum C_iz_i^2}\nonumber$
In the ionic strength equation, µ represents the ionic strength of the solution (mol L-1) , Ci represents the concentration (mol L-1)of each ionic species i and zi represents the charge of each ionic species i.
Q19. What effect would raising the ionic strength of a solution have on the solubility of a sparingly soluble salt? Think about what relatively high concentrations of other ions (e.g., Na+ and Cl-) might have in a solution containing smaller amounts of Ca2+ and CO32-.
In the case of the solubility of a sparingly soluble salt, the analyte ions not only interact with the solvent via ion-dipole interactions, but they also experience coulombic interactions with the supporting electrolyte. The ions from the supporting electrolyte partially screen the charge of each analyte ion, which is shown in Chapter 6, Section 9 of Analytical Chemistry 2.0. These interactions decrease the “effective concentration” or activity of the analyte ions. The relationship between the analytical concentration and the effective concentration (activity) is shown below:
$a_x=γ_x [X]\nonumber$
In the equation above, ax represents the activity of an analyte ion in mol L-1, [X] is the analytical concentration in mol L-1, and γx is the activity coefficient. As the ionic strength of a solution increases, the activity coefficient decreases, and so does the activity of an analyte.
The relationship between activity and concentration is important because chemical equilibria are based on analyte activity and not analyte concentration. Therefore, the equilibrium constant expression for the general reaction:
$\mathrm{aA (aq) + bB (aq) \rightleftharpoons cC (aq) + dD (aq)}\nonumber$
is:
$\mathrm{K_{eq}=\dfrac{a_C^c a_D^d}{a_A^a a_B^b}=\dfrac{γ_C^c γ_D^d}{γ_A^a γ_B^b}\dfrac{[C]^c [D]^d}{[A]^a [B]^b}}\nonumber$
The value Keq is called the thermodynamic equilibrium constant because the constant is based on activities, which are only equal to concentrations under infinite dilution conditions (µ = 0). Many analytical methods determine concentrations, not activities, and the ionic strength of the solutions analyzed tends to be significantly greater than zero. Rearranging the equation above results in the following equilibrium constant expression:
$\mathrm{{K}'_{eq}=K_{eq} \dfrac{γ_A^a γ_B^b}{γ_C^c γ_D^d}=\dfrac{[C]^c [D]^d}{[A]^a [B]^b }}\nonumber$
In the equation above, K’eq is called the concentration-based equilibrium constant. This equilibrium constant describes the favorability of a reaction in a system with an ionic strength greater than zero.
To calculate the concentration-based equilibrium constant, the activity coefficient for each analyte ion is needed. A common approach for the estimation of activity coefficients is the Debye-Hückel equation:
$\mathrm{\log γ_x= \dfrac{-0.51 z^2 \sqrt μ}{1+3.3α_x \sqrt μ}}\nonumber$
In the Debye-Hückel equation, z is the charge on the analyte ion, αx is the hydrated ion diameter of the analyte in nanometers, and µ is the ionic strength in mol L-1. Once the activity coefficient for each analyte ion is calculated, the concentration-based equilibrium constant may be calculated and used to solve chemical equilibrium problems.
Most measurements made in the analytical chemistry laboratory detect the concentration of the analyte. A spectroscopic measurement based on the Beer-Lambert law is an example of the direct proportionality between analyte concentration and absorbance. Interestingly, ion selective electrodes (ISEs) detect the activity of an analyte and not its concentration. Although the effect of ionic strength on the activity coefficient, and therefore on the analyte activity, is expressed through the Debye-Hückel equation, it is common practice to add a high ionic strength solution to the calibration standards and samples so that the ionic strength is constant for all the measurements. Under these conditions, the activity coefficient for each solution will be the same, and the calibration plot can be used to determine the analyte concentration.
Q20. Calculate the molar solubility of AgCl under infinite dilution conditions (i.e. µ = 0) with its solubility in 0.10 M NaNO3. The hydrated ion diameters and thermodynamic equilibrium constants may be obtained from Analytical Chemistry 2.0.
Q21. Calculate the concentration-based equilibrium constant for CaCO3, H2CO3, and HCO3-, and use these constants to predict the calcium concentration in Lithia water at pH 6.4. The ionic strength of Lithia water may be calculated from the solute molarities reported on the plaque from the Major Inorganic Constituents section, and the hydrated ion diameters and thermodynamic equilibrium constants may be obtained from Analytical Chemistry 2.0.
Problem 3 – Effect of dissolved CO2
Air contains carbon dioxide gas that dissolves to some extent in water. The Lithia spring water also contains some dissolved carbon dioxide that comes from carbon dioxide gas entrained in the rocks that feed the springs. When carbon dioxide dissolves in water, a chemical reaction occurs that results in the formation of a weak acid, carbonic acid.
$\ce{CO2 (aq) + H2O (l) ⇌ H2CO3 (aq)}\nonumber$
Q22. Remembering that the Lithia water contains carbonate ion, use LeChatelier’s principle to explain what effect the addition of carbonic acid has on the concentration of carbonate ion in Lithia water?
Q23. Considering your answer to the preceding question, what effect would the addition of carbonic acid have on the solubility of calcium carbonate and other sparingly soluble carbonate salts?
Q24. The concentration of calcium ion based on the 1915 analysis of Lithia water is 347 mg L-1. Assuming that the calcium ion concentration in Lithia water is controlled by the carbonate ion concentration in Lithia water, calculate the carbonate ion concentration in Lithia water using the concentration-based equilibrium constant for calcium carbonate. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/02_Chemical_Equilibria_and_Sample_Preparation.txt |
Learning Outcomes
The purpose of this assignment is to learn how to determine whether all major inorganic constituents have been determined in Lithia water by applying the principle of electroneutrality.
At the end of this assignment, students will be able to:
1. Apply the principle of electroneutrality toward calculating cation and anion concentrations in natural waters
2. Understand the purpose of univariate statistical tests in comparing concentrations of ionic solutes
3. Perform appropriate chemical and statistical calculations for the comparison of ionic solute populations.
Preparation
To ensure that each student has received a proper introduction to the statistical fundamentals of chemical analysis problems, everyone will complete a laboratory exercise using the following ASDL resource:
1. Harvey, D.; Otto, W. Introduction to Data Analysis. JASDL [Online], 2007, article 10042. http://asdlib.org/onlineArticles/ecourseware/Harvey/DataAnalysisHome.html (accessed November 29, 2013).
The use of univariate statistics is critical in evaluating the results of chemical determinations. In addition to completing the ASDL laboratory exercise entitled “Introduction to Data Analysis,” to review basic statistical methods and measures of accuracy and precision, students should also read Chapter 4, “Evaluating Analytical Data,” in David Harvey’s textbook Analytical Chemistry 2.0, which is available at the Analytical Sciences Digital Library (ASDL) website (https://www.asdlib.org/onlineArticles/ecourseware/Text_Files.html).
Characterization of Inorganic Ions in Lithia Water
Assessing whether the complete analysis of the major inorganic species in a Lithia water spring has been accomplished requires a statistical evaluation of the cation and anion concentrations determined in the sample. This assessment is based on the hypothesis that, using analytical techniques encountered in the traditional quantitative analysis laboratory, all detectable cations and anions have been determined. To test this hypothesis, one first must ask how the concentrations of all cationic and anionic species are related to one another in solution. Subsequently, one must then use this information to determine whether these two quantities, the concentration of cations in Lithia water and the concentration of anions in Lithia water, are statistically equivalent at a specified confidence level or whether there is a significant difference between the concentration of cations and anions determined in a Lithia water sample.
Lithia water, as with all matter, should follow the principle of electroneutrality, which states that all pure substances have no net charge. Remember that a solution, such as Lithia water, is simply a mixture of pure substances. Mathematically, one can express this principle in the following manner:
$\sum_{i=1}^{N}n_+ = \sum_{i=1}^{N}n_-\nonumber$
$\sum_{i=1}^{N}c_{i,+}z_{i,+} = \sum_{i=1}^{N}c_{i,-}z_{i,-}\nonumber$
In the first equation above, n+ and n- represent the moles of positive and negative charges, respectively. The moles of positive charge are calculated by multiplying the molarity of each cation (ci,+) by the charge on each cation (zi,-) and summing across all cationic charges. An identical calculation is performed to determine the moles of negative charge in a solution. This process is identical to writing a charge balance equation.
Q1. Once you have completed your Lithia water project experiments in lab, list the cations and anions that were determined. If you are performing these modules as part of a dry lab, please use the 1915 analysis plaque (Figure 2 – Major Inorganic Constituents in Statement of Problem module) to create your list of cations and anions.
Q2. Based on the principle of electroneutrality, what should be true about the concentration of the cationic and anionic charges in Lithia water?
Q3. Assuming that the inorganic constituents of Lithia water have been completely characterized, write down an appropriate charge balance equation for all dissolved ions in Lithia water.
Statistical Analysis of Electroneutrality
Although solutions theoretically have equivalent concentrations of positive and negative charge, experimental error in the determination of solute concentrations will result in numerical differences between the total concentrations of cations and anions. Often times, it is unclear whether these concentration differences are real (aka bias or determinate error) or due to random (indeterminate) error. Assuming that the experimental error is normally distributed, Chapter 4 in Analytical Chemistry 2.0 provides an overview of how to calculate the uncertainty of measured quantities and to compare the means of two populations to determine if they are statistically significant for a given confidence level.
Q4. Use the error propagation equations described in Chapter 4 of Analytical Chemistry 2.0 to propagate the experimental error in the following analytical determinations
1. A solution with a volume of 1.000 ± 0.012 mL was determined to have a mass of 1.008 ± 0.023 g. What is the density of the solution and its uncertainty?
2. The volume delivered by a buret (V) is the difference between the initial volume (Vi) and the final volume (Vf) read on the scale of the buret. If Vi is (5.36 ± 0.02) mL and Vf is (15.68 ± 0.02) mL, calculate the volume delivered and its uncertainty.
3. The density of a liquid was determined by measuring the mass of a 100-mL volumetric flask filled with the liquid, subtracting the mass of the empty 100-mL volumetric flask, and dividing the mass difference by the volume of the 100-mL volumetric flask. If the mass of the full flask is 248.3 ± 0.1 g, the mass of the empty flask is 45.5 ± 0.1 g, and the volume of the flask is 100.0 ± 0.08 mL, calculate the density of the liquid and its uncertainty.
Q5. Based on your reading of Chapter 4 in Analytical Chemistry 2.0, what statistical approach would you use to determine whether equivalent amounts of cationic charge and anionic charge were determined in Lithia water?
Q6. The Nutrition Facts label on a bottle of local mineral water contains the following information for selected inorganic solutes (standard deviations provided by author, assume each result based on triplicate readings (n=3)):
Solute
Concentration (mg L-1)
Bicarbonate
466 (± 24)
Magnesium
124 (± 5)
Calcium
2.67 (± 0.10)
Sodium
3.31 (± 0.15)
Zinc
4.45 (± 0.10)
At the 95% confidence level, determine whether all of the major inorganic solutes have been reported on the Nutrition Facts label from this bottle of mineral water.
04 Assessment Question Set 1 Solubility Equilibria
Q1. Write the balanced reaction for the dissolution of iron (III) hydroxide in water.
Q2. Write the equilibrium constant expression for the solubility of iron (III) hydroxide.
Q3. Calculate the molar solubility of iron (III) hydroxide in pure water.
Q4. Calculate the molar solubility of iron (III) hydroxide at pH 6.4, and compare it to the calculated molar solubility of iron (III) hydroxide in Q3.
Q5. Calculate the molar solubility of iron (III) hydroxide when the ionic strength of the pH 6.4 buffer is 0.10 mol L-1.
Q6. To prevent precipitation of iron (III) hydroxide in household water systems, hydrogen peroxide is often added as an oxidizing agent to municipal water sources containing iron (II) ion concentrations above 0.3 mg L‑1. Compare the effectiveness of this treatment at pH 6.4 versus pH 2.0 at a dissolved iron (i.e. Fe2+) concentration of 0.3 mg L-1. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/03_Principle_of_Electroneutrality.txt |
This document serves as a solution manual for the questions posed in the Lithia Water Springs Project. Instructors should feel free to use individual questions or modules in their entirety as they fit into the course material being covered. These modules may also be adapted to other samples such as wellwater or natural waters, such as freshwater and saltwater sources.
Answers to Questions:
05 Solutions Manual
Q1. Which cations and anions are represented by a single salt and which are represented by multiple salts?
From the summary of the 1915 Lithia water analysis on the engraved stone:
Cations: Na+ is represented by multiple salts, and K+, Li+, Ca2+, and Mg2+ are represented by single salts.
Anions: HCO3- is represented by multiple salts, and Cl-, SO42-, and BO2- are represented by single salts.
Q2. Of those analytes listed on the plaque, are there any cation or anion concentrations that are unknowns based on the information provided? How can you tell?
Iron and aluminum oxides are represented by a single concentration, and they would both be unknowns because it is impossible to calculate an analyte concentration when there are two unknowns and one concentration. Although it is not indicated on the plaque, based on recent analyses for the City of Ashland, instructors can assume that aluminum does not contribute significantly to the ionic content of Lithia water, and iron represents the vast majority of this Lithia water constituent. Instructors may also assume that all measured silica exists in Lithia water as SiO32-, the silicate ion.
Q3. Based on the analysis engraved on the plaque above, how would you calculate the concentration of each ion represented by a single salt? Complete these calculations, and express the concentration in milligrams per liter (mg L-1) and in moles per liter (M).
Listed below are the conversions of single salt concentrations to concentrations of individual cations and anions in Lithia water. Instructors may use these questions to provide students with examples of stoichiometric calculations.
1. Potassium
1. $\left(\dfrac{279.5 \: mg \: KHCO_{3}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: {KHCO}_{3}}{100.12\: {g}}\right)\left(\dfrac{1\: {mol}\: K^{+}}{1\: {mol}\: {KHCO}_{3}}\right)=2.792 \times 10^{-3}\: {M}\: {K}^{+}$
2. $\left(2.792 \times 10^{-3}\: M\: K^{+}\right)\left(\dfrac{39.098\: g\: K^{+}}{mol\: K^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{109.1\: mg\: K^{+}}{L}$
2. Lithium
1. $\left(\dfrac{153.82\: {mg}\: {LiHCO}_{3}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {LiHCO}_{3}}{67.96\: {g}}\right)\left(\dfrac{1\: {mol}\: {Li}^{+}}{1\: {mol}\: {LiHCO}_{3}}\right)=2.263 \times 10^{-3}\: {M}\: Li^{+}$
2. $\left(2.263 \times 10^{-3}\: M\: L i^{+}\right)\left(\dfrac{6.941\: g\: L i^{+}}{mol\: Li^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{15.7\: mg\: Li^{+}}{L}$
3. Calcium
1. $\left(\dfrac{1404\: mg\: Ca\left(H C O_{3}\right)_{2}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: C a\left(HCO_{3}\right)_{2}}{162.12\: g}\right)\left(\dfrac{1\: mol\: C a^{2+}}{1\: mol\: Ca\left(H C O_{3}\right)_{2}}\right)= {8.66 \times 10^{-3}\: M\: C a^{2+}}$
2. $\left(8.66 \times 10^{-3}\: M\: Ca^{2+}\right)\left(\dfrac{40.08\: g\: Ca^{2+}}{mol\: Ca^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{347\: mg\: Ca^{2+}}{L}$
4. Magnesium
1. $\left(\dfrac{1153\: {mg}\: {Mg}\left({HCO}_{3}\right)_{2}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Mg}\left({HCO}_{3}\right)_{2}}{146.35\: {g}}\right)\left(\dfrac{1\: {mol}\: {Mg}^{2+}}{1\: {mol}\: {Mg}\left({HCO}_{3}\right)_{2}}\right)=7.88 \times 10^{-3}\: {M}\: {Mg}^{2+}$
2. $\left(7.88 \times 10^{-3}\: M\: Mg^{2+}\right)\left(\dfrac{24.31\: g\: Mg^{2+}}{mol\: Mg^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{192\: mg\: Mg^{2+}}{L}$
5. Chloride
1. $\left(\dfrac{4515\: mg\: NaCl}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: NaCl}{58.44\: g}\right)\left(\dfrac{1\: mol\: Cl^{-}}{1\: mol\: NaCl}\right)=7.726 \times 10^{-2}\: M\: Cl^{-}$
2. $\left(7.726 \times 10^{-2}\: M\: Cl^{-}\right)\left(\dfrac{35.45\: g\: Cl^{-}}{mol\: Cl^{-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2739\: mg\: Cl^{-}}{L}$
6. Metaborate
1. $\left(\dfrac{321.3\: {mg}\: NaBO_2}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {NaBO}_{2}}{65.80\: {g}}\right)\left(\dfrac{1\: {mol}\: {BO}_{2}^{-}}{1\: {mol\: NaBO}_{2}}\right)=4.883 \times 10^{-3}\: {M}\: {BO}_{2}^{-}$
2. $\left(4.883 \times 10^{-3}\: M\: BO_{2}^{-}\right)\left(\dfrac{42.81\: g\: BO_{2}^{-}}{mol\: BO_{2}^{-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{209\: mg\: BO_{2}^{-}}{L}$
7. Sulfate
1. $\left(\dfrac{3.895\: {mg}\: {Na}_{2} {SO}_{4}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Na}_{2} {SO}_{4}}{142.04\: {g}}\right)\left(\dfrac{1\: {mol}\: {SO}_{4}^{2-}}{1\: {mol}\: {Na}_{2} {SO}_{4}}\right)=2.742 \times 10^{-5}\: {M}\: {SO}_{4}^{2-}$
2. $\left(2.742 \times 10^{-5}\: M\: SO_{4}^{2-}\right)\left(\dfrac{96.06\: g\: SO_{4}^{2-}}{mol\: SO_{4}^{2-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2.63\: mg\: SO_{4}^{2-}}{L}$
8. Silica (as silicate)
1. $\left(\dfrac{94.9\: {mg}\: {SiO}_{2}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: {mol}\: {SiO}_{2}}{60.08\: g}\right)\left(\dfrac{1\: {mol}\: {SiO}_{3}^{2-}}{1\: {mol}\: SiO_2}\right)=1.58 \times 10^{-3}\: {M}\: {Si}O_{3}^{2-}$
2. $\left(1.58 \times 10^{-3}\: M\: SiO_{3}^{2-}\right)\left(\dfrac{76.08\: g\: SiO_{3}^{2-}}{mol\: SiO_{3}^{2-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{120.2\: mg\: SiO_{3}^{2-}}{L}$
9. Iron (assuming aluminum content is negligible)
1. $\left(\dfrac{12.5\: {mg}\: {Fe}_{2}{O}_{3}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Fe}_{2} {O}_{3}}{159.69\: {g}}\right)\left(\dfrac{2\: {mol}\: {Fe}^{3+}}{1\: {mol}\: {Fe}_{2}{O}_{3}}\right)\left(\dfrac{1\: {mol}\: {Fe}^{2+}}{1\: {mol}\: {Fe}^{3+}}\right)=1.566\times10^{-4}\:M\:Fe^{2+}$
2. $\left(1.566 \times 10^{-4}\: M\: Fe^{2+}\right)\left(\dfrac{55.845\: g\: Fe^{2 +}}{mol\: Fe^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{8.74\: mg\: Fe^{2+}}{L}$
Q4. How does your approach to calculating ion concentrations represented by multiple salts differ from expressing ion concentrations represented by a single salt? Complete these calculations, and express the solute concentration in milligrams per liter (mg L-1) and in moles per liter (M).
Multiple sources of a common ion need to be converted from mg L-1 to mol L-1 and then added together. Sodium ion and bicarbonate ion have a common salt so the molarity of sodium bicarbonate may first be calculated. After this calculation, the molarity of the common ions can be calculated using stoichiometric relationships and the concentrations from the previous question.
1. Sodium bicarbonate
1. $\left(\dfrac{2456\: mg\: NaHCO_{3}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: NaHCO_{3}}{84.01\: g}\right)\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: NaHCO_{3}}\right)=2.923 \times 10^{-2}\: M\: Na^{+}$
2. $\left(2.923 \times 10^{-2}\: M\: Na^{+}\right)\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: Na^{+}}\right)=2.923 \times 10^{-2}\: M\: HCO_{3}^{-}$
2. All sources of sodium ion
1. $\left(2.923 \times 10^{-2}\: M\: Na^{+}\right)+\left(7.726 \times 10^{-2}\: M\: Cl^{-}\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: Cl^{-}}\right)\right)+ \left(4.883 \times 10^{-3}\: M\: BO_{2}^{-}\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: BO_{2}^{-}}\right)\right)+\left(2.742 \times 10^{-5}\: M\: SO_{4}^{2-}\left(\dfrac{2\: mol\: Na^{+}}{1\: mol\: SO^{2-}_{4}}\right)\right)= {0.1114\: M\: Na^{+}}$
2. $\left(0.1114\: M\: Na^{+}\right)\left(\dfrac{22.99\: g\: Na^{+}}{mol\: Na^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2562\: mg\: Na^{+}}{L}$
3. All sources of bicarbonate ion
1. $(2.923 \times 10^{-2}\: M\: HCO_{3}^{-})+\left(2.792 \times 10^{-3}\: M\: K^{+}\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: K^{+}}\right)\right)+\left(2.263 \times 10^{-3}\: M\: Li^{+}\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: Li^+}\right)\right)+\left(8.66 \times 10^{-3}\: M\: Ca^{2+}\left(\dfrac{2\: mol\: HCO_{3}^{-}}{1\: mol\: Ca^{2+}}\right)\right)+\left(7.88 \times 10^{-3}\: M\: Mg^{2+}\left(\dfrac{2\: mol\: HCO_{3}^-}{1\: mol\: Mg^{2+}}\right)\right)=0.06736\: M\: HCO_{3}^{-}$
2. $\left(0.06736\: M\: HCO_{3}^{-}\right)\left(\dfrac{61.02\: g\: {HCO}_{3}^{-}}{mol\: {HCO}_{3}^{-}}\right)\left(\dfrac{1000\: {mg}}{g}\right)=\dfrac{4110\: {mg}\: {HCO}_{3}^{-}}{L}$
Q5. Determine the number of milligrams of each analyte present in 25.00 mL of Lithia water from the 1915 analysis, assuming that the composition of Lithia water has not changed since 1915. Using these analyte quantities, determine appropriate analytical techniques based on the classification scheme in Table 1.
Students should recognize that multiplying the analyte concentration by a given volume will result in the mass (or moles) of analyte in that given volume. Each analyte concentration was calculated in the previous question in mg L-1, and those concentrations can be used for the determination of milligrams of each analyte in a 25.00 mL sample of Lithia water. The analytical technique(s) listed are suggested techniques based on analyte classification. It is up to the students, in consultation with the instructor, to determine which methods are most appropriate for a given laboratory project.
Table 1. Suggested analyte classification table for Lithia water analysis. See Analytical Chemistry 2.0 (Chapter 3.4, Scale of Operation) on the Analytical Sciences Digital Library website for additional information.
Analyte
Calculation
Analyte Classification (Potential Technique)
Sodium
$\left(\dfrac{2562\: mg\: Na^{+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=64.04\: mg$
Minor (ISE)
Potassium
$\left(\dfrac{109.1\: {mg}\: {K}^{+}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=2.7\: {mg}$
Minor (ISE or Spectroscopy)
Lithium
$\left(\dfrac{15.7\: {mg}\: Li^{+}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=0.4\: {mg}$
Trace (ISE or Spectroscopy)
Calcium
$\left(\dfrac{347\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=8.7\: mg$
Minor (Titrimetry, ISE or Spectroscopy)
Magnesium
$\left(\dfrac{192\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=4.8\: mg$
Minor (Titrimetry, ISE or Spectroscopy)
Iron
$\left(\dfrac{8.74\: mg\: Fe^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=0.2\: mg$
Trace (Spectroscopy)
Chloride
$\left(\dfrac{2739\: mg\: Cl^{-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=68.5\: mg$
Minor (ISE or Titrimetry)
Metaborate
$\left(\dfrac{209\: mg\: BO_{2}^{-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=5.2\: mg$
Minor (Titrimetry, ISE or Spectroscopy)
Sulfate
$\left(\dfrac{2.63\: {mg}\: {SO}_{4}^{2-}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=0.07\: {mg}$
Trace (Spectroscopy)
Bicarbonate
$\left(\dfrac{4110\: {mg}\: {HCO}_{3}^{-}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL}) =102.8\: {mg}$
Minor (Titrimetry)
Silicate
$\left(\dfrac{120.2\: mg\: SiO_{3}^{2-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=3.0\: mg$
Minor (Spectroscopy)
Q6. Once the appropriate analytical technique(s) for each analyte have been identified based on the classification scheme in Table 1, consider, discuss, and identify additional criteria that would be important in the selection of an appropriate method for each analyte in Lithia water. An appropriate starting point for this discussion would be to read Chapter 3.4 in Analytical Chemistry 2.0.
After students have read Chapter 3.4 in Analytical Chemistry 2.0 and calculated the analyte concentrations in Q5, each group should develop a set of criteria for the selection of each analytical method. Appropriate criteria include, but are not limited to:
1. Method Sensitivity. In this project, students are guided in their selection of an analytical technique based on the analyte classification scheme in Table 1.
2. Cost of materials & equipment. In this analytical chemistry course, the selection of an analytical technique that minimizes the cost of reagents and equipment is preferable.
3. Timeframe of analysis. An analyte determination that can be completed within one laboratory period is preferable. However, it may also be acceptable if one or more group members agree to complete a determination by remaining beyond the scheduled lab period. Such arrangements should be made ahead of time, and in such cases, solution and standard preparation should be completed ahead of time if possible.
4. Is the proper equipment for performing the determination and handling any toxic or noxious reagents available in the department?
The most appropriate choice of a proper technique and method for the determination of a chosen analyte depends on the equipment, time and resources available. A student may discover that there are multiple techniques available for the determination of an analyte. If this is the case, the criteria are designed to guide each group toward the least expensive, and presumably, the least complicated analytical technique that can be used for a given determination. However, it is up to the instructor to decide how the criteria are prioritized based on course goals.
In the past, the analytical chemistry laboratory prescribed the following analytical techniques for Lithia water analytes:
• Sodium, potassium: Ion-selective electrodes
• Calcium, magnesium: Complexometric titrations
• Chloride: Precipitation titration
• Bicarbonate: Acid-base titration
• Boron, Iron, Silicate, Sulfate: Molecular spectroscopy
• Lithium: Atomic emission spectroscopy, molecular spectroscopy
Q7. Lithium carbonate is typically prescribed to control manic depression. A typical dosage is 900 mg day-1 of lithium carbonate. How many liters of Lithia water would someone have to drink to ingest an equivalent amount of lithium that is present in a prescribed dose of lithium carbonate? Based on your calculated volume of Lithia water in the previous question, are there health related issues that could arise due to the presence of other ions that are listed on the Lithia water analysis plaque? To help answer this question, perform an internet search on the health effects and the dietary intake of any of the other cations or anions in Lithia water. Cite any sources you use to support your answer.
If a typical daily dosage of lithium carbonate to treat manic depression is 900 mg per day, the volume of Lithia water needed to supply the appropriate dosage of lithium ion is:
$\left(\dfrac{0.900\: g\: Li_{2}CO_{3}}{day}\right)\left(\dfrac{1\: {mol}\: Li_{2}CO_{3}}{73.892\: g\: Li_{2}CO_{3}}\right)\left(\dfrac{2\: mol\: Li^{+}}{1\: mol\: Li_{2}CO_{3}}\right)\left(\dfrac{1\: L\: Lithia\: water}{0.002263\: {mol}\: Li^{+}}\right)=\left(\dfrac{10.8\: L\: Lithia\: water}{day}\right)\nonumber$
Instructors may want to point out to their students that one of the inorganic ions in Lithia water that serves an important role in human health is sodium. Although sodium is necessary for proper electrolyte balance, which controls such functions as nerve impulse conduction, only small amounts of dietary sodium are necessary to maintain such functions. Excess sodium in the diet leads to health problems such as high blood pressure. According to the Harvard School of Public Health website on Salt and Health, U.S. government recommendations are 2300 milligrams of sodium as a maximum daily dietary intake. The sodium concentration in Lithia water, calculated in the previous question in this section, is 2573 milligrams per liter. Therefore, the volume of Lithia water that must be consumed in order to ingest the maximum daily intake of sodium ions is:
$\left(\dfrac{2.300\: g\: Na^{+}}{day}\right)\left(\dfrac{L\: Lithia\: water}{2.573\: g\: Na^{+}}\right)=\left(\dfrac{0.894\: L\: Lithia\: water}{day}\right)\nonumber$
This volume is twelve times less than the volume needed to supply the therapeutic amount of lithium to treat manic depression. Drinking the Lithia water from the Ashland, OR source would not be a healthy or effective way of administering a therapeutic dose of lithium. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/01_Major_Inorganic_Constituents.txt |
Q8. If the Lithia spring water was to leach ions from limestone deposits to form soda springs, what primary cations and anions would you expect to have dissolved in the spring water due to such leaching action?
Calcium ions and carbonate ions.
Q9. Soda springs are characterized by significant levels of CO2. If the pH of a recently obtained sample of Lithia water is 6.4, draw a ladder diagram for carbonic acid, and predict the predominant form of carbonic acid in this sample of Lithia water.
The pH of Lithia water is typically 6.4, and based on the ladder diagram for the carbonic acid system, the predominant form of carbonic acid is the bicarbonate ion and carbonic acid. Since the pH of Lithia water is slightly greater than pKa1, there is slightly more bicarbonate ion present than carbonic acid.
03 Sample Handling and Treatment
Q10. (Sample Handling & Treatment) Answer Q1 – Q13 in the Chemical Equilibria and Sample Preparation section of the Lithia Water Springs document. After completing these questions, go online to the U.S. Environmental Protection Agency website (www.epa.gov) and search for documentation regarding the preservation of inorganic analytes in water samples.
Q11. (Sample Handling & Treatment) Based on your search results on the U.S. EPA website and your answer to question 13 in the Chemical Equilibria and Sample Preparation section, develop a sampling plan for Lithia water so that all significant analytes can be determined.
One general approach to preserving Lithia water without compromising the determination of bicarbonate is to obtain two samples of Lithia water in Nalgene bottles, and acidify the sample that will be used to determine the concentration of all cationic species. A 1.0 mL sample of 6 M HNO3 added to 60 mL of Lithia water should be sufficient to reduce the sample pH below 2.0. The unpreserved sample should be tightly capped until the bicarbonate ion determination is about to be performed.
Activity #1: At this point, if you are going to have your students analyze some of the constituents of a local water source, it would be an appropriate time to have them collect the samples. If students are interested in investigating changes in sample appearance upon standing, students should obtain a small, clean Nalgene bottle (60 – 250 mL capacity) and rinse it thoroughly with the water sample. After rinsing, fill the bottle completely with water and cap it tightly. Students should record any observations regarding the appearance of the water immediately after collection and a day or two afterwards. They should measure the pH of the water sample and record that value in their laboratory notebook. If students have already developed a sampling plan, appropriate steps should be taken to properly clean sample containers and preserve samples as needed.
Instructors should inform their students that any change in the appearance of their natural water sample may indicate the occurrence of a chemical reaction that changes the chemical composition of the water sample. Such changes include the appearance of turbidity or precipitate, gas evolution and/or a color change. In the case of Lithia water, the Lithia water piped into downtown Ashland, Oregon is clear and colorless with a pH of approximately 6.4 and a faint sulfurous smell. After 1 – 2 days, the Lithia water sample becomes slightly turbid with a light orange precipitate at the bottom of the sample container. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/02_Origins_of_Lithia_Water.txt |
Q1. Which of the cationic solutes in Lithia water are main group ions? Which of the cationic solutes are transition metal ions? Evaluate whether main group and transition metal cations have lost all of their valence electrons. Can main group and/or transition metal cations potentially be oxidized by dissolved oxygen?
The only transition metal that is recognized as an inorganic constituent in Lithia water is iron. All of the other inorganic cations of historical significance determined in this course (K+, Na+, Li+, Ca2+, Mg2+) are main group ions. Main group cations tend to exist in their most oxidized state. Since there are no additional valence electrons, main group cations are unable to function as reductants and cannot reduce elemental oxygen to the oxide. Ferrous ion (Fe2+) can theoretically be oxidized to the ferric ion (Fe3+) and act as a reductant for species such as elemental oxygen.
Q2. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ under standard conditions. Predict the predominant oxidation state for iron under these conditions.
For this redox reaction:
\begin{align} &\ce{O2 (g) + 4 H+ (aq) + 4 e- \rightleftharpoons 2 H2O (l)} &&\mathrm{E° = 1.23\: V}\nonumber\ &\ce{Fe^3+ (aq) + e- \rightleftharpoons Fe^2+ (aq)} &&\mathrm{E° = 0.771\: V}\nonumber \end{align}\nonumber
Based on this ladder diagram, the predominant species in this solution under standard conditions would be H2O and Fe3+.
Q3. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ assuming the following non-standard conditions: assume that the dissolved oxygen concentration is 10 mg L-1, the total iron concentration is 10 mg L-1, and that the initial ratio of [Fe2+]/[Fe3+] is 1000. Predict the predominant oxidation state for iron under these conditions and compare the result to your answer in question 2.
For the redox couple of dissolved oxygen and ferric/ferrous ion, the following calculations need to be performed before constructing the ladder diagram:
Potential of the dissolved oxygen half-reaction:
$\ce{O2 (g) + 4 H+ (aq) + 4 e- \rightleftharpoons 2 H2O (l)}\nonumber$
$[H^+ ]=10^{-pH}=10^{-6.4}=3.98 \times 10^{-7}\: M \nonumber$
$[O_2 ]= \left(\dfrac{10.0\: mg\: O_2}{L}\right)\left(\dfrac{1.00\: mmol\: O_2}{32.0\: mg\: O_2}\right)\left(\dfrac{1\: mol\: O_2}{1000\: mmol\: O_2}\right)=3.1 \times 10^{-4}\: M\nonumber$
\begin{align} E_{O_2} &= E_{O_2}^o-\dfrac{0.05915}{n} \log \dfrac{1}{[O_2 ] [H^+ ]^4}\nonumber\ &= 1.23- \dfrac{0.05915}{4} \log \dfrac{1}{(3.1 \times 10^{-4} ) (3.98 \times 10^{-7} )^4}\nonumber\ &= 0.800\: V\nonumber \end{align}\nonumber
\begin{align} E_{Fe^{3+}/Fe^{2+}} &= E_{Fe^{3+}/Fe^{2+}}^o-\dfrac{0.05915}{n} \log \dfrac{[Fe^{2+} ]}{[Fe^{3+} ]} \nonumber\ &= 0.771- \dfrac{0.05915}{1} \log \dfrac{1000}{1}\nonumber\ &=0.594\: V\nonumber \end{align}\nonumber
The predominant oxidation state for iron under these conditions is Fe3+.
Q4. Refer to the plaque shown in Fig. 2 of the Major Inorganic Constituents section. What are the two predominant anions in Lithia water? Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect these anions to react with dissolved oxygen at concentrations expected in Lithia water? Provide appropriate support for your answer.
The two predominant anions in Lithia water are chloride and bicarbonate. Chloride ion may be oxidized to chlorine, and the Nernst equation plus a ladder diagram may be used to determine whether dissolved oxygen has sufficient capability as an oxidant under conditions found in Lithia water. Atmospheric chlorine is extremely reactive and is removed efficiently from the troposphere by chemical reactions. The smell of chlorine can be detected at 3 – 5 ppm. The determination of atmospheric chlorine in Ashland, OR has not been undertaken but a conservative estimate would be 1 pptv (part per trillion by volume) which is 1 pmol Cl2 per mol of air. Assuming air is an ideal gas:
$\ce{Cl2 (g) + 2e- \rightleftharpoons 2 Cl- } \hspace{30px} \mathrm{E^o = 1.396\: V}\nonumber$
$[Cl_2 ]= \left(\dfrac{1 \times 10^{-12}\: mol\: Cl_2}{mol\: air}\right)\left(\dfrac{1\: mol\: air}{22.4\: L}\right)=4.5 \times 10^{-14}\: M\: Cl_2\nonumber$
\begin{align} E_{Cl_2/Cl^-} &= E_{Cl_2/Cl^-}^o-\dfrac{0.05915}{n} \log \dfrac{[Cl^- ]^2}{[Cl_2 ]}\nonumber\ &= 1.396- \dfrac{0.05915}{2} \log \dfrac{(7.777\times10^{-2} )^2}{4.5 \times 10^{-14}} \nonumber\ &=1.067\: V\nonumber \end{align}\nonumber
According to the ladder diagram, dissolved oxygen and chloride ion do coexist in Lithia water. Under these conditions, dissolved oxygen is not a sufficient oxidant to oxidize chloride ion to chlorine.
In the bicarbonate ion, the oxidation number for carbon is +4, which is the highest oxidation state for carbon since elemental carbon has only four valence electrons. Therefore, dissolved oxygen cannot oxidize the bicarbonate ion.
Q5. Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect main group cations to react with transition metal cations present in Lithia water? Provide appropriate support for your answer.
The predominant main group cations in Lithia water are Li+, Na+, K+, Mg2+, and Ca2+. The predominant transition metal in Lithia water is Fe2+. The standard reduction potentials for the main group cations to their zero valent state and the standard reduction potential for the ferric-ferrous redox couple are listed in the table below:
Standard Reduction Half-Reaction
Eo (V)
$\ce{Li+ + e- ⇌ Li (s)}$
-3.04
$\ce{Na+ + e- ⇌ Na (s)}$
-2.713
$\ce{K+ + e- ⇌ K (s)}$
-2.93
$\ce{Mg^2+ + 2e- ⇌ Mg (s)}$
-2.356
$\ce{Ca^2+ + 2e- ⇌ Ca (s)}$
-2.84
$\ce{Fe^3+ + e- ⇌ Fe^2+ (aq)}$
+0.771
All of the main group cations exist in their fully oxidized state and are extremely weak oxidants as evidenced by the extremely negative (unfavorable) standard reduction potentials. The oxidation of ferrous ion to ferric ion is unfavorable except in the presence of a strong oxidant. Although an instructor may want students to construct a ladder diagram and use the Nernst equation to calculate the electrochemical potential for each half-cell, one can determine by inspection that the reduction of Group IA and IIA cations by the oxidation of ferrous ions to ferric ions is extremely unfavorable. Therefore, no redox reaction should take place among the main group ions and ferrous ions.
Q6. Briefly summarize the redox chemistry that can take place among the major inorganic cations in Lithia water.
Dissolved oxygen acts as an oxidant to oxidize ferrous ions to ferric ions. The oxidation of ferrous ions to ferric ions leads to the precipitation of ferric hydroxide, which appears as a fine orange precipitate at the bottom of the Lithia water sample. The main group cations are stable in Lithia water as they have large negative standard reduction potentials, which make them extremely weak oxidants.
Q7. Using the solubility rules provided above and the primary inorganic constituents in Lithia water, which inorganic salts would be most likely to form a precipitate in Lithia water?
Carbonates and hydroxides would be the most likely inorganic salts to form a precipitate in Lithia water.
Q8. Why is there no term for the solid calcium carbonate in the Ksp expression?
Unlike the concentration of dissolved ions in aqueous solution, which is an extensive property and varies based on the number of ions in a given volume of solvent, the concentration of a pure solid (or a pure liquid) is an intensive property and is constant. Equilibrium constant expressions do not contain concentration terms for pure solids or liquids, and they are given a value of 1 in the equilibrium constant expression.
Q9. Write expressions that relate the concentration of calcium ion and the concentration of phosphate ion to the solubility of calcium phosphate in water.
$\begin{array}{llclcl}\ce{ &Ca3(PO4)2 (s) &⇌ &3 Ca^2+ (aq) &+ &2 PO4^3- (aq)\ Initial\: (I) &------ & &0 & &0\ Change\: (C) &-X & &+3X & &+2X\ Equilibrium\: (E)&------ & &3X & &2X} \end{array}\nonumber$
If the molar solubility of calcium phosphate is X, then
$\mathrm{3X = [Ca^{2+}] \hspace{20px} or \hspace{20px} X = \dfrac{[Ca^{2+}]}{3}}\nonumber$
$\mathrm{2X = [PO_4^{3-}] \hspace{20px} or \hspace{20px} X = \dfrac{[PO_4^{3-}]}{2}}\nonumber$
So the molar solubility of Ca3(PO4)2 can be expressed as either ([Ca2+]/3) or ([PO43-]/2).
Q10. Calculate the solubility of calcium hydroxide in a solution that has a pH of 12. Note: The pH has been adjusted to 12 by some other means than the dissolution of calcium hydroxide. If calcium hydroxide is added to the water with a pH of 12, some of the calcium hydroxide will dissolve to satisfy the Ksp equilibrium expression for calcium hydroxide.
The solubility equilibrium established in a saturated solution of Ca(OH)2 is:
$\ce{Ca(OH)2 (s) ⇌ Ca^2+ (aq) + 2 OH- (aq)}\nonumber$
The equilibrium constant expression for Ca(OH)2 is:
$K_{sp} = [Ca^{2+}][OH^-]^2\nonumber$
The molar solubility of Ca(OH)2 can be defined as [Ca2+] because dissolving one mole of Ca(OH)2 provides one mole of calcium ions. Solving for the molar solubility of Ca(OH)2 may be accomplished using the following steps:
$[H^+ ]= 10^{-pH}= 1.0 \times 10^{-12}\: M \nonumber$
$[OH^-] = \dfrac{K_w}{[H^+]}=\dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-12} } = 1.0 \times 10^{-2}\: M \nonumber$
$[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(0.010)^2} =0.065\: M \nonumber$
Q11. What is the molar solubility and mass solubility (i.e. solubility in grams of solid per liter of solution) of each of the metal hydroxides listed in Table 1 at pH values of 10.0, 6.0, and 2.0? How is the solubility of a metal hydroxide affected by pH? Offer an explanation for this behavior using LeChatelier’s principle.
To solve for the molar solubility of each metal hydroxide at each pH, the hydroxide ion concentration must be calculated at each pH and substituted into the equilibrium constant expression to solve for the molar solubility of the salt. The molar solubility can be converted to mass solubility by multiplying the molar solubility by the molar mass of the salt.
Step 1: Calculate the hydroxide ion concentration at pH 10.0, 6.0, and 2.0
$[H^+ ]= 10^{-pH}= 10^{-10.0}= 1.0 \times 10^{-10}\: M \nonumber$
$[OH^-] = \dfrac{K_w}{[H^+]}=\dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-10}} = 1.0 \times 10^{-4}\: M \nonumber$
$At\: pH\: 6.0,\: [OH^- ]=1.0 \times 10^{-8}\: M \nonumber$
$At\: pH\: 2.0,\: [OH^- ]=1.0 \times 10^{-12}\: M \nonumber$
Step 2: Calculate the molar solubility of calcium hydroxide at each pH level
$At\: pH\: 10.0,\:[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-4}\: M)^2} =6.5 \times 10^2\: M \nonumber$
$At\: pH\: 6.0,\:[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-8}\: M)^2} =6.5 \times 10^{10}\: M \nonumber$
$At\: pH\: 2.0,\: [Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-12}\: M)^2} =6.5 \times 10^{18}\: M \nonumber$
Step 3: Calculate the mass solubility (S) of calcium hydroxide at each pH level
\begin{align} At\: pH\: 10.0,\: S_{Ca(OH)_2} &= [Ca^{2+} ] MW_{Ca(OH)_2 }\nonumber\ &= (6.5 \times 10^2\: M\: Ca^{2+} )\left(\dfrac{1\: mol\: Ca(OH)_2}{1\: mol\: Ca^{2+}}\right)\left(\dfrac{74.093\: g\: Ca(OH)_2}{1\: mol\: Ca(OH)_2}\right)\nonumber\ &=\dfrac{4.8 \times 10^4\: g\: Ca(OH)_2}{L\: solution}\nonumber \end{align}\nonumber
$At\: pH\: 6.0,\: S_{Ca(OH)_2} =\dfrac{4.8 \times 10^{12}\: g\: Ca(OH)_2}{L\: solution}\nonumber$
$At\: pH\: 2.0,\: S_{Ca(OH)_2} =\dfrac{4.8 \times 10^{19}\: g\: Ca(OH)_2}{L\: solution}\nonumber$
The table below contains the molar solubility and mass solubility for all the hydroxide salts from Table 1 in the Chemical Equilibria document.
Molar Solubility (M)
Mass Solubility (g/L)
Salt
Ksp
pH 10
pH 6
pH 2
pH 10
pH 6
pH 2
Ca(OH)2
6.5 x 10-6
6.5 x 102
6.5 x 1010
6.5 x 1018
4.8 x 104
4.8 x 1012
4.8 x 1018
Mg(OH)2
7.1 x 10-12
7.1 x 10-4
7.1 x 104
7.1 x 1012
4.1 x 10-2
4.1 x 106
4.1 x 1014
Al(OH)3
4.6 x 10-33
4.6 x 10-21
4.6 x 10-9
4.6 x 103
3.6 x 10-19
3.6 x 10-7
3.6 x 105
Fe(OH)2
8.0 x 10-16
8.0 x 10-8
8.0 x 100
8.0 x 108
7.2 x 10-6
7.2 x 102
7.2 x 1010
Fe(OH)3
1.6 x 10-39
1.6 x 10-27
1.6 x 10-15
1.6 x 10-3
1.7 x 10-25
1.7 x 10-13
1.7 x 10-1
Note that for salts having a relatively large Ksp or under acidic conditions, the predicted molar solubility and mass solubility are unrealistically large. In these cases, the solubility is no longer dependent on the solubility equilibrium but on the number of moles of solvent required to interact with each mole of solute. The molarity of water is 55.6 M, and each ion would be surrounded by several molecules of water. It is, therefore, unreasonable to expect that the solute concentration under low pH conditions would exceed several moles per liter.
The solubility of metal hydroxides increases with decreasing pH. Metal hydroxides that have a cation with an oxidation state of +2 exhibit a one hundred-fold increase in their solubility for every unit decrease in pH. Metal hydroxides that have a cation with an oxidation state of +3 exhibit a one thousand-fold increase in their solubility for every unit decrease in pH.
Q12. Integrating the ideas of redox chemistry and solubility expressed in this module, summarize the chemical and physical changes you would expect over time in a sample of Lithia water? What sample preparation step could be used to prevent such changes from occurring?
Previous questions on redox chemistry (see Q6 in this guide) indicate that dissolved oxygen does not affect the concentration or oxidation state of main group metals, but that dissolved oxygen is predicted to oxidize ferrous ion to ferric ion. Previous questions on analyte solubility as a function of pH indicate that metal hydroxides of interest in the analysis of Lithia water tend to be insoluble under basic to slightly acidic conditions, but that most of these hydroxides are moderately to freely soluble at low pH. The Ksp of ferric hydroxide is orders of magnitude lower than ferrous hydroxide, which results in the precipitation of ferric hydroxide in Lithia water after the oxidation of ferrous ion by dissolved oxygen occurs. Observations in the author’s laboratory indicated that most of the ferrous ion in Lithia water is oxidized and precipitates as ferric hydroxide within the first 24 hours after sample collection. However, at low pH (< 2.0), even the least soluble of the metal hydroxides, ferric hydroxide, is moderately soluble at pH 2.0, with a mass solubility of 170 mg L-1. Therefore, it is suggested that if the pH of the Lithia water is adjusted to a pH of 2.0 or less, analytes that form insoluble hydroxides can be prevented from precipitating by acidification with trace-metal grade or reagent grade acid. Instructors are encouraged to address this question as students are observing changes in their mineral water sample.
Q13. Collection of samples for trace metal analysis in drinking water is described in the Sample Collection, Preservation, and Storage section of EPA method 200.5, Determination of Trace Elements in Drinking Water by Axially Viewed Inductively Coupled Plasma - Atomic Emission Spectrometry (http://www.epa.gov/nerlcwww/ordmeth.htm). After reading this document, suggest a general approach to preserving Lithia water prior to analysis. Would this work for all analytes of interest in Lithia water? Explain.
EPA method 200.5 is a protocol for the determination of a large set of cations in drinking water. In the Collection, Preservation, and Storage section of EPA method 200.5, to properly keep all inorganic cations in solution, the analyst is instructed to acidify a one liter water sample with 3.0 mL of (1+1) nitric acid, and after 16 hours of cold storage, verify that the pH of the acidified sample is below 2.0.
The effect of sample acidification on the solubility of sparingly soluble salts is introduced in the Lithia water module entitled Chemical Equilibria. In the case of Lithia water, students are expected to discover that dissolved oxygen oxidizes the ferrous ion to ferric ion. Because iron (III) hydroxide is significantly less soluble than iron (II) hydroxide, a layer of iron (III) hydroxide settles out of solution as an orange precipitate in 1-2 days unless the sample is acidified. However, while sample acidification keeps metal ions in solution, sample acidification also protonates the bicarbonate ion to form carbonic acid, which subsequently dissociates into carbon dioxide and water. Therefore, students analyzing Lithia water should suggest the collection of two samples; one acidified sample for the determination of iron and another unacidified sample for the remaining analytes. Lithia water students can also suggest, based on EPA method 200.5, that the acidified sample be used for the determination of metal ions, and the unacidified sample be used for the determination of anions. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/04_Chemical_Equilibria_and_Sample_Preparation_%28Q1__.txt |
Q14. Calculate the alpha values for the three carbonic acid species (H2CO3, HCO3-, CO32-) at a pH of 6.4. Do these values agree with your predictions based on the carbonic acid ladder diagram?
The ladder diagram for the carbonic acid system is shown below (Ka1 = 4.47 x 10-7, Ka2 = 4.68 x 10-11). Once students are introduced to alpha values, they should realize that at a given pH, the fraction of each form of the analyte can be calculated using only the Ka values and the pH.
$\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-6.4}=3.98 \times 10^{-7}\: \mathrm{M}\nonumber$
$\begin{array}{rl} \alpha_{\mathrm{H}_{2} \mathrm{CO}_{3}} &=\dfrac{\left[\mathrm{H}^{+}\right]^{2}}{\left[\mathrm{H}^{+}\right]^{2}+\left[\mathrm{H}^{+}\right] \mathrm{K}_{\mathrm{a} 1}+\mathrm{K}_{\mathrm{a} 1} \mathrm{K}_{\mathrm{a} 2}} \nonumber\ &={\dfrac{\left(3.98 \times 10^{-7}\: \mathrm{M}\right)^{2}}{\left(3.98 \times 10^{-7}\: \mathrm{M}\right)^{2}+\left(3.98 \times 10^{-7}\: \mathrm{M}\right)\left(4.47 \times 10^{-7}\right)+\left(4.47 \times 10^{-7}\right)\left(4.68 \times 10^{-11}\right)}} \nonumber\ &=0.471\nonumber \end{array}\nonumber$
$\begin{array}{rl} \alpha_{\mathrm{HCO}_{3}^{-}}&=\dfrac{\left[\mathrm{H}^{+}\right] \mathrm{K}_{\mathrm{a} 1}}{\left[\mathrm{H}^{+}\right]^{2}+\left[\mathrm{H}^{+}\right] \mathrm{K}_{\mathrm{a} 1}+\mathrm{K}_{\mathrm{a} 1} \mathrm{K}_{\mathrm{a} 2}} \nonumber\ &= \dfrac{\left(3.98 \times 10^{-7}\: \mathrm{M}\right)\left(4.47 \times 10^{-7}\right)}{\left(3.98 \times 10^{-7}\: \mathrm{M}\right)^{2}+\left(3.98 \times 10^{-7}\: \mathrm{M}\right)\left(4.47 \times 10^{-7}\right)+\left(4.47 \times 10^{-7}\right)\left(4.68 \times 10^{-11}\right)}\nonumber\ &=0.529\nonumber \end{array}\nonumber$
$\begin{array}{rl} \alpha_{\mathrm{CO}_{3}^{2-}}&=\dfrac{\mathrm{K}_{\mathrm{a} 1} \mathrm{K}_{\mathrm{a} 2}}{\left[\mathrm{H}^{+}\right]^{2}+\left[\mathrm{H}^{+}\right] \mathrm{K}_{\mathrm{a} 1}+\mathrm{K}_{\mathrm{a} 1} \mathrm{K}_{\mathrm{a} 2}} \nonumber\ &={\dfrac{\left(4.47 \times 10^{-7}\right)\left(4.68 \times 10^{-11}\right)}{\left(3.98 \times 10^{-7}\: \mathrm{M}\right)^{2}+\left(3.98 \times 10^{-7}\: \mathrm{M}\right)\left(4.47 \times 10^{-7}\right)+\left(4.47 \times 10^{-7}\right)\left(4.68 \times 10^{-11}\right)}} \nonumber\ &=0.000062\nonumber \end{array}\nonumber$
The calculated fractions of each carbonate species make sense since the pH (6.4) is close to but slightly more basic than the pKa1 (6.35) for carbonic acid. When pH is equal to pKa1, then the fraction of each species involved in the equilibrium described by Ka1 by definition equals 0.50.
Q15. Given that the pH of the Lithia water is 6.4, what is the total concentration of all species in the carbonic acid equilibrium?
The introduction of alpha values enables students to account for the acid-base behavior of ionic species in solution. In the stoichiometric calculations used to determine the concentrations of inorganic constituents in Lithia water (Major Inorganic Constituents – Q3 and Q4), it was determined that the bicarbonate concentration in Lithia water, based on the analysis on the 1915 plaque, was 0.06736 M. Assuming that the pH of the water was 6.4, 52.9% of the species in the carbonic acid equilibrium exist as bicarbonate. Therefore, the total concentration of all species in the carbonic acid equilibrium is:
$c_T=\dfrac{0.06736\: M}{0.529}=0.1273\: M\nonumber$
This concentration represents the sum of all carbonate-based species in the carbonic acid equilibrium (i.e. cT = [CO32-] + [HCO3-] + [H2CO3])
Q16. What is the concentration of carbonate ion?
$\ce{[CO_3^{2-} ]}= \ce{\alpha_{CO_3^{2-}} } c_T=(0.000062)(0.1273M)= 7.92 \times 10^{-6}\: M\nonumber$
Q17. Using this concentration of carbonate, calculate the concentration of calcium that you would expect in Lithia water from the dissolution of calcium carbonate.
$\mathrm{K_{sp} = [Ca^{2+}][CO_3^{2-}]}\nonumber$
$[Ca^{2+} ][CO_3^{2-} ]=K_{sp}=4.5 \times 10^{-9}\nonumber$
$[Ca^{2+} ]= \dfrac{K_{sp}}{[CO_3^{2-} ]} = \dfrac{4.5 \times 10^{-9}}{7.92 \times 10^{-6} }=5.68 \times 10^{-4}\: M \nonumber$
And the mass solubility (S)
$\mathrm{S=(5.68 \times 10^{-4}\: M\: Ca^{2+} )\left(\dfrac{40.08\: g\: Ca^{2+}}{1\: mol\: Ca^{2+}}\right)\left(\dfrac{1000\: mg}{1\: g}\right)=22.8\: mg\: L^{-1}}\nonumber$
Q18. Compare this value to the value of calcium on the plaque.
In the section on Major Inorganic Constituents, the concentration of calcium in Lithia water was determined to be 347 mg L-1. This concentration was based on the analysis engraved on the 1915 plaque. This concentration is approximately fifteen times higher than the calcium ion content calculated in the previous problem, which indicates that the acid-base chemistry of the carbonic acid system is not the only factor that controls the calcium ion concentration in Lithia water. Instructors may want to stress this difference between calculated and determined calcium concentrations as an introduction to the effect of ionic strength on solubility.
Q19. What effect would raising the ionic strength of a solution have on the solubility of a sparingly soluble salt? Think about what relatively high concentrations of other ions (e.g., Na+ and Cl-) might have in a solution containing smaller amounts of Ca2+ and CO32-.
In the case of the solubility of a sparingly soluble salt, analyte ions (Ca2+ and CO32-) experience coulombic interactions with the supporting electrolyte of opposite charge (e.g. CO32- ions interact with Na+, and Ca2+ ions interact with Cl-) . The ions from the supporting electrolyte partially screen the charge of each analyte ion. These interactions decrease the activity, or the effective concentration, of the analyte ions.The solubility equilibrium and equilibrium constant expression of calcium carbonate is expressed by:
$\ce{CaCO3 (s) \rightleftharpoons Ca^2+ (aq) + CO3^2- (aq)} \hspace{20px} K_{sp}=a_{Ca^{2+}}a_{CO_3^{2-}}\nonumber$
As the ionic strength of a solution increases, the screening of the analyte ions increases, and the analyte ion activity decreases. According to LeChatelier’s principle, if the activity of the dissolved ions decreases, then more of the solid should dissolve so that the product of the calcium ion and carbonate ion concentrations once again equals the solubility product. Instructors should keep in mind that students do not need to know the Debye-Hückel equation in order to answer this question.
Q20. Calculate the molar solubility of AgCl under infinite dilution conditions (i.e. µ = 0) with its solubility in 0.10 M NaNO3. The hydrated ion diameters and thermodynamic equilibrium constants may be obtained from Analytical Chemistry 2.0.
$\ce{AgCl(s) \rightleftharpoons Ag+ (aq) + Cl- (aq)} \hspace{20px} \mathrm{K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}}\nonumber$
Under infinite dilution conditions:
$\mathrm{K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}}\nonumber$
Let X be the molar solubility of AgCl
$\mathrm{[Ag^+][Cl^-] = (X)(X) = X^2 = 1.8 \times 10^{-10}}\nonumber$
$\mathrm{X = 1.34 \times 10^{-5}\: M}\nonumber$
Taking ionic strength into account:
• Calculation of ionic strength (µ)
• $\mu= \dfrac{1}{2} \sum c_i z_i^2 = \dfrac{1}{2} \left((0.10\: M\: Na^+ ) (+1)^2+(0.10\: M\: NO_3^- ) (-1)^2 \right) = 0.10$
• Calculation of activity coefficients (γ) using Debye – Hückel equation
$\log γ_x= \dfrac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}\nonumber$
$γ_{Ag^+} = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(1^2 ) \sqrt{0.10}}{1+3.3(0.25\: nm) \sqrt{0.10}}}} = 0.7449 \nonumber$
$γ_{Cl^-} \mathrm{= 10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(-1^2 ) \sqrt{0.10}}{1+3.3(0.3\: nm) \sqrt{0.10}}}}=0.7537\nonumber$
• Calculation of a concentration-based solubility productand molar solubility of AgCl
• $K_{sp}^{'} = \dfrac{K_{sp}}{γ_{Ag^+} γ_{Cl^-} }=\dfrac{1.8 \times 10^{-10}}{(0.7449)(0.7537)} =3.2 \times 10^{-10}$
• $X^2=3.2 \times 10^{-10}$
• $X=1.79 \times 10^{-5}\: M$
Q21. Calculate the concentration-based equilibrium constant for CaCO3, H2CO3, and HCO3-, and use these constants to predict the calcium concentration in Lithia water at pH 6.4. The ionic strength of Lithia water may be calculated from the solute molarities reported on the plaque from the Major Inorganic Constituents section, and the hydrated ion diameters and thermodynamic equilibrium constants may be obtained from Analytical Chemistry 2.0.
For the solubility equilibrium
$\ce{CaCO3 (s) ⇌ Ca^2+ (aq) + CO3^2- (aq)}\nonumber$
Taking ionic strength into account:
• Calculation of ionic strength (µ): This calculation involves the analyte molarities that were calculated from the 1915 analysis of Lithia water.
• $μ= \dfrac{1}{2} \sum c_i z_i^2 =\dfrac{1}{2} \begin{pmatrix} (0.1114\: M\: Na^+ ) (+1)^2 + (0.002792\: M\: K^+ ) (+1)^2 + (0.002263\: M\: Li^+ ) (+1)^2 + \ (0.00866\: M\: Ca^{2+} ) (+2)^2 + (0.00788\: M\: Mg^{2+} ) (+2)^2 + \ (0.0001566\: M\: Fe^{2+}) (+2)^2 + (0.07726\: M\: Cl^- ) (-1)^2 + \ (0.06736\: M\: HCO_3^- ) (-1)^2 + (0.004883\: M\: BO_2^- ) (-1)^2 + \ (0.00002742\: M\: SO_4^{2-} ) (-2)^2 + (0.00158\: M\: SiO_3^{2-} ) (-2)^2 \end{pmatrix} = 0.170$
• Calculation of activity coefficients (γ) using the Debye – Hückel equation for inorganic salt
$\mathrm{\log γ_x= \dfrac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}\nonumber$
$γ_{Ca^{2+}} = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}} = \mathrm{10^{\Large\frac{(-0.51)(2^2 ) \sqrt{0.170}}{1+3.3(0.60\: nm) \sqrt{0.170}}}} =0.3443\nonumber$
$γ_{CO_3^{2-}} = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}} = \mathrm{10^{\Large\frac{(-0.51)(-2^2 ) \sqrt{0.170}}{1+3.3(0.45\: nm) \sqrt{0.170}}}} =0.3008\nonumber$
For the weak acid equilibria
$\ce{H2CO3 (aq) ⇌ H+ (aq) + HCO3- (aq)} \hspace{20px} K_{a1}=4.47 \times 10^{-7}=\dfrac{[H^+ ][HCO_3^- ]}{[H_2 CO_3 ]}\nonumber$
$\ce{HCO3- (aq) ⇌ H+ (aq) + CO3^2- (aq)} \hspace{20px} K_{a2}=4.68 \times 10^{-11}=\dfrac{[H^+ ][CO_3^{2-} ]}{[HCO_3^- ] }\nonumber$
• Calculation of activity coefficients (γ) using Debye – Hückel equation for weak acid
$γ_{H^+} = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}} = \mathrm{10^{\Large\frac{(-0.51)(+1^2 ) \sqrt{0.170}}{1+3.3(0.9\: nm) \sqrt{0.170}}}} =0.8044\nonumber$
$γ_{CO_3^{2-}} = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}} = \mathrm{10^{\Large\frac{(-0.51)(-2^2 ) \sqrt{0.170}}{1+3.3(0.45\: nm) \sqrt{0.170}}}} =0.3008\nonumber$
$γ_{HCO_3^- } = \mathrm{10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}} = \mathrm{10^{\Large\frac{(-0.51)(-1^2 ) \sqrt{0.170}}{1+3.3(0.45\: nm) \sqrt{0.170}}}} =0.7406\nonumber$
• Calculation of a concentration-based acid dissociation constant and the molar concentration of the carbonate ion
$K_{a2}^{'} = \dfrac{K_{a2} γ_{HCO_3^- }}{γ_{H^+} γ_{CO_3^{2-} }} = \dfrac{(4.68 \times 10^{-11} )(0.7406)}{(0.8044)(0.3008)} =1.43 \times 10^{-10}\nonumber$
$1.43 \times 10^{-10}= \dfrac{[H^+ ][CO_3^{2-} ]}{[HCO_3^- ]} \nonumber$
Based on the 1915 analysis, [HCO3-] = 0.06736 M, and assuming that the pH is 6.4 ([H+] = 3.98 x 10-7 M):
$1.43 \times 10^{-10}= \dfrac{(3.98 \times 10^{-7}\: M)[CO_3^{2-} ]}{(0.06736\: M)}\nonumber$
$2.42 \times 10^{-5}\: M=[CO_3^{2-} ]\nonumber$
• Calculation of a concentration-based solubility product and both the molar solubility and mass solubility for calcium carbonate in Lithia water
• $K_{sp}^{'}= \dfrac{K_{sp}}{γ_{Ca^{2+}} γ_{CO_3^{2-}} }=\dfrac{4.5 \times 10^{-9}}{(0.3443)(0.3008)} =4.35 \times 10^{-8}$
• $K_{sp}^{'}=[Ca^{2+} ][CO_3^{2-} ]$
• $4.35 \times 10^{-8}=[Ca^{2+} ](2.42 \times 10^{-5}\: M)$
• $[Ca^{2+} ]=1.79 \times 10^{-3}\: M$
• $S_{CaCO_3} = 1.79 \times 10^{-3}\: M\left(\dfrac{100.09\: g\: CaCO_3}{mol}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{179.6\: mg\: Ca^{2+}}{L}$
Students should be able to communicate that although the mass solubility of calcium carbonate in Lithia water is within a factor of two of the reported solubility from the 1915 analysis, there still may be another factor left unexplained which would account for the additional calcium in the Lithia water.
Q22. Remembering that the Lithia water contains carbonate ion, use LeChatelier’s principle to explain what effect the addition of carbonic acid has on the concentration of carbonate ion in Lithia water?
The carbonate buffer system consists of carbonic acid, hydrogen carbonate ion (bicarbonate), and the carbonate ion, and all three of these chemical species contain the carbonate ion. Carbonic acid is the fully protonated form in the buffer system and carbonate ion is the fully deprotonated form in the buffer system. When carbonic acid is introduced to a solution containing carbonate ion, the dissociation of protons from carbonic acid shifts the carbonate equilbrium in Lithia water favoring the protonation of carbonate ion to form bicarbonate ion:
$\ce{H3O+ (aq) + CO3^2- (aq) ⇌ HCO3- (aq) + H2O (l)}\nonumber$
This reaction decreases the concentration of the carbonate ion in solution.
Q23. Considering your answer to the preceding question, what effect would the addition of carbonic acid have on the solubility of calcium carbonate and other sparingly soluble carbonate salts?
For the solubility equilibrium
$\ce{CaCO3 (s) ⇌ Ca^2+ (aq) + CO3^2- (aq)}\nonumber$
The addition of carbonic acid would reduce the carbonate ion concentration. Based on Le Chatelier’s principle, if the concentration of a product is decreased in a chemical equilibrium, then the equilibrium would shift towards the products to reestablish equilibrium. This shift in the solubility equilibrium of calcium carbonate would increase the solubility of calcium carbonate.
Q24. The concentration of calcium ion based on the 1915 analysis of Lithia water is 347 mg L-1. Assuming that the calcium ion concentration in Lithia water is controlled by the carbonate ion concentration in Lithia water, calculate the carbonate ion concentration in Lithia water using the concentration-based equilibrium constant for calcium carbonate.
From Q21 (Solubility Equilibria):
• $K_{sp}^{'}= \dfrac{K_{sp}}{γ_{Ca^{2+}} γ_{CO_3^{2-}} }=\dfrac{4.5 \times 10^{-9}}{(0.3443)(0.3008)} =4.35 \times 10^{-8}$
• $K_{sp}^{'}=[Ca^{2+} ][CO_3^{2-} ]$
From Q3 (Major Inorganic Constituents):
• $4.35 \times 10^{-8}=(8.66 \times 10^{-3}\: M)[CO_3^{2-} ]$
• $[CO_3^{2-} ]=5.02 \times 10^{-6}\: M$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/05_Solubility_Equilibria__Advanced_Topics.txt |
Q1. List the cations and anions that were determined this term
This answer will differ based on the type of sample available, time spent on the project, etc.In any case, the answer to this question will guide the student toward constructing the appropriate charge balance equation. For Lithia water, a typical list of ions is:
Cations: Na+, K+, Li+, Ca2+, Mg2+, Fe2+
Anions: HCO3-, Cl-, BO2-, SO42-, SiO32-
Q2. Based on the principle of electroneutrality, what should be true about the concentration of cationic charge and anionic charge in Lithia water.
The number of moles of positive charge should equal the number of moles of negative charge.
Q3. Assuming that the inorganic constituents of Lithia water have been completely characterized, write down an appropriate charge balance equation for all dissolved ions in Lithia water.
Based on the list of ions in Q1 of this section, and assuming that silica exists in solution as the silicate ion, the appropriate charge balance equation for the inorganic ions in Lithia water would be:
$\ce{[Na+] + [K+] + [Li+] + 2[Ca^2+] + 2[Mg^2+] + 2[Fe^2+]} = \ce{[HCO3- ] + [Cl- ] + [BO2- ] + [SO4^2- ] + [SiO3^2- ]}\nonumber$
Because the pH of Lithia water is 6.4 and the concentration of the other ions is large, the contribution of H+ and OH- to the charge balance reaction can be neglected. In other cases, such as solutions having low ionic strength, it may be necessary to include these species in the calculation of electroneutrality.
Q4. Use the error propagation equations described in Chapter 4 of Analytical Chemistry 2.0 to propagate the experimental error in the following analytical determinations
1. $D= \dfrac{M}{V}= \dfrac{1.008\: g}{1.000\: mL}=1.008\: g\: mL^{-1}\nonumber$
To estimate the uncertainty of an experimental result that is calculated using solely multiplication and division:
$\dfrac{u_D}{D}= \sqrt{\left(\dfrac{u_M}{M}\right)^2 + \left(\dfrac{u_V}{V}\right)^2 }\nonumber$
$\dfrac{u_D}{1.008\: g\: mL^{-1} }= \sqrt{\left(\dfrac{0.023\: g}{1.008\: g}\right)^2 + \left(\dfrac{0.012\: mL}{1.000\: mL}\right)^2 }\nonumber$
$u_D=0.026\: g\: mL^{-1}\nonumber$
The density of the solution is 1.008 ± 0.026 g mL-1.
2. $\mathrm{∆V = V_f - V_i = 15.68\: mL - 5.36\: mL = 10.32\: mL}\nonumber$
To estimate the uncertainty of an experimental result that is calculated using solely addition and subtraction:
$u_{∆V}=\sqrt{u_{V_f}^2 + u_{V_i}^2 }\nonumber$
$u_{∆V}=\sqrt{(0.02\: mL)^2+(0.02\: mL)^2 }=0.028\: mL\nonumber$
Rounded to the nearest hundredth of a milliliter, the delivered volume is (10.32 ± 0.03) mL.
3. $D_{liq}=\dfrac{(M_{full}-M_{empty} )}{V_{flask}} = \dfrac{(248.3\: g-45.5\: g)}{100.0\: mL}=2.028\: g\: mL^{-1}\nonumber$
To estimate the uncertainty of an experimental result that is calculated using two or more sets of mathematical operations (in this case, addition and/or subtraction followed by multiplication and/or division), one must propagate error through one set of operations (in this case, subtraction of two masses) followed by propagating the error through the remaining set of operations (in this case, dividing volume into mass):
$(M_{full}-M_{empty} )=(248.3\: g-45.5\: g)= 202.8\: g\nonumber$
$u_{∆M}=\sqrt{u_{M_{full}}^2 + u_{M_{empty}}^2 }\nonumber$
$u_{∆V}=\sqrt{(0.1\: g)^2+(0.1\: g)^2} =0.141\: g \nonumber$
(keep additional 1 or 2 digits to minimize rounding errors)
$\dfrac{u_D}{D}= \sqrt{\left(\dfrac{u_M}{M}\right)^2 + \left(\dfrac{u_V}{V}\right)^2} \nonumber$
$\dfrac{u_D}{2.028\: g\: mL^{-1} }= \sqrt{\left(\dfrac{0.141\: g}{202.8\: g}\right)^2 + \left(\dfrac{0.08\: mL}{100.0\: mL}\right)^2 }\nonumber$
$u_D=0.002\: g\: mL^{-1}\nonumber$
The density of the liquid is 2.028 ± 0.002 g mL-1.
Q5. Based on your reading of Chapter 4 in Analytical Chemistry 2.0, what statistical approach would you use to determine whether equivalent amounts of cationic charge and anionic charge were determined in Lithia water?
A t-test for two experimental means would be the appropriate statistical test for determining whether statistically equivalent amounts of cationic charge and anionic charge were determined in Lithia water.
Q6. The Nutrition Facts label on a bottle of local mineral water contains the following information on select inorganic solutes (standard deviations provided by author, assume each result based on triplicate readings (n=3)):
Solute
Concentration (mg L-1)
Bicarbonate
466 (± 24)
Magnesium
124 (± 5)
Calcium
2.67 (± 0.10)
Sodium
3.31 (± 0.15)
Zinc
4.45 (± 0.10)
At the 95% confidence level, determine whether all of the major inorganic solutes have been reported on the Nutrition Facts label from this bottle of mineral water.
Step 1: Convert all solute concentrations and uncertainties into molarities
$\left(\dfrac{466\: mg\: HCO_3^-}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: HCO_3^-}{61.02\: g\: HCO_3^-} \right)=7.637 \times 10^{-3}\: M\: HCO_3^-\nonumber$
$\left(\dfrac{24\: mg\: HCO_3^-}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: HCO_3^-}{61.02\: g\: HCO_3^- }\right)=0.393 \times 10^{-3}\: M\: HCO_3^-\nonumber$
$\left(\dfrac{124\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Mg^{2+}}{24.305\: g\: Mg^{2+}} \right)=5.102 \times 10^{-3}\: M\: Mg^{2+}\nonumber$
$\left(\dfrac{5\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Mg^{2+}}{24.305\: g\: Mg^{2+}} \right)=0.206 \times 10^{-3}\: M\: Mg^{2+}\nonumber$
$\left(\dfrac{2.67\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Ca^{2+}}{40.08\: g\: Ca^{2+}} \right)=6.662 \times 10^{-5}\: M\: Ca^{2+}\nonumber$
$\left(\dfrac{0.10\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Ca^{2+}}{40.08\: g\: Ca^{2+}}\right)=0.250 \times 10^{-5}\: M\: Ca^{2+}\nonumber$
$\left(\dfrac{3.31\: mg\: Na^+}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Na^+}{22.9898\: g\: Na^+ }\right)=1.440 \times 10^{-4}\: M\: Na^+\nonumber$
$\left(\dfrac{0.15\: mg\: Na^+}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Na^+}{22.9898\: g\: Na^+ }\right)=0.065 \times 10^{-4}\: M\: Na^+\nonumber$
$\left(\dfrac{4.45\: mg\: Zn^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Zn^{2+}}{65.37\: g\: Zn^{2+} }\right)=6.807 \times 10^{-5}\: M\: Zn^{2+}\nonumber$
$\left(\dfrac{0.10\: mg\: Zn^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Zn^{2+}}{65.37\: g\: Zn^{2+}}\right)=0.153 \times 10^{-5}\: M\: Zn^{2+}\nonumber$
Solute
Concentration (mg L-1)
Concentration (M x 103)
Bicarbonate
466 (± 24)
(7.637 ± 0.393)
Magnesium
124 (± 5)
(5.102 ± 0.206)
Calcium
2.67 (± 0.10)
(0.066 ± 0.002)
Sodium
3.31 (± 0.15)
(0.144 ± 0.007)
Zinc
4.45 (± 0.10)
(0.068 ± 0.002)
Step 2: Create a charge balance equation for the bottled spring water sample based on the nutrition label.
$[HCO_3^- ]=2[Mg^{2+} ]+ 2[Ca^{2+} ]+ [Na^+ ]+ 2[Zn^{2+} ]\nonumber$
Step 3: Use the charge balance equation and error propagation to determine the total cationic charge and total anionic charge including their associated uncertainties.
Anionic Charge: 7.637 mM
Anionic Charge Uncertainty: $s_-=0.393\: mM\nonumber$
(bicarbonate ion is the only anionic species in this charge balance equation)
Cationic Charge: 2(5.102 mM) + 2(0.066 mM) + (0.144 mM) + 2(0.068 mM) = 10.616 mM
Cationic Charge Uncertainty:
$s_+= \sqrt{(2(0.206\: mM))^2+ (2(0.0025\: mM))^2+ (0.0065\: mM)^2+ (2(0.0015\: mM))^2 }\nonumber$
$s_+=0.412\: mM\nonumber$
Step 4: Use the F-test to see if the variance for the anionic charge is significantly different than the variance of the cationic charge. If so, the t-test for two experimental means with equal variances can be used to compare the concentrations of the anionic charge and cationic charge. If not, the t-test for two experimental means with unequal variances must be used to compare the concentrations of the anionic charge and cationic charge.
$F_{exp}= \dfrac{s_+^2}{s_-^2 }$ (larger variance is always in the numerator)
$F_{exp}= \dfrac{(0.412\: mM)^2}{(0.393\: mM)^2} =1.10\nonumber$
At 95% confidence level: $F_{crit} (0.05,2,3)=16.04$
degrees of freedom: $ν_{num}=2$
(variance in numerator based on triplicate determinations of bicarbonate ion
$ν_{denom}=3 \nonumber$
(there are four cationic analytes)
Since $F_{exp}<F_{crit}$,
the variances of the two groups are statistically equivalent @ 95% confidence level
Step 5: Use the t-test for two experimental means with equal variances to determine if the anionic and cationic solute concentrations are statistically equivalent at the 95% confidence level.
$t_{exp}= \dfrac{|\bar{X}_+ - \bar{X}_- |}{s_{pooled} \sqrt{\dfrac{1}{n_+} +\dfrac{1}{n_-} }}\nonumber$
where $s_{pooled}= \sqrt{\dfrac{(n_+-1) s_+^2 + (n_--1) s_-^2}{(n_++ n_-- 2)}}$
$s_{pooled}= \sqrt{\dfrac{(4-1) (0.412\: mM)^2+(3-1) (0.393\: mM)^2}{(4+ 3 - 2) }}=0.405\: mM\nonumber$
$t_{exp}= \dfrac{|10.616\: mM-7.637\: mM|}{(0.405\: mM)\sqrt{\dfrac{1}{4}+\dfrac{1}{3}}}=9.63\nonumber$
$t_{critical} (0.05,5)=2.571$, and therefore, $t_{expt}>t_{critical}$
the difference in the ionic concentrations is statistically significant at the 95% c.l. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/06_Principle_of_Electroneutrality.txt |
Q1. Write the balanced reaction for the dissolution of iron (III) hydroxide in water.
Q1 is a knowledge-based question for quantitative analysis students. Chemistry students at this level should be able to write the chemical formula of an ionic compound and express the dissociation of an ionic compound in water, which is general chemistry material.
$\ce{Fe(OH)3 (s) ⇌ Fe^3+ (aq) + 3 OH- (aq)}\nonumber$
Q2. Write the equilibrium constant expression for the solubility of iron (III) hydroxide.
Q2 is a comprehension-based question for quantitative analysis students. If a chemistry student can recall the definition of an equilibrium constant expression, then that student can modify the definition of an equilibrium constant expression for any solubility equilibrium.
$K_{sp}= [Fe^{3+}] [OH^- ]^3\nonumber$
Q3. Calculate the molar solubility of iron (III) hydroxide in pure water.
Q3 is an application-based question for determining the solubility of an ionic compound in water.
In pure water, pH = pOH = 7.00. Therefore, in pure water, [OH-] = 1.0 x 10-7 M :
$\begin{array}{lccccc} \ce{&Fe(OH)3 (s) &\rightleftharpoons &Fe^3+ (aq) &+ &3 OH- (aq)}\ \textrm{Initial} &\textrm{-----} & &0 & &1.0 \times 10^{-7}\ \textrm{Change} &\ce{-X & &+X & &+3X}\ \textrm{Equilibrium} &\textrm{-----} & &\ce{X} & &\mathrm{1.0 \times 10^{-7} + 3X} \end{array}\nonumber$
Assume that X represents the molar solubility of iron (III)hydroxide
$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$
$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-7}+3X)^3\nonumber$
assuming $3X≪1.0 \times 10^{-7}\: M$
$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-7} )^3\nonumber$
$1.6 \times 10^{-18}\: M= X\nonumber$
using 5% rule, $3X<(0.05)(1.0 \times 10^{-7}\: M)$
$3(1.6 \times 10^{-18}\: M)<5.0 \times 10^{-9}\: M$, assumption holds
Q4. Calculate the molar solubility of iron (III) hydroxide at pH 6.4, and compare it to the calculated molar solubility of iron (III) hydroxide in Q3.
Q4 is an analysis-based question because it requires that a student differentiate between the solubility of iron hydroxide in a solvent where one of the ion concentrations (i.e. hydroxide) is held constant (i.e. a buffer) and where the concentration is not held constant.
In a pH buffer, the hydronium ion concentration, and therefore the hydroxide ion concentration, is constant. In a buffered solution with a pH of 6.4:
$\mathrm{pH + pOH = 14}\nonumber$
$\mathrm{pOH = 14 - pH = 14 - 6.4 = 7.6}\nonumber$
$\mathrm{[OH^-] = 10^{-pOH} = 10^{-7.6} = 2.5 \times 10^{-8}\: M}\nonumber$
$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$
$1.6 \times 10^{-39}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$
$1.0 \times 10^{-16}\: M= X\nonumber$
since $1.0 \times 10^{-16}\: M > 4.8 \times 10^{-18}\: M$,
the solubility of Fe(OH)3 increases as the solution becomes more acidic
Q5. Calculate the molar solubility of iron (III) hydroxide when the ionic strength of the pH 6.4 buffer is 0.10 mol L-1.
Q5 is a synthesis-based question because it requires a student to understand how ionic strength controls the activity coefficient of an analyte, to understand how this activity coefficient changes the magnitude of the equilibrium constant, and to apply these concepts in the determination of the molar solubility of the ionic compound. In this problem, a student needs to:
• Use the Debye-Hückel equation to calculate the activity coefficient of each ionic species in the iron (III) hydroxide solubility equilibrium
$\mathrm{\log γ_x= \dfrac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}\nonumber$
$\mathrm{γ_{Fe^{3+}} = 10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(+3^2 ) \sqrt{0.10}}{1+3.3(0.9\: nm) \sqrt{0.10}}}=0.1784}\nonumber$
$\mathrm{γ_{OH^-}= 10^{\Large\frac{-0.51 z^2 \sqrt{μ}}{1+3.3α_x \sqrt{μ}}}= 10^{\Large\frac{(-0.51)(-1^2 ) \sqrt{0.10}}{1+3.3(0.35\: nm) \sqrt{0.10}}}=0.7619}\nonumber$
• Calculate the concentration-based equilibrium constant using the thermodynamically-based equilibrium constant and the activity coefficients calculated in the previous step
$K_{sp}^{'}= \dfrac{K_{sp}}{γ_{Fe^{3+} } (γ_{OH^-} )^3 }=\dfrac{1.6 \times 10^{-39}}{(0.1784) (0.7619)^3 }=2.0 \times 10^{-38}\nonumber$
• Calculate the molar solubility of the ionic compound using the concentration-based equilibrium constant
$K_{sp}^{'}= [Fe^{3+} ] [OH^- ]^3\nonumber$
$2.0 \times 10^{-38}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$
$1.3 \times 10^{-15}\: M= X\nonumber$
Q6. To prevent precipitation of iron (III) hydroxide in household water systems, hydrogen peroxide is often added as an oxidizing agent to municipal water sources containing iron (II) ion concentrations above 0.3 mg L‑1. Compare the effectiveness of this treatment at pH 6.4 versus pH 2.0 at a dissolved iron (i.e. Fe2+) concentration of 0.3 mg L-1.
Q6 is an evaluation-based question as it asks the student to use equilibrium chemistry to compare the effectiveness of iron removal during water purification under two sets of conditions.
Since hydrogen peroxide is used as an oxidizing agent to remove iron, the first step for the student is to recognize that there is a large difference in solubility constants between iron (II) hydroxide and iron (III) hydroxide, and that this difference may lead to marked differences in solubility between the two salts. The second step is to use each solubility equilibrium constant expression to determine the molar solubility of each salt at each pH and compare it to the analyte concentration of interest (0.3 mg L-1) to evaluate the effectiveness of hydrogen peroxide.
The critical molarity of dissolved iron in water for this problem is:
$\left(\dfrac{0.3\: mg\: Fe}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Fe}{55.85\: g\: Fe}\right)=5.4 \times 10^{-6}\: M\: Fe\nonumber$
At pH 6.4:
Iron (III) hydroxide
$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$
$1.6 \times 10^{-39}= (X) (2.5 \times 10^{-8}\: M)^3\nonumber$
$1.0 \times 10^{-16}\: M= X\nonumber$
Iron (II) hydroxide
$K_{sp}= [Fe^{3+} ] [OH^- ]^2\nonumber$
$8.0 \times 10^{-16}= (X) (2.5 \times 10^{-8}\: M)^2\nonumber$
$1.3\: M= X\nonumber$
When the solution pH is slightly acidic, the oxidation of ferrous ion by the addition of hydrogen peroxide results in the formation of an iron (III) hydroxide salt which is substantially less soluble than the iron (II) hydroxide salt. Under these conditions, hydrogen peroxide is effective at decreasing the dissolved iron concentration below the 0.3 mg L-1 threshold.
At pH 2.0:
Iron (III) hydroxide
$K_{sp}= [Fe^{3+} ] [OH^- ]^3\nonumber$
$1.6 \times 10^{-39}= (X) (1.0 \times 10^{-12}\: M)^3\nonumber$
$1.6 \times 10^{-3}\: M= X\nonumber$
Iron (II) hydroxide
$K_{sp}= [Fe^{3+} ] [OH^- ]^2\nonumber$
$8.0 \times 10^{-16}= (X) (1.0 \times 10^{-12}\: M)^2\nonumber$
$8 \times 10^8\: M= X$ (unrealistically large-interpret as freely soluble)
Under highly acidic conditions, both iron hydroxide salts are more soluble than the 0.3 mg L-1 threshold for dissolved iron. This example highlights the significance of sample acidification below pH 2.0 as a method of sample preservation for metals analysis. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Lithia_Water_Springs_Project/05_Solutions_Manual/07_Assessment_Question_Set_1__Solubility_Equilibria.txt |
The module is designed as a sequence of class activities and provides a set of experimental data (TEM images, size and size distribution analyses, and absorbance vs. concentration data) that can be downloaded from this site. Alternatively, if the instructor has access to a transmission electron microscope and a UV-VIS spectrophotometer, the module could be used in the context of a lab experiment where students, divided in small groups, can conduct gold nanoparticle syntheses exploring different experimental parameters. The lab outcome is a comparison of optical properties for gold nanoparticles synthesized under different conditions of reagents ratio and pH.
Sections are designed to be modular in their format and can be used relatively independently. They can also be implemented at different levels of guidance to students.
Experimental Data
• Absorbance Data (Excel)
• Particle Size Data (Excel)
• Size Distribution Data (Excel)
• Visible Spectra (Excel)
• 2-1 pH 5.4 (Image)
Optical Properties of Gold Nanoparticles
Background Information
Nanoparticles have diameters of 1-100 nm, so a 1 nm particle has a diameter of 1 x 10-9 m. Nanoparticle research is currently a very active field of investigation due to a large variety of applications spanning from medicine (1), optics and electronics (2) and chemical and biological detection and measurement (3). Nanoparticles are special because they display characteristics that are different from those of the same material in bulk form. In fact, one can fine tune physical and optical properties of nanoparticles by controlling their size and shape.
You are familiar with the shiny, metallic appearance of bulk gold and silver. However, reducing the dimensions of the particle drastically changes the appearance as a result of the way it interacts with light. Gold nanoparticles with diameters of 25 nm absorb green light and appear red in color. Silver nanoparticles absorb violet light and are yellow. The ability of nanoparticles to absorb or scatter light is not new knowledge. Artisans as far back as the times of Mesopotamia used nanoparticles to generate a glittering effect on ceramic pots and artists in medieval times used them for stained glass. Because nanoparticles are stable, the red and yellow color in these windows remains today. Nanostructures like those that produce bright, shimmering colors on butterfly wings are composed of multiple layers with air gaps in between that refract, diffract and reflect light generating luminous colors (4). Nanoparticle shape also affects its optical properties. For example, spherical gold nanoparticles absorb in the 500 nm spectral region while irregularly shaped nanorods and nanostars absorb in the near-infrared (5).
A large area of application of nanoparticles is in the development of optical sensors for detection and measurement of a wide array of analytes (3). Spherical gold nanoparticles change color from red to blue depending whether they are dispersed or aggregated. Thus, any ion, small molecule or even protein that can trigger gold nanoparticles to aggregate or disperse can be detected.
Figure 1. Example of colorimetric sensing of metal ions using gold nanoparticles functionalized with chelating agents.
For example, sensors have been developed that rely on gold nanoparticles coated with chelating agents. In the absence of the target ion, the nanoparticles are in their dispersed state and appear red. In the presence of the specific ion, the interaction between the chelating agent and the ion bring the nanoparticles together, shifting their color to blue. Similar approaches using different functional groups on the nanoparticles have extended this detection approach to small organic molecules, oligonucleotides and proteins (3).
The color change of the gold and silver nanoparticles illustrates an important concept about nanoscale science. Chemical and physical properties such as color, conductivity, and reactivity do not depend on the identity of the substance but on the size of the particle.
01 Investigation of Gold Nanoparticles
Solutions of gold nanoparticles appear colored which means they are absorbing light in the visible range. Since you will be investigating the optical properties of gold nanoparticles using a technique called spectrophotometry, we will briefly review how a spectrophotometric measurement takes place.
Q1. What wavelength range constitutes the visible region in the electromagnetic spectrum?
When a beam of electromagnetic radiation passes through a sample, most of the radiation is transmitted but, at specific wavelengths, the radiation may be absorbed by chemical constituents within the sample. For an atomic or molecular species, the absorption of light causes valence electrons to be excited from lower to higher energy states. For this transition to occur, the energy provided by the radiation has to match the difference in energy (ΔE) between the lower and higher energy states.
The amount of light transmitted through a sample is measured by the transmittance and is represented by the equation:
$\mathrm{T = \dfrac{P}{P_0}}\nonumber$
where P0 represents the radiant power of the beam before the sample and P the power after the sample. Another commonly used measurement is the absorbance which is related to the transmittance through a logarithmic relationship:
$\mathrm{A = - \log T = - \log \dfrac{P}{P_0} = \log \dfrac{P_0}{P}}\nonumber$
Relationship between absorbance and concentration
It is possible to make quantitative measurements using UV-VIS spectrophotometry because of the linear relationship between absorbance and concentration. This relationship is known as Beer’s law and is represented by the equation:
$\mathrm{A = εbc}\nonumber$
where
• A is the absorbance
• ε is the molar extinction coefficient (with units of M-1cm-1)
• b is the path length of the cuvette (cm)
• c is the concentration (Molarity)
The changes in absorbance with changes in concentration of a sample are measured at λmax which is the wavelength of maximum absorption in the spectrum of the compound being analyzed. By plotting A as a function of varying concentrations of the analyte, a regression curve can be established.
Q2. Draw a representative plot of A versus c.
Q3. How could you use this plot to determine the molar extinction coefficient of the analyte being investigated?
For a discussion of fundamental concepts of UV-VIS spectrophotometry check the following links:
The molar extinction coefficient can be linked to the intensity of light absorption. For example, transition metals and dyes appear brightly colored because some transitions in the visible range are highly probable and have very large extinction coefficients. For example, the extinction coefficient of the bright Bromothymol Blue at 615 nm is approximately 3.5 x 104 M-1cm-1. Gold nanoparticles are also intensely colored and their color as well as its intensity is controlled by their size (Figure 2).
Figure 2. Colors of various sized monodispersed gold nanoparticles – Source: http://www.sigmaaldrich.com/material....kwu2f19v.dpuf
Table 1. Effect of particle size on maximum wavelength of absorbance and molar extinction coefficient of gold nanoparticles.
Diameter (nm)
Peak Wavelength (nm)
Molar Extinction Coefficient (M-1cm-1)
5
515-520
1.10 x 107
10
515-520
1.01 x 108
15
520
3.67 x 108
20
524
9.21 x 108
30
526
3.36 x 109
40
530
8.42 x 109
50
535
1.72 x 1010
60
540
3.07 x 1010
80
553
7.70 x 1010
100
572
1.57 x 1011
Q4. What color would a particle that absorbs in the blue-green region of the electromagnetic spectrum appear?
Q5. What trend can be identified in the molar extinction coefficient value as particle size increases? How may this trend affect the choice of particle size in a colorimetric sensor?
In this module we will explore how experimental parameters may affect the size, size distribution and related optical properties of spherical gold nanoparticles with the goal of producing nanoparticles of consistent size with a well characterized molar extinction coefficient. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/01_Investigation_of_Gold_Nanoparticles/01_Exploring_fundament.txt |
There are several different ways by which gold nanoparticles can be synthesized but the most common reaction involves the reduction of tetrachloroauric acid (HAuCl4) to gold (Au) using trisodium citrate (Na3C6H5O7) (6). This is a red-ox reaction where gold in the tetrachloroauric acid is reduced to elemental gold while the citrate (C6H5O7 3–) is oxidized to dicarboxy acetone (C5H4O5) (7). Figure 3 shows the structures of these species. This reaction was first described by John Turkevich in 1951 (8) and the mechanisms of particle formation under different experimental conditions are well discussed in the literature (9).
Q6. What is the oxidation number of gold in HAuCl4?
Q7. Write the half reaction for the reduction of HAuCl4 to Au.
Figure 3. Chemical structure of citric acid (a); trisodium citrate (b); dicarboxy acetone (c)
Excess citrate ions not involved in the red-ox reaction are also adsorbed on the surface of the particles, thus playing a role in stabilizing the nanoparticles. They are called “capping” agents. Adsorption of citrate ions gives the gold particles an overall negative charge. Mutual repulsion of the small, negatively charged particles keeps them suspended in the solution and prevents them from coagulating to form larger particles that might eventually settle out of solution. This suspension of the gold nanoparticles is known as a colloid.
Since citric acid is a triprotic acid with distinct pKa values of 3.2, 4.8 and 6.4, pH will affect the chemical equilibrium involving the dissociation of the three hydrogens (10).
Q8. Above what pH value will citric acid (C6H8O7) be completely dissociated to citrate (C6H5O73)?
From an application standpoint, it is very important that particle size distribution be uniform. For example, if one was to develop a colorimetric sensor for a given analyte based on the color shift of gold nanoparticles, the optical characteristics of the nanoparticles would have to be highly reproducible to ensure repeatability of the analytical method. As you observe from data in Table 1, size strongly affects the particle wavelength of maximum absorption as well as the molar extinction coefficient which, in turns, defines the sensitivity of the analytical method.
Thus it is important to understand how different synthetic conditions influence the size and uniformity of size distribution. Controlling the size and size distribution is necessary to control the optical properties of the nanoparticles.
In the next section of this module you will explore the synthesis of gold nanoparticles using the reduction of tetrachloroauric acid (HAuCl4) by citrate and identify experimental parameters that may affect the size, size distribution and optical properties of gold nanoparticles.
03 Developing an exper
Purpose: The purpose of this exercise is to read literature describing the synthesis of gold nanoparticles using the citrate reduction method in order to identify experimental conditions that yield gold nanoparticles of different sizes.
Learning Outcomes:
At the end of this assignment you will be able to
1. Adapt experimental procedures from scholarly articles pertaining to the synthesis of gold nanoparticles by the citrate reduction method.
2. Understand how different experimental parameters affect gold nanoparticle size.
Assignment:
Q9. What are some experimental parameters you think may affect the growth of nanoparticles?
There are a number of literature articles that describe how to synthesize gold nanoparticles. The following articles may be a good start but many others are available:
• Size Control of Gold Nanocrystals in Citrate Reduction: The Third Role of Citrate. Am. Chem. Soc. 2007, 129, 13939-13948.
• Turkevich in New Robes: Key Questions Answered for the Most Common Gold Nanoparticle Synthesis. ACS Nano, 2015, 9 (7), pp 7052–7071.
Read the literature and generate a list of experimental parameters that can be varied in the course of the reaction.
Some variables that you may want to consider are:
• Ratio of reactants
• pH
• Rate of citrate addition
• Temperature
• Stir rate
In designing the experiment, remember it is best to change only one variable at any given time. Let’s, for example, focus on ratio of reactants and pH. You could design an experiment in which you vary the ratio of tetrachloroauric acid to citrate at a controlled pH value. Or you could choose a set ratio of reagents and examine the effect that pH has on the products.
A possible experimental grid is presented in Figure 4. This experiment would allow to explore nine possible combinations of reagent ratios and pH conditions. Consult your instructor on the number of combinations to use.
Initial pH
Citrate to Auric Acid Ratio
2.0:1
4.0:1
7.0:1
4.2
5.4
7.0
Figure 4. Example of experimental grid. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/01_Investigation_of_Gold_Nanoparticles/02_Synthesis_of_gold_n.txt |
Purpose: The purpose of this exercise is to determine the average diameter of gold nanoparticles using a Transmission Electron Microscope (TEM). In addition, the size distribution will be evaluated.
Learning Outcomes:
At the end of this assignment you will be able to
1. Analyze TEM images to determine the average particle diameter.
2. Determine the size distribution for each nanoparticle preparation previously synthesized.
Assignment:
Figure 5. HRTEM images of gold nanoparticles synthesized using a pH of 5.4 and a (a) 2:1 and (b) 7:1 ratio of citrate to tetrachloroauric acid.
The images shown in Figure 5 were collected using a High Resolution Transmission Electron Microscope on nanoparticles prepared using different citrate to tetrachloroauric acid ratio, at pH 5.4. Both images use the same scale of 200 nm.
Q10. Looking at the images on a qualitative basis, what differences do you observe? What can be said about the size of the particles and their size distribution at the two different experimental conditions?
To quantitatively analyze the images and extract information about particle size and size distribution, you will use a free software called ImageJ (http://imagej.nih.gov/ij/). The first task will be to determine the Feret’s diameter of the particles. Follow these instructions on how to use ImageJ.
Go to the “Experimental Data” link and download the image labeled ‘2-1 pH 5.4’. Next, load the image in ImageJ and determine the Feret’s diameter.
Q11. What is the value of the Feret’s diameter you obtained from the analysis of the image showed in Figure 5a?
In the data in Table 2, three different ratios of citrate to tetrachloroauric acid were investigated. In addition, for each ratio, three different values of pH were explored. Use the data in Table 2 to answer the following questions.
Table 2. Summary of experimental results
Citrate/Au
pH
Particle
Feret’s Diameter
2.0:1
4.2
20.00 ± 0.50
5.4
21.73 ± 0.40
7.0
21.22 ± 1.00
4.0:1
4.2
23.30 ± 0.78
5.4
26.52 ± 1.57
7.0
26.76 ± 0.99
7.0:1
4.2
31.87 ± 2.45
5.4
33.91 ± 4.25
7.0
No particles formed
Q12. What is the effect of changing the reagents molar ratio on particle size?
Q13. What is the effect of varying the pH on particle size?
Q14. What statistical tests could you apply to determine whether differences in particle diameter are statistically different?
Q15. Summarize your findings. Is there a relationship between molar ratios of reagents and particle size?
Q16. If you wanted to synthesize particles with a diameter of approximately 20 nm, which experimental conditions would you use?
Next, you will explore the particle size distribution using again the image labeled ‘2-1 pH 5.4’. The output from the ImageJ analysis will list each detected particle and its diameter. Count all the particles of a given diameter (you may want to group them in brackets within 0.5 or 1 nm) and, using a spreadsheet, compile a chart reporting the percentage of occurrence of each particle of a given diameter.
Q17. What trends emerge from your analysis? Is there a set of experimental conditions that yield a more uniform particle distribution?
Use the particle size distribution data (see experimental data link) to continue.
Table 3 summarizes the percentage particles of a given diameter observed for gold nanoparticles synthesized using a 2:1 citrate: HAuCl4 molar ratio at pH 5.4. As seen in Table 2, these experimental conditions yield particles with diameter 21.73 ± 0.40.
Table 3. Particle size distribution for gold nanoparticles synthesized using a 2:1 citrate: HAuCl4 molar ratio at pH 5.4.
Particle Diameter (nm)
Percentage
12
0.104058
13
0.208117
14
0.104058
15
0.104058
16
0.780437
17
1.040583
18
2.549428
19
6.919875
20
13.99584
21
21.12383
22
20.91571
23
15.08845
24
9.313215
25
4.942768
26
2.289282
Examine the data in Table 3. You may want to plot a graph to better visualize the data. Remember, we are trying to identify which experimental conditions yield a monodisperse (one uniform size) product.
Q18. What would you conclude about the distribution obtained by the specific set of synthetic parameters shown in Table 3?
In order to put into context the results of your experiment, it may be useful to take a look at the product specifications for Gold Nanoparticle Reference Materials sold by the National Institute of Standards and Technology (NIST). Go to http://www.nist.gov/pml/div683/gold_010808.cfm
Download the RM 8012, Gold Nanoparticles, Nominal 30 nm Diameter Report of Investigation which contains information about the average particle size as determined by different techniques as well as size distribution histograms.
Q19. If you purchased 30 nm gold nanoparticles from NIST, what particle size diameter would you get? Are all the particles of the same size?
Obviously, there is a relatively large variation in the particle size distribution even for a standard material sold by NIST. Continue analyzing particle distribution data (link) for nanoparticles synthesized using the experimental design described in Figure 4.
Q20. Is there an effect on the particle size distribution when changing the reagents molar ratio?
Q21. Is there an effect on the particle size distribution when varying the pH?
Q22. Summarize your findings. Is molar ratio or pH the controlling factor in ensuring a yield of uniform particles? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/01_Investigation_of_Gold_Nanoparticles/04_Determining_nanopar.txt |
Purpose: The purpose of this exercise is to estimate the molar concentration of nanoparticles based on the particle diameter measured by TEM.
Learning Outcomes:
At the end of this assignment you will be able to
1. Calculate the molar concentration of any nanoparticle preparation using the Feret’s diameter estimated through TEM image analysis.
Assignment:
The first step in estimating the molar concentration of nanoparticles is to calculate the average number of gold atoms per nanoparticle. In the specific case of gold, Liu and coworkers (11) determined the following relationship between the average number of gold atoms (N) per nanoparticle and the particle diameter (D):
$\mathrm{N = \dfrac{π \left(19.3 \dfrac{g}{cm^3}\right) \mathit{D}^3}{6\left(197 \dfrac{g}{mol}\right)}} \label{eq. 1}$
This equation assumes a spherical shape and a uniform face-centered cubic (fcc) structure. In equation $\ref{eq. 1}$, 19.3 g/cm3 is the density for fcc gold and 197 g/mol is the gold atomic mass.
Q23. According to data from Table 2, nanoparticles synthesized with a 2:1 citrate to tetrachloroauric acid ratio and pH 5.4 have a Feret’s diameter of 21.7 nm. What is the value of N?
The next step is to calculate the molar concentration of a nanoparticle solution. This can be done by dividing the total number of gold atoms (Ntotal) equivalent to the amount of tetrachloroauric acid added to the reaction volume by the average number of gold atoms (N).
$\mathrm{Concentration\: (mol/L) = \dfrac{N_{Total}}{N \times V \times N_A}} \label{eq. 2}$
where V is the reaction volume in liters and NA is Avogadro’s number. For example, for the 2:1 citrate to tetrachloroauric acid ratio, assume you reacted all the gold contained in 50.0 mL of 0.25 mM HAuCl4.
Q24. What is the total number of gold atoms (NTotal) in 50.0 mL of a 0.25 mM solution of HAuCl4?
Q25. What is the molar concentration of nanoparticles in this solution?
Now you can estimate the nanoparticle concentration in each of your experimental preparations. Complete the data in Table 4 by estimating the nanoparticle concentration for each experimental combination. The concentration value obtained in the case of reacting 2:1 citrate to tetrachloroauric acid at pH 5.4 is provided as an example of order of magnitude to be expected.
Table 4. Summary of experimental results – use this table to report nanoparticle concentration.
Citrate/Au
pH
Particle
Feret’s Diameter
Nanoparticle concentration (M)
2.0:1
4.2
20.00 ± 0.50
5.4
21.73 ± 0.40
3.16 x 10-9
7.0
21.22 ± 1.00
4.0:1
4.2
23.30 ± 0.78
5.4
26.52 ± 1.57
7.0
26.76 ± 0.99
7.0:1
4.2
31.87 ± 2.45
5.4
33.91 ± 4.25
7.0
No particles formed
06 Determining the mol
Purpose: The purpose of this exercise is to experimentally estimate the molar extinction coefficient of gold nanoparticles and determine what effect size has on the magnitude of the coefficient.
Learning Outcomes:
At the end of this assignment you will be able to:
1. Select the appropriate wavelength to collect spectra of gold nanoparticles.
2. Evaluate the molar extinction coefficient of different nanoparticle preparations by applying the Lambert-Beer law.
3. Determine experimentally the effect that size plays on the magnitude of the molar extinction coefficient.
Gold nanoparticles display very intense colors in the visible range. As seen in Figure 2, color varies from deep red to blue depending on size and state of aggregation. These unique optical properties are due to a phenomenon called Surface Plasmon Resonance (SPR) (12). At specific wavelengths the collective oscillation of electrons on the surface of the particle results in strong absorption (and scattering) which means that the color generated by the particle can be seen at very small concentrations. For this reason, gold nanoparticles are widely used for the development of colorimetric sensors. In order to ensure maximum sensitivity in colorimetric detection, it is important to use gold nanoparticles with high molar extinction coefficients.
Because nanoparticles appear colorful, this means that they are absorbing specific wavelengths of light while reflecting others. This property makes them excellent candidates for spectrophotometric analysis. As explained earlier in the module, a spectrophotometric analysis is performed by illuminating the sample with radiation in the visible or ultraviolet range and detecting the amount of light that reaches the detector. Depending on the chemical properties of the species in the sample, some light will be absorbed while the remainder will be transmitted to the detector where it is converted to an electrical signal that is proportional to the amount of light absorbed, called the absorbance. By constructing a standard calibration curve using known concentrations of nanoparticles under investigation, a relationship can be developed between the absorbance of the analyte at a specific wavelength and its concentration. If the nanoparticle concentration is expressed in molarity, the slope of that plot is the molar extinction coefficient.
An example of visible spectra recorded on gold nanoparticles synthesized using a citrate to tetrachloroauric acid ratio of 2:1 and pH ranging from 4.2 to 7.0 is shown in Figure 6.
Figure 6. Visible spectra recorded on gold nanoparticles synthesized using a citrate to tetrachloroauric acid ratio of 2:1 and pH ranging from 4.2 to 7.0.
Q26. At what wavelength does each nanoparticle solution exhibit the maximum absorbance? This wavelength is referred to as λmax.
Q27. What wavelength would you choose to quantitatively determine the concentration of the nanoparticles? Why did you choose this wavelength?
Q28. What is the value of the absorbance at λmax for each nanoparticle solution? Complete Table 5 by reporting the nanoparticle concentration from Table 4 and λmax estimated from the spectra found in the link to the experimental data.
Table 5. Summary of experimental results – complete this table by reporting the nanoparticle concentration from Table 4 and the value of λmax
Citrate/Au
pH
Particle
Feret’s Diameter
Nanoparticle concentration (M)
λmax (nm)
2.0:1
4.2
20.00 ± 0.50
520
5.4
21.73 ± 0.40
3.16 x 10-9
520
7.0
21.22 ± 1.00
525
4.0:1
4.2
23.30 ± 0.78
5.4
26.52 ± 1.57
7.0
26.76 ± 0.99
7.0:1
4.2
31.87 ± 2.45
5.4
33.91 ± 4.25
7.0
No particles formed
Q29. How could you use a plot of absorbance vs. concentration at the wavelength chosen above to determine the molar extinction coefficient of a gold nanoparticle solution? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/01_Investigation_of_Gold_Nanoparticles/05_Estimating_the_conc.txt |
You are now going to prepare a set of dilutions and use them to calculate the molar extinction coefficient of a gold nanoparticle solution. Use the concentrations you calculated and recorded in Table 4 along with the absorbance values recorded on serial dilutions of gold nanoparticle preparations (see experimental data link).
Q30: Consider the 2:1 pH 5.4 gold nanoparticle preparation. Given that the concentration of the stock solution is of the order of 3 x 10-9 M, how many milliliters of this solution will you have to pipette to prepare 5.00 mL of the following dilutions?
Nanoparticle concentration (M)
Milliliters of stock solution to pipette
0.00
5 x 10-11
1 x 10-10
2 x 10-10
4 x 10-10
8 x 10-10
1.5 x 10-9
3 x 10-9
Q31. What is the purpose of preparing a 0.00 M solution? How is it used in the analysis?
The following absorbance values were recorded when analyzing the eight solutions prepared above using the 2:1 pH 5.4 stock solution:
Nanoparticle concentration (M)
Absorbance
0.00
0
5 x 10-11
0.0501
1 x 10-10
0.0913
2 x 10-10
0.1778
4 x 10-10
0.3367
8 x 10-10
0.6364
1.5 x 10-9
1.172
3 x 10-9
2.622
Use Excel or equivalent software to plot the data above.
Q32. What relationship do you observe between the absorbance and concentration of nanoparticles?
Q33. Do any of the data points deviate from the general behavior observed in the plot?
Q34. What parameter allows us to determine whether there is a good fit between absorbance and gold nanoparticle concentration?
Q35. What is the molar extinction coefficient for this specific gold nanoparticle preparation? How does this value compare to the extinction coefficient reported in Table 1 for a gold nanoparticle of similar size?
Q36. How can you estimate the uncertainty on the molar extinction coefficient from the regression analysis?
08 Putting it all toge
The goal of this module was to explore how experimental parameters may affect the size, size distribution and related optical properties of gold nanoparticles with the goal of producing nanoparticles of consistent size with a well characterized molar extinction coefficient.
It is time to put all the information together to draw some conclusions. Complete the data in Table 6 by filling in the nanoparticle concentration and λmax as reported in Table 5 and estimating the molar extinction coefficient from the Absorbance vs. concentration plots for each experimental combination.
Table 6. Final summary of experimental results
Citrate/Au
pH
Particle
Feret’s Diameter
Nanoparticle concentration (M)
λmax (nm)
Molar extinction coefficient
2.0:1
4.2
20.00 ± 0.50
5.4
21.73 ± 0.40
3.16 x 10-9
7.0
21.22 ± 1.00
4.0:1
4.2
23.30 ± 0.78
5.4
26.52 ± 1.57
7.0
26.76 ± 0.99
7.0:1
4.2
31.87 ± 2.45
5.4
33.91 ± 4.25
7.0
No particles formed
Q37. Based on the analysis of the data you summarized in Table 6, how does particle size affect the molar extinction coefficient?
Q38. How does the citrate to tetracholoroauric acid ratio affect the molar extinction coefficient?
Q39. Does pH play a role in the value of the molar extinction coefficient? Justify your answer.
Q40. If you were to develop a colorimetric sensor, what experimental conditions would you choose to synthesize nanoparticles of consistent size with a large molar extinction coefficient?
09 References
1. Guanying Chen, Indrajit Roy, Chunhui Yang, and Paras N. Prasad. Nanochemistry and Nanomedicine for Nanoparticle-based Diagnostics and Therapy. Chem. Rev., 2016, 116 (5), pp 2826–2885.
2. Reiss, Gunter; Hutten, Andreas (2010). "Magnetic Nanoparticles". In Sattler, Klaus D. Handbook of Nanophysics: Nanoparticles and Quantum Dots. CRC Press.
3. Saha, K.; Agasti, S.; Kim C., Li, X.; Rotello, V. Gold Nanoparticles in Chemical and Biological Sensing. Chem. Rev. 2012, 112, 2739-2779
4. Starkey, A. The butterfly effect. New Scientist, 2005, 187(2518), 46.
5. Wenxin Niu, Yi An Alvin Chua, Weiqing Zhang, Hejin Huang, and Xianmao Lu. Highly Symmetric Gold Nanostars: Crystallographic Control and Surface-Enhanced Raman Scattering Property. J. Am. Chem. Soc. 2015, 137, 10460−10463.
6. Ji, X.; Song, X.; Li, J.; Bai, Y.; Yang, W.; Peng, X. Size control of Gold Nanocrystals in Citrate Reduction: The Third Role of Citrate. J. Am. Chem. Soc. 2007, 129, 13939-13948.
7. Wuithschick, M., Birnbaum, A., Witte, S., Sztucki, M., Vainio, U., Pinna, N., Rademann, K., Emmerling, F., Kraehnert, R., and Polte, J. Turkevich in New Robes: Key Questions Answered for the Most Common Gold Nanoparticle Synthesis. ACS Nano, 2015, 9 (7), pp 7052–7071.
8. Turkevich, J,. Stevenson, P. C., Hillier, J.A Study of the Nucleation and Growth Processes in the Synthesis of Colloidal Gold Discuss. Faraday Soc. 1951, 11, 55 – 75
9. Polte, J; Ahner, T.T.; Delissen, F.; Sokolov, S.; Emmerling, F.; Thünemann, A.F.; Kraehnert R. Mechanism of gold nanoparticle formation in the classical citrate synthesis method derived from coupled in situ XANES and SAXS evaluation. J. Am. Chem. Soc. 2010, 132, 1296.
10. Ojea-Jiménez, I.; Campanera, J. M. Molecular Modeling of the Reduction Mechanism in the Citrate-Mediated Synthesis of Gold Nanoparticles J. Phys. Chem. C, 2012, 116 (44), pp 23682–23691.
11. Liu, X., Atwater, M. Wang, J., Huo, Q. Extinction coefficient of gold nanoparticles with different sizes and different capping ligands. Colloids and Surfaces B: Biointerfaces. 2007, 58, 3-7.
12. Zeng, S.; Baillargeat, D., Ho, Ho-Pui, H., Yong, K. Nanomaterials enhanced surface plasmon resonance for biological and chemical sensing applications. Chemical Society Reviews 2014, 43 (10): 3426–3452. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/01_Investigation_of_Gold_Nanoparticles/07_Preparing_gold_nano.txt |
The following procedure outlines a standard approach to using Image J with TEM images. It is recommended that images be saved in .TIFF format.
Additional information, including examples and tutorials, can be found on the ImageJ homepage (https://imagej.nih.gov/ij/) in the Documentation link.
1. Double click on the ImageJ shortcut on the desktop to open the image processing software.
2. Load the TEM image from the desktop that corresponds to the synthesis conditions used by your lab group. Do this by selecting “File” and then “Open”.
3. The software from modern TEM instruments will embed a scale bar on the image in order to give a reference to the size of objects in the image. We will use this to tell the software how big each pixel in the digital image is, which the software can then use to calculate sizes in a more automated manner.
4. Use the zoom tool (looks like a magnifying glass) or the + and – keys on the keyboard to enlarge the image until the scale bar in the image almost fills the screen. Pan around the image using the tool that looks like a hand.
5. Select the line tool from the tool bar at the top of the screen and draw a line over the scale bar (the line should extend exactly the same length as the scale bar).
6. Select “Analyze” and then “Set Scale”.
7. Remove any prior measurements by selecting “Click to remove scale”.
8. In the box labelled “Known distance”, enter in the value on the scale bar in your image. The distance in pixels should already be filled from the line drawn. In the units box, enter the appropriate units, in this case nanometers or nm. Leave the pixel aspect ratio as the default value of 1 (this just means that the horizontal and vertical sizes of the pixels are the same). Click “OK”.
9. Next be sure to zoom back out until you can see the entire image (hold control while clicking with the magnifying glass to zoom out), and then select the region of interest (where you have nanoparticles, but excluding the margins and scale bar). Do this by clicking on the rectangle tool and surrounding the area you want to use.
10. Select “Image” and then “Crop”.
11. You next need to adjust the image so that the software can readily distinguish between a nanoparticle and the background. To do this, you will first adjust the brightness/contrast of the image. Select “Image” followed by “Adjust” and then “Brightness/Contrast”. Sometimes the “Auto” feature is sufficient, but if not you can also play around with the sliding bars. Try to adjust them so that the nanoparticles really stand out and the background is mainly white, but make sure the size of the dots doesn’t change.
12. Select “Image” followed by “Adjust” and then “Threshold”. Move the sliders until the particles you wish to analyze are red, but none of the background is red. This will set the background to be completely white to create a firm visual boundary between the nanoparticle and the background. Click “Apply”.
13. Select “Process”, followed by “Binary” and then “Make Binary”.
14. We are about to measure the size of the particles, but first, we need to tell the software what measurements we want. Go to “Analyze” followed by “Set Measurements”. Check the box “Feret's Diameter”, but none of the others.
15. Finally, we are ready to have the software measure the size of all the nanoparticles. To do this, select “Analyze” followed by “Analyze Particles”.
16. Check the boxes “Display Results” and “Exclude on Edges” and “Summarize”. The default size from 0 to infinity is a good starting point, although it can be beneficial to put an upper and lower limit to prevent errant inclusions. For circularity, a range of 0.6 to 1 is typically acceptable.
17. Select OK, and two tables with your results will appear, a summary and complete table of each particle.
18. Look through the results table and watch for unrealistic particle sizes (for example if the area is on the order of 1x10-4 nm2, this makes no physical sense, it results from background noise being counted as a particle). You can remove these errant points by right clicking on them and selecting “clear”.
19. For each window, select “File” followed by “save as” to save the table as a spreadsheet. End your filename with “.txt”. For the summary table, either print the spreadsheet or copy the results into your lab notebook.
20. While the complete data table is nice, it is challenging to visualize the results in table form. Instead, we will create a histogram which shows the size distribution of our particles. In the Results window, select “Results” followed by “Set Measurements”. Select the Feret’s Diameter box (you can uncheck all the others). Then select “Results” followed by “Distribution”. Under “Parameter”, select “Feret”. Uncheck “Automatic Binning” and set the number of bins to 10, and then select “OK”. Note that the binning determines how many bars are displayed on the histogram.
21. Save your data for future use. You can also copy data (edit>select all>copy) and paste in Excel or comparable software to analyze the data and create graphs. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/02_Image_J_Processing.txt |
Gold Nanoparticles – Example of Lab Experiment
The following experiment provides an example of how some of the activities described in the module may be implemented in a lab environment under the supervision of the instructor. This specific experiment was developed for a General Chemistry lab as a guided inquiry but can be modified and expanded for upper division courses. It can also be used as a dry lab. Experimental data are available (see experimental data link in this module). Several sections describe equipment specific to the chemistry department at Eastern Oregon University that will likely need to be revised for the equipment present at the user’s institution.
03 Laboratory Experiment
• Wear gloves and eye protection.
• Sodium citrate may cause irritation to skin, eyes, and respiratory tract.
• Tetrachloroauric acid causes eye and skin burns. Harmful if swallowed or inhaled.
• All reactions should be conducted under a hood.
• All leftover chemicals must be collected in the waste bottle in the waste hood.
All glassware for the AuNP synthesis must be free of contaminants and should be washed in aqua regia prior to use and rinsed thoroughly with nanopure water. Aqua regia, once prepared, can be used for approximately 4 weeks. If nanopure water is not available, commercially purchased distilled water can be substituted. When weighing gold chloride and sodium citrate, it is common practice to use a folded weighing paper and a spatula as the folded weighing paper gives better control for transferring small quantities. Students should receive instruction in proper use of pipets. Chemical waste should be properly disposed in heavy metal hazardous waste containers.
2. Reagents and Equipment
Hydrogen tetrachloroaurate(III) trihydrate (2.5 mM)
Sodium citrate
Sodium hydroxide (0.02 M)
Hydrochloric acid
UV-VIS Spectrophotometer
NanoGrids TEM grids (Dune Sciences, Inc.) for sample preparation if a Transmission Electron Microscope is available.
3. Introduction
Synthesis and Analysis of Gold Nanoparticles by Citrate Reduction
Experimental Objective:
In this experiment, you will synthesize gold nanoparticles and explore the influence of the ratios of reactants and pH on the size of the nanoparticles as determined by transmission electron microscopy. You will also explore the influence of the size of the nanoparticles on the surface plasmon resonance measured using visible spectroscopy.
Learning Objectives:
1. Explain the basic red-ox chemistry of gold involved in nanoparticle synthesis.
2. Explain the relationship between reaction conditions and nanoparticle size.
3. Describe the relationship between the size of nanoparticles and their visual properties.
4. Manipulate data sets in order to display them in relevant and meaningful ways.
5. Collaborate with other groups to expand the context of your results.
Theory:
Most chemicals have properties determined almost entirely by their chemical composition. For example, elemental gold always has the same general properties. As the size of a solid decreases into the nanoscale regime, this can begin to change however. Many of the properties begin to depend on the size of the material, including the optical, electrical, chemical, and catalytic properties. Gold for example is typically a yellowish colored, lustrous material. As the size of a particle of gold shrinks below around 100 nm, it gives rise to blue to red solutions when suspended in water. The size and shape of the particle both influence the exact color which is observed. This is due to a phenomenon referred to as surface plasmon resonance. This basically refers to free electrons in the gold moving as a result of their interaction with the light shining on the particles. This leads to some of the wavelengths of light being absorbed while others are transmitted. The exact wavelengths which are absorbed are very sensitive to the exact size and shape of the particles. This effect is also highly dependent on what is surrounding the particles. Switching the solvent from water to something else like ethanol significantly alters the color of the particles. As a result, there is significant interest in using gold nanoparticles in sensing applications (among many others). In this lab, you will synthesize gold nanoparticles and explore the relationship between reaction conditions and the size of the particles as well as the influence of the size of the particles on the absorbance of light. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/03_Laboratory_Experiment/1._Safety_and_Waste_Disposal.txt |
(Note - The following procedure was developed using the MicroLab Interface. Proceed with calibrating a pH electrode according to the manufacturer protocol).
1. Turn on the MicroLab interface and select the Titration tab. Choose “Hand entered titration”.
2. Select the pH sensor and choose “Edit”.
3. Click on “Perform New Calibration”.
4. Fill three disposable plastic cups approximately 2/3rds full with pH 4, 7, and 10 buffer solutions.
5. In Microlab, select “Add Calibration Point”.
6. In the window which opens, enter the actual pH of the buffer in the box labelled “Actual Value”, and wait for the meter to stabilize. You will know it is stable when the red bar stays within the green region. Don’t worry about the graph that you see because it will autorange, meaning that once it is stable a very small range will be used, and it will appear to still be fluctuating significantly even though only very small changes are taking place.
7. Rinse the pH electrode with deionized water into a beaker, and then place it into one of the buffer solutions. Ensure that the tip is completely immersed in the solution. Allow the reading to stabilize, and then enter the value of the buffer and click “OK”.
8. Repeat the previous step for the other two buffer solutions.
9. At this point you should have three points on the graph that form a straight line. Now select “First Order” under curve fit options, and print the resulting graph. This is your only opportunity to print your calibration curve! Click “Accept and Save This Calibration”, for units enter “pH”, and assign a filename. Be sure to record the filename and directory in your lab notebook. After saving the calibration, click “Finish”.
5. Gold Nanoparticle Synthesis
1. Prepare 50 mL of 0.25 mM (millimolar) HAuCl4. A 2.5 mM stock solution is provided. What calculation do you need to perform to prepare the new solution? (Check with instructor!).
2. Note the initial pH and Citrate to Auric Acid Ratio from the table below assigned to your group.
3. Add your HAuCl4 solution to a 100 ml beaker with a magnetic stir bar and place it on a stir plate under mild stirring.
4. Place a pH electrode in your solution, and add sufficient 0.02 M NaOH dropwise to achieve the desired pH. (What can you use to correct pH if you add too much NaOH? (Check with instructor!)
Initial pH
Citrate to Auric Acid Ratio
2.0:1
4.0:1
7.0:1
4.2
Group 1
Group 2
Group 3
5.4
Group 4
Group 5
Group 6
7.0
Group 7
Group 8
1. Based on the 0.25 mM concentration of HAuCl4, how many moles of citrate will be needed to achieve the selected ratio? Perform the calculations and check with instructor! Now convert the number of moles into grams.
2. Transfer your solution into a round bottom flask with a few boiling chips, place it in a heating mantle and connect it to a condenser column.
3. Turn on the heating mantle to medium setting and bring the solution to gentle boiling, then add the calculated number of grams of sodium citrate to the solution.
4. Continue boiling until the reaction is observed (an obvious color change will take place).
5. Remove the solution from heat and allow it to cool back to room temperature (be careful of the hot glass).
6. Once the solution is cooled, measure and record the final pH.
6. Visible Spectroscopy
(Note - The following procedure was developed using a Varian CARY 50 Bio Spectrophotometer).
1. Use a UV-VIS spectrophotometer to measure the absorbance spectrum between 300 and 700 nm.
1. In order to collect a UV-VIS spectrum, you will need a blank. What can be readily used to serve as your blank? (Check with instructor!)
2. Double click on the Cary WinUV shortcut key.
3. Double click on the "Scan" shortcut key
4. Click on "Setup" and under X Mode change the start to 650 nm and the stop to 400 nm and then click "OK".
5. Insert your blank, and then click "Zero".
6. Click "OK" on the popup screen.
7. Fill the cuvette with your nanoparticle solution and place it in the spectrophotometer.
8. Click "Start" to collect the scan
9. Navigate to an appropriate file folder, assign a filename and click "Save". Note that this saved file will contain all of the scans you make until you close the program or start a new file. You will next have the opportunity to assign a Sample ID. Once again, choose a descriptive name. Be sure to record both the filename and the Sample ID in your lab notebook.
10. Look at the absorbance spectrum to make sure that the maximum absorbance does not exceed 1. If the absorbance does exceed 1, think of ways you could fix this.
11. Once a suitable spectrum is collected, right click on the background of the graph and select "peak labels".
12. Select the box “X label”.
13. Print the graph and include it in your lab notebook.
14. We will now export the spectrum so that we can share the raw data with others and plot it together with the spectra from other groups. To do this, go to “File” and select “Save Data As”.
15. From the “Files of Type” pull down menu, select “Spreadsheet Ascii(*.CSV)”. CSV stands for comma separated value, which simply means that the data will be stored as a text file, with columns of data separated by commas. Save this file to one of the provided flash drive with a name that describes your synthesis parameters.
16. Dispose of the sample in the appropriate waste container.
17. Upload your spectrum file to a shared file folder. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/03_Laboratory_Experiment/4._Calibrating_a_pH_electrode.txt |
a) Nanoparticle size from ImageJ
Your next task is to determine the average size of the gold nanoparticles in the solution you have synthesized. Since we do not have access to a Transmission Electron Microscope, we will utilize images collected ahead of time on nanoparticle solutions previously synthesized under identical conditions.
1. Double click on the ImageJ shortcut on the desktop to open the image processing software.
2. Load the TEM image from the desktop that corresponds to the synthesis conditions used by your lab group. Do this by selecting “File” and then “Open”.
3. The software from modern TEM instruments will embed a scale bar on the image in order to give a reference to the size of objects in the image. We will use this to tell the software how big each pixel in the digital image is, which the software can then use to calculate sizes in a more automated manner.
4. Use the zoom tool (looks like a magnifying glass) or the + and – keys on the keyboard to enlarge the image until the scale bar in the image almost fills the screen. Pan around the image using the tool that looks like a hand.
1. Select the line tool from the tool bar at the top of the screen and draw a line over the scale bar (the line should extend exactly the same length as the scale bar).
1. Select “Analyze” and then “Set Scale”.
2. Remove any prior measurements by selecting “Click to remove scale”.
1. In the box labelled “Known distance”, enter in the value on the scale bar in your image. The distance in pixels should already be filled from the line drawn. In the units box, enter the appropriate units, in this case nanometers or nm. Leave the pixel aspect ratio as the default value of 1 (this just means that the horizontal and vertical sizes of the pixels are the same). Click “OK”.
2. Next be sure to zoom back out until you can see the entire image (hold control while clicking with the magnifying glass to zoom out), and then select the region of interest (where you have nanoparticles, but excluding the margins and scale bar). Do this by clicking on the rectangle tool and surrounding the area you want to use.
1. Select “Image” and then “Crop”.
2. You next need to adjust the image so that the software can readily distinguish between a nanoparticle and the background. To do this, you will first adjust the brightness/contrast of the image. Select “Image” followed by “Adjust” and then “Brightness/Contrast”. Sometimes the “Auto” feature is sufficient, but if not you can also play around with the sliding bars. Try to adjust them so that the nanoparticles really stand out and the background is mainly white, but make sure the size of the dots doesn’t change.
1. Select “Image” followed by “Adjust” and then “Threshold”. Move the sliders until the particles you wish to analyze are red, but none of the background is red. This will set the background to be completely white to create a firm visual boundary between the nanoparticle and the background. Click “Apply”.
1. Select “Process”, followed by “Binary” and then “Make Binary”.
1. We are about to measure the size of the particles, but first, we need to tell the software what measurements we want. Go to “Analyze” followed by “Set Measurements”. Check the box “Feret's Diameter”, but none of the others.
1. Finally, we are ready to have the software measure the size of all the nanoparticles. To do this, select “Analyze” followed by “Analyze Particles”.
2. Check the boxes “Display Results” and “Exclude on Edges” and “Summarize”. The default size from 0 to infinity is a good starting point, although it can be beneficial to put an upper and lower limit to prevent errant inclusions. For circularity, a range of 0.6 to 1 is typically acceptable.
1. Select OK, and two tables with your results will appear, a summary and complete table of each particle.
2. Look through the results table and watch for unrealistic particle sizes (for example if the area is on the order of 1x10-4 nm2, this makes no physical sense, it results from background noise being counted as a particle). You can remove these errant points by right clicking on them and selecting “clear”.
3. For each window, select “File” followed by “save as” to save the table as a spreadsheet. End your filename with “.txt”. For the summary table, either print the spreadsheet or copy the results into your lab notebook.
4. While the complete data table is nice, it is challenging to visualize the results in table form. Instead, we will create a histogram which shows the size distribution of our particles. In the Results window, select “Results” followed by “Set Measurements”. Select the Feret’s Diameter box (you can uncheck all the others). Then select “Results” followed by “Distribution”. Under “Parameter”, select “Feret”. Uncheck “Automatic Binning” and set the number of bins to 10, and then select “OK”. Note that the binning determines how many bars are displayed on the histogram.
b) Nanoparticle Size Distribution in Excel
One shortcoming of ImageJ is that there is very limited ability to control the appearance of the graphs generated. As a result, it is often beneficial to instead work with the actual data in a more flexible program like Excel.
1. Open Excel, and the select “File → Open” and the “Browse”.
1. Navigate to the folder where you saved your data, and switch the menu from Excel Files to “All Files”, at which point, you should see your results. Select the file, and click “Open”.
1. Because your data isn’t an actual spreadsheet file, you need to give Excel some guidance so that it can open it correctly. Check the “Delimited” bullet option, and the check box “My data has headers”. Click “Next”
1. Check the “Tab” box under Delimiters, and click “Finish”.
1. Browse through your data to again verify that all of the data look reasonable.
2. Highlight the data under “Feret”, and then go to the “Insert” tab, and click on “Recommended Charts”.
1. Click the tab labelled “All Charts”, and then select histogram.
1. A histogram should appear. Right click on the x-axis, and select “Format Axis”.
1. A window will open on the right side of the screen. Set the number of bins to the same number of bars as was generated in ImageJ previously.
2. Give the histogram an appropriate title, and print it.
c) Combining UV-Vis Spectra in Excel
1. Go to the shared google drive folder where you previously uploaded your exported visible absorbance spectrum and download all the spectra in the folder that have been collected by your classmates.
2. Open each file in Excel, and copy and paste the wavelength and absorbance data into a single spreadsheet. You should have something that looks similar to the data shown below.
1. Highlight the wavelength and absorbance columns for group 1, and select the “Insert” tab, and click on “Scatter with smooth lines”.
1. A chart should appear on your spreadsheet. Now we need to add the rest of the data. Right click on the chart, and click on “Select Data”.
1. Click on “Add”
1. Under Series name, type in “Group 2 Abs”.
2. Under Series X values, click on the box next to the text space.
1. A new window should appear. Highlight the wavelengths for Group 2 and click the box next to the text space that is now filled in.
1. Repeat this for the “Series Y values”.
2. Repeat steps 5-9 for each additional data set and click “OK”
3. You should now have all of the plots on one graph. The last thing we need to do is to make sure everything is labelled. Left click on the graph and a + should appear. Click on the + and additional menu items should appear.
1. Check the options “Axes”, “Axis Titles”, “Chart Title” and “Legend”.
2. Label the Y-axis “Absorbance”.
3. Label the X-axis “Wavelength (nm)”.
4. Give the chart an appropriate title.
5. Finally, we are going to rescale the x-axis to only cover the range of interest. Select the x-axis by clicking on it, and then right click on it and select “Format Axis”.
1. Change the minimum bound to 400 and the maximum bound to 650.
1. Print a copy of the combined graph.
8. Discussion Questions
1. What is the oxidation number of gold in HAuCl4?
2. What do we mean when we say that gold is being reduced?
3. Obtain the particle size results from your classmates and construct a table reporting the gold nanoparticle diameter for each synthetic condition. What trends can be observed?
4. Analyze all the visible spectra collected in the shared folder and expand the table constructed in part 3 to include the wavelength of maximum absorbance (λmax). What relationship do you observe between λmax and the particle size?
5. What information do you obtain from a histogram of your particle diameter data that you would miss if you just looked at the average diameter?
6. What are the benefits and limitations of characterizing nanoparticles with UV-VIS spectroscopy? What about with transmission electron microscopy? If you had access to both techniques, what would you use to best characterize the size of the nanoparticle and why? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/03_Laboratory_Experiment/7._Data_Analysis.txt |
Overview on the Use of This Module
This module is designed to guide students to investigate gold nanoparticles, their synthesis and optical properties, with the goal of identifying experimental conditions that lead to the synthesis of nearly monodisperse gold nanoparticles for sensor development applications.
The module is designed as a sequence of class activities and provides a set of experimental data (TEM images, size and size distribution analyses, and absorbance vs. concentration data) that can be downloaded from this site. Additional TEM images are available upon request from the instructor. Alternatively, if the instructor has access to a transmission electron microscope and a UV-VIS spectrophotometer, the module could be used in the context of a lab experiment where students, divided in groups of 2 or 3, can conduct gold nanoparticle syntheses exploring different experimental parameters. The lab outcome would be a comparison of optical properties for gold nanoparticles synthesized under different conditions of reagents ratio and pH.
Whether conducted as class activity or actual experimental lab, the instructor should ensure that students receive instructions for chemical safety and hazardous material handling (refer to SDS for each chemical). Safety goggles, lab coats and gloves should be worn when performing the activity in a lab setting.
The materials are designed to be modular in their format and used relatively independently. They can also be implemented at different levels of guidance to students. For example, instructors in a general chemistry course could provide students with more detailed instructions on the synthesis and characterization of gold nanoparticles while students in an upper division course may be asked to explore the literature, develop the experimental procedure for the synthesis and fine tune experimental conditions and instrumental parameters if a TEM and/or a UV-VIS spectrophotometer are used.
Students are first introduced to the nano dimension and relevance of nanoparticles in the area of optical sensors development. They are then guided to explore fundamental relationships in spectrophotometry which will be used later in the module to determine the molar extinction coefficient of different sizes of nanoparticles. Depending on the level of preparation, the instructor may choose to provide additional information on gold nanoparticle structure and properties and familiarize students with UV-VIS spectroscopy and use of instrumentation, if available.
Answers to Questions
The following section provides answers to the questions interdispersed in subsections of the module. A brief introduction to the nano dimension and relevance of nanoparticles in the area of optical sensors development is provided. Depending on the level of preparation, the instructor may choose to provide additional information on gold nanoparticle structure and properties.
04 Instructors Guide
This section of the module is designed to provide a review of basic concepts of spectrophotometry. If students have never been introduced to spectrophotometry, a more in depth coverage may be necessary to bring them up to speed, particularly if the module is being used as a lab experiment.
Q1. What wavelength range constitutes the visible region in the electromagnetic spectrum?
The visible region corresponds to wavelengths between 400 and 700 nm.
Q2. Draw a representative plot of A versus c.
Students should draw a typical plot reporting the absorbance on the y-axis and the concentration on the x-axis. A generic plot is presented below.
Q3. How could you use this plot to determine the molar extinction coefficient of the analyte being investigated?
According to Beer’s law, A = εbc, where A is the absorbance, ε is the molar extinction coefficient, b is the path length of the cuvette and c is the concentration. Thus, the molar extinction coefficient can be obtained by calculating the slope of the absorbance vs. concentration plot. Since in most instances the path length b of the cuvette is equal to 1 cm, the slope is the same value as ε.
Q4. What color would a particle that absorbs in the blue-green region of the electromagnetic spectrum appear?
The particle would appear red. As showed in Figure 2, gold nanoparticles can appear anywhere from red to purple/blue depending on their size.
Q5. What trend can be identified in the molar extinction coefficient value as particle size increases? How may this trend affect the choice of particle size in a colorimetric sensor?
To answer these questions students should be given time to analyze data provided in Table 1. The table shows that the molar extinction coefficient increases by four orders of magnitude as particle size increases from 5 to 100 nm. Answering the second part of the question may be more challenging. First, the instructor may want to clarify what a colorimetric sensor is. A pregnancy test is a good example. To further guide students, the instructor may want to discuss the significance of the molar extinction coefficient in terms of color intensity. The larger the extinction coefficient, the larger the absorbance for the same concentration of sample. Thus, when developing a colorimetric sensor, particles with larger molar extinction coefficient will produce larger changes in absorbance for the same concentration change, resulting in a more sensitive sensor. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/04_Instructors_Guide/01_Exploring_fundamental_relationships_i.txt |
This section of the module first provides some background information about the most common method used for the synthesis of gold nanoparticles which involves the reduction of tetrachloroauric acid (HAuCl4) by small amounts of citric acid. This method was introduced by Turkevich et al. in 1951 and further refined by Frens in 1973. In this method, hot tetrachloroauric acid reacts with a small amount of sodium citrate solution. The citrate ions act as both a reducing agent, and a capping agent. Reduction occurs according to the reactions:
$\ce{Na3C6H5O7 \:(sodium\: citrate) + 3H+ → C6H8O7 \:(citric\: acid) + 3Na+}\nonumber$
$\mathrm{2HAuCl_4 + 3C_6H_8O_7 \:(citric\: acid) → 2Au + 3C_5H_6O_5 \:\textrm{(3-ketoglutaric acid)} + 8HCl + 3CO_2}\nonumber$
In the traditional Frens’ method, small citrate to tetrachloroauric acid ratios are used without particular attention to pH and the typical trend observed is that particle size decreases as the ratio of citrate to tetrachloroauric acid increases. This is consistent with the mechanism where part of the sodium citrate is used to reduce Au3+ to Au and the remaining sodium citrate ions are available for stabilizing the particles (capping agent). Nanoparticles continue to aggregate until the total surface area of all particles becomes small enough to be covered by the existing citrate ions. Therefore, higher concentrations of sodium citrate result in less particle aggregation and the final size of the particles decrease. However, Ji et al. (reference 7) observed that if the ratio is increased beyond 3.5:1, the particle size actually increases and eventually levels off for very large ratios. This can be attributed to the effect the citrate plays on the final pH of the solution. Upon addition of larger amounts of Na3Ct to HAuCl4, the pH increases and shifts the gold complex equilibrium toward more hydrolyzed forms. In addition, oxidation products of citrate are formed. All these processes create a continuously changing Turkevich reaction system with a great complexity and a variety of possible pathways for the reduction of Au3+. pH also affects the ionization of citrate depending where the pH falls with respects to three pKa values of 3.2, 4.8 and 6.4. As more citric acid is in the form of citrate (with a -3 charge), the ionic strength of the solution increases and the colloidal stability of the seed particles decreases. As consequence, the seed particle size increases leading to less particles and larger final sizes.
Students are directed to investigate both the effect of changing the Na3Ct to HAuCl4 ratio as well as the pH of the solution at three different values slightly above each respective pKa. These values were chosen because of the larger particle size observed which yield larger extinction coefficients and are more suitable for sensor application. However, to thoroughly investigate the role that pH plays in the synthesis of gold nanoparticles, the instructor will have to widen the pH range from 1 to 8 (see reference 7).
Q6. What is the oxidation number of gold in HAuCl4?
The oxidation number is +3.
Q7. Write the half reaction for the reduction of HAuCl4 to Au.
$\ce{HAuCl4 + 3e- → Au + H+ + 4Cl-}\nonumber$
Q8. Above what pH value will citric acid (C6H8O7) be completely dissociated to citrate (C6H5O73)?
Since citric acid is a triprotic acid with pKa values of 3.2, 4.8 and 6.4, the acid will be completely dissociated to citrate at pH values above 6.4.
03 Developing an experimental design to
In this part of the module students are asked to research and read scholarly articles pertaining to the synthesis of gold nanoparticles and come up with an experimental procedure. They should be referred to the basic approach developed by Turkevich (reference 8) as well as more recent papers (reference 6 and 7). Most articles do not address pH as a variable, so the instructor should make sure that students are aware of these two references.
Q9. Which experimental parameters can be varied that could potentially affect the growth of nanoparticles?
The most obvious parameters are the relative concentrations of citrate to auric acid. Students can explore any ratio but data are available for 2:1, 4:1, and 7:1 ratio of citrate to auric acid. pH is also an important variable and gold nanoparticle synthesis could be explored at pHs slightly above each pKa value of citric acid. Data are included for pH values of 4.2, 5.4 and 7.0. Other possible variables to explore include rate of citrate addition, order of reagents, temperature, stirring rate, etc. However, these variables have not been investigated in our study and no data are available to illustrate any potential effects. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/04_Instructors_Guide/02_Synthesis_of_gold_nanoparticles.txt |
In this part of the module students first visually examine Transmission Electron Microscopy (TEM) images of two different nanoparticle solutions synthesized under different conditions and qualitatively estimate differences in size and size distribution. Next, they can use the provided image files and a free software called ImageJ (http://imagej.nih.gov/ij/) to calculate the Feret’s diameter and particle size distribution for each synthetic condition. Instructions on how to use ImageJ are provided in the module.
Q10. Looking at the images on a qualitative basis, what differences do you observe? What can be said about the size of the particles and their size distribution at the two different experimental conditions?
There are distinct differences between the two high resolution transmission electron microscopy images. The gold nanoparticles synthesized at pH 5.4 with a 2:1 ratio of citrate to tetrachloroauric acid (Figure 5a) appear smaller and much more uniform than those synthesized at pH 5.4 with a 7:1 ratio (Figure 5b). The larger citrate to auric acid ratio definitely generates particles of larger size although a wide distribution of sizes is apparent.
Q11. What is the value of the Feret’s diameter you obtained from the analysis of the image showed in Figure 5a?
The average diameter of particles synthesized at pH 5.4 with a 2:1 ratio of citrate to tetrachloroauric acid is 21.73 ± 0.40 (estimated through ImageJ analysis). Note the small standard deviation indicating a homogeneous distribution of sizes.
Q12. What is the effect of changing the reagents molar ratio on particle size?
As the ratio of citrate to tetrachloroauric acid increases from 2:1 to 7:1, the average diameter of the gold nanoparticles increases from approximately 21 to 34 nm.
Q13. What is the effect of varying the pH on particle size?
Although within the range of pH values explored in this study the pH only slightly affects the particle size, overall pH plays a major role on particle size. Students will observe later that pH also has a dramatic effect on particle size distribution. The traditional conditions of gold nanoparticle synthesis happen to be in a fortuitous pH window (2.7 – 4, see reference 7) that promotes seed-mediated growth mechanism with particles of diameter around 10 nm. However, outside this window, the synthesis outcome is very different. For pH < 1.5, no particles are observed because the protonated citric acid is incapable of reducing Au3+ since the reduction mechanism requires one deprotonated carboxy group. For pH > 4, which is the pH range covered in this experiment, the initial amount of reactive [AuCl4]- is too low to trigger the fast formation of seed particles. As a consequence of the changed growth mechanism, the reduction of Au3+ can occur unselectively leading to nonuniform particles for very acidic or neutral reaction conditions.
Q14. What statistical tests could you apply to determine whether differences in particle diameter are statistically different?
Since the data presented in Table 2 are averages of independent measurements from 15 different images collected on each gold nanoparticle preparation, students could apply a t-test comparing two different means.
Q15. Summarize your findings. Is there a relationship between molar ratios of reagents and particle size?
The molar ratio of the reagents definitely affects the particle size. As the concentration of citrate increases compared to the concentration of gold, so does the particle size.
Q16. If you wanted to synthesize particles with a diameter of approximately 20 nm, which experimental conditions would you use?
A smaller citrate to tetrachloroauric acid ratio yields particles with an approximate diameter of 20 nm.
Q17. What trends emerge from your analysis? Is there a set of experimental conditions that yield a more uniform particle distribution?
If students don’t have access to their own experimental data, they can download the ‘Size-distribution-data’ Excel file (see experimental data section) which provides particles sizes determined on multiple images of the same particle solution. They can then generate a graph showing the percentage of occurrence of each particle of a given diameter.
By comparing particle size distributions for the different sets of experimental conditions, they should observe that lower pH values yield more homogeneous distributions. This can also be observed in Table 2 which shows smaller standard deviations for nanoparticles synthesized using the same citrate to tetrachloroauric acid ratio at lower pH values. As the pH increases the seed-mediated growth mechanism, which is predominant at lower pH values, is not retained and the final particles are rather polydisperse. The initial concentration of [AuCl4]- is too low to trigger the fast formation of seed particles. Seed particle formation might still occur but simultaneously uncontrolled growth of existing and formation of further particles can take place, resulting in a larger dispersion of particle sizes.
Q18. What can be said about the distribution obtained by this specific set of synthetic parameters?
The distribution is relatively uniform with 87% of all particles falling within approximately 2.5 nm from the average diameter.
Q19. If you purchased 30 nm gold nanoparticles from NIST, what particle size diameter would you get? Are all the particles of the same size?
This is a great opportunity for students to discover that gold nanoparticles synthesized as standard material are not truly monodisperse and have their own size distribution. Students can be asked to research the NIST website for RM 8012 gold nanoparticles (nominal 30 nm diameter). These particles have a diameter of 27.6 nm with a standard deviation of 2.1 nm. This deviation is comparable to particles of similar diameter synthesized as part of this experiment using a 7:1 citrate to tetrachloroauric acid ratio at pH 4.2.
Q20. Is there an effect on the particle size distribution when changing the reagents molar ratio?
No, the molar ratio only minimally affects the size distribution. The molar ratio plays a major role in the actual size of the particles.
Q21. Is there an effect on the particle size distribution when varying the pH?
Yes, pH has a dramatic effect on particle size and size distribution. The traditional conditions of gold nanoparticle synthesis happen to be in a fortuitous pH window (2.7 – 4, see reference 7) that promotes seed-mediated growth mechanism with particles of diameter around 10 nm. However, outside this window, the synthesis outcome is very different. For pH < 1.5, no particles are observed because the protonated citric acid is incapable of reducing Au3+ since the reduction mechanism requires one deprotonated carboxy group. For pH > 4, which is the pH range covered in this experiment, the initial amount of reactive [AuCl4]- is too low to trigger the fast formation of seed particles. As a consequence of the changed growth mechanism, the reduction of Au3+ can occur unselectively leading to nonuniform particles for very acidic or neutral reaction conditions.
Q22. Summarize your findings. Is molar ratio or pH the controlling factor in ensuring a yield of uniform particles?
pH is the parameter that more dramatically affects the size distribution. For each citrate to tetrachloroauric acid ratio explored, more monodisperse particles were obtained at lower pH values. This is in agreement with findings by Wuithschick et al. (see reference 7). The paper contains an excellent summary in figure 9 which shows the dependence of polydispersity on pH. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/04_Instructors_Guide/04_Determining_nanoparticles_size_and_si.txt |
In this section of the module students calculate the nanoparticle concentration for each synthetic condition. First, they calculate the average number of gold atoms per nanoparticle (N), then the total number of particles based on the moles of gold used in the reaction and from there, the actual nanoparticle concentration expressed in molarity.
Q23. According to data from Table 2, nanoparticles synthesized with a 2:1 citrate to tetrachloroauric acid ratio and pH 5.4 have a Feret’s diameter of 21.7 nm. What is the value of N?
The relationship between the average number of gold atoms (N) per nanoparticle and the particle diameter (D) is provided by equation $\ref{eq. 1}$:
$\mathrm{N = \dfrac{π \left(19.3 \dfrac{g}{cm^3}\right) \mathit{D}^3}{6\left(197 \dfrac{g}{mol}\right)}} \label{eq. 1}$
This equation assumes a spherical shape and a uniform face-centered cubic (fcc) structure. In equation $\ref{eq. 1}$, 19.3 g/cm3 is the density for fcc gold and 197 g/mol is the gold atomic mass. For D = 21.7 nm,
$\mathrm{N = \dfrac{π (19.3\: g/cm^3)\: D^3}{6 (197\: g/mol)}= 316,946\: atoms/nanoparticle}\nonumber$
Q24. What is the total number of gold atoms (NTotal) in 50.0 mL of a 0.25 mM solution of HAuCl4?
The total number of gold atoms in 50.0 mL of a 0.25nM solution of HAuCl4 can be estimated by calculating the number of moles of gold and multiplying by Avogadro’s number:
$\mathrm{0.050\: L \times \left (0.00025\: \frac{mol}{L} \right ) \times \left (\dfrac{1\: mol\: Au}{1\: mol\: HAuCl_4} \right ) \times \left (6.02 \times 10^{23}\: \dfrac{atoms}{mole} \right ) = 7.5 \times 10^{19}\: atoms}\nonumber$
Q25. What is the molar concentration of nanoparticles in this solution?
The molar concentration of a nanoparticle solution can be estimated by dividing the total number of gold atoms (Ntotal) equivalent to the amount of auric acid added to the reaction volume by the average number of gold atoms (N).
$\mathrm{C =\dfrac{N_{total}}{NVN_A}}\nonumber$
This is equivalent to the equation
$\mathrm{C = \dfrac{moles\: from\: reaction}{N \times Volume}}\nonumber$
$\mathrm{C = \dfrac{1.25 \times 10^{-5}\: mol\: Au}{316,946\: atoms/NP \times 0.05\:L} =7.9 \times 10^{-10}\: mol/L}\nonumber$
Following the example provided above, students should repeat the calculations for all particles and complete all data in Table 4.
06 Determining the molar extinction coef
Using the data sets made available in the experimental data, students can first estimate the wavelength of maximum absorbance and then estimate the absorptivity coefficient for each nanoparticle preparation by calculating a linear regression equation based on the absorbance vs. concentration data.
Q26. At which wavelength does each nanoparticle solution exhibit the maximum absorbance? This wavelength is referred to as λmax.
Students can estimate λmax by examining Figure 6 which shows spectra recorded on nanoparticles synthesized under different pH conditions or analyzing the visible spectra provided in the experimental data. As the pH increases, a red shift is observed in agreement with increased particle size. This shift is also portrayed in Table 1 in the module.
Q27. What wavelength would you choose to quantitatively determine the concentration of the nanoparticles? Why did you choose this wavelength?
Students should answer this question by reporting λmax. This wavelength is chosen because it provides the maximum sensitivity for the determination of the concentration according to Beer’s law.
Q28. What is the value of the absorbance at λmax for each nanoparticle solution? Complete Table 5 by reporting the nanoparticle concentration from Table 4 and λmax estimated from the spectra found in the link to the experimental data.
Table 5 can be completed by inserting the nanoparticle concentrations calculated in Table 4 and by reporting the λmax derived by analyzing the visible spectra provided in the experimental data.
Q29. How could you use a plot of absorbance vs. concentration at the wavelength chosen above to determine the molar extinction coefficient of a gold nanoparticle solution?
According to Beer’s law, A = εbc, where A is the absorbance, ε is the molar extinction coefficient, b is the path length of the cuvette and c is the concentration. Thus, the molar extinction coefficient can be obtained by calculating the slope of the absorbance vs. concentration plot. Since in most instances the path length b of the cuvette is equal to 1 cm, the slope is the same value as ε. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/04_Instructors_Guide/05_Estimating_the_concentration_of_nanop.txt |
Q30: Consider the 2:1 pH 5.4 gold nanoparticle preparation. Given that the concentration of the stock solution is of the order of 3 x 10-9 M, how many milliliters of this solution will you have to pipette to prepare 5.00 mL of the following dilutions?
Students will use the dilution formula: M1 x V1 = M2 x V2 where M1 and V1 are the molarity and volume of the stock and M2 and V2 are the molarity and volume of the dilute solution.
For example, if we wish to prepare 5.00 mL of a 1.5 x 10-9 M solution from a 3 x 10-9 M stock:
$\mathrm{3 \times 10^{-9}\: M \times V_x = 1.5 \times 10^{-9}\: M \times 5.00\: mL}\nonumber$
Solving for Vx:
$\mathrm{V_x = \dfrac{1.5 \times 10^{-9}\: M \times 5.00\: mL}{3 \times 10^{-9}\: M} = 2.5\: mL}\nonumber$
To prepare the dilution, 2.5 mL of the stock solution (3 x 10-9 M) would have to be pipetted using a volumetric pipette into a 5.00 mL volumetric flask and brought to volume with water.
The best approach will be to make a number of serial dilutions as preparation of the more diluted solutions from the stock solution is not feasible in light of very small volumes that would have to be measured. Such small volumes would increase the inaccuracy of the analysis.
Q31. What is the purpose of preparing a 0.00 M dilution? How is it used in the analysis?
A solution containing zero concentration of analyte is considered a blank and it is used to zero the instrument to eliminate interferences from any reagents used in the analysis.
Q32. What relationship do you observe between the absorbance and concentration of nanoparticles?
There is a very linear relationship between the absorbance and the concentration for any of the nanoparticles preparations.
Q33. Do any of the data points deviate from the general behavior observed in the plot?
In general, there is very small deviation in any of the plots. The absorbance of nanoparticles solutions displays a high linearity with changes in concentration.
Q34. What parameter allows us to determine whether there is a good fit between absorbance and gold nanoparticle concentration?
The correlation coefficient is a measure of good fit. For most solutions examined in this study the correlation coefficient is 0.999 or better.
Q35. What is the molar extinction coefficient for this specific gold nanoparticle preparation? How does this value compare to the extinction coefficient reported in Table 1 for a gold nanoparticle of similar size?
Students can analyze the data provided in the experimental results. Based on the absorbance vs. concentration plot, the slope of the curve is 831284607.8 or 8.3 x 108 M-1cm-1. By comparison, the extinction coefficient for a 20 nm particle is approximately 9 x 108 M-1cm-1 as reported in Table 1.
Q36. How can you estimate the uncertainty on the molar extinction coefficient from the regression analysis?
The Summary Output from the regression analysis is displayed for each set of absorbance vs. concentration data. The Standard Error associated with X Variable 1 is the uncertainty on the slope. In this particular case it is 4289294.315. Thus the extinction coefficient can be reported as (8.3 ± 0.4) x 108 M-1cm-1.
08 Putting it all together
Q37. Based on the analysis of the data you summarized in Table 6, how does particle size affect the molar extinction coefficient?
In general, the molar extinction coefficient increases with increasing particle size.
Q38. How does the citrate to tetrachloroauric acid ratio affect the molar extinction coefficient?
Since the citrate to tetrachloroauric acid ratio affects the particle size, the ratio does affect the molar extinction coefficient. As the particle size increases, so does the molar extinction coefficient.
Q39. Does pH play a role in the value of the molar extinction coefficient?
In the pH range explored in this experiment, pH does not dramatically affect particle size and, therefore, molar extinction coefficient. However, for pH values lower than 4, pH plays a dramatic effect on particle size, thus affecting the molar absorptivity coefficient.
Q40. If you were to develop a colorimetric sensor, what experimental conditions would you choose to synthesize nanoparticles of consistent size with a large molar extinction coefficient?
This is an open ended question. Looking at the overall results, particles with an approximate diameter of 20 nm can be synthesized with relatively small polydispersion using a 2:1 citrate to auric acid ratio at pH values between 4 and 5. By increasing the citrate to auric acid ratio to 4:1, the particle size increases along with increased polydispersion. This trend continues for the higher ratio (7:1) where particle size is above 30 nm with wider particle distribution.
For sensor application, probably the 2:1 citrate to auric acid at pH values between 4 and 5 would yield the best conditions. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/04_Instructors_Guide/07_Preparing_gold_nanoparticle_dilutions.txt |
This module introduces students to the topic of sample preparation using a variety of real-world examples from pharmaceutical analysis to environmental analysis. Liquid-liquid extraction, solid phase extraction, QuEChERS, derivatization in chromatography and solid-phase microextraction methods are discussed in a discovery-based and contextual manner.
Sample Preparation
Overview
This module introduces some of the basics of sample preparation, and the ways in which samples must be handled and modified to make them amenable to a particular instrumental method of analysis. Included are theory and applications of liquid-liquid and solid phase extraction and microextraction techniques, and derivatization methods for GC and HPLC. Throughout the module, the primary goal is to present thought-provoking questions that lead to critical thinking about which of the available sample preparation methods is appropriate for a given sample scenario. Supporting materials detailing these techniques can be found in Chapter 7 in the online textbook Analytical Chemistry 2.01 by David Harvey available from the Analytical Sciences Digital Library (www.asdlib.org ). The final section of Chapter 7 contains key terms and links to further discussion. If you need to review any terminology, the final section of Chapter 3 in Harvey also defines many of the key terms used in this module.2
Learning Objectives
After completion of this learning module, you will be able to:
• Describe why many samples need some form of preparation before analysis.
• Describe the properties of an analyte that can be used to separate it from a mixture.
• Describe the general theory of the following extraction techniques
• Liquid-liquid extraction (LLE)
• Solid-phase extraction (SPE)
• Adsorption
• Charge (anion/cation exchange)
• Mixed modes
• QuEChERS
• Solid-phase microextraction
• Describe common derivatization methods used in sample preparation methods for gas chromatography
• Describe common derivatization methods that “tag” an analyte, allowing it to be detected spectrophotometrically, electrochemically or using fluorescence
• Predict the best extraction method for isolating the analyte given a sample scenario.
Introduction
Consider your sample as shown in Figure 1, comprised of your analyte (triangles) in a matrix of various interferents (circles and moons). Consider someone trying to measure neonicotinoid pesticides in a plant sample collected in an area with a noticeable bee decline. These pesticides have in recent years come under scrutiny for contributing to the decline in honeybee populations.3 A field is sprayed with a neonicotinoid pesticide. You take some plant samples back to the lab and homogenize for subsequent preparation and analysis.
Q1. Do an Internet search to find structures of nicotine and imidacloprid (one of the neonicotinoids). Do these structures indicate to you how this class of insecticides was named?
Q2. In the plant sample, state the identity of your analyte and list some possible interferents.
The interferents can cause problems with your eventual measurement by doing things such as
• Adding to the analyte signal
• Masking the analyte signal
• Quenching the analyte’s signal
• Wreaking havoc on your instrument (e.g. clogging a chromatography column)
Figure 1: Shows the removal of interferents (circles and moons) from the analyte (triangles) in a complex sample (A). After isolation (B), the sample may be ready for analysis, or additional preparation steps (like concentration or derivatization) may be required.
Q3. List at least four chemical properties of the analyte that may allow it to be separated from interfering components in its matrix.
Keep your list in mind as you read through the next sections describing various extraction methodologies.
Sample preparation
A) How do you get your sample from this…
B) To this?
Figure 2. Photographs of samples before (A) and after (B) the preparation steps for injection onto a gas chromatograph.
As shown in Figure 2, in most chemical analyses, the sample as collected is not suitable for direct analysis, but must be converted into a proper form through a process known as sample preparation. For example, a solid may have to be dissolved in water or digested in acid, and then filtered to remove insoluble components. Compounds that could interfere with the analysis must be removed or masked during sample preparation; a process that typically involves a separation step. Also, within this step it may be necessary to dilute or concentrate the sample to place the analyte within the appropriate concentration dynamic range of the measurement.
In this module several aspects of sample preparation will be considered. The focus of the first section is directed at liquid-liquid extraction (LLE) and solid phase extraction (SPE). Though other aspects of sample preparation such as digestion, precipitation, and pre-concentration are critical to a successful analysis, they are typically specific for the sample and analyte to be determined. On the other hand, methods of extraction can be considered in more general terms, and their successful exploitation often determines the success of a validated analytical method. Figure 3 below outlines the extraction methods discussed in this series of modules.
Figure 3. Diagram of the extraction methods discussed in this module series.
Additionally, even after purification, samples may not be in a form that is amenable to determination by typical instrumental methods. In many cases, the sample must be converted by reaction to a form containing an additional functionality detectable by the instrument of choice. This conversion, referred to as derivatization will be discussed in sections that address sample analysis by GC and by HPLC. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/01_Introduction_to_Sample_Preparation.txt |
Liquid-liquid extraction (LLE)
Liquid-liquid extraction (LLE) is based on the transfer of a solute from one liquid phase into another immiscible liquid phase according to differences in solubility. A common analytical challenge is measuring the level of a dilute analyte in a complex aqueous sample matrix, for example blood plasma or wastewater. Extraction of the analyte (A) from the aqueous sample into an immiscible organic solvent can both reduce matrix interferences and facilitate concentration of the analyte. When two immiscible phases are in contact, an analyte (A) that is soluble in the two phases will exchange between them. In this system, phase 1 could be blood plasma and phase 2 an immiscible solvent like dichloromethane. The equilibrium concentrations of A in the two phases will depend upon its relative solubility in each phase defined by the chemical equilibrium where [A]1 represents the concentration of species A in phase 1 and [A]2 represents the concentration of A in phase 2:
$\mathrm{\color{firebrick}{[A]_1} \color{black}\leftarrow \rightarrow \color{navy}{[A]_2}}\nonumber$
The analyte is said to partition between the two phases and the equilibrium constant for this reaction is called the partition coefficient, KP or P.
$\mathrm{K_P = P= \dfrac{\color{navy}[A]_2}{\color{firebrick}[A]_1}}\nonumber$
As you have probably learned when studying equilibrium, KP depends on activity, a thermodynamic parameter related to concentration. Because it can be challenging to measure or calculate analyte activities, it is common to substitute molar concentrations for activities in the calculation of KP, as is shown in the equation above. One of the most important partition equilibria involves water and octanol. The log P for a molecule partitioning between octanol (phase 2) and water (phase 1) is commonly used as a surrogate for lipid solubility and is an important determinant in the uptake and metabolism of pharmaceutical agents or toxic substances.
An essential analytical consideration in LLE is the extraction efficiency, defined as the fraction of analyte A transported from the aqueous phase into the organic phase by the partition equilibrium. To increase the extraction efficiency, it is often desirable to perform multiple extractions, combining the organic volumes. By careful selection of the volumes used for the aqueous and organic phases of the LLE, the analyte can be concentrated substantially prior to analysis, a process facilitated by the use of a volatile organic solvent that can be easily reduced in volume after it has been separated from the aqueous phase.
As described above, partition coefficients are thermodynamic constants based on the activities of the neutral solute in each phase. However, many common analytes are weak acids or bases that exist in equilibrium in aqueous solution between charged and neutral species. Therefore, equilibria involving the acid and base forms of the analyte must be considered to fully exploit its partitioning behavior in LLE. Similar to KP, we can also define a constant, D, the distribution coefficient that describes the partitioning of a solute between two phases based on its analytical (total or formal) concentration in the two phases:
$\mathrm{D = \dfrac{A_{total}\: in\: phase\: 2}{A_{total}\: in\: phase\: 1}}\nonumber$
For example, suppose the analyte is an amine, B, and that phase 1 is aqueous and phase 2 is an immiscible organic solvent such as toluene or dichloromethane. As seen in Figure 1, the protonated form of the amine (BH+) will generally be soluble only in the aqueous phase, but the neutral form will be soluble in both phases. Therefore, the partition coefficient, KP, will be based on the neutral form. Assuming that concentration is a good approximation for activity, the partition coefficient KP is defined by
$\mathrm{K_P = \dfrac{[B]_2}{[B]_1}}\nonumber$
However, the distribution coefficient, D, will be based on the total concentration of B in the aqueous phase (B + BH+) which will depend on the pH and the pKa value of BH+. As a consequence, it is often necessary to adjust the pH of the aqueous phase appropriately to maximize the equilibrium concentration of the neutral species B and efficiently extract it into the organic phase.
Figure 1. Partitioning of the basic analyte B between organic (yellow) and aqueous (green) phases.
Q1. You are asked to determine the optimum pH for the extraction of a weak base (pKa of BH+ = 8.54) into toluene. What pH range for the aqueous sample: acidic, neutral or basic, do you expect will be most suitable for this extraction? Why?
Q2. Suppose the analyte is a weak acid (HA, pKa =4.25).
1. Write the expression for KP.
2. Which form of the analyte, HA or A- do you expect to be extracted into the organic phase? Why?
3. What pH range for the aqueous sample: acidic, neutral or basic, do you expect will be most suitable for this extraction? Why?
Application of LLE
Drug development in the pharmaceutical industry requires several physical and chemical properties of the new drug to be evaluated in the drug substance and in formulations to satisfy FDA regulations. The FDA guidelines require structural identification of impurities and degradation products present in the formulated drug above the identification threshold, 0.1%. The compound 4-aminophenol (PAP, Figure 2) is a common impurity in acetaminophen tablets. There are two possible sources of this impurity. It can be a present as a synthesis impurity; essentially starting material that did not react completely to give the amidated product, acetaminophen (Figure 2). An additional source of this impurity is thermal degradation of acetaminophen.
Figure 2. Chemical structures of acetaminophen and 4-aminophenol (PAP).
Pubchem gives the following information (Table 1) for PAP1 and acetaminophen2. Note that the logP is the partition coefficient (KP) for PAP between water and octanol.
PAP
acetaminophen
log P
0.04 at pH 7.4
0.46
pKa1
5.48 (at 25 °C)
9.38 (at 25 °C)
pKa2
10.46 (at 25 °C)
--
Table 1. Partition coefficient and pKa data for PAP and acetaminophen.
Q3. Think about the chemical structures of acetaminophen and PAP (Figure 5). What chemical property is different? Suggest an alternate extraction method from LLE that could be used to separate PAP and acetaminophen. Could pH be useful here?
In this analysis of a Tylenol sample, a very small amount of PAP (an impurity) is expected to be present, and the differences in the chemical properties of the two species are not sufficient to lead to an efficient LLE. As a sample preparation method, LLEs are also labor intensive, time consuming and require large amounts of organic solvents. The next sections of this module will discuss the other extraction techniques listed in Figure 1 in the Introduction, which may be able to overcome these limitations. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/02_Liquid-Liquid_Extraction.txt |
Solid Phase Extraction (SPE)
Solid phase extraction1 (SPE) is a sample preparation technique using a solid adsorbent contained most commonly in a cartridge device (Figure 1), or on a disk to adsorb select species from solution. SPE is used to isolate a species in a sample or to clean-up a sample before analysis.
Figure 1. Photograph of an SPE cartridge (top).
As the sample is slowly passed through the SPE cartridge or disk, the analyte and some of the sample matrix compounds may be retained on the SPE material. Depending on the properties of the analyte and the SPE sorbent, a wash solvent can be chosen to selectively remove (elute) components from the SPE sorbent while retaining others. The ultimate goal is to remove interferents present in the matrix from the analyte, producing a solution containing primarily analyte. The following different scenarios can be utilized to achieve this goal.
1. Bind and elute strategy: The analyte(s) are retained and matrix components do not retain. The analyte subsequently is washed and eluted with a strong solvent. (Figure 2A)
2. Removal/trapping strategy: The analyte(s) are not retained and elutes, while the matrix components are retained. (Figure 2B)
The captured analyte elution solvent can be further reduced in volume to concentrate the analyte prior to analysis.
SPE is used to:
• simplify complex sample matrices
• purify compounds of interest
• reduce the ion suppression in mass spectrometry applications
• fractionate complex mixtures for analysis by classification
• concentrate analytes present at low levels.
Figure 2. SPE bind and elute strategy (A) where the analyte is selectively retained on the adsorbent, while interferents pass through (a-b). Next, the analyte is eluted (c); and removal/trapping strategy (B) where the analyte is not retained and elutes, while the interferents are retained on the sorbent.
How does SPE compare to liquid-liquid extraction?
Liquid-liquid extraction (LLE) separates an analyte through partitioning between two immiscible solvents. LLE suffers from several disadvantages that can be overcome using SPE, including the need for expensive glassware (separatory funnels), the large and costly amount of organic solvents needed for separation, as well as the less than quantitative recoveries that can stem from low values of KP or incomplete phase separations. For a review of LLE, refer to Section 2 of this module.
The general steps involved in performing SPE are:
1. Pre-treat the sample (e.g. dilution, adjustment of pH)
2. Condition the cartridge (run water or solvent through it)
3. Load the sample
4. Elute the fractions
SPE methods start with a conditioning step (Step 2 above). This consists of rinsing the SPE cartridge with a solvent similar to the matrix of the sample you want to load. For an aqueous sample, we would choose an aqueous solvent such as water for conditioning. Once the cartridge has been conditioned, the sample is then loaded on the cartridge (Step 3; note Steps 3 and 4 can be viewed in Figure 3). The amount of sample loaded will vary depending on the purchased cartridge. The cartridges can be loaded on a vacuum manifold to apply a vacuum or centrifuged to speed up the separation process. This separation can also be done using gravity, but will take more time. Once the sample has been collected on the adsorbent, the cartridge can be rinsed with 100% H2O to remove un-retained compounds. More rinses can then be done with increasing percentages (ex.: 5%, 10%, 20%, 100%) of an organic solvent, such as isopropanol, to remove compounds in terms of decreasing polarity.
Q1. Explain the difference between liquid-liquid extraction and solid-phase extraction.
Figure 3 shows an SPE procedure on a grape Kool-Aid sample. In this example, a polar solvent is passed through a less polar adsorbent material. This is referred to as reversed-phase (RP) SPE. For this example, let’s say we are interested in determining the amount of red and blue dye in the sample. (Please note that the examples in this module involve separating dyes, as the separation can be easily visualized. However, SPE also can be used to extract and separate colorless species.)
Q2. List one or two species that you might find in the Kool-Aid matrix. Do you expect these species to be water soluble?
Q3. Consider Figure 3. In steps 1 and 2, grape Kool-Aid is passed through the cartridge, followed by water. Which species are retained on the adsorbent and which species are eluted with the water?
Figure 3. SPE of a grape Kool-aid sample. Looking left to right, the following are eluted through the cartridge: grape Kool-aid, 100% water, 95% water/5% isopropanol, 90% water/10% isopropanol, 80% water/20% isopropanol, and 100% isopropanol. (Photos are courtesy of Cynthia Larive.)
Q4. After the water is passed through the cartridge in Figure 3 (step 2), water/isopropanol mixtures with increasing amounts of isopropanol are passed through the cartridge. Is the polarity of the solvent increasing or decreasing?
If you said “decreasing” you are correct. This is termed making the solvent “stronger”, i.e. more like the adsorbent, which in this case is non-polar.
Q5. Consider the two scenarios at the beginning of this section for retaining and eluting analyte (dye molecules) and interferences. Which scenario is being used in Figure 3?
Types of SPE
In the Kool-Aid example, we used a SPE cartridge where the adsorbent was less polar than the solvent. We called this reversed phase SPE. The first step in using SPE is to select the correct type of adsorbent for the type of separation that is needed. The type of SPE sorbent is chosen based on the chemical properties of the species we wish to separate. It is important to have an understanding of the sample matrix, and the analytes of interest, when selecting the type of SPE. For example, if the analyte of interest is charged or ionizable, you would select an ion exchange cartridge, either anion or cation depending on the charge of your analyte.
The types of SPE fall into several categories including, reversed phase, normal phase, ion exchange (anion/cation), as well as mixed-mode phases, which have the properties of more than one type of SPE material. Figure 4 describes the properties of these phases.
Figure 4. Various types of SPE with details about their separation mechanisms.
Adsorption (reversed phase)
Reversed phase SPE techniques are optimal for analytes with moderate to low polarity, and separate analytes based on hydrophobicity with the most polar compounds eluting first. Because many analyses involve analytes dissolved in an aqueous sample (e.g. surface waters, wastewater, urine or plasma), SPE using reversed phase sorbents are commonly employed.
Figure 5. Red and blue dyes of Kool-Aid separated using reverse phase SPE.2
Let’s think back to the Kool-Aid example. Here we want to again separate the red and blue dyes that form the purple colored Kool-Aid solution. Figure 5 shows a cartoon similar to the process in Figure 3.
The process of separating the red and blue dyes within the purple Kool-Aid sample will start by loading an aliquot of sample onto a reversed phase cartridge that has been pre-conditioned with water. The first fraction collected will contain mostly water and compounds of high polarity and the dye will be retained in the adsorbent. Rinse steps with increasing concentrations of isopropyl alcohol (IPA) will be introduced to elute the red and blue dyes separately. Once the dyes have been collected, the IPA is increased to elute non-polar species within the sample.
Q6. Figure 5 depicts the separation of dyes within the Kool-Aid sample. The red dye appears to elute before the blue dye with only a small addition of IPA. Please explain why the red dye elutes first.
Adsorption (normal phase)
Normal phase SPE techniques are commonly used when the analyte of interest has low to high polarity, or is neutral.3 The cartridge contains a polar adsorbent, such as silica. This separation will be based on polarity, with the least polar components eluting first. The sample is usually in a non-aqueous matrix. For example, a researcher wishes to determine the concentration of polycyclic aromatic hydrocarbons (PAH) in coffee samples. Many PAH are carcinogenic and can be introduced during the roasting process. The coffee beans are ground up, dissolved and the oily residue extracted into the organic solvent 1-pentane. This extraction sample contains the PAH but also potential interferents. Next, SPE is used to clean up the sample and/or concentrate the PAH. A “bind and elute” (Figure 2A) sample cleanup strategy is employed. A PAH, such as benzo(k)fluoranthene shown in Figure 6, is dissolved in the 1-pentane phase, but the pi-electrons can interact with the polar functional group (e.g. silica) on the solid phase. When the sample is loaded onto the SPE cartridge, the PAH are retained onto the stationary phase. Next, the solvent is changed and the PAH are eluted from the cartridge.
Q7. In order to elute the PAH from the cartridge, how would you alter the polarity of the solvent?
Figure 6. Chemical structure of the PAH benzo(k)fluoranthene.
Q8. A pesticide has been extracted from a soil sample. The pesticide is relatively non-polar and soluble in organic solvents. However, it contains a polar functional group. Describe a procedure to separate the pesticide from potential interferents (such as hydrocarbons) in the sample.
Q9. If we wanted to use SPE in the preparation of soda and tea samples, which type of adsorption would we use (Normal or Reversed phase) and why?
Ion Exchange (cation/anion)
Ion exchange SPE lets us separate compounds based on charge. Let us look back to the example of 4-aminophenol and acetaminophen (Section 1 on liquid-liquid extraction). Notice that the 4-aminophenol has an –NH2 group that will be positively charged at an acidic pH, while the acetaminophen will not. We would conclude that the best way to separate these two compounds would be by charge using ion exchange.
There are two types of ion exchange SPE, cation exchange and anion exchange. For this example, we would select cation exchange because the analyte we wish to retain is cationic (positively charged). This separation will be a basic 3-step approach to SPE (Figure 2A), starting with the first step of loading the sample onto the cartridge. The cation exchange cartridge will retain the positively charged 4-aminophenol by electrostatic interaction. The first wash step will elute the negatively charged and neutral species such as acetaminophen. To elute the 4-aminophenol that has been retained in the cartridge, a solvent with high ionic strength can be used. This will introduce a large amount of positively charged ions to compete for binding of the substrate, causing the 4-aminophenol to elute from the cartridge.
Q10. Given the example of cation exchange, can you explain a scenario where anion exchange SPE would be necessary?
Q11. Assuming you have two analytes of interest (refer to Figure 7), using ion exchange SPE, please explain how you could isolate the two analytes separately.
Figure 7. Process of eluting two analytes of interest separately.2
Mixed mode
In a perfect example of bind and elute SPE, only the analyte of interest would be retained while all interferents would be rinsed off leaving a pure, concentrated sample with little to no loss. Unfortunately, this scenario is unlikely with the use of a single SPE cartridge. Many times, conditions strong enough to remove impurities will also remove some analyte, reducing the recovery. Mixed-mode SPE can enhance analyte retention and/or reduce matrix impurity contamination.
Mixed-mode SPE combines the capabilities of ion-exchange and reversed-phase SPE for an enhanced separation. Examples of mixed-mode SPE include:
• Reversed-phase/strong cation-exchange
• Reversed-phase/strong anion-exchange
• Reversed-phase/weak cation-exchange
• Reversed-phase/weak anion-exchange
The ion-exchange component of the cartridge allows the appropriately charged compounds to be retained for further extraction. Once the appropriately charged compounds are retained, the SPE can be rinsed to remove any impurities. The pH of the eluent can then be adjusted to reduce the charge on the analyte molecules, which will release them from the ion-exchange component of the cartridge causing them to elute. The mixed-mode SPE system also offers a reverse-phase mechanism. This will allow you to change the organic concentration of the eluent to achieve improved selective elution.
Mixed-mode is commonly used in the separation of basic, acidic, and neutral compounds in a mixture. This is widely used in the concentration, purification, and analysis of pharmaceutical compounds and antibacterial agents found in wastewater.
Summary
Use the table below to answer the following questions about an SPE method.
The compound punicalagin (Figure 8), is present in pomegranate juice. Punicalagin analysis in pomegranate juice can be a method of quantifying how much pomegranate juice is in a juice blend. Note that punicalagin is colorless and it is anthocyanin compounds that give pomegranate juices their red color.
Figure 8. Chemical structure of punicalagin.
Q12. Consider the various SPE modes we discussed. Choose and defend your choice for the most appropriate methods to extract punicalagin from pomegranate juice. Suggest both an adsorbent material and eluents.
Table 2: Summary of SPE Modesa
Mode
Adsorbent
Retention Mechanism
Sample Matrix
Analyte Properties
Elution Scheme
Example Applications
Reversed Phase
Non-polar
• C2
• C8
• C18
• Phenyl
Non-polar or hydrophobic interactions
Aqueous samples
• Biological fluids
• Aqueous extracts
• Environmental water samples
• Wine or beer
Exhibits non-polar functionalities
• Most organics
• Alkyl, aromatic, alicyclic functional groups
Disrupt reversed-phase interaction with solvent or solvent mixtures of sufficient non-polar character
• Methanol, acetonitrile, dichloromethane
• Buffer/solvent mixtures
• Drugs and metabolites in biological fluids
• Environmental pollutants in water
• Aqueous extracts of tissues and solids
Normal
Polar
• Alumina
• Silica (Si-OH)
• Diol
• cyano
• amino
• florisil
Polar Interactions
• H-bonding
• pi-pi
• dipole-dipole
• induced dipole
Non-polar
• Organic extracts of solids
• Very non-polar solvents
• Fatty oils, hydrocarbons
Exhibits polar functionalities
• Hydroxyl, carbonyls, amines, double bonds
• Hetero atoms (O,N,S,P)
• resonance
Polar interactions disrupted with a more polar solvent or solution
• acetonitrile, methanol, isopropanol
• combination of solvent /solvent mixtures
• Clean-up of organic extracts of soils and sludge
• Fractionation of petroleum hydrocarbons
• PCB’s in transformer oil
• Isolation of compounds in cosmetics
Ion Exchange
• Quaternary amine (anion)
• Sulfonic acid (cation)
Electrostatic attraction of charged functional groups of the analyte to oppositely charged functional groups on the sorbent
Aqueous or organic samples of low walt concentration (<0.1M)
• Biological fluids
• Solution-phase synthesis reactions
Cation exchange – for basic compounds such as amines
Anion exchange – for acidic compounds
Electrostatic interactions disrupted via:
• pH modification to neutralize compound and/or sorbent functional groups
• increase salt concentration (>1M); or use a more selective counter-ion to compete for ion-exchange binding sites
• Pharmaceutical compounds and drugs in biological fluids
• Fatty acid removal from food or agricultural samples
• Clean-up of synthetic reactions
• Organic acids in urine
• Herbicides in soil
aTable adapted from Sigma Aldrich website4 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/03_Solid-Phase_Extraction.txt |
QuEChERS and the Dirty Dozen
You’re walking through the produce aisle at the grocery store and wonder: “Which organic fruits, veggies are worth spending more on?” This was the title of a recent Omaha World Herald article.1 You can also find similar articles on the web at organic.org2 (The Dirty Dozen) and Good Housekeeping3 (The New Dirty Dozen: 12 Foods to Eat Organic.)
The terms “The Dirty Dozen Plus” and “The Clean Fifteen” have been trademarked by the Environmental Working Group (EWG) in their Shopper’s Guide to Pesticides in Produce.4 Articles such as these are geared toward the general public. As scientists, let’s take a closer look at these articles and determine from where they obtained their information and how this information is being utilized.
Q1. EWG compiles these lists every year. Find a recent list and write down the top five produce types most likely to contain pesticides. Note: Reference 4 is a great place to start your research. If you scroll down and click “see the list” you should be able to access a recent list without registering at the website.
Q2. How do you suppose the EWG came up with the “Dirty Dozen” list? At the top of the page at the site ( http://www.ewg.org/foodnews/ ) click on “Read the Report” and you should be provided with the Executive Summary. Read through the first few paragraphs of this document.
A) What group carried out the testing of produce for pesticides?
B) Read through the document and scroll down to the “Methodology” section. What metrics did they use to rank the foods? You should find six metrics.
C) There are some references to the pesticide data from the United States Department of Agriculture (USDA) at the end of the Executive Summary. List the relevant references along with brief descriptions of what they contain.
Q3. Find any recent USDA Pesticide Data Report online. The website for the Pesticide Data Program (PDP) is www.ams.usda.gov/AMSv1.0/pdp (Databases and Annual Summary Reports). Note, you should scroll to the middle of the page to “PDP Annual Summary Reports” and click on a specific report. In the 2010 Report, Table 1 (page 15 of the pdf file) lists the commodities sampled and tested. Did they test any produce that made your top 5 list in question 1?
Q4. In the report, Refer to Appendix B: Distribution of Residues by Pesticide in Fruit and Vegetables (or search the pdf for the fruits or vegetables you listed in Q3). Which pesticides were detected on these? Make a table structured like the one below. Note, some fruits or veggies may have more than one pesticide detected on them. Choose two produce items for your table and list 3-5 pesticides for each. Also find the percent of samples that contained each pesticide along with the concentration range and tolerance limit.
Table 1: Pesticides Found on Fruit and Veggie Samples
Fruit/Veggie
pesticide
% of samples
Conc. Range (ppm)
EPA tolerance (ppm)
A review of years of PDP data would reveal that apples are consistently at the top of the “Dirty Dozen” list.
Q5. Look at your table from Q4 and list three to four pesticides most detected on apples. Did any samples contain levels above the EPA tolerance?
Q6. To get an idea of the chemical and physical properties of these four pesticides most detected on apples, look up their chemical structures on Pubchem (https://pubchem.ncbi.nlm.nih.gov/ ) and draw below or insert into your data table. Additionally, in the Chemical and Physical Properties section, find information on the pH stability and the pKa (dissociation constant) of each of the four pesticides.
You’ve now seen that pesticide residues have been detected and quantitated on many of the fruits and vegetables you find at the grocery store. As a student studying analytical chemistry, you are probably wondering what technique was used to identify and quantitate the pesticide residues. Many of these measurements were made using some form of gas chromatography coupled with mass spectrometry (GC-MS) or liquid chromatography with MS detection. You already know that you can’t inject an apple on a GC and that some form of sample preparation is needed. So once again . . .
A) How do you get your sample from this form…
B) To a form suitable to inject on the GC . . .
It may be wise to change the caption in part A) above to “How do you get 1000’s of samples from this form . . . “ Looking through the PDP literature, you can see that an extensive amount of samples were collected, prepared and analyzed throughout the United States. From the dates of the reports, you can also see that this is done annually. So at this point, you know we have a complex sample and multiple analytes. Note that the pesticides are also referred to as “residues” in the pesticide analysis lingo. An analysis that determines the concentration of multiple analytes is called a “multi-residue” analysis.
Q7. Name both practical obstacles and requirements for this type of multi-residue sample preparation and quantitation on fruits and veggies. Especially think about the sample preparation and extraction step.
Q8. Name some important analytical requirements for this type of multi-residue sample preparation and pesticide quantitation on fruits and veggies. Especially think about the sample preparation and extraction step.
Q9. Consider the sample preparation and extraction techniques you have learned about so far in this module. Are there any that may be applicable to multi-residue pesticide analysis? Are there any that would be a poor choice?
Sample Preparation Background
As you have seen in the previous sections of this module, some drawbacks of conventional extraction techniques, particularly LLE, are the use of large volumes of organic solvents and the long sample preparation times. In fact, Waters Corporation estimates that 60% of analytical work activity and operating cost can be spent on sample preparation procedures.5 In the solid phase extraction (SPE) section, you learned how SPE uses less solvent and makes sample preparation more efficient. However, SPE can be both costly and complex to implement, and method development can take a long time.
The sample preparation approach of QuEChERS (pronounce “ketchers”) was originally designed to improve the efficiency of pesticide extraction from fruits and vegetables.6 The method has been improved upon7, and there are now many other applications of the QuEChERS process, particularly for samples containing little to no water. The acronym stands for Quick, Easy, Cheap, Effective, Rugged, and Safe. It has been estimated that a single analyst can prepare 8 samples in 45 minutes using disposable materials worth only a few dollars or euros.8 QuEChERS uses much less solvent than the other extraction methods and uses no chlorinated solvents.9
How QuEChERS works
A QuEChERS extraction involves two steps.
1) The first step is a liquid-liquid extraction via a “salting out” step in acetonitrile.
2) The second step is a dispersive SPE (d-SPE) cleanup step to remove polar interferents.
In the previous sections of this module, you have learned about the theory behind both LLE and SPE. To see how these are combined in the QuEChERS process, click here10 to watch a video showing the procedure. The video lasts approximately 6 minutes. Watch the entire video and answer the following questions to make sure you understand the process. The webpages at references 9 – 12 are also helpful for understanding the specific steps in the process.
Q10. Find the structure of acetonitrile and look up the density. Water and acetonitrile are generally miscible, but the high concentration of salt added causes a phase separation between acetonitrile and the salt water.11 Would you expect acetonitrile to be the top or bottom layer in a LLE with the salt water?
Q11. Fill in the second column of the table below with the description of each step.
Table 2: Steps of a QuEChERS Procedure
Step
Description
Additional Information
1
---------
2
3
4
5
6
7
-------
Q12. Add the following details to the “Additional Information” column.
A. What mass of sample should be added to the 15 mL centrifuge tube prior to adding extraction solvent?
B. What volume of extraction solvent is added for each gram of sample?
C. What are the identities of two types of buffering salts?
D. After the extraction into acetonitrile, what volume of supernatant is added to the 2 mL centrifuge tube? What volume is added to the 15 mL centrifuge tube?
E. In the centrifuge cleanup tube, which layer contains the analyte? Which layer contains the sorbent?
Q13. According to the video, do you expect the majority of your interferents to be polar or non-polar? Think about the compounds you may find in a fruit or veggie sample. What might be the identity of an interferent in a fruit sample?
Q14. What is the purpose of adding the buffer salts?
Q15. After the first QuEChERS tube is centrifuged, there are four layers. Label them below and note which layer contains your analyte.
Table 3: Layers formed in the first QuEChERS Tube
Top layer =
Middle layer =
Middle layer =
Bottom layer =
Q16. The next step is to remove an aliquot of the top (liquid) layer and transfer to a subsequent centrifuge tube. The tube size choices are 15 mL and 2 mL. What analytical figure of merit can the 15 mL cleanup tube improve?
Q17. The cleanup step consists of adding “sorbents” to the analyte-acetonitrile aliquot. Why is this called dispersive SPE (d-SPE)? Specifically explain the identity of the sorbent and its purpose.
Q18. Another supplier of QuEChERS, Teledyne Tekmar, notes that C18 is incorporated into the sorbent for higher fat samples. Explain.
Q19. As you know many fruits and vegetables are highly colored. What additional sample preparation step can be applied to samples such as peppers, carrots and spinach, which contain large amounts of chlorophyll and carotenoids.
Q20. During the cleanup step, MgSO4 is also added with the “sorbent”. What it its purpose?
Summary
We’ve just gone through a lot of information. If you need to, fill out the table below to keep track of which sorbent removes which interferents.
Table 4: QuEChER d-SPE sorbent and interferents removed
sorbent
Interferent removed
Summary Questions
Q21. Propose a procedure for extracting pesticides from a sample of peaches (88% water). Also indicate at which step you would add internal standard. Consult one of the USDA Pesticide data reports (Appendix B) and also provide expected pesticide levels. Write out the general steps below. Include the identities of three pesticides you expect to detect on over 20% of your samples and cite your reference. Note, that the “find” function in your pdf reader is helpful for searching a specific commodity.
Q22. How would you modify your procedure in Q21 if you were analyzing oats for pesticides?
Q23. You learned about determining the percent recovery of an extraction procedure in the SPE section. Propose an experiment for determining the percent recovery of a QuEChERS extraction of pesticides from grapes.
Calculation Questions
Use the following procedure to answer the next two questions.
A lab analyst followed the following general procedure while performing a QuEChERS sample preparation for validating a method for determining pesticides in apples. Note that the QC (quality control) samples will be used to determine the % recovery of the extraction. For determination of the %recovery, organic apples presumed to contain no pesticide residues were used.
Extraction: A 15.0 g previously homogenized sample was placed into a 50 mL centrifuge tube. QC samples were fortified with 100 μL of appropriate pesticide QC spiking solution yielding QC samples. One hundred microliters of IS spiking solution (15 μg/mL of TPP) were added to all samples except the control blank. Note, the two pesticides in which we are interested are diphenylamine (DPA) and thiabendazole (TBZ). The solution concentration and aliquot volumes are shown in Table 5.
Tubes were capped and vortexed for 1 min. Next, 15.0 mL of 1% acetic acid in acetonitrile (ACN) were added to each tube using the dispenser. To each tube, a buffered extraction packet from the kit containing 6 g of anhydrous MgSO4 and 1.5 g of anhydrous sodium acetate, was added directly to the tubes. The tubes were shaken vigorously for 1 min by hand to ensure that the solvent interacted well with the entire sample and crystalline agglomerates were broken up. Sample tubes were centrifuged at 4000 rpm for 5 min.
Table 5: QuEChER Sample Preparation for Method Validation and Unknown Sample
Sample
TPP spike conc. (μg/mL)
TPP vol. (μL)
Conc. of 100 μL DPA
Conc. of 100 μL TBZ
Conc. (ng/g) DPA/TBZ
Conc. (ng/g) TPP
QC 1
(spiked organic)
15.0
100
1.50 μg/mL
1.50 μg/mL
QC 2
(spiked organic)
15.0
100
7.50 μg/mL
7.50 μg/mL
QC 3
(spiked organic)
15.0
100
30.0 μg/mL
30.0 μg/mL
Control Blank
--
--
--
--
--
--
Unknown
sample
15.0
100
--
--
D-SPE Sample Cleanup. A 1.00 mL aliquot of the upper ACN layer was transferred into a QuEChERS AOAC 2 mL dispersive SPE tube. This 2 mL tube contained 50 mg of PSA and 150 mg of anhydrous MgSO4. The tubes were tightly capped and vortexed for 1 min. The 2 mL tubes were centrifuged with a micro-centrifuge at 13000 rpm for 2 min. An aliquot from the extract (500 μL) was transferred into an autosampler vial. The samples were then ready for GC/MS analysis.
Q25. Use the appropriate information in Table 5 to calculate the concentration (ng/g) of each compound (TPP, DPA and TBZ) spiked into each QC apple sample and the concentration of internal standard TPP in the unknown apple sample in units of ng compound /g apple. You may enter your results in the table.
Calibration Curves
The quantitation method commonly used for these methods is internal standard method, using the compound triphenyl phosphate (TPP) as an internal standard. Calibration curves, spiked in matrix blanks, were made at levels of 5, 10, 50, 100, 200, and 250 ng/g. (Table 6) The TPP (IS) was used at the level you calculated in Q25. The calibration curves were generated by plotting the relative responses of analytes (peak area of analyte/peak area of IS) to the relative concentration of analytes (concentration of analyte/concentration of IS).
Table 6: Calibration Curve Data for Pesticide Analysis
Analyte
1 mL dSPE regression equation
R2
Thiabendazole
y = 0.0132x + 0.0002
0.9968
Diphenylamine
y = 0.0110x + 0.0003
0.9995
Table 7: GC/MS Peak Area Data for Pesticide Analysis
Sample
DPA/TPP Peak Area
TBZ/DPA Peak Area
Conc. of 100 μL DPA
Conc. of 100 μL
TBZ
Conc. (ng/g) DPA or TBZ
Conc. (ng/g) TPP
QC 1
(spiked organic)
0.1032
0.1264
1.50 μg/mL
1.50 μg/mL
QC 2
(spiked organic)
0.5032
0.6120
7.50 μg/mL
7.50 μg/mL
QC 3
(spiked organic)
2.313
2.678
30.0 μg/mL
30.0 μg/mL
Control Blank
--
--
--
--
--
--
Unknown
sample
1.564
3.123
--
--
Q26. With any sample preparation procedure, an analyst is concerned about potentially losing sample during the process. The purpose of the QC samples was to determine the % recovery of the procedure. Use the calibration curve data in Table 6, the spike concentration data you determined in Q25, and the peak area data in Table 7 to determine the % recovery of DPA and TBZ at each concentration level.
Level (ng/g)
%DPA
%TBZ
10.0
50.0
100
Q27. After the % recovery of the method was determined, it was used to determine the concentration of DPA and TBZ (ng/g of apple) in an unknown apple sample (conventionally grown) obtained from the grocery store. Use the calibration curve data and peak area ratio data in the table to determine the amounts of DPA and TBZ in this apple sample. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/04_QuEChERS.txt |
I. Headspace Sampling for Volatile and Semi-volatile Analytes
Introduction. Imagine you are a member of the highway patrol, and observe a car veering across the center and side lines of the interstate at 2:30 am on a Sunday morning. Suspecting that the operator is driving under the influence of alcohol, you pull the car over, and get the driver out of the car to perform a field sobriety test. After these are failed, you ask the driver to submit to a breathalyzer, an electrochemical test for alcohol in the expelled breath of the driver, to determine if their blood alcohol level exceeds the legal limit in most states of 0.08% w/v. If the driver fails or refuses to submit to this test, it is common practice for you as the law enforcement officer to place the suspect under arrest, and transport the suspect to the nearest medical facility or police forensic laboratory for a blood test to determine the level of alcohol content (BAC). If the driver refuses this test, the law generally requires that the suspect be charged as if the maximum level has been exceeded.
Figure 1. Field Sobriety Test1
Q1. You probably have seen enough crime shows to predict how the laboratory analysis for blood alcohol content will be done. Whether you know or not, you should be able to describe the physical characteristics of both the sample (blood) and the material to be analyzed (ethyl alcohol) that might have to be considered in the analysis. Describe if you can what part of the sample will be used, and what method of analysis might be employed.
Blood is a complex matrix, containing hundreds of substances, including red and white blood cells along with large molecular weight proteins and enzymes. The primary component is water (92%), which we know boils at about 100 oC (depending on pressure), and has a vapor pressure of about 3 kPa near room temperature. Ethanol boils at a lower temperature than water (about 78 oC) and has a higher vapor pressure, about twice that of water (6 kPa at 20 oC).
Recall that vapor pressure is defined as the pressure exerted by a gas in a closed container when at equilibrium with the corresponding liquid phase. This means that at a fixed temperature, the amount of substance in the gas phase relative to the liquid phase increases with vapor pressure.
Substances with high values of vapor pressure are considered to be more volatile than those that do not. Considering that the larger molecular weight components of blood have much smaller vapor pressures than water and ethanol, one might consider the vapor phase above the liquid (blood) sample as a good place to look for ethanol in the absence of the very large number of possible interferents.
The equilibrium is illustrated in Figure 2. A sealed vial contains, in this case, equal volumes of air space and liquid. The substance of interest, say ethanol, is dissolved in the complex matrix, say blood, at some level. Once equilibrium is reached between the gas and liquid phases, ethanol (and other volatile substances) will be found at concentrations described by the partition coefficient (K) for each analyte:
$\mathrm{K = \dfrac{C_{liquid}}{C_{gas}}} \label{Eq. 1}$
Figure 2. Equilibrium between liquid and gas phases in sealed vial.
In addition, each analyte will have a propensity to enter the gas phase that is dependent upon the relative volumes of the gas and liquid phases. This is described by the phase ratio, β, given by
$\mathrm{β = \dfrac{V_{gas}}{V_{liquid}}} \label{Eq. 2}$
Combining the equations for K and for β yields an expression2 that describes the concentration of analyte in the gas phase as
$\mathrm{C_{gas} = \dfrac{C_{liquid}}{K + β}} \label{Eq. 3}$
Q2. Knowing that we are interested in analyzing the amount of alcohol in blood, you would rightly assume that the larger the concentration of the analyte in the gas phase, the easier it would be to evaluate. Using Le Chatelier’s principle as a guide, list and explain ways you might increase the concentration of ethanol in the gas phase over the liquid blood sample.
Q3. What analytical method(s) do you know about that are amenable to volatile substances in the gas phase?
Static headspace sampling with gas chromatographic analysis. The most common test performed by forensic laboratories is that for blood ethanol.3 Typically this involves the use of gas chromatography following headspace sampling. A gas-tight syringe is most often used to collect a sample from the headspace of a sealed vial, and that gas sample is injected into the gas chromatograph. In most cases, the vial is heated, but no purge gas is introduced into the sample vial during static headspace analysis (more on this later). Sampling and sample introduction can be done manually, but more often an autosampler is used. If the autosampler employs a syringe, it is generally heated to prevent the condensation of volatiles prior to their injection. Heated transfer lines that directly connect the sample vial and the injector on the chromatograph are also frequently employed.
Figure 3. Headspace Sampling.4
Q4. What advantages can you see for chromatographic analysis when sampling only from the volatile gas phase above a complex matrix?
The headspace technique samples only the volatile substances at equilibrium in the gas phase, and eliminates the need for sample cleanup while preventing the injection of non-volatile materials that can often contaminate the chromatographic system. While this introduction has focused on blood as a sample matrix, headspace analysis is possible for solid or liquid samples.
Improving the amount of volatile analyte in the gas phase. The two most important parameters affecting the distribution between two phases are temperature and solubility. The effect of temperature on the distribution coefficient, K, is given by
$\mathrm{\ln K = \dfrac{A}{RT} – B} \label{Eq. 4}$
where A and B are thermodynamic constants, R is the ideal gas constant, and T is the temperature in Kelvin.2 Thus, increasing the temperature will decrease the value of K. Recall that K = Cliquid/Cgas, which means that Cgas increases with T. For ethanol in water, the values for K are 1355 at 40 oC and 328 at 80 oC.4
Increasing the temperature to increase Cgas works best for analytes that have a high solubility (high K value) in the sample matrix. For analytes with low solubility in the sample matrix (low K value), an increase in the volume of sample relative to the headspace volume (decreasing β) will be more effective at increasing Cgas.5 The addition of large concentrations of an inorganic salt like sodium chloride can also result in a reduction of matrix solubility (“salting out”) for many analytes, especially polar ones, leading to an enhanced analyte concentration in the headspace.4
Q5. To this point in our discussion of static headspace analysis, we have considered the GC analysis of a gaseous sample taken from above the sample matrix. The concentration of analyte in the gas sample is dependent upon temperature and solubility in the sample matrix. Instead of just sampling the gas directly, can you think of a way to physically enhance the amount of analyte you obtain from the gas sample prior to its introduction into the GC? (Hint: Brita® filter.)
Dynamic headspace sampling with gas chromatographic analysis. The Brita® filter, and water purification devices like it, contain activated carbon, a form of carbon that has been treated to yield very high surface area-to-mass ratios which remove impurities by adsorption, and ion exchange resins, which remove impurities based on their charge. A cartoon6 (from Brita®) of the filter is shown below. The measure of impurities that can be removed is a function of the surface area in contact with the water sample being filtered, and the extent of charge interaction with the ion exchangers.
Figure 4. Cutaway View of a Brita® Filter.6
In dynamic headspace sampling, the analyte is allowed to equilibrate into the headspace above the sample matrix as before, but then a flow of inert gas (usually GC carrier) is introduced into the headspace through a needle and allowed to flow out of the vial via a second needle. This gas flow is directed either onto a cold trap, or more commonly onto a trap containing a solid adsorbent. As analyte is removed from the headspace, it is replenished according to its value of K, the partition coefficient. Over time, the concentration in the sample matrix is reduced, allowing for an exhaustive, or near-exhaustive collection of the analyte on the trap. Dynamic headspace can be used for either solid or liquid samples. A schematic of the method is given in Figure 5.
Figure 5. Dynamic Headspace GC.
The concentration of analyte in the headspace is reduced over time according to
$\mathrm{C_t = C_o e^{-(F\, t\, /\, V)}} \label{Eq. 5}$
where Ct is the concentration of analyte in the headspace after time t, Co is the initial equilibrium concentration in the headspace, F is the flow rate of purge gas, t is the purge time, and V is the headspace volume.7
Q6. Rearrange Equation \ref{Eq. 5} to solve for the time, t, required to remove 99% of the analyte from a sample if the flow rate of helium is 40 mL/min and the volume of the headspace is 10 mL.
Commonly used sorbent materials for dynamic headspace analysis include silica gel, activated carbon, and Tenax®, a porous polymer prepared from 2,6-dipheny-p-phenylene oxide.8 These can be used alone, or in series as a multi-layer trap, depending upon the complexity of the sample being collected. Tenax® has a low affinity for water, and works well for nonpolar volatiles. Silica gel is highly polar and retains water and other polar analytes well. Activated carbon is hydrophobic and is very good at trapping highly volatile compounds.5
Following extraction, the trap is heated and the sample thermally desorbed onto the column of a gas chromatograph. Reverse inert gas flow is directed through the trap so that the least volatile analytes, which were trapped first, are prevented from contacting the strongest of the adsorbents, which could lead to slow desorption.5
Purge and trap. If instead of directing the inert gas flow through the headspace above the sample, the gas is bubbled through an analyzed liquid sample, the technique is referred to as purge and trap. This is most often applied to volatile analytes in a water matrix, as for example in EPA 524.2, a method for the analysis of purgeable organics in drinking water.9 In general, a fritted sparging tube, either 5 or 25 mL in volume, as shown in Figure 6 (from Restek10), is used for purge and trap. The inert gas is introduced through a needle placed within the solution volume, while the sample rests on a glass frit. As in dynamic headspace analysis, the purge gas is directed onto a sorbent trap, where it is concentrated prior to desorption onto a gas chromatograph. Detection limits for many volatiles can be increased more than 1000x with the use of purge and trap compared with static headspace.2
Figure 6. Sparge Tube for Purge and Trap10
II. Solid Phase Microextraction (SPME)
Introduction. Headspace GC has been demonstrated to be an effective technique for the analysis of volatile compounds in liquid or solid matrices. More of a challenge is encountered with semi-volatile or non-volatile compounds in these matrices. Consider atrazine, one of the most commonly used pesticides in the United States, and a major contaminant of ground and surface waters. Atrazine causes disruption of hormonal levels in humans, and the EPA considers a safe level in finished waters to be only 3 ppb for a yearly average.11
Figure 7. Structure of Atrazine
Q7. Consider the structure of atrazine. Would you expect it be water soluble? Why? What other characteristics of the molecule might be important to consider when performing an analysis?
Historically, the most common method for the analysis of atrazine in natural waters involved a liquid-liquid extraction (LLE), a sample volume of 1 L (adjusted to pH 7.0) and multiple volumes (100 mL minimum) of an organic solvent like dichloromethane. The organic extracts were combined and reduced in volume prior to analysis by GC or HPLC.12
Q8. What would you consider to be the drawbacks to the LLE method detailed above?
Q9. If you were to design an “ideal” method for sampling and preconcentration of an environmental contaminant like atrazine in natural waters, what characteristics would you build in to the method?
Much effort is being expended in the analytical community to develop methods for environmental contaminants that replace techniques like LLE that require large sample volumes, often unfriendly chlorinated solvents, and long prep times. One technique that has proven to be quite capable of addressing most, if not all, of these concerns is solid phase microextraction (SPME).
Solid phase microextraction (SPME) utilizes a small length (1-2 cm) of fused silica fiber that is coated with a sorbent material allowing to selectively preconcentrate analyte from either the gas or liquid phase. The technique allows for easy, one-step sample collection without the need of solvents or long sample preparation time. It can be used for direct sample collection from air or water in the field, and has even been applied to in vivo analysis.13 After sample exposure, the fiber is introduced into the heated injector of a gas chromatograph where the analyte is desorbed. With an appropriate interface allowing for removal of the analyte into a solvent, the technique can also be applied to HPLC analysis.14
The SPME fiber, which is fragile, is housed inside of a syringe needle to protect it before and after sampling. In headspace or direct immersion sampling, the fiber is exposed to the sample only after the vial septum has been punctured. Following extraction, the fiber is retracted before insertion into the GC, and once inside the injection liner, the fiber is extended for desorption. The fiber apparatus is shown in Figure 8.
Figure 8. SPME Fiber Assembly. At top, the fiber is retracted inside the syringe needle. Below, the inner sheath and the fiber have been extended out of the needle for sample extraction or desorption.
Sorbent materials may be either liquid polymers, like polydimethylsiloxane (PDMS), polyacrylate (PA) and polyethyleneglycol (PEG), or solid particles, like CarboxenTM (porous carbon) and divinylbenzene (DVB), that are suspended within a liquid polymer. The mechanism for extraction is different for the two materials. The liquid polymers are referred to as absorbent coatings, into which analyte molecules diffuse, attracted primarily by the polarity of the polymer. Retention within the coating is a result of both these attractive forces and the thickness of the polymer, much like that seen for chromatographic stationary phases. Porous particles extract analyte molecules on the basis of adsorption, the result of $\pi$-$\pi$ or H-bonding, and van der Waals type interaction.12 Porous materials are generally characterized by pore size and extent of porosity, along with total surface area, with the strength of adsorption dependent upon analyte particle size.
SPME fibers are available commercially from Suplelco/Sigma-Aldrich. In general, the choice of fiber is dictated by the properties of the analyte, notably polarity, molecular weight, volatility, and concentration range.14 PDMS coatings are available in a variety of thicknesses, dictated by sample volatility, and exhibit best sensitivities toward non-polar analytes. PA coated fibers are suited primarily for polar analytes in polar media. Mixed phase coatings are the best choice for smaller, volatile, and polar analytes.16 Detailed selection guides are available from the manufacturer, and in references 15 and 16.
SPME fibers may be used in either the headspace above a sample matrix (HS-SPME), or directly immersed (DI-SPME) into a liquid. As for static headspace, the method relies on establishing equilibrium between two phases, and does not exhaustively remove the analyte from the sample, generally extracting between 2 – 20%.16
Figure 9. SPME fiber in vial headspace.
Q10. Predict how the process of equilibrium between a SPME fiber in the gas phase above a sample and one immersed in a liquid sample might differ. Which of the two would you predict would occur faster?
Q11. What complications would you expect from direct immersion (DI) SPME that you expect to be absent in HS-SPME?
The distribution coefficient for an analyte in a sample matrix and the SPME fiber within that matrix is given by
$\mathrm{K_{fs} = \dfrac{C_f}{C_s}} \label{Eq. 6}$
where Cf is the concentration of analyte within the fiber and Cs is the concentration of analyte in the sample matrix.16 Until an equilibrium condition is reached, the concentration within the fiber will increase over time. If the fiber is in solution, it is important the solution (or the fiber) is agitated during the concentration step as the analyte concentration in the solution layer close to the fiber will be depleted over time. In the gas phase above the sample matrix, the description of the distribution coefficient is more complicated, involving three phases instead of two. Additionally, as the analyte is desorbed from the headspace, more analyte will be pulled into the gas phase from the sample matrix.
The equilibrium concentration of analyte into the fiber will be identical using extractions from either of the phases, as long as the volumes of the two are equivalent.15 Also, the amount of analyte extracted into the fiber is independent of the sample volume if KfsVf << Vs, where Vf is the volume of the fiber coating, and Vs is the volume of the sample.16
Advantages of HS-SPME over DI-SPME are similar to those described for HS-GC in the previous section. Sampling in the gas phase prevents fouling of the fiber with high molecular weight impurities, or those that might irreversibly adsorb to the fiber surface. In the absence of these complex matrix effects, DI extraction works best for analytes with lower volatility and polarity. Conversely, HS extraction is better suited for analytes with higher volatilities and polarity.5 Because of higher analyte diffusion rates in the gas phase relative to the liquid phase, equilibrium occurs faster for HS than for DI. Further, because of the relationship between large sample volume and amount of extracted analyte, SPME is ideal for field sampling, allowing direct analyte extraction in large volumes of air or water with no required sample pretreatment.13
Optimization of Conditions for SPME. In addition to sorbent type and thickness, other factors to consider in the optimization of analyte extraction by SPME include temperature, sample volume, pH and ionic strength, agitation conditions, and desorption parameters.5
Temperature. Increasing the temperature at which extraction is carried out increases the diffusion coefficients for the analyte, thus leading to a more rapid extraction. However, these kinetic effects are countered by a decrease in Kfs and consequently a lower amount of extracted analyte once equilibrium is achieved.15 If the analyte concentration is high in the sample, and sensitivity is not an issue, higher temperatures allow for shorter analysis times. Lower temperatures and longer extraction times may be required if sensitivity is a problem. Extraction efficiencies may also be increased in some instances by the use of a cold fiber technique, in which the fiber is cooled either by introduction of liquid CO2 via an inner capillary, or by Peltier cooling.15
Sample Volume. The equilibrium amount of analyte extracted from a sample is described by
$\mathrm{n = \dfrac{K_{fs}\, V_f\, V_s\, C_o}{K_{fs}\, V_f + V_s}} \label{Eq. 7}$
where n is the number of moles extracted, Co is the analyte concentration in the sample, and other variables are as previously defined.16 In general, if KfsVf > Vs, then the amount of analyte extracted will increase with sample volume.15 In most laboratory settings, sample vials for use with SPME have volumes from 1.5 to 20 mL, with extraction amounts normally increasing with Vs over this volume range.15 As described previously, for very large values of Vs (direct air or water sampling, eg), the amount of extracted analyte will depend only upon Co for set values for Kfs and Vf. For low concentrations of volatile analytes (<50 ppb), the amount of extracted analyte may be observed to be independent of sample volume, while exponentially increasing calibration curves are often seen for large samples (>5 mL) containing high concentrations of analyte.13
pH and Ionic Strength. Commercially available SPME fibers employ sorbents that are neutral, and thus basic or acidic analytes must be converted to neutral species prior to extraction. Care must be exercised in pH adjustment of the sample matrix, as some fibers may be degraded at very high or very low pH. The addition of an inert salt, as described for HS-GC, will many times serve to enhance the extraction efficiencies for certain analytes.13
Agitation. Samples for SPME are typically agitated during extraction to decrease the time required to achieve equilibrium with the fiber, and to increase the amount of analyte extracted. There are many ways to achieve proper agitation, including magnetic stirring, sample vial movement, or movement of the fiber within the sample vial. The choice of agitation may be dictated by autosampler make/model, or selected according to specific method requirements for a particular analyte.
Desorption. For GC analysis, samples are normally desorbed from the SPME fiber by insertion into the injection port of the chromatograph. Controllable parameters include inlet temperature, depth of fiber insertion, carrier gas flow at inlet, and desorption time. Typically, a small bore inlet liner (ca. 0.75 mm) is used without split flow during injection to increase the flow rate of carrier gas around the fiber and thereby the efficiency of extraction for adsorbed analytes.13 Inlet temperatures, fiber depth, and desorption times are optimized for rapid, but complete extraction of analyte. The Supelco website (among others) provides a large number of SPME applications that provide a convenient starting point for methods development.17
Q12. Access the Supelco SPME applications website (reference 17), and obtain suggested experimental conditions for the analysis of atrazine (a triazine environmental pesticide) from water samples.
Stir Bar Sorptive Extraction (SBSE). SBSE, a variation of the SPME method involves the coating of sorbent material directly onto glass surrounding a magnetic stir bar with a length of 10-20 mm. The coating, PDMS or PDMS/ethlylene glycol (EG) copolymer, has a greater thickness than those available for SPME fibers (between 0.5-1.0 mm compared to 30-100 μm for SPME), allowing much larger sorptive capacities, and up to a 1000 times more sensitivity than SPME.18 The stir bar can be used either in the headspace above a sample, or more commonly within a liquid matrix with magnetic stirring.14 Vial size is 10-20 mL. Once extraction has been accomplished, the stir bar is transferred to a thermal desorption unit, where it is desorbed and introduced into the inlet of a gas chromatogram. Figure 10 (from Gerstel18) shows a stir bar inside of a sample vial (left) and inside of a desorption tube which is part of the thermal desorption unit (right). A library of SBSE applications can be found on the Gerstel website.19
Figure 10. Stir Bar Sorptive Extraction (SBSE).18 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/05_Sampling.txt |
Consider the analysis of alcohol in blood you studied earlier in this module. It was determined that a good choice of analytical method was headspace gas chromatography (HS-GC).
Q1. List two or three of the reasons that HS-GC was an ideal method for the quantification of ethanol in the complex sample matrix of blood.
Though not discussed in this module, the basics of gas chromatography can be found in the learning module Environmental Analysis – Lake Nakuru Flamingos: Pesticides1, with additional information available in the online textbook Analytical Chemistry 2.1.2 In general, substances in a mixture are volatilized in a heated injector port and swept onto a capillary column using an inert carrier gas like helium, where the components are separated from one another by taking advantage of differing physical interactions (structure dependent) with the column material and then individually passed through a detector, which produces an electrical signal proportional to the concentration of the individual component. A simplified diagram of a GC is shown below.
Figure 1. Block diagram of a gas chromatograph. 3
In a typical GC analysis of ethanol, the choice of detector is frequently a flame ionization detector (FID), which oxidizes the sample in an air/hydrogen flame to produce cations that are then attracted to a cathode where a current is measured. This type of detector responds to certain functional groups present on the measured species, in this case the alcohol moiety.4
Q2. Consider the methods of analysis you are familiar with or may have used in the past. For each of the methods, describe the physical process responsible for the production of an analytical signal.
In your consideration of various methods of analytical measurement, you probably noticed that for most, if not all the methods, the detection of analytes involved a physical response to some sort of external stimulation; for example, the absorption (or emission) of radiation (chromophore or fluorophore), or the oxidation or reduction of a functional group (electrophore).
Additional considerations in the choice of an acceptable method for a particular analyte must include such things as volatility and stability at elevated temperatures (most notably in GC), polarity and pH effects, solubility, chirality, poor chromatographic behavior, or just plain old lack of sensitivity sufficient for detection of the level of analyte present in the sample.5,6
Q3. What would you do if a molecule did not contain a chromophore, a fluorophore, or a redox active center, or lacked sufficient sensitivity at the chosen detector?
Many of you (hopefully) answered Q3 with some semblance of a response that included, simply stated, “change the molecule in such a way as to allow its detection”. This process whereby a molecule is chemically altered to enhance its detection by a particular analytical method is called derivatization.7
Case Study: Glyphosate (aka “Roundup”)
Glyphosate (N-(phosphonomethyl) glycine) is the most widely used herbicide in the world, with nearly 0.5 pounds/acre applied on cropland worldwide in 2014 (0.8 lb/acre in US).8,9 It was classified as a “probable human carcinogen” by the International Agency for Research on Cancer in 2015, with a “safe” level for human exposure set at 1.75 mg/kg/day by EPA.9 The structure of glyphosate is shown below.
Figure 2. Structure of Glyphosate.10
Q4. The physical properties of glyphosate are listed in Table 1. Analyze the structure of glyphosate in Figure 2, along with the physical properties given in Table 1 to predict some of the considerations that must be made when choosing an analytical method for the compound dissolved in surface water.
Table 1. Physical Properties of Glyphosate.11,12
Molecular Weight
169.07
Physical form
White solid, Zwitterion structure
Melting Point
184 oC (decomposes 187 oC)
Density
1.705 (pH = 1.9, 20 oC)
Dissociation Constants
pKa1,2 = 2.0, 2.6 (phosphate); pKa3 = 5.6 (amine); pKa4 = 10.6 (carboxylic acid)
log Kow
-3.40
Solubilities
10.5 g/L in water (pH = 1.9, 20 oC);
practically insoluble in organic solvents
Analysis of the structure and physical properties of glyphosate should lead to some generalizations as to its amenability to direct chromatographic or spectroscopic determination. It turns out that the analysis of glyphosate is complicated by the following factors (among others).8,13
• poor solubility in organic solvents, making it difficult to extract from water samples, in which it is quite soluble
• high polarity
• lack of volatility
• absence of chromophore, fluorophore, or electrophore
• generally poor chromatographic behavior (bad peak shape, adsorption, etc)
A combination of all these factors in a single molecule makes glyphosate the poster child for derivatization! Specific methods of analysis for glyphosate have been reviewed.8,13 The remaining discussion here will be of a general nature, and will deal with commonly used methods of chemical derivatization for analytical measurements that involve chromatography, including many that are applicable to glyphosate.
Common Methods of Derivatization
Gas Chromatography. Derivatization is generally necessary in the analysis by GC of compounds that exhibit low volatility, have poor thermal stability, contain “active” groups that can lead to loss of sample to intermolecular hydrogen bonding or adsorption on the inlet or column, and/or demonstrate poor sensitivity at the detector. Functional groups that are commonly derivatized for GC analysis include alcohols, carboxylic acids, amines, and sulfhydryls. The three most widely used methods of derivatization in GC are silylation, acylation, and alkylation.14
Silylation. In this reaction, active hydrogens are displaced by a silyl group, most often tetramethylsilane (TMS). The general reaction scheme is illustrated for TMS reacting with an alcohol below.
$\ce{R-OH + (CH3)3SiX → R-O-Si(CH3)3 + HX}\nonumber$
The reaction is applicable to active hydrogens from acid, alcohol, thiol, amine, and amide groups.6 The resulting derivatives have increased thermal and chemical stability, and increased volatility.15 Further, the presence of the silyl group improves the response in mass spectral analysis, supplying an additional characteristic fragments in the mass spectrum.15,16
Q5. The form of silane most commonly used for GC derivatization is trimethyl-chlorosilane, (CH3)3SiCl. What side reaction might you need to protect your reagent against in normal use?
Q6. Derivatization for chromatography can be performed either pre-column or post-column (prior to the detector). Which technique do you expect to be preferable for GC analysis? Explain.
In addition to trimethylchlorosilane (TMCS), a large number of silylation reagents exist that have differing reactivities and specificities. Many classes of compounds, like steroids, may possess more than one type of functional group that requires derivatization by more than one reagent.14 More information on the specifics of silylation can be found in review articles6,16-17 or from vendors like Sigma-Aldrich who have a large collection of application notes and product information available on their web sites.15,18
Acylation. In this process, compounds with active hydrogens (OH, SH, NH) are reacted with carboxylic acid or derivatives of carboxylic acids to form esters, thioesters, and amides, respectively.6 The general reaction for the acylation of an alcohol by an acid halide to form an ester is shown below.
$\ce{R-C}\textrm{(:O)}\ce{-X + R’-OH → R-C}\textrm{(:O)}\ce{-O-R’ + HX}\nonumber$
In addition to acid halides, acylation reagents include acid anhydrides and reactive acyl derivatives like acylated imidizoles and perfluorinated acyls.15 As was the case for silylation, acylation produces derivatized species that are more volatile, generally less polar, and less prone to adsorption than the original compound. Derivatization with fluorinated acyl groups leads to enhanced detectability at the electron capture (EC) detector for GC, as well as increased structural information from mass spectrometric detection.16 In general, acylated derivatives are more stable than their silylated counterparts.15
Popular acylation reagents include acetic anhydride (AA) trifluoroacetic acid (TFA), trifluoroacetic anhydride (TFAA), trifluoroactylimidazole (TFAI), pentafluoropropionic anhydride (PFAA), and N-methylbistrifluoroacetamide (MBTFA).6 Reagents useful for the derivatization of carbohydrates, amino acids, and drugs of abuse are also commercially available. The reader is once again directed to vendor web sites for information specific to a particular analysis.15,18-19
Q7. Do a Google search for 2,4,6-trichloranisole (TCA). What is it? Draw its structure, and predict whether the molecule would be amenable to derivatization by acylation. Explain your answer.
Q8. Resveratrol is an anti-oxidant commonly found in red wine. Click on the link given in reference 15, which will take you to information on derivatization reagents available on the Sigma-Aldrich website. Locate the application note for SPME analysis of resveratrol (page 26). Give the structure of resveratrol. What SPME fiber is used and what derivatization of TCA is required for analysis?
Alkylation. Derivatization by alklylation involves the replacement of active hydrogens of the type described previously with an aliphatic or aliphatic-aromatic (like benzyl) group. The most common application of the technique is the formation of esters from carboxylic acids. The esters formed in this way are less polar, more volatile and display improved chromatographic behavior than the original compound. In general, these derivatives are more stable than the corresponding silylation product.15 Esterification is most often accomplished by reaction of an alcohol with the carboxylic acid in the presence of an acid catalyst. This reaction is shown below.20
Rapid esterification (~2 min) can be accomplished at 100 oC (water bath) with BF3-methanol solution.6
A more general alkylating agent is N,N-dimethylformamide dimethylacetal (DMFDA), whose structure is given in Figure 3.
Figure 3. Structure of DMFDA.21
In addition to carboxylic acids, DMFDA rapidly reacts with amines, phenols, thiols, and amino acids to give the products in Table 2.6
Compound
Derivative
R-COOH
R-COOCH3
R-NH2
R-NH-(CH2)2-N(CH3)2
R-OH
R-O-CH3
R-SH
R-S-CH3
-O-C(O)-CH(R)-NH3+
CH3-O-C(O)-CH(R)-NH-(CH2)2-N(CH3)2
Table 2. Reaction Products for DMFDA Derivatization.
Derivatization with DMF-Acetals containing longer side chains (between 2 and 4 carbons) are also possible, with the added benefit of controlling analyte retention times.16
As with previous derivatizations, the addition of halogens in the alkylation reaction will enhance the response at the EC detector for derivatized analyte. One commonly used reagent for this purpose is pentafluorobenzyl bromide (PFB-Br), which is capable of adding the PFB group to carboxylic acids, alcohols, sulfonamides, and thiols.6
Liquid Chromatography. High performance liquid chromatography (HPLC) is preferred for separations involving chemical species that do not exhibit sufficient volatility or stability to allow the use of gas chromatography (GC). Applications of HPLC routinely include drug analysis, measurement of biologically important species like amino acids, and pesticide analysis, among others. Often these analytes lack a necessary structural component that makes them easily detectable using common techniques of HPLC detection like spectrophotometry, fluorimetry, or electrochemistry. As in GC analysis, methods to improve the detectability of a target analyte by reaction with a suitable derivatizing agent are available for HPLC. In addition to detectability, the derivatization of polar compounds can increase their retention in reverse-phase HPLC by making the derivatized product more hydrophobic.15
Q9. Consider the HPLC analysis of glyphosate in groundwater utilizing spectrophotometric detection. Due to the lack of a chromophore or fluorophore in the structure of the herbicide, it is necessary to perform a derivatization reaction before the modified analyte can be detected (see Figure 4). Speculate on the stages in the sample prep/analysis where a conversion can be achieved. Are there advantages and disadvantages that you recognize with each of your choices?
Figure 4. Reaction of Glyphosate with 9-Fluorenyl Methoxycarbonyl Chloride (FMOC-Cl)13
Q10. Identify the structural characteristic(s) of the derivatized glyphosate that now allow for spectrophotometric quantification. Specifically, how would you now expect to quantify the molecule with the maximum amount of sensitivity? Explain your answer.
While derivatization in GC occurs nearly always pre-injection, derivatization reactions in HPLC are generally completed either pre-column, where sample and reagent are mixed prior to injection, or post-column, where separation of the sample components is achieved prior to reaction with derivatizing agent.
With appropriate instrumentation, both techniques can be carried out in an on-line mode, where the addition of reagent and subsequent mixing is performed by the instrument.22 In some instances, pre-column derivatization is best done off-line, allowing for more complicated sample manipulations, and removal of excess reagent or interferents prior to injection. Often the choice is dependent upon the speed with which a derivatization reaction can be made to occur, the stability of the derivatized product, or whether automation is available to accomplish the reaction.23
Important distinctions between pre- and post-column derivatization are summarized below.22
• Pre-column
• no restrictions on reaction conditions
• quantities of derivatizing agents can be minimized
• provides generally better analyte sensitivity due to lower backgrounds
• requires reaction products that are stable over time
• Post-column
• reaction must be rapid and compatible with eluent
• derivatization reagent mixed with chromatographically pure analyte after elution
• unstable products can be analyzed immediately
• constant flow of reagent stream requires large quantities of derivatizing agent
Derivatization for HPLC most often involves the introduction of a “tag” to an analyte molecule that imparts to the derivatized product a physical property lacking in the underivatized analyte, which can then be exploited in the detection process. Three of the most important functional groups in derivatization for HPLC are chromophores, fluorophores (or a species capable of forming a fluorophore with the analyte), and redox-active centers.23
Classes of molecules that frequently require derivatization to allow detection at desired concentration levels include aliphatic amines, carboxylic acids, and alcohols/thiols.6 For HPLC, derivatization reactions can be easily accomplished in either aqueous or non-aqueous solvent, leading to multitudinous methods for both pre- and post-column reactions.6 Review articles are available that provide detailed coverage of these methods.5-6, 22-24 Some of the more frequently encountered methods will be presented here.
UV-Visible Detection. Though the most widely utilized type of detection for HPLC, UV-Vis detection lacks the necessary sensitivity/selectivity for many analytes. Derivatization generally involves the introduction of highly conjugated aromatic moieties to the analyte, giving the product high absorptivity and improved detection.15
Examples of derivatizing agents commonly employed for HPLC/UV-VIS are given in Table 3.
Derivatizing Agent
Group(s) Derivatized
1-Fluoro-2,4-Dinitrobenzene (FDNB)
RNH2, RNHR’, R-OH
4-N,N-Dimethylaminoazobenzene-4’-Sulfonyl Chloride (DABS-Cl)
RNH2, RNHR’, R-SH, R-OH
p-Bromophenacyl Bromide (BPB)
R-COOH, R-SH
Benzoyl Chloride
RNH2, RNHR’, R-SH, R-OH
9-Fluorenylmethoxycarbonyl Chloride (FMOC-Cl)
RNH2, RNHR’, R-OH
2,4-Dinitrophenylhydrazine (2,4-DNPH)
RC(O)R’, RC(O)H
Table 3. Common Derivatizing Agents for HPLC/UV-VIS Analysis.
Specific applications can be used to illustrate the usefulness of these reagents.
FDNB reacts with primary and secondary amines, and aliphatic hydroxyl groups.23 For example, the tricyclic amine memantine hydrochloride, useful in the treatment of Parkinson’s disease and other movement disorders, was analyzed in finished dosage forms following pre-column derivatization with FDNB. UV detection of the product was accomplished at 360 nm.25
Figure 5. Structures of Memantine hydrochloride (left) and FDNB (right).25
Q11. What do you predict the reaction product to be for the reaction between memantine HCl and FDNB? Draw its structure.
The sulfonyl group in DABS-Cl, whose structure is shown in Figure 6, can react with primary and secondary amines, alcohols, and thiols to form derivatives with strong visible absorption between 425 and 468 nm.24 Recently, the complete amino acid analysis of octreotide, used clinically to treat endocrine and carcinoid tumors, among other things, was accomplished using DABS-Cl derivatization followed by visible detection at 436 nm.26
Figure 6. Structure of DABS-Cl.
Q12. Do an internet search for octreotide and locate its structure. Considering only the primary amine sites on the molecule, predict how many sites will be derivatized with DABS-Cl. Next, using “R” as the rest of the molecule, show the structure of the reaction product of DABS-Cl and a single primary amine.
Aromatic halides like p-bromophenacyl bromide (BPB) are effective derivatization agents for carboxylic acids, producing strongly absorbing phenacyl ester derivatives.15 Pre-column derivatization of fatty acid with BPB in the presence of crown ether to aid in product extraction into non-aqueous solvent allowed detectability limits of 50 ng of derivatized acid containing 20 carbons. UV detection was carried out at 254 nm.27
BPB has been used to derivatize the thiol moiety of tiopronin, a drug used to prevent kidney stones and in the treatment of rheumatoid arthritis, in blood plasma. The derivatization reaction is given in Figure 7. UV detection was done at 236 nm, with a reported limit of detection for tiopronin in plasma of 12 ng/mL.28
Figure 7. The Derivatization Reaction between Tiopronin (center) and BPB (left).
Q13. BioRad maintains a free database of spectra, where you can obtain the UV spectra of 4-bromophenacyl bromide (https://spectrabase.com/), shown below. Speculate as to why the derivatized fatty acid and derivatized tiopronin above were detected at 254 and 236 nm.
Figure 8. UV Spectrum of 4-Bromophenacyl Bromide.
Hydroxy compounds, including alcohols and phenols, carbohydrates and steroids, can be derivatized by reaction with acyl chlorides like benzoyl chloride that convert the hydroxyl group into an ester.15 Additionally, this reagent can react with primary and secondary amines, thiols, and phenols to introduce a spectroscopically useful phenyl group. A recent report demonstrated the utility of benzoyl chloride in the derivatization of 70 neurologically relevant compounds, including catecholamines and their metabolites prior to analysis by HPLC/MS/MS.29
Figure 9. Derivatization of Dopamine (left) with Benzoyl Chloride.
FMOC-Cl, to which you were introduced earlier in this module, has been used to derivatize primary, secondary, and to some extent tertiary amines, as well as for some alcohols. It has been shown to be useful for enhancement of both UV detection (at 265 nm) and fluorescence detection (excitation at 265 nm and emission at 340 nm).24 Its reaction with glyphosate was seen in Figure 4. FMOC-Cl has been widely used for analysis of amino acids. It can be used in conjunction with o-phthalaldehyde (OPA), selective for primary amines, to doubly derivatize amino acids and biogenic amines in biological tissues.31 The structures of OPA and FMOC-Cl are given in Figure 10.
Figure 10. Structures of OPA (left) and FMOC-Cl (right).
Q14. Look up the structures of putrescine and spermidine, and determine how they differ. Explain what advantage double-derivatization might offer in the identification of these molecules in a mixture.
2,4-DNPH is one of the most widely used derivatizing agents in analytical analysis, converting carbonyl compounds to their corresponding 2,4-dintrophenylhydrazone derivatives.32 The general reaction scheme is given in Figure 11.
Figure 11. Reaction of 2,4-DNPH (left) with Carbonyl to Form DNP-hydrazone.
Important applications of this method can be found in the analysis of carbonyl compounds arising from motor vehicles and industrial emissions, as well as “sick building syndrome” involving airborne volatile molecules like formaldehyde.33 2.4-DNPH-coated silica gel cartridges have been used in the collection of formaldehyde from ambient air, with extraction into organic solvent and detection at 366 nm.34 A procedure for the analysis of 20 carbonyl compounds in aqueous, soil, waste, and stack samples by UV absorption at 360 nm following 2,4-DNPH derivatization can be found in EPA Method 8315A.35
Q15. Use the link in Reference 35 to obtain a copy of EPA Method 8315A. Outline the procedure necessary to quantify carbonyl compounds (other than formaldehyde) in aqueous samples. Next, determine how the MDL values for hexanal and decanal isolated from solid samples differs from that for liquid samples.
Fluorescence Detection (FD). Owing to low cost and high availability in many laboratories, UV-VIS detection is the most widely used method for the determination of derivatization products in HPLC.28 Fluorescence detection, which is less commonly available, offers lower detection limits and wider linear range than does UV-VIS.15 The technique has better inherent specificity than absorption alone since both excitation and emission wavelengths are controlled. Emission is generally measured at 90o to exciting radiation to minimize stray light.36
Many of the derivatization agents for enhanced UV-VIS detection in HPLC form products that are also amenable to fluorescence detection. Table 4 lists some of the more commonly used compounds for the formation of fluorescent derivatives.
Derivatizing Agent
Group(s) Derivatized
o-Phthalaldehyde, OPA
RNH2, RNHR’
Dimethoxyanthracene sulfonate, DMAS
RNH2, RNHR’, R3N
2,3-Naphthalene dialdehyde, NDA
RNH2, RNHR’
Monobromobimane, MBB
R-SH
9-Fluorenylmethoxycarbonyl chloride, FMOC-Cl
RNH2, RNHR’, R3N , -OH
5-Dimethyl amino naphthalene-1-sulphonyl chloride (Dansyl chloride), DNS-Cl
RNH2, RNHR’, $\Phi$-OH
5-(Dimethylamino)-1-naphthalenesulfonic hydrazide, DH (Dansyl hydrazine)
RC(O)R’, RC(O)H, RCOOH
Table 4. Common Derivatizing Agents for HPLC/FL Analysis.
Examples of methods utilizing these fluorescence derivatizing agents are given below.
o-Phthaladehyde. Perhaps the most commonly used derivatizing agent in the production of fluorescent products is o-phthalaldehyde (OPA). OPA reacts with primary amines and amino acids in the presence of thiol or sulfite to produce isoindole derivatives.23 Secondary amines can be also derivatized following the addition of NaOCl.
Gabapentin has been used to treat epilepsy and is known to increase the production of $\gamma$-aminobutyric acid (GABA) in the brain.37 A methanolic solution of OPA/mercaptoethanol was used to derivatize gabapentin in blood plasma after clean-up on a C18 SPE column. Excitation frequency was 350 nm, with emission monitored at 450 nm. Quantitation of the drug was possible down to a level of 0.3 mg/L, with the therapeutic range reported to be 2-20 mg/L. The formation of the fluorescent derivative of gabapentin is shown below.
Figure 12. Reaction of o-Phthalaldehyde with Gabapentin.
Q16. What properties are associated with fluorescent molecules?
9,10-Dimethoxy-2-Anthracene Sulfonate. Dimethoxy anthracene sulfonate (DMAS) is capable of forming fluorescent ion pairs with amines and has been successfully used in the quantification of drug substances such as chloropheniramine and ephedrine, which contain teritary amine groups.6 The spectrofluorometric determination of the β-blockers arotinolol, atenolol and labetalol derivatized with DMAS has been reported by Abdine, et al.38 The formation in acidic medium of the fluorescent charge transfer complex between DMAS and arotinolol, which has a 2:1 stoichiometry, is shown below.
Figure 13. Reaction of Arotinolol with Dimethoxyanthracene Sulfonate.
The ion pairs for each of the β-blockers were extracted from tablets and from serum with chloroform, excited at 385 nm and measured at 452 nm. The reported LOD for arotinolol was 0.12 μg/mL.
Q17. What is a charge transfer complex?
2,3-Naphthalene Dialdehyde. While OPA is a very popular choice as a fluorescent tag for amines and amino acids, its use suffers from several drawbacks, including short derivative lifetimes and unfavorable side reactions with imidazole-containing compounds like histamine. The naphthalene derivative of OPA, 2,3-naphthalene dialdehyde (NDA), has been shown to ameliorate these problems, leading to more stable derivatives and more sensitive detection for many applications.39
Li, et al. studied the reaction of glutathione with NDA and two newly synthesized derivatizing agents containing methoxy and fluorine groups respectively at the 6-position of the naphthalene ring of NDA.40 Of the three probes, NDA showed the best results with glutathione, with a calculated detection limit of 6.4 x 10-8 M. The formation of the fluorescent derivative of glutathione (GSH) with NDA is shown below.
Figure 14. Reaction of Glutathione with 2,3-Naphthalene Dialdehyde.
Maximum excitation of the derivatized product was observed at 450 nm, with detection being done at an emission wavelength of 531 nm.40 The probes were then used in fluorescence microscopy measurements of GSH in live cells.
Q18. Naphthalene, the base compound in the formation of NDA, is one of a class of compounds called polycyclic aromatic hydrocarbons (PAH) which are known or suspected carcinogens. An interesting analysis scheme for these compounds involves the use of cyclodextrins to enhance the energy transfer between PAH and the strong fluorophore acceptors BODIPY and Rhodamine 6G. Use the supplied link to access the report, find the structures of the two fluorophore acceptors, and the mechanism by which cyclodextrin compounds contribute to the analysis (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4376008/ ).
Monobromobimane. MBB reacts with compounds containing reactive sulfur or thiol groups, including H2S, glutathione, proteins, and nucleotides.41 While the probe itself has little fluorescence, highly fluorescent and extremely stable thioether derivatives are formed quickly at room temperature.
Shen, et al. have reported the sensitive measurement of hydrogen sulfide in plasma by HPLC following derivatization with MBB.42 H2S is produced in mammalian tissue during the metabolism of cysteine. It is involved as a gaseous mediator in the cardiovascular and other systems, and is linked to a number of disorders, including Alzheimer's disease. In this study, plasma samples were incubated with MBB in Tris buffer at pH 9.5 for 30 minutes, and the fluorescent product quantified following RP-HPLC at an excitation wavelength of 390 nm and an emission wavelength of 475 nm. The detection limit for hydrogen sulfide was found to be 2 μM. The derivatization scheme is shown below.
Figure 15. Reaction of Monobromobimane with Hydrogen Sulfide.
Q19. Locate the structure of cysteinylglycine, and its biological function. What product would you expect in the reaction of cysteinylglycine with MBB?
9-Fluorenylmethoxycarbonyl chloride. As was mentioned in the UV-VIS section, FMOC is an effective derivatizing agent for primary, secondary, and tertiary amines, as well as some alcohols. Its reaction with glyphosate was given in Figure 4.
Q20. Access EPA Method 547 for glyphosate at https://www.o2si.com/docs/epa-method-547.pdf, and give a short summary of the method. What is the HPLC column specified in the method, and what is the reported MDL for glyphosate in ground water?
The tricyclic antidepressant drug, nortriptyline (NT), a non-selective serotonin uptake inhibitor, was quantified in plasma samples by FMOC-Cl derivatization.43 The reaction was carried out at room temperature for 30 minutes, and the product was stable for at least 48 hours. A wide calibration range between 5-5000 ng/mL was obtained, and an LOD of 2 ng/mL was reported. Excitation wavelength was 260 nm with emission detection at 310 nm. The derivatization scheme for nortriptyline is shown below.
Figure 16. Reaction of 9-Fluorenylmethoxycarbonyl Chloride with Nortriptyline.
5-Dimethyl amino naphthalene-1-sulphonyl chloride. Dansyl chloride (DNS-Cl), like OPA, has found widespread use in the analysis of amino acids at very low levels, in the nanomole to picomole range.44 An advantage of DNS-Cl over OPA is that the derivatized products are more stable, and give better results at the lower concentrations.
DNS-Cl can be successfully applied to the derivatization of polyamines to form fluorescent products for HPLC analysis. Molin-Legua, et al. describe the use of DNS derivatization in the quantification of putrescine, cadaverine, spermidine, and spermine in urine following purification by solid phase extraction (SPE).45 After pre-concentration of the analytes of interest on the SPE packing, DNS-Cl reagent in buffer at pH 9.5 was introduced to the column where the derivatization reaction was allowed to take place for 30 minutes at room temperature. The fluorescent derivatives were desorbed using acetonitrile and introduced into the HPLC. Excitation was accomplished at 252 nm, while emission was measured at 500 nm. Detection limits for the four polyamines in urine was 2 ng/mL. The reaction scheme is illustrated below for putrescine.
Figure 17. Reaction of Putrescine with 5-Dimethyl Amino Naphthalene-1-Sulphonyl Chloride.
5-(Dimethylamino)-1-naphthalenesulfonic hydrazide. Dansyl hydrazine (DH), reacts with ketones (including ketosteroids), aldehydes, and reducing sugars to form intensely fluorescent hydrozones.46 The reagent itself is highly fluorescent and can lead to interference if not removed from the reaction mixture prior to analysis.
2,5-hexanedione (2,5-HD) is a metabolite of hexane exposure in mammals. It is an active neurotoxin and has been related to testicular cancer in mice. The presence in urine can signal hexane exposure, often as a result of workplace use, gasoline, or from voluntary “sniffing” of solvents.47-48
Maestri, et al. have reported a sensitive method for determining 2,5-HD in urine by HPLC after derivatization with dansyl hydrazine.49 Urine samples were purified with C18 cartridges prior to derivatization. The hydazone derivative of 2,5-HD was detected by fluorescence, with an excitation wavelength of 340 nm and an emission wavelength of 525 nm. The reported detection limit for 2,5-HD in urine was 5 μg/L. The reaction between 2,5-HD and DH is shown below.
Figure 18. Reaction of 2,5-Hexanedione with 5-(Dimethylamino)-1-Naphthalenesulfonic Hydrazide.
Q21. What is a ketosteroid? Give the reaction product of a typical 17-ketosteroid with DH.
Electrochemical Detection (ECD). Electrochemical detection for HPLC first came into prominence in the 1970’s as the detector of choice in measurement of endogenous catecholamines and indoleamines, providing extremely low levels of detection (down to femtomolar range) for these substances in neurochemical analysis.50-53 The catechols and indoles lent themselves well to the technique, being inherently active electrochemically at relatively low applied potentials.
Electrochemically active compounds include aromatic alcohols and amines, nitroaromatics, heterocyclics, aliphatic amines, and aliphatic polyenes.54 By varying the potential applied in ECD, a certain degree of specificity can be achieved for some analytes in the presence of others.
Q22. Access the Basic Principles document on electrochemical detection at the BASi website (www.basinc.com/manuals/LC_epsilon/Principles/Basic/basic#hdv), and scroll down to the section on “Hydrodynamic Voltammograms” containing Figure 1.7. Explain in a few sentences how applied detector potential allows you to choose which analytes you wish to quantify.
Two factors limit the usefulness of electrochemical detection for the majority of analytes: 1) lack of an electroactive functional group that can be oxidized or reduced at a reasonable potential, or 2) presence of an electroactive group whose redox potential lies outside the working window of the device.55-56
Figure 19 presents approximate ranges for the oxidation/reduction potentials for many common organic functional groups.57
Figure 19. Oxidation/Reduction Potentials for Common Organic Functional Groups
(adapted from Fuchigami, Reference 57)
Q23. The structure of dopamine, a neurotransmitter in mammalian brain known to be involved in a variety of motivated behaviors, was given in Figure 9. It can be easily oxidized at a carbon electrode at potentials less than +0.5 V vs SCE. What functional group(s) would you expect to oxidize at these potentials, and what do you think is a reasonable product if the oxidation mechanism involves 2 electrons and 2 protons?
Q24. In the Basic Principles document accessed in Q22, scroll down to the section on “Background Current” containing Figure 1.14. What are the approximate potential windows, both anodic and cathodic, for routine ECD? What common sources of background current can limit the extent of the available potential window?
As was the case previously for spectroscopic detection in HPLC, the introduction of a derivatization reagent can often allow the analyst to take advantage of the enhanced sensitivities and limits of detection available in ECD for substances that are not naturally amenable to analysis by this method.54,58 Many of the same reagents used to produce fluorescent derivatives can also be used to introduce an electrophore to the molecule, though fluorescence detection is used more extensively than electrochemical detection.59
Q25. Based on the potential windows arising from background current found in Q24, which of the groups in Figure 19 would likely require derivatization to allow successful use of electrochemical detection in HPLC?
Two types of electrochemical detectors are in common use.54,60-62 In an amperometric detector, the effluent from the chromatographic system is passed over a smooth electrode surface at which the potential is controlled and the current resulting from oxidation or reduction of the analyte is measured. This type of detector is characterized by low noise and high sensitivity, but only about 5-15% of the total amount of analyte comes in contact with the electrode, which limits the limit of detection. In a coulometric detector, a high surface area porous electrode is utilized through which the column effluent flows, and at which close to 100% of the analyte is oxidized/reduced. This results in a higher signal and in many cases lowered detection limits.
The use of the amperometric detector is limited primarily to isocratic elution for HPLC as changes in eluent composition lead to changes in the observed background current. The use of the coulometric detector reduces the effect of background current changes and can be used for gradient elution.54,56
Examples of derivatization agents used in the production of electroactive products for HPLC are given in Table 5.
Derivatizing Agent
Group(s) Derivatized
o-Phthalaldehyde, OPA
RNH2, RNHR’
2,3-Naphthalene dialdehyde, NDA
RNH2, RNH2R’
Ferrocene derivatives
RNH2, RNH2R’, R-COOH,
R-OH , R-SH, RC(O)R’, RC(O)H
1-(2,5-dihydroxyphenyl)-2-bromoethanone, 2,5-DBE
R-COOH
4-NPH
RC(O)R’, RC(O)H
Table 5. Common Derivatizing Agents for HPLC/ECD Analysis.
o-Phthalaldehyde. An example of the use of OPA in ECD derivatization is for histamine, which in its native form requires an applied potential of >1.2 V vs Ag/AgCl for detection. The OPA/2-mercaptoethanol derivative of histamine can be detected at a modest applied potential of +0.4 V.63
Figure 20. Reaction of o-Phthalaldehyde with Histamine.
Q26. Using the potential ranges in Figure 19 as a guide, predict which of the functional groups on OPA-derivatized histamine leads to the observable signal observed near +0.4 V vs Ag/AgCl.
Reaction times are on the order of 2 minutes for OPA, making its use as a post-column derivatizing agent possible. The reaction product is highly unstable, though the use of sulfite in place of thiol has been shown to produce a more stable derivatized product.58 Chen, et al. have reported the detection of histamine in brain diasylate after OPA/sodium sulfite derivatization using dual detectors.64 The first was poised at a potential of +0.25 V vs Ag/AgCl (where the histamine product did not oxidize) to reduce noise from other chemicals present, with a second detecting derivatized histamine at +0.55 V vs Ag/AgCl.
2,3-Naphthalenedialdehyde. Another frequently encountered fluorescent derivatizing agent, 2,3-naphthalenedialdehyde (NDA), reacts with primary amines and amino acids in the presence of cyanide ion to produce electroactive products that are detectable at ca. +0.75 V vs Ag/AgCl.58,65 Reaction times are greater than those observed for OPA (15-60 minutes), but the reaction products are more stable, making pre-column derivatization easier.23 The reaction between NDA and a primary amine in the presence of cyanide to form a cyano[f]benzoisoindole (CBI) is shown below.
NDA CBI
Figure 21. Reaction of 2,3-Naphthalenedialdehyde with Primary Amine.
Ferrocene. Ferrocene (bis-cyclopentadienyl Fe(II)) and its substituted derivatives display highly reversible electrochemical behavior. Fe(Cp)2 has an accessible redox potential in organic solvents near +0.5 V vs. SCE, with the moiety retaining its excellent electrochemical performance even when attached to more complex molecules.66 This makes it an exceptional choice as a derivatizing agent for HPLC-ECD.
Figure 22. Structure of Ferrocene.
Q27. Predict what might happen to the redox potential of ferrocene as the cyclopentadienyl groups are modified by the addition of different functional groups like methyl, alkylamine, or carboxylic acid. Explain.
Ferrocene-based derivatizing agents have been applied most widely to the analysis of amine containing compounds including amino acids, peptides and proteins.56 By altering the reactive groups on the ferrocene, derivatives have been formed with oxidation potentials generally in the range of +0.20 to +0.60 V vs Ag/AgCl.56,59 Ferrocene carboxaldehyde and ferrocene carboxylic acid chloride (FAC) have been used to derivatize primary amines, with FAC also reacting with secondary amines to form stable products that are easily oxidized.65,67 The reaction between FAC and amino acid to form electroactive product is shown below.
Figure 23. Reaction of Ferrocene Carboxylic Acid Chloride with Amino Acid.
In addition to amines, ferrocene containing reagents have been used to derivatize alcohols, thiols, ketones, aldehydes, and carboxylic acids for analysis by HPLC-ECD resulting frequently in limits of detection in the femtomolar range.56,59,65 The sulfhydryl group of cysteine can be selectively reacted with ferrocene derivatives, as shown below for the reaction of glutathione (GSH) with ferrocenyl ethylmaleimide (FEMI).68
Figure 24. Reaction of Glutathione with Ferrocenyl Ethylmaleimide.
Other derivatizing agents. Carboxylic acids have been derivatized using substituted phenacyl bromides, as illustrated below for the reaction between 1-(2,5-dihydroxyphenyl)-2-bromoethanone (2,5-DBE) and quinoxaline-2-carboxylic acid (QCA). The resulting ester derivatives were oxidized between 0.4 – 0.6 V vs Ag/AgCl.69
Figure 25. Reaction of 1-(2,5-Dihydroxyphenyl)-2-Bromoethanone with Quinoxaline-2-Carboxylic Acid.
Alcohols and ketones can be derivatized using aromatic nitro-derivatives.63 As an example, Shimada reported the electrochemical detection of the hydrazone derivative of 17-ketosteroids at +0.8 V vs Ag/AgCl following reaction with 4-nitrophenylhydrazine (NPH).70 This reaction is shown below for the derivatization of androsterone (AND).
Figure 26. Reaction of 4-Nitrophenylhydrazine with Androsterone.
Q28. The use of 4-nitrophenylhydrazine introduces more than one electrochemical active functionality to androsterone in the reaction from Figure 26. What two groups in the derivatized molecule would you would expect to be addressable via ECD? Approximately what detector potential would be needed for each? Which group do you think is reacting at +0.8 V vs Ag/AgCl?
Detailed derivatization procedures, conveniently organized by both functional group and derivatizing agent, are collected in Handbook of Derivatization Reactions for HPLC, edited by Lunn and Hellwig.71 In addition, tables of derivatizing agents and citations to primary literature can be found in references 54, 56, 59, and 65. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Sample_Preparation/06_Derivatization.txt |
This module provides an introduction to the measurement technique of potentiometry. It is intended to be a primary learning tool for a student in a Quantitative Analysis or Analytical Chemistry course and as a review resource for a student in Instrumental Analysis. It could also serve as a beginning resource for new practitioners.
If you have ever used a pH meter, then you have already performed potentiometry, an electrochemical method in which the potential of an electrochemical cell is measured while little to no current is passed through the sample. In a potentiometric measurement, an indicator electrode responds to changes in the activity, or “effective concentration” of the analyte. A potential, or voltage, that develops at the interface between the electrode and the analyte solution is measured relative to a reference electrode. This potential will be proportional to the amount of analyte in the sample. An illustration of one type of indicator electrode, the hydrogen ion or pH electrode, is shown below.
Ion-selective electrodes, or ISEs, are probably the most important class of indicator electrodes. As indicated by the name, ISEs possess a high degree of selectivity. These electrodes are routinely used in clinical laboratories to determine various ion concentrations (such as calcium) in blood samples. In fact, millions of measurements are made each year with the valinomycin K+ electrode in clinical assays. However, the applications of ISEs are not limited to clinical applications. ISEs are also regularly used in environmental analysis, such as in a water treatment plant to monitor nitrate levels.
The instrumentation used to perform potentiometry is straightforward, consisting of an indicator electrode, a reference electrode, and a potential measuring device. This simplicity makes potentiometry an inexpensive technique compared to atomic spectroscopy or ion chromatography. When making a potentiometric measurement, the sample typically experiences minimal perturbation. Additionally, measurements at ISEs are not affected by the color or turbidity of a sample, making them applicable to clinical and environmental applications, as noted above.
Before we get into more details about potentiometry and ion-selective electrodes, we will review some basic electrochemical theory that is needed to thoroughly understand potentiometric measurements. These topics include junction potentials, reference electrodes, and the Nernst equation. If you would also like a refresher on electrochemical cells and standard potentials, a companion concepts module is available.
Contributors and Attributions
• Erin M. Gross (Creighton University), Richard S. Kelly (East Stroudsburg University), and Donald M. Cannon, Jr. (University of Iowa)
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.5 License
Analytical Electrochemistry: Potentiometry
After completion of this e-Module, you should be able to:
• Describe the basic concepts of making a potentiometric measurement.
• Name some applications of potentiometry.
• Know the difference between a reference electrode and an indicator electrode.
• Describe the reactions of the typical reference electrodes.
• Define liquid junction potential and boundary potential.
• Describe how ion-selective electrodes (ISEs) function.
• Describe how both a pH electrode and a pH meter work.
• Describe the errors involved in pH measurements.
• Perform basic troubleshooting while making a pH measurement.
• Use the Nernst equation to perform calculations for potentiometric measurements.
02 Potentiometry Timeline
Shown below are major milestones in the development of potentiometry. Additional information is available in the references cited.
Adapted from references 1- 7. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/01_Goals_and_Objectives.txt |
The origin of the measured potential at an indicator electrode is most generally the separation of charge across an interface between solutions of differing ionic strengths (an inner solution at fixed analyte activity and an outer solution with variable analyte activity).
The mechanism leading to this charge separation varies with electrode type. After defining what is meant by a junction potential, we will consider two types of indicator electrodes:
1. the metallic direct indicator electrode, whose response involves a surface or solution redox reaction, and
2. the membrane electrode, or ion-selective electrode (ISE).
Click here to learn about junction potentials.
03 Potentiometric Theory
A potential develops at any interface, or junction, where there is a separation of charge. For example, a potential can develop when a metal electrode comes in contact with a solution containing its cation. A potential of this type can be described using the Nernst Equation.
A potential can also develop when electrolyte solutions of differing composition are separated by a boundary, such as a membrane or a salt bridge (a gel-filled tube containing an inert electrolyte that connects half-cells to allow charge neutrality to be maintained).
The two solutions may contain the same ions, just at different concentrations or may contain different ions altogether. These ions have different mobilities, which means that they move at different rates.
For example, a porous glass frit may separate two solutions of NaOH:
NaOH (0.01 M) // NaOH (0.001 M)
OH- moves approximately 5X faster than Na+ . . .
. . . developing of a potential at the interface, or boundary, between the two solutions.
So now for an electrochemical cell containing a salt bridge, the cell potential is actually:
\[E_\ce{cell} = E_\ce{cathode} - E_\ce{anode} + E_\ce{junction} \tag{4}\]
Typical values for a liquid junction potential range from a few mV up to ~40 mV, depending on the identities and concentrations of the electrolyte solutions.
For a simple cell such as the one mentioned on the previous page, the junction potential can be calculated from the ion mobilities. However, most practical electrochemical cells are more complicated. To correct for or minimize errors due to a junction potential you can use the standard addition quantitation method. Here a potential measurement is made on a solution containing just the analyte ion. Next this solution is “spiked” with a known volume of standard ion solution and a second potential measurement is made. This procedure can be continued for multiple spikes (multiple standard addition method). You can assume that the addition of the standard does not alter the junction potential. You can read further about this quantitation method in the Chloride Experiment in the Experimental section of this module.
02 Direct Indicator Electrodes
The simplest type of direct indicator electrode is a metal, M, in contact with a solution containing its own cation, M+. At the metal-solution interface, a potential develops that is proportional to the activity of the metal ion in solution. The potential can be measured directly with respect to a reference electrode using the simple arrangement shown in Figure 1.
Figure 1. Electrochemical cell for a potentiometric measurement with a metallic indicator electrode.
The voltage measured is simply the difference between the potential at each electrode:
$E_\ce{cell} = E_\ce{indicator} - E_\ce{reference} \tag{1}$
Inert metal electrodes like Pt or Au can be used as indicator electrodes for ions involved in redox reactions that occur in solution but do not include the metallic form of the analyte.
Let’s discuss how a simple metal electrode can be used to measure the activity of its cation. This type of measurement is based on the fact that when a metal is in contact with a solution containing its cation, a potential develops. For example, when a piece of copper is placed in a copper sulfate solution:
The half-reaction is
$\ce{Cu^2+}(aq) + 2e^- ↔ \ce{Cu}(s) \tag{2}$
And the cell notation is
$\ce{Cu}(s) / \ce{Cu^2+}(aq) \tag{3}$
These types of electrodes develop an electrical potential in response to the equilibrium between the cation and metal surface, as shown in equation 2. The potential developed is proportional to analyte activity ( ACu2+ ) as determined from the Nernst Equation. For example, for the Cu(s) / Cu2+(aq) half cell:
$E_{ind} = E^0_\mathrm{Cu^{2+}/Cu} − \dfrac{0.05916}{2} \log \dfrac{1}{A_\mathrm{Cu^{2+}}} = E^0_\mathrm{Cu^{2+}/Cu} + \dfrac{0.05916}{2} \log A_\mathrm{Cu^{2+}} \tag{4}$
With a logarithmic dependence of potential on activity, it is convenient to express the analyte activity as a p-function (pCu = -log ACu2+ ). Substituting this relationship into equation 4 gives
$E_{ind} = E^0_\mathrm{Cu^{2+}/Cu} - \dfrac{0.05916}{2} \mathrm{pCu} \tag{5}$
In dilute solution of fixed ionic strength, we can assume that the analyte activity is approximately the same as its molar concentration, i.e. ACu2+ = [Cu2+] and that pCu is related to [Cu2+].
Let’s work on an example problem that shows the relationship between potential and analyte concentration.
Example Problem: What’s the relationship between concentration and cell potential?
Let’s put a metal electrode in solution with its mono-valent cation:
$\ce{M+} (aq) + e^- ⇋ \ce{M}(s) \hspace{40px} E_{ind} = E^0_{ind} - 0.05916 \log(1/[\ce{M+}])$
We can use a saturated calomel electrode (SCE) as a reference electrode:
$\ce{\frac{1}{2} Hg2Cl2}(s) + e^- ⇋ \ce{Hg}(l) + \ce{Cl-} (aq) \hspace{40px} E_{ref} = \mathrm{0.241\: V}$
The cell potential (or voltage you measure) is related to M+ concentration as defined by the Nernst Equation:
$E_{cell} = E_{ind} - E_{ref} = E^0_{ind} - \mathrm{0.05916\log \left( \dfrac{1}{[M^+]} \right) - 0.241\: V}$
Which reduces to:
$E_{cell} = \left\{E^0_{ind} - 0.241\: \ce V\right\} + 0.05916 \log [\ce{M+}] = \left\{E^0_{ind} - 0.241\: \ce V\right\} - 0.05916\: p\ce{M}$
What voltage change (in mV) would you expect for a 10X increase in cation concentration? What about a 100X increase?
Make your prediction and continue onto the next steps to see if you are correct . . .
Problem Exercise
Open up the following spreadsheet and follow the instructions below to see if your predictions are correct. Also try to answer the questions. Click here to open up the spreadsheet.
1. Make sure that the sheet for the monovalent tab is showing.
2. Enter a value for E0ind in the pink box. For example, E0 = +0.799 V for the Ag+/Ag system and E0 = +0.337 V for the Cu2+/Cu system.
3. Enter a range of concentration values for the cation in the yellow boxes. Use the range from 5.00 x 10-6 to 1.00 x 10-3M.
4. Inspect the graph that is generated from your values and answer the following questions.
5. What is the voltage change for in 10X increase in cation activity? Was your prediction correct?
6. What is the voltage change for in 100X increase in cation activity? Was your prediction correct?
7. What do you predict is the slope of the line in the plot?
8. Create a trendline by right-clicking on the data points and choosing “Add trendline” in the menu. Under the Type tab, make sure that “linear” is highlighted. Under the Options tab, click on the box next to “Display equation on chart”. Click OK. Was your prediction correct for the slope of the line?
9. Repeat steps 2-8 for creating a plot in the “divalent” tab.
Summary
When making a potentiometric measurement, we do not want the measurement itself to alter potential. Therefore the following conditions must be met:
1. Little to no current is passed through the cell.
2. There is no net reaction.
3. An equilibrium (static) measurement is made.
Direct indicator electrodes can experience interference from more easily reduced cations, thus decreasing their selectivity. Some metals used as the indicator electrode can be oxidized. Therefore, they are not as widely used as the next type of indicator electrodes – membrane electrodes, also referred to as ion-selective electrodes (ISE’s). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/03_Potentiometric_Theory/01_Junction_Potentials.txt |
So far you have learned that during the technique of potentiometry, the potential, or voltage, of an electrochemical cell is measured. The cell consists of both an indicator and reference electrode. Since the potential of the reference electrode is constant, it is the potential developed at the indicator electrode that contains information about the amount of analyte in a sample. During the measurement, there is little to no current flow. An electrochemical cell for making a potentiometric measurement with a membrane electrode (also known as an ion-selective electrode, ISE) is shown in Figure 1. As you can see the main difference between an ISE and the direct indicator electrode is in the ISE's composition.
Figure 1. Electrochemical cell for making a potentiometric measurement with an ISE.
As indicated by their name, ion-selective electrodes possess a high degree of selectivity. The selectivity of the ISE is determined by the composition of the membrane. Ideally the membrane allows the uptake of only one specific ion into it. The analyte ion may be a cation or an anion. The three main components of making a measurement at an ISE are an inner reference, or standard, solution and an outer analyte, or sample, solution separated by a thin membrane. These components are shown in Figure 1. Redox processes do not occur at ISEs. The potential developed at the membrane is the result of either an ion exchange process or an ion transport process occurring at each interface between the membrane and solution. The basics of ion exchange and ion transport are reviewed in the next sections.
Ion Exchange Process
In the following ion exchange process, a lithium cation displaces a potassium cation from the organic anion, R- :
$\mathrm{ {\color{Magenta} K}{\color{Blue} R} + {\color{Magenta} Li^+} \leftrightharpoons {\color{Magenta} Li}{\color{Blue} R} + {\color{Magenta} K^+}}$
We can imbed the lipophilic R- in a membrane, as shown in Figure 2, and place it in a solution of Li+
$\mathrm{ {\color{Magenta} K}{\color{Blue} R_{(mem)}} + {\color{Magenta} Li^+_{(aq)}} \leftrightharpoons {\color{Magenta} Li}{\color{Blue} R_{(mem)}} + {\color{Magenta} K^+_{(aq)}}}$
Figure 2. Ion-exchange process.
In order to construct an ion-selective electrode, we would add an inner reference solution to the other side of the membrane. This solution would contain a fixed concentration of the ion of interest, Li+ in this example. This is typically accomplished by placing a thin membrane at the end of the plastic tube and filling the tube with a standard (known concentration) solution of the analyte. As shown in Figure 1, a reference electrode is placed in the inner solution and a second reference electrode is in contact with the analyte (outer) solution. At each solution-membrane interface, an ion-exchange equilibrium is established. The partitioning of Li+ between the aqueous solution phase and the membrane phase depends on its activity, or concentration. The resulting charge separation at each interface results in a phase-boundary potential.8-10 If the analyte ion concentration on each side of the membrane was equal, the potential difference across the membrane would be zero. However, if the concentrations are not equal, a membrane potential will develop. The two reference electrodes (see Figure 1) measure the potential difference across the membrane.
Ion Transport with an Ionophore
Now that you understand the basics of ion exchange, let’s put a membrane, containing an ionophore, between an “unknown” analyte solution and a “known” reference solution (Figure 3). In this example, the analyte is A+. The ionophore is a neutral “carrier” molecule represented by the blue oval. Figure 4 shows the chemical structure of two ionophores. The ionophore cannot diffuse out of the membrane and but can “trap” the analyte ion (A+) at the interface between the solution and membrane. Without the ionophore, the analyte would be unable to partition into the organic membrane.
Figure 3. Development of a potential at an ISE.
Figure 4. Chemical structures of a crown ether (left) and of valinomycin (right), two ionophores.
As with the ion-exchange process, equilibrium is established at both solution-membrane interfaces. The resulting charge separation at each interface leads to a phase-boundary potential.
Now that we have developed an electrical potential across the membrane, we need to find a way to measure it. As before, we put an internal reference electrode in the internal reference solution and an external reference electrode in the analyte solution, as shown in Figure 5. The potential difference measured at these two electrodes is the membrane potential.
Figure 5. Electrochemical cell for a potentiometric measurement with an ISE.
The Membrane
By now you have learned that the identity of the membrane determines the selectivity of the electrode. In other words the type of membrane used dictates which analyte you can detect. Therefore different electrodes are used for different ions. The membrane should also have low electronic conductivity. The membrane must have low solubility in the analyte solution – we don’t want it to dissolve!!
There are three main types of membranes. The electrodes are classified by the membrane material.
1. Glass membrane electrodes: The most famous glass electrode determines H+ activity or pH (click here for the pH electrode section). The membrane is composed of a silicate glass. Glass electrodes can also be constructed that are sensitive to other cations such as sodium.
2. Single crystal LaF3 is widely used to determine F-. The crystal is usually doped with europium to improve the conductivity. At each membrane-solution interface, the following equilibrium takes place:
$\ce{LaF3} (s) \leftrightharpoons \ce{LaF2+} (s) + \ce{F-} (aq)$
You can see that the formation of LaF2+ creates a charge at the surface. The equilibrium will be shifted to the right for the solution with a smaller F- concentration, and the potential will become more positive relative to the other side of the membrane. It is this potential difference across the LaF3 crystal membrane that is measured and related to F- concentration. The fluoride electrode is extremely selective for F- but can experience interference from OH- above pH 8. This electrode is used in one of the experiments listed at the end of this learning module. Click here to learn more.
3. One of the most famous liquid membrane electrodes has been used for calcium determination. Initially, researchers attempted to use glass membrane electrodes (which had been successful for monovalent cations such as H+ and Na+) to detect divalent cations, such as Ca2+. When this was determined to be unfeasible, liquid membranes were developed. This electrode works by an ion-exchange process. The cation-exchanger is an aliphatic diester of phosphoric acid, (RO)2PO2-, where each R group is an aliphatic hydrocarbon chain containing between 8 and 16 carbons. The phosphate group can be protonated, but has a strong affinity for Ca2+. The cation exchanger is dissolved in an organic solvent and held in a porous compartment between the analyte solution and internal reference calcium chloride solution. The ion-exchanger uptakes Ca2+ into the membrane by the following mechanism, forming a complex with the structure shown in Figure 6:
$\ce{Ca^2+}\, (aqueous) + \ce{2(RO)2PO2-}\, (organic) \leftrightharpoons \ce{[(RO)2PO2]2Ca}\, (organic)$
Figure 6. Calcium dialkyl phosphate complex.
Calcium ISEs are commonly used to measure calcium ion activity in biological fluids, as calcium ion is important in many physiological processes, such as bone formation.
4. Ionophores, or chelating agents, that selectively complex ions include crown ethers and the antibiotic valinomycin (see Figure 4). The important feature of the neutral carrier molecule is its cavity which has dimensions approximately that of a molecule or ion. The valinomycin electrode was one of the first polymer membrane electrodes and is routinely used to determine potassium. The electron-rich center of valinomycin efficiently extracts K+ ions due to the similarity between the diameter of K+ and the inner diameter of the valinomycin molecule. The outer lipophilic part of the valinomycin molecule allows it to remain in the polymeric membrane. In the United States alone, nearly 200 million measurements are made annually of blood potassium levels using this electrode.11
Relationship Between Potential and Concentration
For all of the ISEs described above, the same equations can be used to predict the relationship between potential and analyte activity (A) or concentration.
The potential measured, Emeas, is the potential difference between the analyte (Eouter) side of the membrane and the reference (Einner) side of the membrane:
$E_{meas} = E_\ce{outer} - E_\ce{inner} \tag{1}$
The potential of each side is related to activity (A) or concentration as described by the Nernst equation, where z is the charge of the ion of interest:
$E_{outer} = E^o − \dfrac{0.05916}{z} \log \dfrac{1}{A_{unk}} \tag{2}$
$E_{inner} = E^o - \dfrac{0.05916}{z} \log \dfrac{1}{A_{ref}} \tag{3}$
Since Aref and Eo are both constant, Einner (equation 3) is constant. If equations 2 and 3 are plugged into equation 1, they combine to give the following:
$E_{meas} = const + \dfrac{0.05916}{z} \log A_{unk}\tag{4}$
Note that the “const” term contains Einner and the E0 from the Eouter term. Also, don’t forget the that –log(1/a) = log(a).
You can see that ISEs should exhibit a Nernstian response, as you previously learned for the direct indicator electrodes.
Selectivity of ISEs
You have learned that one of the most important analytical characteristics of ISEs is their selectivity; i.e. a specific ion electrode will only respond to the presence of one species. In reality, ion-selective electrodes can experience interferences by responding to the presence of other ions. Although the fluoride electrode comes close, no membrane is 100% specific for only one ion. Equation 4 assumes that all of the electrode response (Emeas) is due to one ion. Let’s call this analyte ion i and its activity Ai (note that the charge is now zi). We will call the interfering ion j with corresponding activity Aj and charge zj. We can account for the lack of 100% specificity by incorporating the activity of j and a selectivity coefficient (kij) into equation 4. This new equation is called the Nikolskii-Eisenman equation:
$E_{meas} = const + \dfrac{0.05916}{z} \log(A_i + k_{ij} A_j^{\Large\frac{z_i}{z_j}}) \tag{5}$
The selectivity coefficient is a numerical measure of how well the membrane can discriminate against the interfering ion. To put this in perspective, if an electrode has equivalent responses to the two ions, then kij = 1.0. As you can see from the equation, the smaller the kij values, the less impact the interfering ion will have on the measured potential. When kij values are less than 1, the ISE is more responsive to the analyte ion and when kij values are greater than 1, the ISE is more responsive to the interfering ion. For example, a kij value of 0.01 means that the electrode is 100 times more responsive to ion i over j.
Selectivity coefficients can be experimentally determined. Selectivity coefficients for some of the electrodes previously discussed are listed below.
Analyte ion (i) Interfering Ion (j) kij
K+ (valinomycin)a
Na+
Ca2+, Mg2+
10-4
10-7
Ca2+ b
Mg2+
K+
0.02
0.001
areference 12. breference 11.
Continue onto the next page to work on some problems about ISEs.
Problems
1. Describe the difference between an ion exchange process and how an ionophore functions.
2. The following calibration data was collected at a Pb2+ ISE. Assuming that the ionic strength remains constant and that the activity coefficient is 1.0, what is the Pb2+ concentration of a solution that gives a potential reading of 145 mV?
For some help with this problem, open up the following spreadsheet. Make sure that you are on the Pb2+ tab.
[Pb2+] (mol/L) E (mV)
5.00 x 10-5 92
5.00 x 10-4 121
5.00 x 10-3 151
? 145
3. A fluoride ISE gives the following calibration data. What potential (mV) reading do you expect for the last solution?
For some help with this problem, open up the following spreadsheet. Make sure that you are on the F- tab.
[F- ] (mol/L) E (mV)
1.00 x 10-5 235
5.00 x 10-5 195
1.00 x 10-4 175
5.00 x 10-4 ?
Further Reading on ISEs
For further reading on ISEs, refer to references 8 through 13 and the following online resources:
http://www.nico2000.net/Book/Guide1.html
Section on ISEs found at Chemistry Hypermedia Project
ISE products from GlobalSpec (an engineering search engine with free registration)
Point of care diagnostics: “Clinical Instrumentation Refresher Series: Ion Selective Electrodes” by William R. Hliwa, Western New York Microcomputer, Inc., April, 1998: available for purchase through Med TechNet Online Services. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/03_Potentiometric_Theory/03_Ion-Selective_Electrodes.txt |
It should be clear by now that at least two electrodes are necessary to make a potential measurement. As Kissinger and Bott have so perfectly expressed, “electrochemistry with a single electrode is like the sound of one hand clapping” (http://currentseparations.com/issues/20-2/20-2d.pdf). In potentiometry, those two electrodes are generally called the indicator electrode and the reference electrode. The indicator electrode possesses some characteristic that allows it to selectively respond to changes in the activity of the analyte being measured. For the measured potential to have meaning in this context, the reference electrode must be constructed so that its composition is fixed and its response is stable over time, with observed changes in measured potential due solely to changes in analyte concentration.
You are probably familiar with tables of standard reduction potentials from a general chemistry course. The standard reduction potential, or E0, allows you to predict the ease with which a half-cell reaction occurs relative to other half-reactions. (For a review of standard potentials and electrochemical cells, you can access the companion “Concepts” module). Values of E0 are most often reported as the potential measured in an electrochemical cell for which the standard hydrogen electrode is used as a reference.
The standard hydrogen electrode, or SHE, is composed of an inert solid like platinum on which hydrogen gas is adsorbed, immersed in a solution containing hydrogen ions at unit activity. The half-cell reaction for the SHE is given by
\[\ce{2H+ (aq) + 2 e- \rightleftharpoons H2 (g)}\]
and the half-cell potential arbitrarily assigned a value of zero (E0 = 0.000 V).
Practical application of the SHE is limited by the difficulty in preparing and maintaining the electrode, primarily due to the requirement for H2 (g) in the half-cell. Most potentiometric methods employ one of two other common reference half-cells – the saturated calomel electrode (SCE) or the silver-silver chloride electrode (Ag/AgCl).
1. Saturated Calomel Electrode (SCE)
The SCE is a half cell composed of mercurous chloride (Hg2Cl2, calomel) in contact with a mercury pool. These components are either layered under a saturated solution of potassium chloride (KCl) or within a fritted compartment surrounded by the saturated KCl solution (called a double-junction arrangement). A platinum wire is generally used to allow contact to the external circuit. The half reaction is described by
\[\ce{Hg2Cl2 (s) + 2 e- \rightleftharpoons 2 Hg} (l ) + \ce{2 Cl-} \textrm{(sat’d)}\]
with an E0 value of +0.244 V. A common arrangement for the SCE is shown below, left side. In this arrangement, a paste is prepared of the calomel and solution that is saturated with KCl.
The solution over the paste is also saturated with KCl, with some solid KCl crystals present. Contact to the measurement cell is made through a porous glass frit or fiber which allows the movement of ions, but not the bulk solution. In many electrodes designed for potentiometry, the reference half cell is contained within the body of the sensing electrode. This arrangement is referred to as a “combination” electrode.
2. Silver/Silver Chloride (Ag/AgCl)
The silver/silver chloride reference electrode is composed of a silver wire, sometimes coated with a layer of solid silver chloride, immersed in a solution that is saturated with potassium chloride and silver chloride. The pertinent half reaction is
\[\ce{AgCl (s) + e- ⇔ Ag (s) + Cl-} \textrm{(sat’d)}\]
with a value for E0 of +0.222 V. The actual potential of the half-cell prepared in this way is +0.197 V vs SHE, which arises because in addition to KCl, AgCl also contributes to the chloride activity, which is not exactly unity. A schematic of the Ag/AgCl reference electrode is shown at right in the previous figure.
Both the SCE and the Ag/AgCl reference electrodes offer stable half-cell potentials that do not change over time or with temperature. In addition, the loss of electrolyte to evaporation does not change the saturated nature of the solution, nor the potential. One must be aware that the contact junctions of the half cells by nature slowly leak fill solution into the external solution in which they are found. As such, there are instances where measurements of certain ions, like chloride, might be affected by the ions introduced to the measurement solution by leakage. The doublejunction design prevents this problem by placing a second solution between the reference half cell and the measurement solution. This of course adds a second junction potential, a topic covered elsewhere in this module. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/03_Potentiometric_Theory/04_Reference_Electrodes.txt |
The technique of potentiometry involves the measurement of cell potentials under conditions of no current flow. In the electrochemical cell, if a high impedance device like a voltmeter, is placed between the indicator and reference electrodes, no current will flow between the two compartments. As we have seen, it is possible under these conditions to measure the potential difference that exists between the two electrodes. For cells with all reactants present at unit activity, the measured cell potential (neglecting junction potential effects) will be the standard cell potential, E0cell. As a reminder, activity defines an “effective concentration” for a particular species, and takes into account all the interionic interactions that the ions experience in the solution, not just a count of the number of molecules of a species per liter (molarity). In real applications of potentiometry, reactant activities are seldom (read never) equal to unity, and measured cell potentials move away from those that result from the tabulated values of E0.
A fundamental expression for characterizing redox systems under equilibrium conditions is the Nernst equation. One usually has encountered this expression early in their study of electrochemistry, perhaps in a general chemistry course long ago. The Nernst equation allows the calculation of relative activities of the species in a redox reaction as a function of the measured electrode potential (E) and the standard reduction potential (E0) for the half reaction. For the general redox reaction (written as a reduction)
$aA + n e^- ⇔ bB$
the Nernst equation takes the form
$\mathrm{E = E^0 - (RT/nF) \log [(\mathcal{A}_B)^b / (\mathcal{A}_A)^a ]}$
where R is the gas constant (8.314 (V . C)/(K . mol)), T is temperature in K, n is the stoichiometric number of electrons involved in the process, F is the Faraday constant (96,485 C/equivalent) and $\mathcal{A}_B$ and $\mathcal{A}_A$ are the activities of the reduced and oxidized members of the redox pair, respectively. At 25 oC, the value of (RT/F) is equal to 0.0592, and the Nernst equation becomes
$\mathrm{E = E^0 - (0.0592/n) \log [(\mathcal{A}_B)^b / (\mathcal{A}_A)^a ]}$
In practical usage, it is convenient to replace activities in the Nernst equation with concentrations. Because E0 values are standard reduction potentials they are not appropriate when concentration values are used. (For a review of standard potentials and electrochemical cells, you can access the companion “Concepts” module). Instead, it is proper to use formal potentials, E0’ , which are half-cell potentials measured when the concentration (not activity) ratio of [B]/[A] is unity and concentrations of other solution species are specified and fixed. These values provide corrections for activity effects under the cell conditions for which they have been measured, and should be used whenever they are available. Tables containing formal potentials for many redox reactions are readily available. If an E0’ value is unavailable, the use of E0 generally results in a reasonably small error.
Substituting E0’ and concentration terms for activities, the Nernst equation for the general reaction above (at 25 oC) becomes
$\mathrm{E = E^{0’} – (0.0592 / n) \log ([B]^b / [A]^a )}$
where square brackets represent concentrations in moles/L.
The Nernst equation can be applied to each of the half reactions in an electrochemical cell in turn, allowing the calculation of cathode and anode potentials at varying values of concentrations. The two half reaction potentials are then combined as before to determine the value of the overall cell potential.
Consider the following half reactions for an electrochemical cell:
$\ce{Zn^{+2} + 2e- ⇔ Zn^0}$
$\ce{Cu^{+2} + 2e- ⇔ Cu^0}$
If instead of unit activities, assume the cell concentrations of Zn+2 and Cu+2 were set at 0.00100 M and 0.00500 M respectively by dissolving the appropriate amount of each nitrate salt in the respective half-cells. The Nernst equation can now be used to calculate each of the half cell potentials at this concentration. (E0 values will be used in this example instead of E0’.) As the solid electrodes are both the elemental form of the element, the activity for each (and hence the concentration term) is reduced to 1.0.
\begin{alignat}{3} &\mathrm{\underline{Anode}}\textrm{:} &&\mathrm{E_a} &&= \mathrm{E^0 - (0.0592 / n) \log ([Products] / [Reactants])}\ & && &&= \mathrm{- 0.763\: V - (0.0592/2) \log ([0.00100] / [1])}\ & && &&= \mathrm{- 0.674\: V}\ &\mathrm{\underline{Cathode}}\textrm{:}\hspace{40px} &&\mathrm{E_c} &&= \mathrm{0.339\: V - (0.0592/2) \log ([0.00500] / [1])}\ & && &&= \mathrm{0.339\: V - (0.0592/2) \log ([0.00500] / [1])}\ & && &&= \mathrm{0.407\: V} \end{alignat}
The overall cell potential (now Ecell instead of E0cell) is calculated as follows:
$\mathrm{E_{cell} = E_{cathode} - E_{anode} = 0.407\: V - (- 0.674\: V) = +1.08\: V}$
In practice, potentiometric measurements involve at least one reference electrode whose composition and potential are fixed, and the concepts of anode and cathode are of only peripheral importance. Most measurements will include the use of a calibration plot for which only the magnitude of measured potential is related directly to analyte activity.
04 Instrumentation
You have realized by now that potentiometric measurements are fairly easy to make from the standpoint of instrumentation. In addition to the indicator electrode and the reference electrode, the only remaining component is a device used to measure the potential difference that exists between the two electrodes. If you have been with us to this point, you should remember that potentiometric measurements are ideally made under conditions of very little current flow. This means that the resistance (impedance to current flow) in the electrochemical cell must be very high (up to 100 MΩ). This is usually not a problem due to the nature of the indicator electrode, but the measurement of potential under these conditions requires the use of a device whose input resistance is even larger than the cell resistance.
When a voltmeter is introduced into the circuit, it brings with it a certain resistance, which changes the overall characteristics from those of the cell alone. That is, the very act of trying to make a measurement changes the quantity you set out to measure. In this context, the potential difference produced in the cell is not the quantity measured by the voltmeter. This perturbation is called the loading error, ER, and its value is given as % relative error by
$E_R = \dfrac{-R_C}{R_M + R_C} \times 100$
where RC is the resistance of the cell, and RM is the input resistance of the meter. In the diagram below, the circuit diagram on the right is a representation of the physical cell, shown on the left . The true cell potential arising from the analyte and its interaction with the sensing electrode is given by VCell.
To achieve a loading error of - 0.1% for a cell resistance of 10 MΩ, an input impedance of 9990 MΩ (~10 GΩ) is required. Modern digital voltmeters utilizing operational amplifiers, which are high input resistance devices, are capable of easily achieving this level of accuracy in potential measurement.
Most of us recognize a “pH meter” when we see one. The measurement of pH is actually a measurement of potential at a glass membrane which is then converted to the pH scale. Most of these meters will also display potential in mV. These devices can be used with most types of the electrodes used in potentiometry, including other ion-selective electrodes, and electrodes utilizing enzyme membranes in their construction. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/03_Potentiometric_Theory/05_The_Nernst_Equation.txt |
The most widely used ion-selective electrode is the glass pH electrode, which utilizes a thin glass membrane that is responsive to changes in H+ activity. F. Haber, in 1901, was the first person to observe that the voltage of a glass membrane changed with the acidity of a solution. In 1906, M. Cremer observed the pH dependence of measured potential across a thin glass membrane.
Today, pH sensitive glasses are manufactured primarily from SiO2 which are connected via a tetrahedral network with oxygen atoms bridging two silicon atoms (see an interactive 3d structure at http://www.geo.ucalgary.ca/~tmenard/...al/quartz.html). In addition, the glasses are made to contain varying amounts of other metal oxides, like Na2O and CaO. Oxygen atoms within the lattice that are not bound to two silicon atoms possess a negative charge, to which cations can ion pair. In this way, ions (primarily Na+) are able to diffuse slowly in the lattice, moving from one charge pair site to another. While the membrane resistance is very high (~100 MΩ), this movement of cations within the glass allows a potential to be measured across it.
If glasses of this type are placed in an aqueous solution containing H+, the glass surface in contact with solution becomes hydrated as water enters a short distance into the crystal lattice and causes it to swell. The “interior” of the glass remains dry. Some of the metal ions within the glass close to the solution boundary are able to diffuse into the solution, and some H+ ions are able to charge pair with oxygen near the glass surface. In this way, ion exchange equilibrium is established between the fixed negative sites on the glass surface and H+, with an increasing number of charge pairs with H+ occurring as its activity in the contacting solution increases. This equilibrium can be expressed by
$\ce{H^{+/-}O-Si}- ⇔ \ce{H^+} + {^-\ce{O-Si}-}$
where H+/-O-Si- and -O-Si- represent oxygen sites at the glass membrane ion paired with H+ and unpaired, respectively.
To function as a pH sensor, a layer of pH sensitive glass is placed between two solutions containing H+. As only the glass closest to solution becomes hydrated, two individual equilibria are established that are dependent upon the respective H+ activity on either side of the layer. This situation is illustrated below.
A difference in the H+ activities on either side of the glass membrane leads to a difference in the number of ion pairs that exist, and an imbalance in the surface charge between the hydrated layers. This results in a membrane potential that is pH dependent, described according to the Nernst equation
$\mathrm{E_{membrane} = E_{inner} - E_{outer} = 0.0592 \log [(\mathcal{A}_{inner}) / (\mathcal{A}_{outer})]}$
where Einner and Eouter are the surface potentials on either side of the membrane, and $\mathcal{A_\mathrm{inner}}$ and $\mathcal{A}_\mathrm{outer}$ represent the H+ activities of the inner and outer solutions, respectively.
Most commonly, pH electrodes are of a combination design, in which the glass membrane and the necessary reference electrodes are incorporated into the same electrode body. A figure of a typical combination pH electrode is shown below. (http://upload.wikimedia.org/wikipedi...ode_scheme.jpg)
In this design, the inner fill solution contacting the glass membrane contains a fixed activity of H+. Typically, a Ag/AgCl reference electrode is in contact with this inner solution, and the solution contains 0.1 M HCl saturated with AgCl. A second Ag/AgCl reference electrode is located within a solution compartment surrounding the inner solution compartment. This solution is in contact with the external solution of unknown H+ activity through a porous frit on the side of the electrode barrel.
The combination electrode allows the measurement of both the inner and the outer membrane surface potentials, which as we saw above, is related to the solution pH by the Nernst equation. The theoretical potential across the glass membrane changes by 59.2 mV for each unit change in solution pH. A combination electrode cell can be represented by the shorthand notation below.
Practical considerations
The potential measured by a pH indicator electrode includes not only the desired membrane or boundary potential, but also small contributions from what are known as junction potentials and asymmetry potentials. Junction potential has been described in a separate section of this module, and for the specific case of the combination pH electrode is that which develops across the porous frit separating the second reference electrode from the external measured solution. Asymmetry potentials result from physical differences between the inner and outer surfaces of the glass membrane, leading to different inner and outer potentials for the same H+ activity. Corrections for these small potential errors can be made by frequent calibration of the glass electrode in standard solutions covering the pH range in which measurements are desired.
Users of glass pH electrodes should also be aware of alkaline and acid errors that limit the pH range over which effective measurements can be made. At very high pH, generally > 10, most glass electrodes become responsive to both H+ and Na+, with the measured pH being lower than actual. At low pH, typically < 1, glass membranes are susceptible to saturation by H+, and produce pH readings that are higher than actual. Specialty glasses are available that minimize these errors if routine measurements must be made at these pH extremes.
Solid state pH sensors
For many applications, such as those requiring ruggedness and miniaturization, a thin glass membrane is impractical as a pH sensor. Alternatives to the glass electrode include the ISFET, or ion selective field effect transistor. More information on solid state electrodes can be found at http://www.ph-meter.info/pH-electrode-solid-state, http://ieee-sensors.org/wp-content/u...T-Bergveld.pdf, and http://csrg.ch.pw.edu.pl/tutorials/isfet (accessed 7/28/2008). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/05_pH_Electrodes.txt |
Introduction
The purpose of this experiment is to perform a potentiometric titration of an acid using a glass electrode and a pH meter. The potential that develops across the glass membrane of the electrode is proportional to the concentration (more exactly, the activity) of H+ in solution. In this laboratory, you will be given a sample of a weak acid, and with information obtained from your titration curve, you will be able to calculate the molecular weight and the pKa of the unknown. This should allow you to identify the unknown, and to calculate a theoretical titration curve which you can match to your experimental result.
The dissociation of a weak acid (represented by HA) in water can be described by
$\ce{HA \rightleftharpoons H+ + A-}$
where H+ represents the acidic proton and A- represents the conjugate base of the weak acid. The equilibrium constant for the dissociation is given by Ka which can be expressed as
$\mathrm{K_a = \dfrac{[H^+][A^-]}{[HA]}}$
This equation can be rearranged in terms of [H+]
$\mathrm{[H^+] = \dfrac{Ka [HA]}{[A^-]}}$
At a volume of titrant equal to half that required to completely neutralize the weak acid present, the remaining concentration of HA will be equal to that of A- formed during the neutralization reaction.
$\mathrm{[HA] = [A^-]}$
and
$\mathrm{[H^+] = K_a}$
By converting [H+] to pH, a useful relationship between the pH at the midpoint of the titration and the pKa is obtained
$\mathrm{pH = -\log [H^+] = -\log K_a = pK_a}$
Procedure
You will given an unknown sample, and you should ask the instructor the maximum sample size to weigh out. Weigh the sample by difference into a clean 100 mL beaker and dissolve it with about 50 mL of H2O. If the sample appears to be insoluble, start again with a clean beaker and add the minimum amount of ethanol needed to dissolve, then add 50 mL of H2O. Quantitatively transfer the dissolved sample to a clean 200.00 mL volumetric flask and dilute to the mark. Mix thoroughly. When the actual titration is to begin, pipette 25.00 mL of H2O (unless you have solubility problems, in which case you should use either ethanol or a water/ethanol mixture) and 50.00 mL of the sample solution (only 1/4 of the total mass!) into a 150 mL beaker.
Make sure you understand the operation of the particular pH meter you will use for your measurements. Standardize the pH electrode using pH 4.00 and pH 7.00 standard buffer solutions. Place a stir bar in the bottom of the beaker containing the unknown, and the beaker on the stir plate. Place the pH electrode into the solution, taking care that it is well away from the stirring magnet (which should stay in the center of the beaker). Run the stirring magnet at a slow speed throughout the titration. You will be measuring pH as a function of standard NaOH solution (prepared and standardized previously) added from a 50.00 mL buret (Class A tolerance). The most helpful information in the identification of your unknown will occur in those regions where the pH changes are great for small aliquots of base. Therefore, in these regions you should collect more data points than in the “buffer” regions.
It is advisable to titrate one sample quickly in one- or two-mL steps, taking the pH reading after it becomes stable (generally no more than 30 seconds). Prepare a table in which you record your data, with columns for “titrant added” and “pH”. Read the buret and the pH meter to two decimal places. This “quickie” will allow you to find the endpoint region, and to be prepared for its arrival during the next titration.
Repeat the process of pipetting 25.00 mL of H2O and 50.00 mL of sample solution. Do a second titration, being careful to take enough points to define the endpoint well. When the change in pH value upon addition of 1-mL of titrant exceeds 0.10 pH units, start adding 0.5-mL aliquots, and read the buret and the pH meter as accurately as possible. When the change exceeds 0.10 pH units each time an aliquot is added, start adding 0.1-mL aliquots (or smaller with “split drops”). When the endpoint is passed, continue to add titrant, gradually increasing the size of each aliquot added up to 1.0 - 2.0 mL. Continue titrating until the pH is well above 10 - 11.
Results
Plot your titration curve using a program like Excel. Since mL added is the independent variable, it goes on the x-axis. Next, calculate and plot the first derivative of your data, which usually is peak-shaped, and gives the endpoint as the maximum value. The moles of titrant used to reach the endpoint will be equal to the moles of unknown acid in the solution titrated. Use this information to calculate the molecular weight of the unknown. Since pH = pKa at a volume of titrant corresponding to half of the endpoint volume, you may easily calculate the Ka for your unknown.
Your unknown is one of the compounds in the following list. Once you have identified it, use its actual value for Ka and calculate a theoretical titration curve. This should have pH values that are calculated for each of the following: initial pH for the concentration of unknown you started with; pH at 1/4, 1/2, and 3/4 of the way to the endpoint; pH at equivalence point; at least two points past the equivalence point. Plot the actual titration curve on the same page as the theoretical curve for an easy comparison. If the two do not match very well, you may need to choose another possibility for the identity of your unknown.
Turn in these plot(s) with your report, and include all of your calculations. In the discussion section, give all the reasoning you used to identify the unknown. Ideal titration curves will not be obtained in all cases but enough information will be available to complete the experiment successfully. You may wish to look for outside confirmation such as information in textbooks, Merck index, CRC, etc. For example, some of your materials will be granular or flakes, colored or colorless, very soluble in water or fairly insoluble. This type of information may help you choose between more than one likely candidate. Be sure to include this information in your discussion.
Data for Acid Unknowns
Name Formula pKa Value Formula Weight
acetic acid HC2H3O2 4.74 60.05
anilinium hydrochloride C6H5NH3+Cl- 4.60 129.59
benzoic acid C6H5COOH 4.20 122.12
boric acid H3BO3 9.24 61.84
chloroacetic acid ClCH2COOH 2.82 94.50
m-chlorobenzoic acid ClC6H4CO2H 3.82 156.6
o-chlorobenzoic acid ClC6H4CO2H 2.92 156.6
p-hydroxybenzoic acid HOC6H4CO2H 4.58 138.12
hydroxylammonium hydrochloride HONH3+Cl- 5.96 69.49
mandelic acid C6H5CH(OH)CO2H 3.40 152.15
m-nitrobenzoic acid O2NC6H4CO2H 3.50 167.12
p-nitrobenzoic acid O2NC6H4CO2H 3.44 167.12
potassium bitartrate KHC4H4O6 4.54 188.18
potassium hydrogen phthalate KHC8H4O4 5.41 204.23
salicyclic acid HC7H5O3 2.98 138.1
sodium bicarbonate NaHCO3 10.36 84.01
sodium citrate, dibasic Na2HC6H5O7 6.40 236.1
sodium malonate NaHC3H2O4 5.66 126.1
sodium phosphate, dibasic Na2HPO4 12.32 143.0 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/06_Experiments/01_Potentiometric_Titration_of_an_Unknown_Weak_Mon.txt |
1. Purpose
This procedure will determine the concentration of chloride ion with a chloride specific ion electrode using potentiometry.
2. Background
Potentiometry is an electrochemical method in which the potential of an electrochemical cell is measured while little to no current is passed through the sample. In titrimetric methods, this measurement can be used to indicate the end point. When an ion selective electrode is used, the measured potential is related to the ion concentration in solution and a quantitative determination can directly be made. The instrumentation used to perform potentiometry is straightforward and relatively inexpensive, consisting of an indicator electrode, a reference electrode, and a potential measuring device. One of the most common applications of potentiometry is in the measurement of pH.
As indicated by the name, ion-selective electrodes (ISE’s) possess a high degree of selectivity. These electrodes are routinely used in clinical laboratories to determine various ion concentrations (such as calcium ion) in blood samples. In this laboratory, the electrode used is specific for chloride ion. Because specific ion electrodes measure activity and not concentration, a large amount of an inert strong electrolyte (e.g. nitrate ion) can be added to fix the ionic strength to a constant value. When the ionic strength is constant, the activity is constant and concentration can be accurately measured. In this laboratory you will fix the ionic strength with an ionic strength adjustment buffer (ISAB). You will be asked to determine the concentration of chloride in an unknown sample at the ppm level.
The chloride ion selective electrode you will use is a crystalline solid-state electrode that contains a membrane, as shown in the diagram below (Figure 1). The membrane consists of a solid salt of silver sulfide / silver chloride (Ag2S / AgCl). The membrane must be insoluble in the analyte solution and contain the analyte ion of interest. The membrane is placed at the end of a solid plastic tube. This membrane is in contact with the analyte solution during the measurement. Inside of the tube is a reference solution, which contains a known and fixed concentration of analyte (Cl-) solution. The concentration difference between this inner solution and analyte solution causes the migration of charged species across the membrane. This ion exchange process at the surface of the membrane causes a potential to develop. Since the potential of both the reference electrode and the inner reference (immersed in the standard solution) are constant, any change in measured potential is caused only by a change in potential across the membrane and is a function of the analyte chloride ion activity (or concentration).
It is also important to note that Figure 1 depicts a reference electrode external to the ISE (indicator electrode). However, a reference electrode can also be placed inside of the tube containing the ISE. This is called a “combination electrode” and is the type that you will be using in this laboratory experiment.
Figure 1. Schematic of a potentiometric measurement.
3. Materials and Equipment
• Sodium chloride
• Ionic strength adjustment buffer (ISAB)
• 1 L volumetric flask
• 250 mL beaker
• 250 mL volumetric flask
• 50 mL pipette
• 1 L plastic bottle
• Potentiometer
• Eppendorf micro pipette
• Magnetic stir bar
4. Safety / Special Handling Procedures
Protective eyewear must be worn at all times.
5. Experimental Method
5.1. Preparation of the standard solution
Accurately weigh pure sodium chloride to give about 2 grams of chloride ion (show calculations). Record this mass to 4 decimal places in your laboratory notebook and quantitatively transfer the NaCl with a wash bottle to a one liter volumetric flask. Dissolve, dilute to the mark with nanopure water, and mix. Calculate the concentration of chloride ion in parts per million. Transfer to a plastic bottle and label it "chloride working standard".
5.2. Preparation of the unknown
5.2.1. Obtain an unknown chloride sample and quantitatively transfer it into a 250 mL volumetric flask. Dilute to the mark with nanopure water and mix. Transfer to a plastic bottle and label appropriately.
5.2.2. Obtain the large bottle marked "ISAB" (ionic strength adjusting buffer). It contains approximately 0.2 M sodium nitrate solution. If the solution level is very low or the bottle is empty, prepare additional solution. If there is sufficient solution, go to step 5.2.3.
5.2.2.1.Weigh out pure sodium nitrate on the top loader scale. Add this to the large 2 liter bottle. Add water to mark on top shoulder of the bottle and mix. Note: This solution is not considered a standard solution.
5.2.3. You will prepare three samples of the unknown using three separate 250-mL beakers. To do so, pipette a 50.00 mL aliquot of ISAB into three different clean and dry 250 mL beakers. Then pipette 50.00 mL aliquots of the unknown solution into each beaker. These three diluted samples will be individually analyzed by the instrument.
5.3. Use of the Accumet Model 15 Specific Ion Potentiometer:
Figure 2. Accumet Model 15 Potentiometer
5.3.1. Bring the following to the instrument room:
1. Beaker with the sample
2. Bottle with the chloride standard
3. Eppendorf micro pipette (blue top. set to deliver 1000 μL with a new plastic tip
4. Wash bottle
5. Kim Wipes (if none are already there)
6. Empty beaker for rinsing purposes
7. Magnetic stir bar
5.3.2. Remove the combination chloride/reference electrode from the storage vial and set it in the holder. Refer to Figure 2 for a picture of the potentiometer and Figure 3 for an image of the electrode.
5.3.3. The bottom of the electrode is a flat crystal surface and must not be touched with fingers (only Kim wipes). Open the fill hole at the top of the electrode. See Figure 3 below.
Figure 3. Chloride electrode (left) with the fill hole closed (middle) and open (right).
5.3.4. Rinse the electrode with nanopure water. Collect the rinses in the empty beaker and blot the electrode dry with a Kim wipe.
5.3.5. Place the sample beaker on the magnetic stirrer and carefully drop in the magnetic bar without splashing.
5.3.6. Insert the electrode so that not only the electrode but also the salt bridge is immersed in the liquid. See Figure 4 below.
Figure 4. Proper positioning of electrode in beaker.
5.3.7. Adjust the stirrer to a slow speed (note: clockwise increases the speed). Do not allow the stirrer to touch the electrode.
5.3.8. Make sure that the potentiometer is plugged in and turned on. Press the "meas/monitor" button so that the instrument is actively reading the potential of the solution (if you see a symbol in the shape of a key on the screen, it is in the locked mode and so press the meas/monitor button again to unlock the instrument). Verify that the meter reads "mV" at the bottom of the screen. The accuracy of the instrument is to the nearest tenth of a millivolt (0.1 mV). When a steady reading is produced, record the millivolt reading in your notebook.
5.3.9. Spike the sample with 1000 μL of the standard chloride solution. Again, when the instrument produces a steady reading, record the millivolt value in your notebook.
Note: For a review of proper micropipette usage see Appendix A of this procedure.
5.3.10. Repeat steps 5.3.1 through 5.3.9 for a total of three samples prepared. Rinse the electrode and dry between each sample reading (post spike addition).
5.3.11. Rinse the electrode with nanopure water and place in the storage flask. Make sure that the fill hole is closed. Refer to Figure 4 for placement.
6. Data Analysis / Calculations
Note: These calculations are quite long. It is a good idea to write them out on scrap paper, double check them, and then record the calculations neatly in your notebook.
The calculation for the standard addition method in potentiometry uses the Nernst equation, where the potential is measured in volts (what were the units you recorded??). Note that while the Nernst Equation typically involves the molarities (or, more accurately, the activities) of ions, it is perfectly acceptable to use ppm values in the following calculations:
(before the spike): $E' = k + \dfrac{0.05916}{z} \log[\ce{Cl-}]_{unk}$
(after the spike): $E'' = k + \dfrac{0.05916}{z} \log ([\ce{Cl-}]'_{unk} + [\ce{Cl-}]'_{std})$ where $\mathrm{[Cl^-]’ = DF \times [Cl^-]}$,
Note that in the second equation, the addition of the standard spike changes the volume of solution in the beaker and so both the standard and unknown concentrations are diluted from their original values. You must account for this by determining the different dilution factors (i.e., [Cl-]’ = DF x [Cl-], where DF is the dilution factor) for the standard and unknown in the second equation.
The other variables in the equations are potential (E, in volts) and charge (z) of the chloride ion. The “k” is a constant, therefore, identical in both equations. This information should allow you to manipulate the equations and solve for [Cl-]unk. Once you have determined [Cl-]unk in your unknowns diluted by the ISAB, you will need to use an aliquot factor to determine the original concentration of chloride in your unknown Perform separate calculation for each individual trial then calculate the average chloride concentration in ppm. DO NOT average the mV readings and perform the calculation on the average readings.
7. Reporting Requirements
In your conclusion, report the individual results for each trial and the average concentration (in ppm) chloride in the original unknown (the 250 mL solution) to one decimal place. Also calculate the percent relative deviation of your trials.
8. Waste Disposal
Discard all sodium chloride solutions down the drain.
02 Determination of Chloride using Potentiometry
A1. Volume Setting
The volume is adjusted by pressing down the lateral catch and turning the control button at the same time.
It is advisable to carry out volume setting from the higher down to the lower value. i.e. first go above the desired volume and then return to the lower value.
A2. Pipette tips
Typically the color of the control button will correspond to the color of the eppendorf tip or tip rack. For the best precision and accuracy, pre-wet all new tips by aspirating and dispensing liquid 2-3 times before pipetting.
A3. Aspirating liquid
• Attach suitable pipette tip to the pipette firmly.
• Press down the control button to the first stop (measuring stroke).
• Immerse the pipette tip vertically ~3 mm into the liquid.
• Allow the control button to slide back slowly.
• Pull the tip out of the liquid slowly.
• To remove any remaining droplets, dab with non-fibrous cellulose material, ensuring that liquid does not come out of the tip. You can also dab on the side of the beaker containing the liquid you are pipetting.
A4. Dispensing liquid
• Hold the tip at an angle against the inside wall of the tube/flask.
• Press down the control button slowly to the first stop (measuring stroke) and wait until the liquid stops flowing.
• Press down the control button to the second stop (blow-out) until the tip is completely empty.
• Hold down the control button and pull the tip out of the inner wall of the tube/flask.
• Allow the control button to slide back slowly.
• Tip is ejected by pressing the control button to the final stop.
Do not lay down the pipette when a filled pipette tip is attached as this may result in liquid entering the pipette.
A5. Verification of pipette
You can verify that the pipette is performing accurately by dispensing nanopure water from a prewetted tip into a tared flask or tube onto an analytical balance. Typically, one might test full volume and ½ volume, e.g. 500 and 1000 μL for a 1000 μL pipette.
Convert the mass to volume by dividing by the density at room temperature. For example, the density of water is 0.9982 mg/μL at 20°C. This number is the volume actually delivered by the pipette. Determine the error relative to the set value. Repeat a few times to verify that the pipette is accurately delivering water. If not, consult your instructor. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/06_Experiments/02_Determination_of_Chloride_using_Potentiometry/A.txt |
The Fluoride Ion Selective Electrode Experiment
Direct Potentiometry and Standard Addition Methods
Dr. David L. Zellmer
Revised February 15, 1999
The fluoride ion selective electrode experiment consists of the following parts:
• Choose a real world unknown and prepare it for analysis. Suggestions for sample preparation may be found in Skoog, West, and Holler, Fundamentals of Analytical Chemistry, 7th Ed., Saunders, 1996, Section 36I-5 "The Direct Potentiometric Determination of Fluoride Ion.", pp. 850-852. Note how Total Ionic Strength Adjustment Buffer (TISAB) is used to maintain constant ionic strength and to remove certain interferences.
• Obtain background information on probable fluoride levels in your unknown. Toothpaste and mouthwash must contain fluoride concentrations on their labels. For levels in drinking water you could try 1 mg F/L as a starting point, or check web sources such as Water Fluoride, or use a web search engine for additional information. Call your local water district if you plan to test your own drinking water.
• Prepare a series of standard fluoride solutions with a constant amount of TISAB in each. Use these solutions to determine if your electrode is operating properly by plotting E vs. log C and checking for Nernstian behavior.
• Using direct potentiometry, measure the potential of your unknown and read its provisional concentration from your E vs. log C graph. You will need both the slope of the electrode and the provisional concentration to do the standard addition portion of the experiment.
• Use standard addition to make a more careful measurement of your unknown. The mathematical details are given below.
Direct Potentiometry
In this method a series of fluoride standards are prepared in a background matrix of TISAB (Total Ionic Strength Adjustment Buffer). The unknown is prepared using TISAB in the hope that the matrix will be similar to the standards.
Since E = K + S log C
A plot of E vs. log C should yield a working curve that can be used to measure an unknown.
The first thing we learn from this plot is that our electrode is working properly, and has the expected linear response of E vs. Log C with slope measured at -59.4 mV, very near to the theoretical -59.2 mV at 25o C. A Linear Least Squares analysis of this plot allows us to measure the concentration of our unknown directly, and to compute the error of this determination. In the figure above the unknown was found to be 2.28 +or- 0.05 mg/L (one-sigma error).
Standard Addition
The problem with direct potentiometry is the possible matrix effect on the unknown. The presence of iron(III), for example, might complex part of the fluoride in spite of the TISAB which is supposed to combat this. To check for this possibility we will need to use Standard Addition. [Note that the standard addition method described will not correct for the presence of other ions to which the electrode also responds; where E = K + S log(Cunk + kCintf)].
The graphics above show two types of standard addition: constant volume and variable volume. (For constant volume standard addition see the tutorial at http://zimmer.csufresno.edu/~davidz/...n/StdAddn.html).
The discussion below will be for variable volume standard addition. In this method we have a large volume of unknown to which small amounts of a standard solution are added. Although the total volume will not change a great deal, we will have to compensate for the volume changes produced. Note that a special pipet will have to be used that can dispense small volumes with high precision. (As an alternative, you could measure the density of the standard solution, then dispense the standard by mass using a weight buret.)
For an ion selective electrode with slope S that responds only to the analyte ion with concentration C,
Now, if we wish to remove any effect that ties up part of the analyte ion, we need to divide the response with standard addition by the response without standard addition.
The final equation is in the form of y = mx + b, where y is the electrode response to the amount x of standard added. Note that x is not expressed as a concentration, but rather as the total amount CstdVstd of standard added. The slope of this line will be m=1/Cunk, from which we could compute Cunk directly. If we wish to use the more traditional standard addition approach using an x-axis intercept, then when y=0, the x-axis intercept will be at x=-b/m or CstdVstd=-CunkVo. Divide the amount CstdVstd at the x-axis intercept by Vo and you have Cunk.
Confused? The following example may help.
An unknown solution is prepared by working up a sample and buffering it with TISAB. 100.0 mL of this solution is placed in a beaker and its potential measured. This potential is called E1. Then 0.300 mL of a 300 mg/L standard is added, the solution mixed, and the potential measured. This potential is called E2. Additional standard is added, and additional values of E2 are measured. A plot of (Vo+Vstd)*10^((E2-E1)/S) is plotted vs. CstdVstd. The following graph was produced.
We can see that at y=0 a value of around 220 micrograms of fluoride is found. Since this was in 100 mL of solution the unknown must be around 220 micrograms/100 mL = 2.2 micrograms/mL or 2.2 mg/L. We are then told that the unknown solution was actually 2.15 mg/L, so we are pretty close.
Another important feature of this graph is the amount of increase shown in the response function for each addition of standard. Each amount of standard added should increase the fluoride level about 30% above the concentration of fluoride present due to the unknown. If the amount of standard added is too small, we will be left with a very large extrapolation to the x-axis. If the amount added is too large, the size of the x-axis intercept will be very small compared to the rest of the graph. Both of these extremes will produce unacceptable levels of relative error.
The spreadsheet that produced the graph above includes a careful linear least squares analysis of the results.
Some portions of this spreadsheet were blocked out because they represent "secret" knowledge about the unknown or about the response function of the electrode. The remaining numbers are similar to those found in a real experiment, however, and give a final result of 2.16 +or- 0.05 mg/L for the unknown solution. Compare this to the actual value of 2.15 mg/L, which is given as "secret" information in Cell B10. In a real analysis you would not have this value, of course.
Note that the value of "S" given in Cell B6 is the -59.4 mV found from the direct potentiometric method given above. Don't confuse this with the slope of the response function as computed by LLS analysis of the data from the standard addition runs. If a poor value of "S" is chosen, the Response function will become nonlinear, making it useless for a standard addition determination.
Note that the formula in Cell D14 is the Response function (Vo+Vstd)*10^((E2-E1)/S) which should be linear with respect to the number of micrograms of fluoride added to the system. When Vstd=0, the exponential term contains (E1-E1)=0 which makes the Response function equal to Vo or 100.0 mL. When filled down, the value of E1 stays fixed on Cell C14, while the value of E2 changes to the new potentials measured as each standard addition is made.
When doing an x-axis intercept at y=0, the value of M is assumed to be infinite because the value of y=0 is assumed to be perfectly known. The traditional formula for "s sub c" is modified by removing the 1/M term. The formula given in row 38 for Cell C30 contains an error which has confused a few people. The reference to cell C19 points to a blank cell. This happens to work out OK, since C19 refers to the y-value of the "unknown" which in this case of standard addition would be zero (the X-axis intercept is at y=0). The calculation comes out OK, since Excel assigned a value of zero to this blank cell. To correct the equation shown in row 38, replace "C19" with 0.
(See http://zimmer.csufresno.edu/~davidz/...modelSp99.html for details about the "traditional" LLS formulas.)
In Cell C33 the final one-sigma reported error for the unknown concentration is computed by multiplying the computed concentration of the unknown by the relative error of the x-axis intercept. The ABS() function is used to remove any minus signs that may result.
In case you may not have noted the pertinent information needed in your report to describe the electrodes used, the following images may prove useful: Corning1, Corning2, Orion1, Orion2, OrionSJR1, and OrionSJR2.
For current information, try websites for Corning and for Orion.
Contributors and Attributions
• David L. Zellmer, Ph.D. (California State University, Fresno) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/06_Experiments/03_Fluoride_Ion_by_Direct_Potentiometry_Standard_A.txt |
While understanding the underlying general concepts of potentiometry is a useful first step at becoming a regular "potentiometric practitioner", experience is also a great resource for effectively conducting these types of measurements. Through experience comes familiarity with common "problem areas" of this field. This page is intended to present some troubleshooting tips. It is not our intention to replace recommendations outlined in manufacturer literature.
Before specific discussion on common problem areas, the subtle nuance differences between efforts in calibration methods and quality control (QC) must be highlighted. Calibration and QC methods are complementary to one another and are often integrated into a method validation program that defines the overall reliability. Calibrations give analytical methods an initial quantitative starting point, whereas QC validates the developed calibration model.
Many classic questions arise, such as, "How often does calibration need to be performed?" and "What are some examples that require re-calibration?" QC should be performed on a daily basis in an industrial setting, if not more often. Various companies are required to maintain SOPs (standard operation procedures), among other validation protocol required for state and federal regulations. The generic answer to when the analyst should consider re-validating the method is when conditions change. Some examples of change include hardware, consumables (change in vendors and even a change in lots from the same vendor), sampling methods, and of-course if QC fails.
So, what are some of the specific concerns when using potentiometric electrodes? Faulty measurements, long response times, and unstable values can typically be traced back to problems at the liquid junction. One of the most important tasks in maintaining a trouble-free potentiometric analysis is electrode conditioning, more specifically "membrane conditioning" when using Ion-Selective Electrodes (ISE). In addition to providing selectivity through specific interactions, these membranes will also serve as salt bridges when an internal reference electrode setup is being used. This salt bridge is necessary to provide connection between the internal reference electrode and the unknown solution, and therefore these membranes must be maintained for proper use. The old adage of "an ounce of prevention is worth a pound of cure," justly applies here.
Membrane Conditioning
Initial conditioning is extremely important and commonly overlooked. For example, prior to using pH electrodes, the exterior surface of the glass membrane must be fully hydrated. Ion exchange at the hydrated glass surface then leads to the establishment of a potential across the membrane, its magnitude being determined by the concentration difference between the internal reference solution and the external analyte solution.
Routine maintenance of electrodes is also necessary. For combination electrodes containing a reference electrode separated from the external solution by a porous frit or plug, the level of internal electrolyte must be regulated appropriately. Proper operation requires that the drainage hole, typically located at the top of the electrode, be opened during measurements, allowing electrolyte solution to slowly flow through the porous junction into the external solution. The level of the internal fill solution should be kept above that of the measured analyte solution. One concern here is to avoid the species in the solution forming a precipitate with the exiting electrolyte species. To ensure that the sample solution does not alter critical parameters of the analyte solution, constant ionic strength buffers are frequently used. Electrodes with a second electrolyte junction (double junction) generally have longer lifetimes than those with a single junction.
Matrix Effects
Matrix effects can have a substantial impact on the response function (e.g. sensitivity) of potentiometric electrodes. Standard addition methods are recommended for nonideal solutions, such as lake-water sampling. Standard addition methods are recommended procedure for determining unknown analyte concentrations. Simple calibration might not account for response of background, or interfering species. Internal standard methods, used for compensating for variations in experimental parameters, are not widely used for potentiometric determinations, unless extensive sample preparation methods are necessary and one would want to account for sample loss during preparation.
Calibrations
• total ionic strength adjustor buffer (TISAB) to ensure standards and samples have similar ionic strength and to reduce interference from other ions
• calibration best performed with standards that bracket expected unknown concentration…especially if outside linear dynamic range
• standard addition – limited samples, high ionic strength, or complicated background – must be on linear part of calibration curve
• titrations – quantitative measurement of species concentration via addition of reagent that reacts with species of interest…~10x more precise than direct, but very time consuming
Other On-line Resources
For a more detailed trouble shooting guide (in PDF format) please visit http://www.metrohm.co.uk/elcat_e_10_basic_pot.pdf. Common problems with pH glass electrodes are listed in Table 3. Metal electrode troubleshooting is presented in Table 5. Interfering ions for specific ISEs are listed is Table 6. Lastly, ISE troubleshooting tips are detailed in Table 7 of the above link.
08 References
1. Cremer, M,. Z. Biol. 1906, 47, 562.
2. Buck, R.P. and Lindner, E. Anal. Chem. 2001, 73, 88A.
3. Frant, M.S., Analyst, 1994, 119, 2293.
4. Frant, M.S. and Ross, J.W., Science, 1966, 154, 1553.
5. Ross, J.W., Science, 1967, 156, 1378.
6. Simon, W., Swiss Pat., 479870, 1969.
7. Frant, M.S. and Ross, J.W., Science, 1970, 167, 987.
8. Bakker, E. and Pretsch, E., Anal Chem., 2002, 74, 420A.
9. Meyerholf, M.E. and Opdycke, W.N. In Advances in Clinical Chemistry, vol. 25; Spiegel, H.E., Ed.; Academic Press, Inc., Orlando, 1986, pp. 1-47.
10. Bakker, E.; Buhlmann, P.; Pretsch, E. Talanta, 2004, 63, 3.
11. Wang, J. Analytical Electrochemistry, 3rd ed.; Wiley: Hoboken, NJ 2006, pp. 165-200.
12. Skoog, D.A., Holler, F.J., Crouch, S.R., Principles of Instrumental Analysis, 6th ed.; Thomson Brooks/Cole: Belmont, CA, 2007.
13. Buhlmann, P.; Pretsch, E.; Bakker, E. Chem Rev. 1998, 98, 1593. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_Potentiometry/07_Common_Troubleshooting_Tips.txt |
Learning Objectives
Each part of this module describes certain aspects of the experience that is analytical electrochemistry. The goal is to include enough information that you can understand and use electrochemical techniques effectively without getting bogged down in theory (or lost in the double layer). After study of this module, you should be able to:
• Discuss the physical basis underlying voltammetric methods.
• Describe an electrochemical cell and its contents.
• Visualize how a potentiostat works.
• No longer cringe at the mention of electrochemistry!
This module gently introduces a topic whose mere mention often strikes fear in the hearts of students and teachers alike. Whether because of some intimidating mathematical aspects of the subject, or a lack of time to provide an adequate basis for understanding, it is often left off of course syllabi in analytical chemistry, This learning module is designed as an introduction to the underlying theory and general practice of several common techniques in analytical electrochemistry. It is intended to contain sufficient background material so that educators or new practitioners in the field can use the material in a "stand-aIone" way. References and links are provided so that users who so desire can pursue a more extensive study of each topic presented. Sufficient coverage is provided such that the module could serve as a two- to three-week segment of a course in analytical chemistry or instrumental methods.
A. Electrochemical Therm
Electrochemistry can be defined as the study of phenomena at electrified interfaces.1 As such, two things are fundamental to electrochemical measurements:
1. Interface: boundary between two distinct, chemically different phases
2. Electric field: existing across the interface
Generally, the interface occurs between a metallic conductor (the electrode) and a fluid, ionic conductor (the solution). The electric field usually results as a consequence of contact between a solid electrode and the solution. In electrochemistry, the electric field is most often one that is under user control, for example by utilizing a device called a potentiostat, which will be discussed in a later section.
B: The Electrode Proce
As you have probably realized, the fundamental basis for most electrochemical techniques is the measurement of current or voltage changes between two electrodes in solution. Given that basis, it follows that analytical electrochemistry is broken down into two major categories: 1) techniques that measure current following a change in potential, and 2) techniques that measure potential under conditions of no current flow. Techniques of the first type are known as voltammetric methods, while those in the second are referred to as potentiometric methods. We will concern ourselves here with voltammetry, leaving a discussion of potentiometry including measurement of pH for a separate learning module.
Voltammetry is defined as the measurement of current which flows at an electrode as a function of the potential applied to the electrode. Current-potential curves are the electrochemical equivalent of absorbance-wavelength curves recorded in spectrophotometric experiments. Once you gain an understanding of the voltammetric process (hopefully by the time you finish this module), you will recognize that voltammograms have the potential (no pun intended) to yield qualitative, quantitative, thermodynamic, and kinetic information about redox active species.
01 Potentia
Methods of this type involve stepping the electrode potential from a value where little or no faradaic current is observed to one sufficient to oxidize or reduce an electroactive species in the vicinity of the electrode. Large amplitude step methods described in this section involve a change in potential capable of instantly converting essentially all of the electroactive material at the electrode surface to its redox partner. In contrast, small amplitude potential methods, described elsewhere in this module, involve small changes in surface redox concentrations as a function of changing potential, with the concentrations of the members of the redox pair described by the Nernst Equation.
We will include here two of the most frequently used large amplitude potential step (potentiometnc) methods, chronoamperometry and chronocoulometry.
Chronoamperometry involves the measurement of current passing in the electrochemical cell at a fixed potential as a function of time (i vs. t).
Chronocoulometry is the measure of total charge (the integrated i-t response, Q), also as a function of time.
02 Potentia
Potential sweep methods are those utilizing an applied potential that changes with time as the excitation signal. The most common waveform is that of a linear sweep, beginning at a potential far removed from the E0’ for the species of interest and increasing in magnitude at a constant rate. The current passing at the working electrode is measured as a function of the applied potential, with electron transfer accomplished by scanning the potential through the regions on either side of the E0’. The most common potential sweep methods are linear sweep voltammetry (LSV), cyclic voltammetry (CV), and anodic stripping voltammetry (ASV). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/01_Module_Description.txt |
Most voltammetric measurements make use of a device called a potentiostat, which is capable of applying a controlled potential to a working electrode and measuring the current that passes as a result of electron transfer to solution species of interest. The working electrode, along with a reference electrode and an auxiliary electrode are commonly placed in an eIectrochemical cell, with a fill solution containing an inert electrolyte and the analyte. Cell designs can be quite simple to quite complex, with solution volumes ranging from a few microliters to many milliliters.
05 Experimental Hardware
1. Defining the cell
The voltammetric methods we have described thus far involve the application of a potential from an external power source to a working electrode (WE) relative to that of a reference electrode (RE), and the measurement of current that flows as a result of that applied potential. In most instances, the electrochemical cell in which this process is carried out involves a solid working electrode immersed in a solution of charged electrolyte along with a reference half-cell of fixed composition.
As was discussed in the Electrochemical thermodynamics section of this module, current passing in the cell as a result of applied potential can be of two basic types:
1. nonfaradaic, in which a change in the potential at the WE causes a rearrangement of electrolyte charges at each electrode/solution interface to counter the charge at the electrode surface;
2. faradaic, in which electrons are transferred across the electrode/solution interface to a solution species which is either oxidized or reduced.
The two-electrode cell shown in Figure 32 is suitable as long as very little current passes during the experiment. This is the case for very low concentration solutions, or for very small working electrode dimensions (ultramicroelectrodes). In most practical applications, however, the circuit is designed to include a third electrode called the auxiliary electrode (AE). The device applying the potential and measuring the current (the potentiostat) electronically isolates the reference electrode so that little or no current passes through it, while allowing current to pass through the working and auxiliary electrodes. In this way, the activities of the reference half-cell components and thus the reference potential are prevented from changing during the experiment (a desirable characteristic!). The potentiostat will be discussed in a later section.
Figure 32
2. Positioning of the electrodes
For the remainder of our discussion, we will consider only the three-electrode electrochemical cell. The primary consideration for the relative location of the three electrodes in a cell is the minimization of the solution resistance. This is most usually accomplished by keeping the tips of the three electrodes as close together as feasible, while not interfering with the current paths between one another. Close approach is especially critical for the reference and working electrodes. The solution resistance between these electrodes leads to an iR drop that manifests itself as an error in the measured potential difference between them. In addition, the potentiostat is normally unable to electronically compensate for this resistance, as it can for the resistance between the reference electrode and the auxiliary electrode. A Luggin capillary arrangement is often employed, in which the RE is placed inside a tube drawn at the end to a fine capillary allowing very close positioning relative to the WE.
A second important consideration involves the shape and size of the auxiliary electrode relative to the working electrode. The AE should be at least as large in area as the WE, and positioned symmetrically with respect to the WE so that the current density and potential experienced along its entire length is constant.
In most voltammetric experiments, the three electrodes can be placed together in the same solution. Under certain conditions, however, either the AE or the RE (or both) may need to be physically isolated from the solution containing the WE. For example, if the goal of the experiment is the total electrochemical conversion of the bulk electroactive material to product (bulk electrolysis), products produced at the AE may be detrimental to the reaction occurring at the WE. Other times, small amounts of leakage of RE components, like Ag+ or even water from a Ag/AgCl reference, may cause undesirable reactions in the electrolyis solution, especially in experiments that involve non aqueous solvents. These problems can generally be avoided by the use of a glass frit or Vycor membrane (which allow charge to pass but not mixing of electrolyte) to separate the compartments of the electrochemical cell.
3. Other considerations
Most experiments require that dissolved oxygen be removed from the cell. Not only is it electrochemically active across the cathodic potential range, it is very likely to react with products formed by electron transfer. Typically, the solution is sparged with an inert gas like nitrogen or argon for 5 – 10 minutes prior to the experiment, and a “blanket” of inert gas maintained above the solution during the experiment.
Cell volumes are quite variable and range from microliters to many mL, depending upon the goal of the experiment. Designs for cells can be quite simple – a glass vial with a screw cap drilled with holes for each electrode – to quite complex, requiring significant glass-blowing and/or machining skills. Cells are available from numerous electrochemical suppliers, some with very specific applications.
Two common cell configurations for quiescent solution voltammetry are shown in Figure 33. On the left, the AE has been fashioned to be symmetrical to the WE. At right, a cell is shown with RE isolated from the solution containing the WE using a Luggin capillary. The AE is also isolated in a fritted compartment. Both cells have a fritted sparge tube to allow deoxygentation of the solution with inert gas.
Figure 33 | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/05_Experimental_Hardware/A._Electrochemical_Cells.txt |
1. Reference Electrodes
Voltammetric methods are those in which current passing in an electrochemical cell is measured as a function of the potential applied to the working electrode. Potential, by definition, is not something that can be directly measured. Rather, the measurement of applied potential requires that a reference point first be established, and individual potentials be measured relative to that reference point. This is accomplished by placing a second electrode, called the reference electrode, in the cell and measuring potential as the energy difference between the two electrodes. As Kissinger and Bott have perfectly expressed, “electrochemistry with a single electrode is like the sound of one hand clapping”.21
Reference electrodes should be constructed using half-cell components that are stable over time and with changing temperature, present at well-defined values of activity. They should possess fixed, reproducible electrode potentials. The reference half-cell with which most of us are familiar is the standard hydrogen electrode (SHE), composed of an inert solid like platinum on which hydrogen gas is adsorbed, immersed in a solution containing hydrogen ions at unit activity. The half-cell reaction for the SHE is given by
$\mathrm{2H^+ (aq) + 2 e^- \rightleftharpoons H_2(g)}$
with a half-cell potential arbitrarily assigned a value of zero (E0 = 0.000 V). Tables of electrode potential values for many redox couples relative to the SHE are commonly available.
Practical application of the SHE is limited by the difficulties in preparing solutions containing H+ at unit activity and maintaining unit activity for H2 (g) in the half-cell. Most experiments carried out in aqueous solutions utilize one of two other common reference half-cells – the saturated calomel electrode (SCE) or the silver-silver chloride electrode (Ag/AgCl).
a) Saturated Calomel Electrode (SCE)
The SCE is a half cell composed of mercurous chloride (Hg2Cl2, calomel) in contact with mercury metal, either as a pool or as a paste with calomel. These components are either layered under a saturated solution of potassium chloride (KCl), or within a fritted compartment surrounded by the saturated KCl solution (called a double-junction arrangement). A platinum wire is generally used to allow contact to the external circuit. The half reaction is described by
$\mathrm{Hg_2Cl_2(s) + 2 e^- \rightleftharpoons 2 Hg (\mathit{l}) + 2 Cl^- (sat’d)}$
with a potential of 0.241 V with respect to the SHE at 25 oC. The double-junction arrangement of the SCE is illustrated in Figure 34, at left. Contact to the electrochemical cell is made through a porous glass frit or fiber, which allows ions to cross but not bulk mixing of the electrolytes.
b) Silver/Silver Chloride Electrode (Ag/AgCl)
The silver/silver chloride reference electrode is composed of a silver wire (Ag) that has been coated with a layer of solid silver chloride (AgCl), immersed in a solution that is saturated with KCl and AgCl. The pertinent half reaction is
$\mathrm{AgCl (s) + e^- \rightleftharpoons Ag(s) + Cl^-(sat’d)}$
with a potential of 0.197 V with respect to the SHE at 25 oC. This value differs slightly from the E0 for the couple (0.222 V) because both KCl and AgCl contribute to the chloride activity, which is not exactly unity. A schematic of the Ag/AgCl reference electrode is shown in Figure 34, at right.
Figure 34
Both the SCE and the Ag/AgCl reference electrodes offer stable half-cell potentials that do not change over time. Only a slight temperature dependence of the potential is observed in these electrodes, changing by approximately 0.5 – 1.0 mV/oC.22 In addition, the loss of electrolyte to evaporation does not change the saturated nature of the solution, nor the potential.
One must be aware that the contact junctions of the half cells by nature slowly leak fill solution into the external solution in which they are found. For example, a small amount of soluble AgCl2-1 ion in the internal KCl solution of the Ag/AgCl reference can find its way into the analyte solution over time, where it can lead to silver deposition on the working electrode at relatively low applied potentials. Silver deposited in this way can contaminate the surface and change the properties of the working electrode. Additionally, there are instances where certain measurements can be affected by the ions, like chloride ion, introduced to the measurement solution by leakage. A double-junction design can reduce the problem of contamination by placing a second solution between the reference half cell and the measurement solution.
c) Non-aqueous reference electrodes
In many applications, even a small amount of electrolyte solution leaking from the reference electrode can immediately compromise the electrochemical reactions occurring in the analyte solution. Primary among these applications is non-aqueous electrochemistry. In these applications, it may be possible to use what is called a pseudo-reference electrode. The simplest pseudo-reference electrode is a metal wire, like platinum, inserted directly into the analyte solution. A reference potential will develop that is strictly due to the composition of that solution. Though this half-cell will supply a constant reference potential during a single experiment, any changes in the cell solution will result in a change in the reference potential. It is accepted practice when employing such a reference that an internal reference redox compound with well-defined potentials, like ferrocene, be added (generally at the end of an experiment), and observed experimental potentials adjusted to the known potential of the standard.
A slightly better approach is to use a reference half-cell that includes the solvent system employed in the analyte solution. One such half-cell that can be used with most organic solvents is the silver/silver ion (Ag/Ag+) pseudo-reference electrode. In this electrode, a silver wire is immersed in a fill solution of the desired organic solvent that is saturated with a silver salt and which contains the same supporting electrolyte contained in the analyte solution. A porous glass or Vycor frit is used to separate the inner solution from that containing the working electrode. The half reaction for this electrode is given by
$\mathrm{Ag^+ + e^- \rightleftharpoons Ag^0}$
with the reference potential being a function of the given electrolyte solution.
Lastly, there are commercially available reference electrodes that advertise themselves as “no-leak” which are suitable for many non-aqueous applications. Users should test these under their particular cell conditions before accepting them for routine use. They should also be aware that many of the materials which constitute the body of the reference electrode may not hold up well in the solvent of choice.
2. Auxiliary electrodes
The purpose of the auxiliary electrode (AE) is to provide a pathway for current to flow in the electrochemical cell without passing significant current through the reference electrode. There are no specific material requirements for the electrode beyond it not adversely influencing reactions occurring at the working electrode (WE). Remember that if a reduction occurs at the WE, there must be an oxidation that takes place at the AE. Care should be taken that electrode products formed at the AE do not interfere with the WE reaction. The AE can be physically separated from the WE compartment using a fritted tube, but one should be aware that under certain circumstances this can have a deleterious effect.
The most commonly used material for the auxiliary electrode is platinum, due to its inertness and the speed with which most electrode reactions occur at its surface. Other, less expensive materials may also be used as auxiliary electrodes. Choices include carbon, copper, or stainless steel if corrosion is not an issue for a particular electrolyte solution or reaction.
As discussed in the previous section on electrochemical cells, the AE should supply current density and potential that is constant across the length of the WE. Many times this means fashioning the AE such that it “mirrors” the shape of the WE, as is the case for a rectangular WE being located near a rectangular AE of similar area. Wire is convenient in that it can be coiled in a symmetrical arrangement around the WE. In some instances, the electrochemical cell can be fashioned from the material chosen for the AE, and the cell itself can serve that purpose. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/05_Experimental_Hardware/B._Reference_and_Auxiliary_Electrod.txt |
1. Electrode types
The working electrode (WE) represents the most important component of an electrochemical cell. It is at the interface between the WE and the solution that electron transfers of greatest interest occur. The selection of a working electrode material is critical to experimental success. Several important factors should be considered. Firstly, the material should exhibit favorable redox behavior with the analyte, ideally fast, reproducible electron transfer without electrode fouling. Secondly, the potential window over which the electrode performs in a given electrolyte solution should be as wide as possible to allow for the greatest degree of analyte characterization. Additional considerations include the cost of the material, its ability to be machined or formed into useful geometries, the ease of surface renewal following a measurement, and toxicity.
The most commonly used working electrode materials are platinum, gold, carbon, and mercury. Among these, platinum is likely the favorite, demonstrating good electrochemical inertness and ease of fabrication into many forms. The biggest disadvantage to the use of platinum, other than its high cost, is that the presence of even small amounts of water or acid in the electrolyte leads to the reduction of hydrogen ion to form hydrogen gas (hydrogen evolution) at fairly modest negative potentials (E = -0.059 x pH). This reduction obscures any useful analytical signal.
Gold electrodes behave similarly to platinum, but have limited usefulness in the positive potential range due to the oxidation of its surface. It has been very useful, however, for the preparation of modified electrodes containing surface structures known as self-assembled monolayers (SAMs).
Carbon electrodes allow scans to more negative potentials than platinum or gold, as well as good anodic potential windows. The most common form of carbon electrode is glassy carbon, which is relatively expensive and difficult to machine. Carbon paste electrodes are also useful in many applications. These electrodes are made from a paste of finely granulated carbon mixed with an oil substrate like Nujol. The paste is then packed into a cavity in an inert electrode body. Carbon paste electrodes have the disadvantage of being prone to mechanical damage during use.
Mercury has historically been a widely used electrode material, primarily as a spherical drop formed at the end of a glass capillary through which the liquid metal is allowed to flow. It displays an excellent potential window in the cathodic direction, but is severely limited in the anodic direction by its ease of oxidation. A dropping mercury electrode (DME), in which drops are formed and fall off repeatedly during a potential scan, being replaced by a “fresh” electrode every second or so, was commonly in past years the first electrode many students encountered in their studies. The toxicity of mercury has lead to a limited use these days, though it still is a very useful surface in methods that involve the preconcentration of a metallic analyte prior to potential scan, such as is done in anodic stripping voltammetry (ASV). Many practitioners now make use of mercury films formed on the surface of solid electrodes rather than the pure metal. Under these conditions, the small volume of the film allows analyte to concentrate at large values, with rapid diffusion times.
2. Advantages and limitations
Table 1 lists the commonly used electrode materials and summarizes the advantages and limitations of each.
Table 1
Material Advantages Limitations
Pt
available wire, flat plate & tube
large range of sizes
Pt-Rh alloy for rigidity
low hydrogen overvoltage so cathodic potential range limited
expensive
Au
configurations same as Pt
larger cathodic potential range
larger cathodic potential range
anodic window limited by surface oxidation
expensive
Carbon
many types and configurations
good cathodic potential range
quality varies greatly
hard to shape
C-paste
wide potential range
low background current
inexpensive
unstable in flow cells
cannot use in organic solvents
Hg
excellent cathodic window
easy to “refresh”
forms amalgams
limited anodic window due to mercury oxidation
toxic
The approximate useful potential ranges for platinum, mercury, and carbon electrodes in aqueous electrolyte solutions along with those for platinum in a number of non-aqueous systems can be found as an appendix in Reference 2.
Solid electrodes for voltammetric measurements are most often fabricated by encapsulating the electrode material in a nonconducting sheath of glass or inert polymeric material like Teflon, Kel-F (poly-chlorotrifluoroethylene ) or PEEK (poly-etheretherketone). Most commonly, the exposed electrode material is in the form of a disk. Common commercially available disk diameters are 1.0, 3.0 and 10.0 mm. Electrodes of this size generally produce measured currents in the μA to low mA range for analytes at concentrations near 1 mM. Two common commercial sources of working electrodes, are ESA, Inc. (www.esainc.com)23 and Bioanalytical Systems, Inc. (www.bioanalytical.com)24.
Figure 35 shows examples of working electrodes from these two manufacturers. At left are macro-sized disk electrodes enclosed in non-conducting polymer, while at right, microelectrodes (see below) are shown that were fabricated by sealing wires of inert metals in glass insulating bodies.
Figure 35
Electrodes with diameters less than about 25 μm, termed microelectrodes or ultramicroelectrodes, advanced by R. M. Wightman and coworkers offer unique electrochemical characteristics. These include, in addition to their extreme small size, minimization of solution resistance effects and rapid response times. Important applications of these electrodes are in very high speed voltammetry (>10,000 V/s), electrochemistry in highly resistive solvents, and in vivo voltammetry. Electrodes can be fabricated in the low μm diameter range by sealing platinum or gold microwires, or carbon fibers in glass. Electrodes at these dimensions can also be produced by metal deposition or photolithography.
The electrochemical behavior of microelectrodes can appear markedly different from that observed at conventionally sized electrodes. To illustrate the difference, we will first consider the case of a planar electrode of mm dimensions in a cell with a volume of several mL. If a change is made in the applied potential, either a step or a sweep, from a value where no electron transfer occurs to a redox active solution species, to one in which electron transfer occurs, the concentration of the redox active species will be lowered at the electrode surface, leading to the formation of a concentration gradient (see chronoamperometry, Section II A, part 1-a for review). The longer the electrode is at a potential sufficient to allow electron transfer, the further away from the electrode into solution the concentration gradient extends.
The existence of the gradient induces diffusion of electroactive material from regions of high concentration (the bulk of the solution) to regions of low concentration (near the electrode surface). This diffusion can be described by Fick’s Laws, which take slightly different forms for changing electrode geometries. For the large planar electrode described here, the diffusion layer moves very quickly far out into the solution, exceeding the distance a molecule can diffuse on the time scale of a typical experiment. Under these conditions, diffusion from the bulk of solution where concentration is constant to the electrode surface is nearly all linear in nature, in a direction perpendicular to the electrode surface.
In cyclic voltammetry (see Section II A, part 2-b) these conditions typically give rise to the traditional peak-shaped voltammogram. A CV recorded for ferrocene at a 3 mm diameter glassy carbon disk electrode is shown at the left side of Figure 36. Ferrocene was present at a concentration of 0.6 mM in acetonitrile with 0.1 M tetrabutylammonium hexafluorophosphate as supporting electrolyte. The scan rate was 0.10 V/s.
Figure 36
Next consider a planar microelectrode with micrometer or smaller dimensions. At the right side of Figure 36, a voltammogram for 0.6 mM ferrocene recorded at a glassy carbon electrode with a diameter of only 10 μm is shown. With all of the other experimental conditions remaining the same, a sigmoidal rather than a peak-shaped voltammogram was observed. This was the result of a steady-state condition between diffusion and electron transfer, where the rate of diffusion matches the rate of electron transfer. Why the difference? Because of the small size of the electrode, the contribution to the current by diffusion from the edges of the electrode becomes important in the total mass transport of electroactive species. This edge effect, or radial diffusion is generally very small at large electrodes relative to the linear diffusion mentioned above. For microelectrodes, the flux per unit time and area is greater than for large electrodes because the region from which electroactive species diffuses to the surface is in essence hemispherical in shape.
It is important to understand that the voltammograms in Figure 36 are for one set of conditions. There are conditions in which a CV recorded at a large planar electrode will demonstrate steady-state behavior, and conditions for which peak-shaped voltammograms are seen at microelectrodes. The normalizing factor for all conditions is the experimental time. The approximate distance (in cm) a diffusing molecule can travel during a time period t (in seconds) is given by
$\mathrm{d = (2\, D_0\, t)^{1/2}}$
where D0 is the diffusion coefficient (cm2/s). When d is small relative to the radius of the electrode, linear diffusion will predominate, and the observed voltammogram will be peak shaped. For small electrode dimensions, d will frequently be large relative to the electrode radius, and a steady-state voltammogram will result.
While this has been only a cursory look at the differences between macro- and micro-sized electrodes, there are many excellent review articles available for readers desiring more in-depth information on this topic.25-28
3. Electrode Preparation
Ideally, a working electrode should behave reproducibly each time that it is used. Factors that affect the electrochemical behavior of a surface are its cleanliness, the kind and extent of chemical functionalities (including oxides) that are present, and the microstructure of the electrode material itself. Generally, a pretreatment step or steps will be carried out prior to each experiment to assure that the surface going into the electrochemical cell can be reproduced experiment-to-experiment. These may be as simple as mechanical polishing, and may include pre-scanning across a certain potential range or exposure to a solvent or chemical species to “activate” the electrode. Specific procedures for different electrode materials can be found in References 29-30 and those contained therein. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/05_Experimental_Hardware/C._Working_Electrodes.txt |
A potentiostat is a device that controls the potential applied to a working electrode (WE), relative to a reference electrode (RE), and measures the current flowing between the working electrode and a third electrode, called the auxiliary electrode (AE). These devices can be constructed using operational amplifiers (OAs), which are circuit components having dual inputs of very high impedance. One of the inputs inverts the polarity of the applied signal (the inverting input, -), while the other is noninverting (+) with respect to the input polarity.
Most practical applications of OAs use some form of signal feedback between the output and one of the inputs. The simplest such device is called a voltage follower, illustrated in Figure 37. OAs are represented in circuit diagrams as triangles.
Figure 37
The voltage follower detects any imbalance between the two inputs signals, and works diligently to maintain a null difference between them by adjusting the output voltage to an appropriate value. For the voltage follower illustrated above, eout = ein. When used in the design of a potentiostat, the high input impedance of the voltage follower is used to isolate the reference electrode and prevent appreciable current from passing through it.
A basic potentiostat constructed from three OAs is shown in Figure 38, with connections illustrated to the three electrodes located in a typical electrochemical cell.
Figure 38
In simple terms, this circuit is designed to “sense” the potential difference between the working electrode (WE) and the reference electrode (RE). This difference is measured at the voltage follower OA-2. The applied potential of the WE is changed by increasing or decreasing the input voltage (ein) at the control amplifier OA-1. Since the output of OA-2 is the feedback signal to OA-1, a change in ein away from the fixed potential of the reference electrode (eref when R1 = R2) causes OA-1 to force current into the electrochemical cell between the auxiliary electrode (AE) and the WE to balance the difference in the inputs at OA-1. The current flowing in the cell as a function of the changing potential output of the control amplifier can be measured as a voltage drop across the feedback resistor in OA-3 (i = eout / R3). OA-3 is configured as a current follower, whose output is generally amplified before it is sent to a recording or display device. Potential during a scan can be monitored at the output of OA-2. This monitored potential is the negative of the actual WE potential, as the WE is held at virtual ground when the current follower is used.
The input voltage, ein, to the control amplifier can be supplied by an external potential source or waveform generator. Commonly used input signals include potential steps, and either triangle or staircase type waveforms that increase/decrease in a linear fashion.
In the arrangement shown above, no current is allowed to pass to the RE because of the high input impedance of OA-2. This allows the activities of the reference half-cell components, and thus the reference potential, to remain unchanged during the experiment.
More information on the fabrication and operation of potentiostats can be found in References 31-32.
06 References and Links
1. Bockris, J. O’M.; Reddy, A. K. N. Modern Electrochemistry I; Plenum Press: New York, 1970; p 2.
2. Bard, A. J., Faulkner, L. R. Electrochemical Methods: Fundamentals and Applications, 2nd Edition; John Wiley & Sons, Inc.: New York, 2001.
3. Maloy, J. T. J. Chemical Education, 1983, 60(4), 285.
4. Gosser, D. K., Jr. Cyclic Voltammetry: Simulation and Analysis of Reaction Mechanisms; Wiley-VCH, Inc.: New York, 1993; p 46.
5. Rudolph, Manfred DigiElch, v. 3.1; ElchSoft: Kleinromstedt, Germany, 2007; www.elchsoft.com (accessed January 5, 2009).
6. Efstathiou, C. E. Educational Applets, University of Athens. http://www.chem.uoa.gr/applets/Apple...l_Diffus2.html (accessed January 5, 2009).
7. Nicholson, R. S. Analyt. Chem. 1965, 37, 1351.
8. Nicholson, R. S., Shain, I. Analyt. Chem. 1964, 36, 706.
9. op. cit, Gosser, p 50.
10. Bott, A. W. Current Separations, 1999 18:1, 9; http://currentseparations.com/issues/18-1/cs18-1b.pdf (accessed January 5, 2009).
11. Center for Research in Electrochemical Science and Technology, University of Cambridge; www.cheng.cam.ac.uk/research/...5html/cvr.html (accessed January 5, 2009).
12. Department of Biological Sciences, University of Paisley; www-biol.paisley.ac.uk/marco/...ltammetry1.htm (accessed January 6, 2009).
13. Heineman, W. R., Kissinger, P. T. Large Amplitude Controlled Potential Techniques in Laboratory Techniques in Electroanalytical Chemistry, 2nd Edition; Kissinger, P. T. and Heineman, W. R., Eds.; Marcel Dekker, Inc.: New York, 1996; pp 51-125.
14. Bott, A. W., Howell, J. O. Current Separations, 1991, 11:1/2, 21.
15. Branina, K., Neyman, E. Electroanalytical Stripping Methods; John Wiley & Sons: New York, 1993; pp 1-3.
16. Tercier, M. –L., Buffle, J. Electroanalysis, 1993, 5, 187.
17. Wang, J. Analytical Electrochemistry; Wiley-VCH: New York, 1994; p 45.
18. Cypress Systems, Inc. Anodic Stripping Voltammetry – Detection and Quantitation of Metal Contaminants; http://www.cypresssystems.com/Experiments/asv.html (accessed January 5, 2009).
19. Bioanalytical Systems, Inc. Pulse Voltammetric Techniques; www.basinc.com/mans/EC_epsilo...lse/pulse.html (accessed January 5, 2009).
20. op. cit, Wang, p 52.
21. Kissinger, P. T., Bott, A. W. Current Separations, 2002, 20, 51; http://currentseparations.com/issues/20-2/20-2d.pdf (accessed January 5, 2009).
22. Bott, A. W. Current Separations, 1995, 14, 64; http://currentseparations.com/issues/14-2/cs14-2d.pdf (accessed January 5, 2009).
23. ESA Bioscience, Inc. http://www.esainc.com (accessed January 5, 2009).
24. Bioanalytical Systems, Inc. www.bioanalytical.com (accessed January 5, 2009).
25. Wightman, R. M., Wipf, D. O. in Electroanalytical Chemistry, Bard, A, J., Ed.; Marcel Dekker: New York, 1989; Vol. 15, 267-353.
26. Howell, J. O. Current Separations, 1989, 8:1/2, 2.
27. Heinze, J. Angew. Chem. Int. Ed. Engl. 1993, 32, 1268-1288.
28. Wightman, R. M. Analyt. Chem. 1981, 53, 1125A-1134A.
29. Kuwana, T. Care and Feeding of Electrodes in ASDL Learning Module Electroanalytical Chemistry: A Laboratory Manual (Tech Notes), 2008, www.asdlib.org.
30. Bott, A. W. Current Separations, 1997, 16, 79; http://currentseparations.com/issues/16-3/cs16-3b.pdf (accessed January 5, 2009).
31. Doelling, Rudolf Potentiostats; Bank Elektronik Intelligent Controls; Pohlheim, Germany, 2000; http://www.bank-ic.de/encms/downloads/potstae2.pdf (accessed January 5, 2009).
32. Kissinger, P. T. Introduction to Analog Instrumentation; in Laboratory Techniques in Electroanalytical Chemistry, 2nd Edition, Kissinger, P. T. and Heineman, W. R., Eds., Marcel Dekker, Inc.: New York, 1996; pp 165-194. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/05_Experimental_Hardware/D._Potentiostats.txt |
CHEM355 is a junior level combined lecture/ laboratory, 4 credit course required for BA Chemistry, BA Biochemistry, BS Chemistry and BS Biochemistry majors. It can also be used towards the chemistry minor. Class sizes is small (typically 8 to 14 students). Analytical chemistry can be defined as the application of chemical principles to the characterization, identification and measurement of chemical species. It involves the study of instrumental and non-instrumental techniques in addition to statistical methods, calibration and validation, sampling strategies, and sample treatment.
Contributors and Attributions
• Neil Fitzgerald Assistant Dean of Science and Associate Professor of Chemistry, Luisa Li Instructional Designer (Marist College)
CHEM355: A Flipped Analytical Chemistry Course (Fitzgerald)
What makes it a flipped classroom course?
The ‘flipped’ refers to requiring students to learn course content at home and doing assigned problems in class. An advantage of a flipped course is that it allows for increased interaction between instructor and students. Students in science courses typically struggle with mathematically problem solving more than course content. The flipped approach allows me to work more with students on areas where they have difficulties.
How is class time organized?
Students are required to view a Prezi presentation (incorporating slides, simple animation, voice over and video) and review the relevant section of the textbook before class. The students are given the opportunity to identify areas of difficulty on a course Facebook page prior to the class. Class starts with a short quiz to test if students have reviewed the material appropriately. I will then spend a couple of minutes to review material that students identified as problematic and stress important concepts. Students then work through assigned problems in groups of three. Questions are provided on an online platform (Sapling Learning). The instructor circulates to help students and give hints to the class when necessary. Most questions are completed in class, any uncompleted questions are the responsibility of the student to complete before the deadline.
View my prezi presentation about the course here.
Pros and Cons of Flipped Courses
The flipped classroom is a form of blended learning which puts the emphasis of learning course material on the students in order to free up time in the classroom for active learning activities. As with any teaching method, there are pros and cons of the approach:
Pros:
• Increased interaction between instructor and students. Traditional lectures tend to be passive learning experiences with little interaction between instructor and students.
• Allows more problem solving and active learning experiences. Education research suggests that these experiences improve student learning.
• More effective use of instructor time. Instructors can spend more time helping students with activities they find challenging and less time teaching content students may find easily understandable.
• Earlier intervention. The flipped classroom allows learning difficulties to be recognized earlier, whether class-wide or individual students.
• Develops teamwork. Students can be organized in groups for classroom activities.
• Peer learning. Students help one another which is beneficial for students of all abilities.
Cons:
• Preparation time. To prepare an effective flipped classroom course can take significantly more time than a traditional lecture.
• Student response. Although I haven’t seen it in my course, the “I had to teach myself” criticism in likely.
• Effectively replacing the lecture. Lectures are effective mechanisms for providing information. The difficulty is how to effectively replace them. Effective use of technology can help here.
• Ensuring compliance. How do we make sure students are completing assigned work in their own time. Quizzes can help. Ultimately students are hurting themselves by not completeting assigned work.
• Class size. My class is relatively small. Effective active learning for large classes may require different strategies
Quick Guide to Prezi
Prezi is cloud-based presentation software using a story telling approach to present ideas on a virtual canvas. The main advantage of prezi is in the ability to present material in a non-linear way (in contrast to Powerpoint). Material can be presented as a concept map and students are able to “dip in” at any point. Prezi allows for voice overs, simple animation, videos, web links and images in addition to text.
Chapter 01: Introduction
An introduction to analytical chemistry and the analytical process by Neil Fitzgerald.
Chapter 02: Basic Tools of Analytical Chemistr
Numbers in analytical chemistry, Concentration units, basic laboratory equipment, and preparing solutions by Neil Fitzgerald.
Chapter 03: The Language of Analytical Chemist
Definitions, selecting an analytical method, developing the procedure by Neil Fitzgerald.
Chapter 04: Evaluating Analytical Data
Descriptive statistics, propagation of uncertainty, distribution of measurements, Hypothesis testing by Neil Fitzgerald.
Chapter 05: Calibrations Standardizations and
External, internal and standard additions methods. Regression analysis and blank corrections by Neil Fitzgerald.
Chapter 06: Equilibrium Chemistry
Reversible reactions and thermodynamics, equilibrium constants for chemical reactions, Ladder diagrams, solving equilibrium problems, buffer solutions, activity by Neil Fitzgerald.
Chapter 07: Collecting and Preparing Samples
Designing a sampling plan, implementing a sampling plan, separations by Neil Fitzgerald.
Chapter 08: Gravimetric Methods
Precipitation Gravimetry, volatilization gravimetry, particulate gravimetry by Neil Fitzgerald.
Chapter 09: Titrimetric Methods
Overview, acid-base titrations, complexation titrations, redox titrations, precipitation titrations by Neil Fitzgerald.
Chapter 10: Spectroscopic Methods
Absorption, emission and scattering methods including UV/Vis, IR, atomic absorption, photoluminescence, atomic emission, turbidimetry and nephelometry by Neil Fitzgerald.
Chapter 11: Electrochemical Methods
Overview, Potentiometric Methods, Coulometric Methods, Voltammetric and Amperometric Methods by Neil Fitzgerald.
Chapter 12: Chromatographic and Electrophoreti
General theory, optimizing separations, GC, HPLC, electrophoresis and other chromatographic methods by Neil Fitzgerald.
Chapter 13: Kinetic Methods
Chemical kinetics, radiochemistry, flow injection analysis by Neil Fitzgerald. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/CHEM355%3A_A_Flipped_Analytical_Chemistry_Course_(Fitzgerald)/01_Prezi_presentation_-_Flipped_Course_Workshop.txt |
Separation Science - Equilibrium Unit
Thomas Wenzel
Department of Chemistry
Bates College, Lewiston ME 04240
[email protected]
The following textual material is designed to accompany a series of in-class problem sets that develop many of the fundamental aspects of chemical equilibrium calculations.
02 Text
Suppose you are a chemist involved in developing a new product for a small manufacturing company. Part of the process leads to the formation of the compound lead phosphate. The lead phosphate will end up in the wastewater from the process. Since you are a small facility, instead of having your own wastewater treatment plant, you will discharge the wastewater to the local municipal wastewater treatment plant. The municipal wastewater treatment plant faces strict requirements on the amount of lead that is permitted in their end products. A wastewater treatment plant ends up with "clean" water and a solid sludge. Most lead ends up in the sludge, and the Environmental Protection Agency has set a limit on how much lead is permitted in the sludge. Most municipalities will require you to enter into a pre-treatment agreement, under which you will need to remove the lead before discharging to the plant. For example, the City of Lewiston, Maine will require you to discharge a material that contains no more than 0.50 mg of total lead per liter.
Lead phosphate is a sparingly soluble material so most of it will actually be a solid in your waste, thereby allowing you to filter it out before discharge to the treatment plant.
What is the concentration of total dissolved lead in the discharge from your facility?
What we need to consider here is the reaction that describes the solubility of lead phosphate. Lead phosphate has the formula Pb3(PO4)2, and the accepted practice for writing the solubility reaction of a sparingly soluble compound that will dissociate into a cation and anion is shown. The solid is always shown on the left, or reactant, side. The dissolved ions are always shown on the product side.
$\ce{Pb3(PO4)2(s) \leftrightarrow 3Pb^2+(aq) + 2PO4^3- (aq)}$
Next, we can write the equilibrium constant expression for this reaction, which is as follows:
$\ce{K_{sp} \:=\: [Pb^2+]^3[PO4^3- ]^2}$
This general equilibrium constant expression for a sparingly soluble, ionic compound is known as the solubility product, or Ksp. Note that there is no term for the solid lead phosphate in the expression. One way to view this is that a solid really cannot have variable concentrations (moles/liter) and is therefore not important to the expression. Ksp values have been measured for many substances and tables of these numbers are available. The Ksp for lead phosphate is known and is 8.1$\times$10-47. What this means is that any solution that is in contact with solid lead phosphate will have a solubility product ($\ce{[Pb^2+ ]^3[PO4^3- ]^2}$) that exactly equals its Ksp (8.1$\times$10-47).
There is a complication to this process though. It turns out that the phosphate ion is a species that appears in the dissociation reactions for a substance known as phosphoric acid (H3PO4). Acids and their corresponding conjugate bases are very important in chemistry and the properties of many acids and bases have been studied. What can happen in this case is that the phosphate ion can undergo a set of stepwise protonations, as shown below.
\begin{align} \ce{Pb3(PO4)2 \leftrightarrow 3Pb^2+ +\: &2PO4^3- \nonumber\ &\:\:\:\Updownarrow\nonumber\ &HPO4^2- \nonumber\ &\:\:\:\Updownarrow\nonumber\ &H2PO4^- \nonumber\ &\:\:\:\Updownarrow\nonumber\ &H3PO4\nonumber} \end{align}
If we wanted to calculate the solubility of lead phosphate in water, we would need to consider the effect of protonation of the phosphate on the solubility. Remember, the Ksp expression only includes terms for Pb2+ and $\ce{PO4^3-}$, and it is the product of these two that must always equal Ksp if some solid lead phosphate is in the mixture. Protonation of the phosphate will reduce the concentration of $\ce{PO4^3-}$. If the concentration of $\ce{PO4^3-}$ is reduced, more of the lead phosphate must dissolve to maintain Ksp.
We can look up relevant equilibrium constants for the dissociation of phosphoric acid. There is an accepted practice in chemistry for the way in which these reactions are written, and the series for phosphoric acid is shown below. This describes the chemistry of an acid and the equilibrium constant expressions are known as Ka values, or acid dissociation constants.
\begin{alignat}{3} \ce{&H3PO4 &&+ H2O &&\leftrightarrow H2PO4- &&+ H3O+} &&\hspace{40px}\mathrm{K_{a1}}\ &\ce{H2PO4- &&+ H2O &&\leftrightarrow HPO4^2- &&+ H3O+} &&\hspace{40px}\mathrm{K_{a2}}\ &\ce{HPO4^2- &&+ H2O &&\leftrightarrow PO4^3- &&+ H3O+} &&\hspace{40px}\mathrm{K_{a3}} \end{alignat}
$\mathrm{K_{a1} = \dfrac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]} \hspace{100px} K_{a2} = \dfrac{[HPO_4^{2-}][H_3O^+]}{[H_2PO_4^-]}}$
$\mathrm{K_{a3} = \dfrac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]}}$
But before we can proceed, there is still one other complication to this process. It turns out that the lead cation has the possibility of forming complexes with other anions in solution. One such anion that is always present in water is hydroxide ($\ce{OH-}$). The hydroxide complex could be another insoluble one with lead. More important, though, is whether lead can form water-soluble complexes with the hydroxide ion. A species that complexes with a metal ion is known as a ligand. It turns out that hydroxide can form water-soluble complexes with lead ions, and that there are three of them that form in a stepwise manner. The equations to represent this are always written with the metal ion and ligand on the reactant side and the complex on the product side, as shown below.
\begin{align} \ce{&Pb^2+ (aq) + OH- (aq) \leftrightarrow Pb(OH)+ (aq)} &&\mathrm{K_{f1}}\ &\ce{Pb(OH)+ (aq) + OH- (aq) \leftrightarrow Pb(OH)2(aq)} && \mathrm{K_{f2}}\ &\ce{Pb(OH)2 (aq) + OH- (aq) \leftrightarrow Pb(OH)3- (aq)} &&\mathrm{K_{f3}} \end{align}
The equilibrium constant expressions are shown below, and these are known as formation constants (Kf).
$\mathrm{K_{f1} = \dfrac{[Pb(OH)^+]}{[Pb^{2+}][OH^-]} \hspace{100px} K_{f2} = \dfrac{[Pb(OH)_2]}{[Pb(OH)^+][OH^-]}}$
$\mathrm{K_{f3} = \dfrac{[Pb(OH)_3^-]}{[Pb(OH)_2][OH^-]}}$
The important thing to realize is that any complexation of lead ions by hydroxide will lower the concentration of Pb2+. Since [Pb2+] is the concentration in the Ksp expression, complexation of lead ions by hydroxide will cause more lead phosphate to dissolve to maintain Ksp. Since all soluble forms of lead are toxic, this increase in lead concentration is a potential problem. We can now couple these reactions into our scheme that describes the solubility of lead phosphate in this solution.
\begin{align} \ce{Pb3(PO4)2 \:\:\:\:\:\:\leftrightarrow\:\:\:\:\:\: &3Pb^2+ \:\:\:\:\:\:+ &&2PO4^3- \nonumber\ &\:\:\:\Updownarrow &&\:\:\:\Updownarrow\nonumber\ &Pb(OH)+ &&HPO4^2- \nonumber\ &\:\:\:\Updownarrow &&\:\:\:\Updownarrow\nonumber\ &Pb(OH)2 &&H2PO4- \nonumber\ &\:\:\:\Updownarrow &&\:\:\:\Updownarrow\nonumber\ &Pb(OH)3- &&H3PO4\nonumber} \end{align}
This is now quite a complicated set of simultaneous reactions that take place. Our goal in the equilibrium unit of this course will be to develop the facility to handle these types of complicated problems.
Before we get started into this process, there are a couple of other general things to know about chemical equilibrium. Consider the general reaction shown below.
$\ce{aA + bB \leftrightarrow cC + dD}$
One way of describing equilibrium is to say that the concentrations do not change. The concentrations of the species in this solution represent a macroscopic parameter of the system, and so at the macroscopic level, this system is static.
Another way of describing equilibrium is to say that for every forward reaction there is a corresponding reverse reaction. This means at the microscopic level that As and Bs are constantly converting to Cs and Ds and vice versa, but that the rate of these two processes are equal. At the microscopic level, a system at equilibrium is dynamic.
Unless you have taken physical chemistry, I am fairly certain that everything you have learned until this point has taught you that the following expression can be used to describe the equilibrium state of this reaction.
$\mathrm{K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}}$
Well it turns out that this expression is not rigorously correct. Instead of the concentrations of reagents, the actual terms we need in an equilibrium constant expression are the activities of the substances. The expression shown below is the correct form of the equilibrium constant, in which aA represents the activity of substance A.
$\mathrm{K=\dfrac{[a_C]^c[a_D]^d}{[a_A]^a[a_B]^b}}$
If you examine the group of As and Bs below, hopefully you can appreciate that the A shown in boldface is “inactive”. For that A species to react with a B, another A species must move out of the way.
A B A A A B B A
If the correct form of the equilibrium constant expression uses the activities of the chemicals, why have you always been taught to use concentrations? It turns out that in most situations we do not have reliable procedures to accurately calculate the activities of substances. If we did, we would almost certainly use the correct form of the expression. Since we do not know how to evaluate the activities of substances under most circumstances, we do the next best thing and use concentrations as an approximation. This means that all equilibrium calculations are at best approximations (some better than others). In other words, equilibrium calculations usually provide estimations of the situation, but not rigorously correct answers. Because the entire premise is based on an approximation, this will often allow us to make other approximations when we perform equilibrium calculations. These approximations will usually involve ignoring the contributions of minor constituents of the solution.
One last thing we ought to consider is when the approximation of using concentration instead of activity is most valid. Perhaps a way to see this is to consider a solution that has lots of A (the concentration of A is high) and only a small amount of B (the concentration of B is low). Inactivity results if a similar species is in the way of the two reactants getting together. Since the concentration of B is low, there is very little probability that one B would get in the way of another and prevent it from encountering an A. For A, on the other hand, there are so many that they are likely to get in each other’s way from being able to encounter a B. Concentration is a better approximation of activity at low concentrations. The example I have shown with A and B implies there is no solvent, but this trend holds as well if the substances are dissolved in a solvent. Notice as well that the activity can never be higher than the concentration, but only lower.
How low a concentration do we need to feel fully comfortable in using the approximation of concentrations for activities? A general rule of thumb is if the concentrations are less than 0.01 M then the approximation is quite a good one. Many solutions we will handle this term will have concentrations lower than 0.01 M, but many others will not. We do not need to dwell excessively on this point, but it is worth keeping in the back of one’s mind that calculations of solutions with relatively high concentrations are always approximations. We are getting a ballpark figure that lets us know whether a particular process we want to use or study is viable. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/01_Overview%3A_Significance_of_Chemical_Equilibrium.txt |
In-class Problem Set #1
Unless specifically told otherwise, whenever a problem lists a concentration, that is the value of material added to solution prior to any reactions occurring to achieve equilibrium. So in the first problem below, 0.155 moles of ammonia were dissolved in 1 liter of solution. The final concentration of ammonia would be something less than 0.155 moles/liter provided some form of equilibration occurred.
1. Calculate the pH of a solution that is 0.155 M in ammonia.
The first step in any equilibrium problem is an assessment of the relevant chemical reactions that occur in the solution. To determine the relevant reactions, one must examine the specie(s) given in the problem and determine which types of reactions might apply. In particular, we want to consider the possibility of acid-base reactions, solubility of sparingly soluble solids, or formation of water-soluble metal complexes.
When given the name of a compound (e.g., ammonia), it is essential that we know or find out the molecular formula for the compound, and often times we have to look this up in a book or table. The molecular formula for ammonia is NH3. Ammonia can be viewed as the building block for a large family of similar compounds called amines in which one or more of the hydrogen atoms are replaced with other functional groups (a functional group is essentially a cluster of atoms - most of these are carbon-containing clusters). For example, the three compounds below result from replacing the hydrogen atoms of ammonia with methyl (CH3) groups.
CH3NH2 Methyl amine
(CH3)2NH Dimethyl amine
(CH3)3N Trimethyl amine
Amines and many other organic, nitrogen-containing compounds constitute one of the major families of bases. Ammonia is therefore a base.
Bases undergo a very specific reaction with water to produce the hydroxide ion. The appropriate reaction needed to describe what will happen when ammonia is mixed with water is shown below.
$\ce{NH3 + H2O \leftrightarrow NH4+ + OH-}$
We can describe this reaction by saying that ammonia reacts with water to produce the ammonium cation and hydroxide anion.
Now that we know the reaction that describes the system, we have to ask what K expression is used to represent that particular reaction. For the reaction of a base, we need an equilibrium constant known as Kb. The expression for Kb is shown below.
$\mathrm{K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]}}$
If we examine the tables of equilibrium constants, though, we observe that the table does not list Kb values, but instead only lists Ka values for substances. A species that is in the reaction that we do find a Ka value for in the table is the ammonium cation. It is important to note that the species ammonia and ammonium differ by only a hydrogen ion.
$\ce{NH3/NH4+}$
Species that differ from each other by only a hydrogen ion are said to be a conjugate pair. A conjugate pair always contains a base (ammonia in this case) and an acid (ammonium in this case). The acid is always the form with the extra hydrogen ion. The base is the form without the extra hydrogen ion.
The Ka reaction is that of the ammonium ion acting as an acid.
$\ce{NH4+ + H2O \leftrightarrow NH3 + H3O+}$
The equilibrium constant expression for Ka is shown below.
$\mathrm{K_a = \dfrac{[NH_3][H_3O^+]}{[NH_4^+]}}$
Furthermore, the Kb and Ka values for the base and acid form respectively of a conjugate pair have a very specific relationship that is shown below.
$\mathrm{K_b \times K_a = K_w = 1 \times 10^{-14}}$
Remember, Kw is the equilibrium expression that describes the autoprotolysis of water.
$\ce{H2O + H2O \leftrightarrow H3O+ + OH-}$
$\mathrm{K_w = [H_3O^+][OH^-] = 1\times10^{-14}}$
The expression below shows that the result of multiplying Kb times Ka is actually Kw
$\mathrm{K_b \times K_a = \dfrac{[NH_4^+][OH^-]}{[NH_3]} \times \dfrac{[NH_3][H_3O^+]}{[NH_4^+]} = [OH^-][H_3O^+] = K_w}$
Now that the Kb value is known, it is finally possible to solve for the pH of the solution of ammonia. A useful way to keep track of such problems is to use the reaction as the headings for columns of values that describe the concentrations of species under certain conditions. The first set of numbers represents the initial concentrations in solution prior to any equilibration.
\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\nonumber\ &\ce{Initial} &&0.155 &&0 &&10^{-7}\nonumber \end{align}
We do not need an initial value for water since it’s the solvent. The hydroxide is given a value of 10-7 M because of the autoprotolysis of the water. The second set of numbers are expressions for the equilibrium concentrations of the species. In this case, we want to keep in mind that the value for Kb is small, meaning we do not expect that much product to form.
\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\nonumber\ &\ce{Initial} &&0.155 &&0 &&10^{-7}\nonumber\ &\ce{Equilibrium} &&0.155 - \ce{x} &&\ce{x} &&10^{-7} + \ce{x}\nonumber \end{align}
If we wanted, these values could now be plugged into the Kb expression and it could be solved using a quadratic. There may be a way to simplify the problem, though, if we keep in mind that Kb is so small. In this case, we expect the value of x to be small and we can make two approximations.
The first is that x $\ll$ 0.155 so that (0.155 – x) = 0.155
The second is that x $\gg$ 10-7 so that (10-7 + x) = x
\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\nonumber\ &\ce{Initial} &&0.155 &&0 &&10^{-7}\nonumber\ &\ce{Equilibrium} &&0.155 - \ce{x} &&\ce{x} &&10^{-7} + \ce{x}\nonumber\ &\ce{Approximation} &&0.155 &&\ce{x} &&\ce{x}\nonumber \end{align}
Now we can plug the approximations in the Kb expression and solve for the value of x.
$\mathrm{K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]} = \dfrac{(x)(x)}{0.155} = 1.76 \times 10^{-5}}$
$\mathrm{x = [OH^-] = 1.65\times 10^{-3}}$
Before we can use this to calculate the concentration of H3O+ and solve for pH, we first must check the two approximations to make sure they are both valid.
$\dfrac{1.65\times 10^{-3}}{0.155} \times 100 = 1.1\% \hspace{80px} \dfrac{10^{-7}}{1.65\times 10^{-3}} \times 100 = 0.0061\%$
It is worth noting that the assumption that the initial hydroxide or hydronium ion can be ignored is almost always made in these problems. The only two instances in which this approximation would break down are if:
1. the acid or base is exceptionally weak so that so little dissociation occurs that the initial amount is significant or
2. the acid or base is so dilute that very little dissociation occurs.
Since both approximations are less than 5%, the concentration of H3O+ can be calculated using the Kw expression and the pH can be calculated.
[H3O+] = 6.31$\times$10-12
pH = 11.2
02 Nomenclature of Acids and Bases
Before continuing on to more problems, it is useful to consider some general rules for the nomenclature of species common to acid-base systems.
The names of species with a positive charge (cations) almost always end with an “ium” ending.
NH3 was ammonia. Its protonated ion (\(\ce{NH4+}\)) is called the ammonium ion.
Earlier the species methyl amine (CH3NH2) was mentioned. The protonated form of this (\(\ce{CH3NH3+}\)) would be the methyl ammonium ion.
When you name the protonated form of a base, the scheme is to remove the last vowel (which is usually an“e”) and replace it with “ium”.
The protonated form of aniline, a base, would be anilinium.
The elements sodium and calcium are found in nature as the Na+ and Ca2+ ion respectively.
We can therefore state that the protonated form of “wenzel” would be “wenzelium”.
The names of most species with a negative charge (anions) end with an “ate” ending.
H2SO4 is sulfuric acid, whereas \(\ce{SO4^2-}\) is the sulfate ion.
Butyric acid (CH3CH2CH2COOH) has the smell of dirty socks. CH3CH2CH2COOis the butyrate ion.
The general rule is to drop the “ic” ending of the name of the acid and replace it with “ate”.
When in doubt, if you need the name of the anion, add an “ate” ending. The anion of “wenzel” is therefore “wenzelate”.
There are other endings in the nomenclature for anions besides the “ate” ending. For example, we are quite familiar with the “ide” ending that occurs with the halides (e.g., fluoride, chloride, bromide, and iodide). There are other anions that are named using an “ite” ending (e.g., nitrite, sulfite). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/01_Solution_of_a_Weak_Base.txt |
2. Calculate the pH of a solution that is 0.147 M in pyridine and 0.189 M in pyridinium chloride.
The first step in any equilibrium problem is to determine a reaction that describes the system. This system has appreciable quantities of both pyridine (Py) and pyridinium chloride. The structure of pyridine is shown below and is a base.
As a base it could undergo the following reaction (note that this is the Kb reaction).
$\ce{Py + H2O \leftrightarrow PyH+ + OH-}$
The structure of pyridinium chloride is shown below. It is important to realize that when added to water, the pyridinium and chloride ions will separate from each other such that the ions will be solvated by water (the pyridinium ion will have the negative oxygen atoms directed toward it, the chloride ion will have the positive hydrogen atoms of the water directed toward it)
We can write potential reactions for both the pyridinium and chloride ions reacting with water as follows.
$\ce{PyH+ + H2O \leftrightarrow Py + H3O+}$
Note that this is the Ka reaction for pyridinium. Looking in the table of values shows a pKa of 5.22. This means that pyridinium is a weak acid.
$\ce{Cl- + H2O \leftrightarrow HCl + OH-}$
Note that this is the Kb reaction for chloride. Chloride is the conjugate base of hydrochloric acid. Looking up hydrochloric acid in the table shows that hydrochloric acid is a strong acid, meaning that it reacts essentially 100% in water to produce Cl and H3O+. Because of this, the reaction above of chloride with water to produce HCl and hydroxide ion will not occur and can be ignored.
At this point it seems we have two reactions (the Kb reaction for pyridine producing pyridinium and hydroxide being one, the Ka reaction for pyridinium producing pyridine and hydronium being the other) that describe the system. As a test, let's do the calculation using both possible reactions.
Using pyridine acting as a base (pKb = 8.78, Kb = 1.66$\times$10-9):
\begin{align} & &&\ce{Py}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{PyH+} \hspace{25px} + &&\ce{OH-}\nonumber\ &\ce{Initial} &&0.147 &&0.189 &&0\nonumber\ &\ce{Equilibrium} &&0.147 - \ce{x} &&0.189+\ce{x} && \ce{x}\nonumber\ &\ce{Approximation} &&0.147 &&0.189 &&\ce{x}\nonumber \end{align}
Note that the initial amount of hydroxide, which is set at 0, assumes that the amount that will be produced is significant compared to 10-7 M. Also, the approximations can be attempted since the value of Kb is small.
The approximations can now be plugged into the Kb expression and x evaluated.
$\mathrm{K_b = \dfrac{[PyH^+][OH^-]}{[Py]} = \dfrac{(0.189)(x)}{0.147} = 1.66\times10^{-9}}$
$\mathrm{x = [OH^-] = 1.29\times10^{-9}}$
However, we must first check the approximation before calculating the pH.
$\dfrac{1.29\times10^{-9}}{0.147} \times 100 = 8.77 \times 10^{-7}\% \hspace{60px} \dfrac{1.29\times10^{-9}}{0.189} \times 100 = 6.83 \times 10^{-7}\%$
These approximations are both valid. However, if you consider that we ignored the initial amount of hydroxide present from the autoprotolysis of water (10-7 M), this would seem to be in error because of the low level of hydroxide (1.29$\times$10-9 M). For the moment, let’s just move ahead assuming it was okay to ignore the autoprotolysis of water, and more will be said later about the appropriateness of this decision. The concentration of hydronium ion and pH can be calculated.
[H3O+] = 7.75$\times$10-6 pH = 5.11
Using pyridinium acting as an acid (pKa = 5.22, Ka = 6.03$\times$10-6):
\begin{align} & &&\ce{PyH+}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{Py} \hspace{25px} + &&\ce{H3O+}\nonumber\ &\ce{Initial} &&0.189 &&0.147 &&0\nonumber\ &\ce{Equilibrium} &&0.189 - \ce{x} &&0.147+\ce{x} && \ce{x}\nonumber\ &\ce{Approximation} &&0.189 &&0.147 &&\ce{x}\nonumber \end{align}
The approximations can now be plugged into the Ka expression and x evaluated.
$\mathrm{K_a = \dfrac{[Py][H_3O^+]}{[PyH^+]} = \dfrac{(0.147)(x)}{0.189} = 6.03 \times 10^{-6}}$
$\mathrm{x = [H_3O^+] = 7.75\times10^{-6}}$
However, we must first check the approximation before calculating the pH.
$\dfrac{7.75\times10^{-6}}{0.147} \times 100 = 0.00527\% \hspace{60px} \dfrac{7.75\times10^{-6}}{0.189} \times 100 = 0.00410\%$
In this case, we can also examine whether it was appropriate to ignore the hydronium ion concentration from the autoprotolysis of water.
$\dfrac{1.0\times10^{-7}}{7.75\times10^{-6}} \times 100 = 1.29\%$
In this case (unlike with the pyridine acting as a base), ignoring the autoprotolysis of water is appropriate.
Since all of the approximations are valid, we can use the hydronium ion concentration to calculate the pH.
[H3O+] = 7.75$\times$10-6 pH = 5.11
What is important to realize that we get the same pH (5.11) using either the Ka or Kb equation. These two answers are reassuring but also problematic. The reassuring part is that a solution can only have one pH. If either of the two reactions can be used to describe the system, then both ought to give the same answer for the pH. But one reaction has pyridine acting as a base, another pyridinium acting as an acid. Which one is actually correct? The way to assess that is to examine the relative values of Ka for the conjugate acid and Kb for the conjugate base. In this case, the Ka for the acid is about 1,000 times larger than the Kb for the base. Because of that, a small amount of the acid would dissociate to the base. And note, we did get a pH that was acidic for the answer in each case. But it really does not matter since the amount of change is so small that it can be ignored.
However, there is something very important to realize about this system. A solution with appreciable concentrations of both members of a conjugate pair is known as a buffer. Buffers are solutions that resist changes in pH. This resistance is created by having both members of the conjugate pair.
If acid is added, the base component of the conjugate pair reacts to form the conjugate acid.
If base is added, the acid component of the conjugate pair reacts to form the conjugate base.
As long as the concentration of the buffer components are not excessively dilute (on the order of 10-6 M or lower), a buffer controls the pH of the system and in buffer solutions we can always ignore the initial concentration of hydronium or hydroxide ion from the autoprotolysis of water. A convenient way to calculate the pH of a buffer is to use what is known as the Henderson-Hasselbalch equation. This equation can be derived from the Ka expression.
Ka expression:
$\mathrm{K_a = \dfrac{[Py][H_3O^+]}{[PyH^+]}}$
Take the negative logarithm of both sides:
$\mathrm{-\log K_a = -\log \left(\dfrac{[Py][H_3O^+]}{[PyH^+]} \right )}$
Rearrange the right hand side using the properties of logs:
$\mathrm{-\log K_a = -\log \left(\dfrac{[Py]}{[PyH^+]} \right ) - \log[H_3O^+]}$
Remember that:
–log(Ka) = pKa
–log[H3O+] = pH
Substituting these in gives:
$\mathrm{pK_a = -\log \left(\dfrac{[Py]}{[PyH^+]} \right ) + pH}$
Rearranging gives the final form of the Henderson-Hasselbalch equation:
$\mathrm{pH = pK_a + \log \left(\dfrac{[Py]}{[PyH^+]} \right )}$
If we substitute in the values for this problem (and note, with a buffer we will be able to ignore any redistribution of the appreciable amount of the two species), we get:
$\mathrm{pH = pK_a + \log \left(\dfrac{[Py]}{[PyH^+]} \right ) = 5.22 + \log\left(\dfrac{0.147}{0.189} \right ) = 5.11}$
This is the same answer we got using either the Ka or Kb expressions.
We can also write two generalized forms of the Henderson-Hasselbalch equation for the two generalized types of weak acid/weak base buffer solutions (the generalized formulas for a weak acid, HA and BH+).
$\ce{HA + H2O \leftrightarrow A- + H3O+}$
$\mathrm{pH = pK_a + \log\left(\dfrac{[A^-]}{[HA]} \right )}$
$\ce{BH+ + H2O \leftrightarrow B + H3O+}$
$\mathrm{pH = pK_a + \log\left(\dfrac{[B]}{[BH^+]} \right )}$
Earlier we said that a buffer is effective at controlling the pH because the acid form of the conjugate pair can neutralize bases and the base form can neutralize acids. Examining the Henderson-Hasselbalch equation also allows us to appreciate from a quantitative sense how buffers are able to control the pH of a solution. If you look at this equation, you notice that the pH is expressed as a constant (pKa) that then varies by the log of a ratio. One thing to note about log terms is that they change rather slowly. Someone who offers you the log of a million dollars is not being very generous with their money. It takes a very large change in the ratio of the two concentrations to make a large difference in the log term. This large change will only occur when one of the two components of the buffer gets used up by virtue of the acid or base that is being added.
04 Solution of a Weak Acid
3. Calculate the pH of a solution that is 0.332 M in anilinium iodide.
The anilinium ion is in the table of Ka values and is a weak acid (pKa = 4.596). Anilinium iodide could be formed by the reaction between aniline (the conjugate base of anilinium) and hydrogen iodide, as shown below.
$\ce{An + HI \leftrightarrow AnH+I-}$
In water, anilinium iodide will dissociate to produce the anilinium cation and the iodide anion. What must now be assessed is whether either of these ions will react with water. The two possible reactions that could occur are shown below.
$\ce{AnH+ + H2O \leftrightarrow An + H3O+}$
$\ce{I- + H2O \leftrightarrow HI + OH-}$
The anilinium ion is behaving as an acid and since it has a pKa value in the table (4.596), this reaction will occur. The iodide ion is acting as a base. To see if this reaction occurs, we would need to look up hydroiodic acid (HI) in the table, and see that it is a strong acid. The important feature of strong acids is that, for all practical purposes, strong acids go 100% to completion. This means that HI in water will dissociate 100% to H3O+ and I. Actually, some amount of undissociated HI must remain, but it is so small that we never need to consider it under normal circumstances in water. Regarding I acting as a base, this means that it will all stay as I and no HI will form as shown above. We can therefore solve the answer to this problem by only using the reaction of the anilinium ion.
The procedure is rather analogous to what we have already used in problems 1 and 2 above. We ought to write a table for initial values, equilibrium values, and then examine whether any assumptions can be made. If x is the amount of AnH+ that reacts, there are two important assumptions that can be made in this problem. One is that, because Ka is so small, very little of the AnH+ reacts so that 0.332 $\gg$ x. However, enough AnH+ reacts to produce a much larger concentration of H3O+ than was initially in solution such that x $\gg$ 10-7 M.
\begin{align} & &&\ce{AnH+}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{An} \hspace{25px} + &&\ce{H3O+}\nonumber\ &\ce{Initial\: amount} &&0.332 &&0 &&10^{-7}\nonumber\ &\ce{Equilibrium} &&0.332 - \ce{x} &&\ce{x} &&\ce{x}+10^{-7}\nonumber\ &\ce{Assumption} &&0.332 \gg \ce{x} && &&\ce{x}\gg10^{-7}\nonumber\ &\ce{Approximation} &&0.332 &&\ce{x} &&\ce{x}\nonumber \end{align}
These values can now be plugged into the equilibrium constant expression for the reaction.
$\mathrm{K_a = \dfrac{[An][H_3O^+]}{[AnH^+]} = \dfrac{(x)(x)}{0.332} = 2.54 \times 10^{-5}}$
$\mathrm{x = [H_3O^+] = 2.9\times 10^{-3}}$
$\mathrm{pH = 2.54}$
Both approximations must now be checked for validity.
$\dfrac{2.93\times 10^{-3}}{0.332}\times 100 = 0.88\% \hspace{60px} \dfrac{10^{-7}}{2.93\times 10^{-3}} \times 100 = 0.0034\%$
Both are okay, so the pH we calculated above is correct. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/03_Solution_Containing_a_Conjugate_Pair_%28Buffer%29.txt |
4. Calculate the pH of a solution that is prepared by mixing 45 ml of 0.224 M 3-chlorobenzoic acid with 30 ml of 0.187 M ethylamine.
Chlorobenzoic acid (Hcba) is a weak acid with a pKa value of 3.824. Ethylamine is not in the table, but ethylammonium, its conjugate acid, is (pKa = 10.63). Therefore ethylamine (EA) is a weak base (pKb = 3.37). This solution consists of a mixture of a weak acid and a weak base.
What happens when we mix an acid with a base? From prior material we should know that an acid and a base react with each other in what is known as a neutralization reaction. The neutralization reaction between chlorobenzoic acid and ethylamine is shown below.
$\ce{Hcba + EA \leftrightarrow cba- + EAH+} \hspace{60px} \mathrm{K_n}$
We can calculate initial amounts of Hcba and EA that exist in solution, but some of these will react according to the neutralization reaction. What we need to know is the extent of the neutralization reaction, in other words the value of K for this reaction. There are no tables of Kn values so what we need to do is see if there is a way to come up with the Kn expression by adding up a series of reactions that we do have K values for.
We do have reactions for Hcba and EA that we can look up in the table. These are as follows:
\begin{align} &\ce{Hcba + H2O \leftrightarrow cba- + H3O+} \hspace{60px} &&\mathrm{K_a\: of\: Hcba}\ &\ce{EA + H_2O \leftrightarrow EAH+ + OH-} \hspace{60px} &&\mathrm{K_b\: of\: EA} \end{align}
Adding these two together produces the following reaction:
$\ce{Hcba + EA + 2H2O \leftrightarrow cba- + EAH+ + H3O+ + OH-} \hspace{40px} \mathrm{K = K_a(acid) \times K_b(base)}$
This almost looks like Kn but it is not exactly the same. The reactant side has two water molecules, and the product side has the hydronium and hydroxide ion. Note that these species do not show up in the neutralization reaction above. As tempting as it might be to say hydronium and hydroxide will react to produce the water molecules (thereby just cancelling these out and ignoring them), they are real species in the reaction that need to be accounted for in the final form of Kn. The way to eliminate these would be to add in the following reaction:
$\ce{H3O+ + OH- \leftrightarrow 2H2O} \hspace{60px} \mathrm{K = \dfrac{1}{K_w}}$
This reaction is the reverse of Kw, a reaction we have seen before. If the direction of a reaction is reversed, its equilibrium constant is just the inverse or reciprocal value, 1/ Kw in this case.
The final expression to calculate the value of Kn then is the following:
$\mathrm{K_n = \dfrac{K_a(acid)\times K_b(base)}{K_w} = K_a(acid)\times K_b(base) \times 10^{14}}$
If we evaluate the value of Kn for the reaction in the problem, we get the following value.
\begin{align} &\mathrm{pK_a\: (Hcba)} = 3.824 &&\mathrm{K_a = 1.5\times 10^{-4}}\ &\mathrm{pK_b\: (EA)} = 3.37 &&\mathrm{K_b = 4.27\times 10^{-4}} \end{align}
$\mathrm{K_n = (1.5\times 10^{-4})( 4.27\times 10^{-4})(10^{14}) = 6.4\times 10^6}$
This Kn value of slightly more than six million is very large. That says that this reaction, for all practical purposes, will go to completion.
Before solving this problem, it would be worth a digression to examine more generally what we might expect for the value of Kn. Can we always expect Kn to be large such that neutralization reactions always go to completion? Or are there occasions when Kn might be relatively small such that the reaction will not go to completion?
One thing to keep in mind is a solution with excessively dilute concentrations of an acid or base. For example, suppose the concentrations of the acid and base are on the order of 10-10 M. In this case, there is so little acid and base, that even if the Kn value were large, the actual extent of reaction could still be small. It is not that common that we would encounter such solutions in a laboratory setting where we usually use much higher concentrations. But this could occur in environmental samples for some species.
Assuming solutions with appreciable concentrations of acid and base, would we ever have a small value of Kn? Recollecting back, we talked about weak acids as having Ka values on the order of, from strongest to weakest, 10-3, 10-5, 10-7, and 10-9. Similarly weak bases had Kb values from strongest to weakest on the same scale (10-3 to 10-9). Remembering the equation for Kn:
Kn = Ka$\times$Kb$\times$1014
We can see that it will take a mixture of an excessively weak acid and base to get a small value for Kn. For example, mixing an extremely weak acid with a Ka of 10-9 with an extremely weak base with a Kb of 10-9 will give a Kn of 10-4, a small number. This neutralization reaction would not proceed much at all. If the acid had a Ka value of 10-7 and the base a Kb value of 10-7, the value of Kn would be 1, an intermediate value. This neutralization reaction would proceed to some extent.
If we considered a neutralization reaction in which either the acid or base was strong (a strong acid or base might have a Ka or Kb value on the order of 106 or higher), you would need the other species to have a K value of 10-20 or lower to get a small value of Kn. Since this is an unreasonably low value for the weak acid or base, any acid-base reaction that involves either a strong acid or a strong base will go to completion.
The first step is to calculate the initial concentrations of Hcba and EA, remembering that mixing the two solutions dilutes each of the species.
$\textrm{[Hcba]: } \mathrm{(0.224\:M)\times\dfrac{45\:ml}{75\:ml} = 0.1344\:M}$
$\textrm{[EA]: } \mathrm{(0.187\:M)\times\dfrac{30\:ml}{75\:ml} = 0.0748\:M}$
The next step, since Kn is large, is to allow the reaction to go to completion. This is a one-to-one reaction, so the species with the lower concentration will be used up and limit the amount of product that forms.
\begin{align} & &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\ &\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\ &\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber \end{align}
Of course, the amount of EA cannot really be zero, since Kn is a finite value and there needs to be some finite amount of EA. The next step in this problem is to think that some small amount of back reaction occurs.
\begin{align} & &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\ &\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\ &\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber\ &\ce{Back\:reaction} &&0.0596 + \ce{x} &&\ce{x} &&0.0748 - \ce{x} &&0.0748 - \ce{x}\nonumber \end{align}
And we can now consider whether there are any approximations that can be made. Considering that Kn is so large, the extent of back reaction is very small. This means that it is likely that x is very small compared to 0.0596 and 0.0748, such that 0.0596 $\gg$ x and 0.0748 $\gg$ x.
\begin{align} & &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\ &\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\ &\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber\ &\ce{Back\:reaction} &&0.0596 + \ce{x} &&\ce{x} &&0.0748 - \ce{x} &&0.0748 - \ce{x}\nonumber\ &\ce{Assumption} &&0.0596 \gg \ce{x} && &&0.0748 \gg \ce{x} &&0.0748 \gg \ce{x}\nonumber\ &\ce{Approximation} &&0.0596 &&\ce{x} &&0.0748 &&0.0748\nonumber \end{align}
Before we go on, it is worth examining these final concentrations. One interesting thing to note is that we have appreciable quantities of Hcba and cba. These two are conjugate pairs, and we know that a solution with appreciable quantities of both members of a conjugate pair represents a buffer. We can therefore use the appropriate Henderson-Hasselbalch equation for chlorobenzoic acid to solve for the pH of this solution.
$\mathrm{pH = pK_a + \log \left(\dfrac{[cba^-]}{[Hcba]}\right) = 3.824 + \log\left(\dfrac{0.0748}{0.0596}\right)=3.92}$
Before assuming that this answer is the correct one, we ought to check our assumptions. Using the Kn expression, we can calculate the value of x.
$\mathrm{K_n = \dfrac{[cba^-][EAH^+]}{[Hcba][EA]} = \dfrac{(0.0748)(0.0748)}{(0.0596)(x)} = 6.4 \times 10^6}$
$\mathrm{x = 1.46\times 10^{-8}}$
This number is very small and obviously less than 5% of 0.0596 and 0.0748. If we assume that 1.46$\times$10-8 is the final value of EA after back reaction, and that 0.0748 is the final value of EAH+, we have final values for both members of a conjugate pair. If we substitute these into the Henderson-Hasselbalch equation of ethylammonium we ought to get the same pH as above. (Note, the EA and EAH+ are not a buffer since the amount of EA is not appreciable. But if you know the concentration of both members of a conjugate pair, you can still use the Henderson-Hasselbalch equation to solve for the pH, since it is just a rearrangement of the Ka expression.)
$\mathrm{pH = pK_a + \log\left(\dfrac{[EA]}{[EAH^+]} \right ) = 10.63 +\log\left(\dfrac{1.46\times10^{-8}}{0.0748} \right ) = 3.92}$
It should not be a surprise that the two values are identical. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/05_Solutions_that_are_Mixtures_of_Acids_and_Bases_Problem_.txt |
5. Calculate the pH of a solution that is prepared by mixing 75 ml of 0.088 M aniline with 50 ml of 0.097 M 2-nitrophenol.
From the table, we can determine that aniline (An) is a base and nitrophenol (HNp) is an acid. This solution consists of a mixture of an acid and a base, so the first thing we must consider is that a neutralization reaction takes place. In this case we also note that aniline is a very weak base (Kb = 3.94$\times$10-10) and nitrophenol is a very weak acid (Ka = 5.83$\times$10-8). The value of Kn is calculated below.
Kn = Ka $\times$ Kb$\times$1014 = (5.83$\times$10-8)(3.94$\times$10-10)(1014) = 2.3$\times$10-3
This value is fairly small, so we cannot assume that this neutralization reaction will go to completion. Instead we anticipate that this reaction will go to a small extent. Since it goes to only a small extent, we can try making the assumption that x is small compared to the initial concentrations of the aniline (0.0528 $\gg$ x) and nitrophenol (0.0388 $\gg$ x).
\begin{align} & &&\ce{An}\hspace{35px} + &&\ce{HNp} \hspace{25px}\leftrightarrow &&\ce{AnH+} \hspace{25px} + &&\ce{Np-}\nonumber\ &\ce{Initial} &&0.0528 &&0.0388 &&0 &&0\nonumber\ &\ce{Equilibrium} &&0.0528 - \ce{x} &&0.0388 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\ &\ce{Assumption} &&0.0528 \gg \ce{x} &&0.0388 \gg \ce{x} && &&\nonumber\ &\ce{Approximation} &&0.0528 &&0.0388 &&\ce{x} &&\ce{x}\nonumber \end{align}
These values can be plugged into the Kn expression to solve for x:
$\mathrm{K_n = \dfrac{[AnH^+][Np^-]}{[An][HNp]} = \dfrac{(x)(x)}{(0.0528)(0.0388)} = 2.3\times10^{-3}}$
$\mathrm{x = 2.17\times10^{-3}}$
Now we could solve the two Henderson-Hasselbalch equations for each of the conjugate pairs, since we know the concentrations of both members of each pair.
$\mathrm{pH = pK_a + \log\left(\dfrac{[An]}{[AnH^+]} \right) = 4.596 + \log\left(\dfrac{0.0528}{0.00217} \right ) = 5.98}$
$\mathrm{pH = pK_a + \log\left(\dfrac{[Np^-]}{[HNp]} \right) = 7.234 + \log\left(\dfrac{0.00217}{0.0388} \right ) = 5.98}$
The two identical values suggest that the pH of the solution will be 5.98. It is interesting to check the assumptions that were used in calculating the values.
$\dfrac{0.00217}{0.0388}\times100=5.6\% \hspace{60px} \dfrac{0.00217}{0.0528}\times 100 = 4.1\%$
One does meet the 5% rule, the other is just a little over. This might suggest that solving a quadratic is in order, however, if you solve the quadratic and substitute in the values, you will still get a pH of 5.98.
07 Acid Base Properties of Amino Acids
It is worth highlighting the acid-base properties of amino acids, since these are so important in biochemistry. The structure of the amino acid alanine, as you would typically see if written, is shown below.
If you look in our chart of pKa values, you would find that two values are given for alanine (pKa1 = 2.34; pKa2 = 9.69), which might surprise you at first. As you look at the structure, remember that the -COOH group is an acid, but also that an amine group (NH2) is a base. There are two pKa values because we can protonate the amine group, as shown below.
Usually the protonated form is prepared by reaction with hydrogen chloride, so instead of referring to the alaninium ion, we would call it alanine hydrochloride.
One last thing to consider about the neutral amino acid. If we have a substance that has an acid (-COOH) and base (-NH2) within the same molecule, we could ask whether this could undergo an internal acid-base neutralization reaction (realize that we would have many of these molecules in solution so we could also view the acid and base functionalities of different alanine molecules neutralizing each other). It turns out that this actually occurs with amino acids in water, leading to an alanine species with two charges that is called a zwitterion.
Note that the zwitterion still has a net neutral charge, so we do not need to distinguish whether its written form is neutral or zwitterionic in equilibrium calculations. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/06_Solutions_that_are_Mixtures_of_Acids_and_Bases_Problem_.txt |
In-class Problem Set # 2
1. Starting with 30 ml of a solution that is 0.1 M in butylamine, calculate the original pH, and then the pH as 5 ml increments of 0.1 M hydrochloric acid are added. Continue the series of calculations until 40 ml of acid have been added. Plot the data (pH on the y axis, volume of added acid on the x).
Has 99.9% of the butylamine been titrated at the equivalence point?
Butylamine is a base (Kb = 3.98$\times$10-4). Hydrochloric acid is a strong acid, so it will convert the butylamine into the butylammonium ion by a neutralization reaction. Remember, the Kn of a neutralization reaction will always be large if one of the species is strong.
Calculating the initial pH of a weak base is something we have done before.
\begin{align} & &&\ce{BNH2}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{BNH3+} \hspace{25px} + &&\ce{OH-}\ &\ce{Initial} &&0.1 &&0 &&0\ &\ce{Equilibrium} &&0.1 - \ce{x} &&\ce{x} &&\ce{x}\ &\ce{Approximation} &&0.1 &&\ce{x} &&\ce{x} \end{align}
$\mathrm{K_b = \dfrac{[BNH_3^+][OH^-]}{[BNH_2]} = \dfrac{(x)(x)}{0.1} = 3.98\times10^{-4}}$
$\mathrm{x = [OH^-] = 6.31\times 10^{-3}}$
If we check the approximation:
$\dfrac{6.31\times10^{-3}}{0.1} \times 100 = 6.31\%$
It’s not quite valid, but we’ll still use this value. Solving the quadratic would only lead to a small change in the initial value, and this is close enough for our purposes now.
pOH = 2.2
pH = 11.8
First thing we ought to ask is whether we think this is a reasonable number. Remembering that it is a base, a pH of 11.8 is basic, so that seems good.
Now we can examine the first 5 ml increment of hydrochloric acid that’s added. Remember that the HCl (shown as H3O+ in the reaction below since all of it is dissociated in water) will convert butylamine to butylammonium as shown below.
$\ce{BNH2 + H3O+ \leftrightarrow BNH3+ + H2O}$
There is something else important to consider about this reaction. If we start with 30 ml of 0.1 M butylamine, that corresponds to 0.0030 moles. If we add 5 ml of 0.1 M hydrochloric acid, that corresponds to 0.0005 moles. The amounts of these initial reagents before and after the reaction are listed below
\begin{align} & &&\ce{BNH2}\hspace{25px} + &&\ce{H3O+} \hspace{25px}\leftrightarrow &&\ce{BNH3+} \hspace{25px} + \hspace{25px}\ce{H2O}\ &\ce{Initial} &&\textrm{0.0030 mol} &&\textrm{0.0005 mol} &&0\ &\ce{After\:neutralization} &&\textrm{0.0025 mol} &&0 &&\textrm{0.0005 mol} \end{align}
There is something very interesting to note about this solution. There are appreciable amounts of butylamine and butylammonium in the final solution. These two are a conjugate pair, so this solution is a buffer. We can solve for the pH of this solution by using the Henderson-Hasselbalch equation for butylamine.
But there is also something else that is interesting about this when you try to solve for the pH using the Henderson-Hasselbalch equation. If we examine the form of the Henderson-Hasselbalch equation, we note that the final term consists of the ratio of the concentrations of the two components of the buffer. Remember that we started with 30 ml of butylamine and added 5 ml of hydrochloric acid. This causes the final solution to have a volume of 35 ml. If we write out the terms in the equation as shown:
$\mathrm{pH = pK_a + \log\left(\dfrac{[BNH_2]}{[BNH_3^+]} \right ) = 10.6 + \log\left(\dfrac{\dfrac{0.0025\: mol}{\cancel{35\:ml}}}{\dfrac{0.0005\: mol}{\cancel{35\:ml}}} \right ) = 11.3}$
What you note is that the two volume terms in the concentration ratio cancel each other out. In other words, the pH of a buffer solution can be calculated either by determining the ratio of the concentrations of the two components, or by determining the ratio of the moles of the two components.
There is another very important outcome of this. The pH of a buffer does not change if the solution is diluted. In other words, suppose we just added 5 ml of water to the above solution. The final volume would now be 40 ml, but the moles of each component would still be 0.0025 and 0.0005. The pH would remain the same because the volumes cancel. Now, does this hold under all circumstances? At some point if we diluted the solution too much, we may start to promote a significant redistribution of the two species in the buffer and this observation would break down. But generally, we find that the pH of a buffer solution stays fixed under dilution with water.
Before we go on to the next increment of hydrochloric acid, let’s consider one other aspect of this initial addition of 5 ml of acid. If we reconsider the initial solution, we found that x, which was the concentration of $\ce{BNH3+}$, was 6.31$\times$10-3 M. If we calculate the moles of that we find out that it is:
$\mathrm{[BNH_3^+] = (6.31\times 10^{-3}\: mol/L)(0.030\: L) = 0.0002\: moles}$
We could write these as the approximate amounts in the initial solution at equilibrium.
\begin{align} &\ce{BNH2}\hspace{25px} + \hspace{25px}\ce{H2O} \hspace{25px}\leftrightarrow &&\ce{BNH3+} \hspace{25px} + &&\ce{OH-}\ &\textrm{0.0028 mol} &&\textrm{0.0002 mol} &&\textrm{0.0002 mol} \end{align}
When we thought about adding the first 0.0005 moles of acid, we thought of it converting butylamine to butylammonium. Does that mean we should have removed 0.0005 moles of the 0.0028 moles that are listed under the reaction above? If so, that would alter the pH we got after the first addition. NO IT DOESN’T. We have to remember that 0.0002 moles of hydroxide are produced by this initial reaction. Hydroxide is a strong base and the first 0.0002 moles of hydrochloric acid will react with the hydroxide ion. The remaining 0.0003 moles of the acid will then start reacting with the butylamine.
The best way to proceed through the other increments of added hydrochloric acid is to construct a chart of the species in solution. This is shown in Table 1 with the first two pH values included.
Table 1. Moles of butylamine and butylammonium in the titration of butyl amine (0.1 M, 30 ml) with hydrochloric acid (0.1 M).
Step # Added HCl (ml) BNH2 (moles) $\ce{BNH3+}$ (moles) pH
1 0 0.0030 0 11.8
2 5 0.0025 0.0005 11.3
3 10 0.0020 0.0010
4 15 0.0015 0.0015
5 20 0.0010 0.0020
6 25 0.0005 0.0025
7 30 0 0.0030
8 35 0 0.0030
9 40 0 0.0030
Examine the table and consider steps 3 through 6 (10 to 25 ml of acid added). In each of these cases, we have an appreciable amount of each of the two components of the conjugate pair and each of these solutions is a buffer. That means we can use the Henderson-Hasselbalch equation to solve for the pH. It is also sufficient to use the mole ratio of the two and not worry about the dilution of the molar concentrations that would occur.
$\textrm{10 ml:} \hspace{60px} \mathrm{pH=10.6 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right )= 10.9}$
$\textrm{15 ml:} \hspace{60px} \mathrm{pH=10.6 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right )= 10.6}$
Note an important point about this one. Here we have equal concentrations, or equal moles of the two components of the buffer, and the log of 1 is 0. At the point at which the concentrations of both members of the conjugate pair are equal, the pH of a buffer equals the pKa.
$\textrm{20 ml:} \hspace{60px} \mathrm{pH=10.6 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right )= 10.3}$
$\textrm{25 ml:} \hspace{60px} \mathrm{pH=10.6 + \log \left( \dfrac{0.0005\: mol}{0.0025\: mol}\right )= 9.9}$
The situation at 30 mL of acid deserves some attention. First note, that this is called the equivalence point. The equivalence point is the point in a titration where the moles of titrant (hydrochloric acid) that have been added exactly equal the moles of analyte (butylamine) that were in the initial solution. It might be tempting to think that, since there are equal moles of acid and base, that the pH of a solution at the equivalence point of an acid-base titration must be 7. Let’s examine the solution that we have at the equivalence point of this titration.
To a first approximation, all of the butylamine has been used up and converted to butylammonium ion. This is the equivalent to saying what would be the pH of a solution prepared by adding some amount of butylammonium to water. If we think about the nature of butylammonium, we realize that it is a weak acid. So the solution at the equivalence point of this titration is a solution of a weak acid. If that’s the case, the pH at the equivalence point ought to be slightly acidic. Also note, that at this point we now have to account for the effects of dilution since we no longer have appreciable amounts of both members of the conjugate pair (we have 0.0030 moles of $\ce{BNH3+}$ in a total volume of 60 ml or 0.060 L).
$\mathrm{(0.0030\: mol/0.060\: L) = 0.050\: M}$
\begin{align} & &&\ce{BNH3+}\hspace{25px} + \hspace{25px}\ce{H2O} \hspace{25px}\leftrightarrow &&\ce{BNH2} \hspace{25px} + &&\ce{H3O+}\ &\ce{Initial} &&0.05 &&0 &&0\ &\ce{Equilibrium} &&0.05 - \ce{x} &&\ce{x} &&\ce{x}\ &\ce{Approximation} &&0.05 &&\ce{x} &&\ce{x} \end{align}
$\mathrm{K_a = \dfrac{[BNH_2][H_3O^+]}{[BNH_3^+]} = \dfrac{(x)(x)}{0.05} = 2.51\times10^{-11}}$
$\mathrm{x = [H_3O^+] = 1.12\times 10^{-6}}$
$\mathrm{pH = 5.9}$
Checking the approximation shows that it is valid.
$\dfrac{1.12\times 10^{-6}}{0.05} \times 100 = 0.00224\%$
Note that the pH at this point, the equivalence point, is slightly acidic, just as we would expect for a solution of a weak acid. However, one other thing to note is that the true initial value of H3O+ at the equivalence point (10-7 M) is more than 5% of the final concentration of H3O+ (1.12$\times$10-6). So if we wanted to be rigorously correct, we would need to account for that in determining the true pH at the equivalence point.
If we now consider the solution where 35 ml of acid have been added, we note that the butylamine/butylammonium system is used up and no more changes can occur here. Instead, what we now have is an excess of strong acid. So this solution consists of a mixture of a weak acid (butylammonium) with some strong acid (hydrochloric). It should seem reasonable that the extra strong acid will be the critical part in determining the pH of the solution. Remember that every mole of HCl will be converted to H3O+.
5 ml of extra acid adds 0.0005 moles of H3O+.
(0.0005 mol/0.065 L) = 7.69$\times$10-3 M [H3O+]
pH = 2.1
And for 40 ml:
10 ml of extra acid adds 0.0010 moles of H3O+.
(0.0010 mol/0.070 L) = 1.4$\times$10-2 M = [H3O+]
pH = 1.8
It’s worthwhile at this point to compile a complete table (Table 2) of this process and examine some of the general trends that occur.
Table 2. Calculated pH values for the titration of butyl amine (0.1 M, 30 ml) with hydrochlorid acid (0.1 M).
Step # Added HCl (ml) BNH2 (moles) $\ce{BNH3+}$ (moles) pH
1 0 ml 0.0030 0 11.8
2 5 0.0025 0.0005 11.3
3 10 0.0020 0.0010 10.9
4 15 0.0015 0.0015 10.6
5 20 0.0010 0.0020 10.3
6 25 0.0005 0.0025 9.9
7 30 0 0.0030 5.9
8 35 0 0.0030 2.1
9 40 0 0.0030 1.8
First, note the large drop in pH between 25 and 30 ml of added acid. At this point we have exhausted the buffer and so it should not be surprising that a small amount of extra acid causes a large drop in pH. Also note, that the pH of 0.1 M hydrochloric acid is 1.0, so that the pH would slowly approach a limit of 1.0 if we continued to add more acid to the solution.
It is also worth examining a plot of the pH during the course of the titration as shown in Figure 1. The relatively flat portion of the plot between 5 ml and 25 ml of acid is known as the buffer region. Notice how the center of the buffer region corresponds to the pKa value. So the butylammonium/butylamine system would be a useful buffer at a pH of around 10.6.
Figure 1. Plot of pH versus ml titrant for titration of butylamine (0.1 M, 30 ml) with hydrochloric acid (0.1 M).
Suppose we had used an identical situation (30 ml of 0.1 M base, add 5 ml increments of HCl) but had a base whose conjugate acid had a pKa of 8. What would that plot look like? The plot is shown in Figure 2 and compared to what we observed with butylamine. Note how the initial pH would be a little less basic, how the buffer region is now centered around 8, how the equivalence point still occurs at 30 ml of acid, but how the pH at the equivalence point is a little more acidic because the weak acid is a little stronger than the butylammonium ion. If we then showed a plot for a species where the acid form had a pKa of 6, we start to note that it becomes more difficult to distinguish the equivalence point in the plot. A concentration of a species like butylamine can be analyzed using an acid-base titration. The concentration of a base whose conjugate acid has a pKa value of 6 could not be analyzed using an acid-base titration.
Figure 2. Plot of pH versus ml titrant for the titration of weak bases (0.1 M, 30 mL, pKa values of 6, 8 and 10.6) with hydrochloric acid (0.1 M).
We can also talk about the qualities that define a “good” buffer. The term that is used here is buffer capacity (a measure of how much acid or base a buffer can neutralize without an appreciable change in pH), and we want a high buffer capacity.
Presumably there is a particular pH that you want to buffer your solution at. The first criteria is to select a buffer that has a pKa close to the pH that you want to buffer at. The “official standard” is that the pKa of the acid must be within +/– one unit of the pH you want to buffer at, but the closer the better. Note from the Henderson-Hasselbalch equation that one unit of change from pKa would correspond to either a 1/10 or 10/1 ratio of the two members of the conjugate pair. If you look at the plots above, note that ratios of 1/10 or 10/1 are out on the extreme end of the buffer region. At these extremes, there is a lot of buffer capacity in one direction, but almost none in the other. It would be risky to use such a solution as a buffer and much better to use a species with a pKa much closer to the pH you need to buffer at.
The second criterion is to have a high concentration of both components of the conjugate pair so that the buffer will neutralize more acid or base. A good buffer therefore is one in which both components of the conjugate pair are highly soluble in water. So buffers usually have high concentrations of species relative to the other species you are studying in solution. The actual concentration of components you use for a buffer depend on the nature of your investigation. In biochemistry, where the concentrations of proteins and nucleic acids are usually quite low, the concentrations of buffer components are relatively low. In chemical analysis procedures where the concentrations of reagents might be fairly high, the concentration of buffer needs to be high as well.
The last criteria in selecting a buffer is to ensure that the buffer components do not interfere in any way with the process being studied. For example, if you want to determine the amount of a substance in solution by measuring its absorption of light, it’s essential that the buffer not absorb at that wavelength. If your procedure involves the formation of a metal complex, it’s essential that the components of the buffer do not complex with the metal ion.
The criteria used in selecting a buffer can be summarized as follows:
1. The buffer substance needs a pKa value as close as possible to the desired pH.
2. The buffer components must have high solubility.
3. The buffer components cannot interfere in any way with the other species in solution or the measurement you want to make.
Now, there is one other component to the question that we have not addressed yet. The question asks whether 99.9% of the butylamine has been titrated at the equivalence point (in other words, has 99.9% of the butylamine been converted to butylammonium). This is an important question in analysis procedures. If we used this titration to determine the concentration of butylamine in the solution, the assumption is that “all” of the butylamine has been converted to butylammonium so that we are getting an accurate measurement. Of course, we can never convert all of the butylamine, since the K values are finite and so there must always be a little bit of butylamine in the solution, but if we can convert at least 99.9%, that’s a high enough degree of accuracy for most purposes. The way we assess this is to compare the concentrations of the two species at the equivalence point. Going back to our pH calculation at 30 ml of acid, we have the following values:
[BNH2] = 1.12$\times$10-6 M
[$\ce{BNH3+}$] = 0.05 M
Admittedly, the $\ce{BNH3+}$ is a little less than 0.05 M, but the approximation we made when solving the problem can still be used. If we evaluate the ratio of BNH2 to $\ce{BNH3+}$, as shown below, we find that 0.002% is BNH2 and 99.998% is $\ce{BNH3+}$ at the equivalence point. That says that this titration procedure would be an effective way to analyze the concentration of butylamine in the initial solution.
$\dfrac{1.12\times10^{-6}}{0.05}\times 100 = 0.00224\%$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/08_Titration_of_a_Weak_Base.txt |
In-class Set #3
1. Calculate the pH of a 0.127 M solution of ascorbic acid (H2asc).
This is a new situation that we have not encountered before, since ascorbic acid (which is vitamin C by the way) has two dissociable hydrogen ions. We can find this out by looking in the table and seeing that two pKa values (4.30 and 11.82) are provided. The relevant reactions needed to describe what happens in a solution of ascorbic acid in water are shown below.
$\ce{H2asc + H2O \leftrightarrow Hasc- + H3O+}\qquad\qquad \mathrm{K_{a1}}$
$\ce{Hasc- + H2O \leftrightarrow asc^2- + H3O+}\qquad\qquad \mathrm{K_{a2}}$
The problem we have is that there are two reactions that can cause production of H3O+. Remember that there is a small amount of H3O+ in solution to begin with (10-7 M from the dissociation of water), but like in other problems of weak acids, we can assume that the H3O+ produced by the dissociation of ascorbic acid will be much larger than the amount there from the dissociation of water.
The first step in understanding how to do this problem is to write an expression for H3O+ in terms of the other species that are produced when the ascorbic acid dissociates. In other words, we ought to be able to write an expression that equates the concentration of H3O+ to the concentrations of Hasc and asc2–.
First consider the Hasc species. If we look at the first reaction above, it should be apparent that one H3O+ occurs for each Hasc found in solution. If this was the only reaction that took place in solution, we could therefore write:
$\mathrm{[H_3O^+] = [Hasc^-]}$
Next consider the asc2 species. One way to think about this is to envision a situation in which all of the ascorbic acid dissociated into the asc2– form. Under this situation, we could write the following reaction to describe this process.
$\mathrm{H_2asc = asc^{2-} + 2 H_3O^+}$
The important thing to note here is that two H3O+ occur for each asc2– found in solution. Similarly, if only a small amount of asc2– is found in solution, as would occur in this solution of ascorbic acid given the pKa values, it would still be the case that two H3O+ ions must occur for each asc2– species found in solution.
$\mathrm{[H_3O^+] = 2[asc^{2-}]}$
Note, there are two H3O+ for every asc2. Substitute a “1” in for asc2 in the equation above and you will get “2” for the H3O+.
We can combine these two into one equation that describes the total concentration of H3O+ in a solution of ascorbic acid as follows:
$\mathrm{[H_3O^+] = [Hasc^-] + 2[asc^{2-}]}$
What this means is that for each Hasc- in solution we have one H3O+, and for each asc2– in solution, we have two corresponding H3O+ ions.
The next critical step is to consider the relative magnitudes of these two terms. If we examine the two pKa values for the ascorbic acid reactions, and convert them to Ka values, note that the second reaction has a Ka value that is about 107 times smaller than the first. This means that the extent of the second reaction is minimal compared to that of the first. In other words, the amount of H3O+ formed by the second reaction is insignificant compared to how much H3O+ is produced by the first. This means that:
$\mathrm{[Hasc^-] \gg 2[asc^{2-}]}$
and we can use the following approximation to describe this solution:
$\mathrm{[H_3O^+] = [Hasc^-]}$
In other words, we only need to consider the first reaction to determine the pH of a solution of ascorbic acid. Even though this looked initially like it might be a complicated system, if we only need to consider the first reaction, solving this problem is identical to what we have done earlier when solving for the pH of a solution of a monoprotic acid. This raises the question of whether we can always simplify such a problem down to only one reaction. The answer depends in part on the relative magnitudes of the two pKa values. If these two differed by two units (pKa1 = 3, pKa2 = 5), this represents a 100-fold difference in the extent of reaction and we can ignore the second reaction. If you were to examine the typical pKa values in the table for polyprotic acids, you would notice that the relative values would almost always allow you to treat this comparable to a monoprotic system. Only in a few instances when the two pKa values are almost identical would you need to treat this in a more complex manner by including both reactions in the problem.
What we will find in general with polyprotic acid/base systems is that we can almost always find one reaction that is significant and ignore the other reactions in the series. As we examine more problems of this variety, we will see how this will apply to other situations.
Now we can solve for the pH of this solution. Note, that there are two important assumptions. Because Ka1 is small, the value of x is small compared to the starting concentration of H2asc (0.0127 $\gg$ x). Also, since H2asc is a weak acid, the amount of H3O+ produced by dissociation of the acid will be much larger than the starting concentration of H3O+ that exists due to the autoprotolysis of water (x $\gg$ 10-7)
\begin{align} & &&\ce{H2asc}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{Hasc-} \hspace{25px} + &&\ce{H3O+}\nonumber\ &\ce{Initial} &&0.127 &&0 &&10^{-7}\nonumber\ &\ce{Equilibrium} &&0.127 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\ &\ce{Assumption} &&0.127 \gg \ce{x} && &&\ce{x} \gg 10^{-7}\nonumber\ &\ce{Approximation} &&0.127 &&\ce{x} &&\ce{x}\nonumber \end{align}
$\mathrm{K_{a1} = \dfrac{[Hasc^-][H_3O^+]}{[H_2asc]}=\dfrac{(x)(x)}{0.127}=5.0\times10^{-5}}$
$\mathrm{x = [H_3O^+] = 2.52\times10^{-3}}$
$\mathrm{pH = 2.6}$
Checking the approximation shows that this is valid.
$\dfrac{2.52\times10^{-3}}{0.127}\times100 = 1.98\%$
As one final check that it was reasonable to ignore the second reaction, we can plug in the values of [Hasc] and [H3O+] calculated above into the Ka2 expression. That leads to an interesting finding as seen below that [asc2–] equals Ka2. Note how small this is, and therefore how little extra H3O+ would come from the second reaction. Ignoring this reaction in the calculation of pH was a valid thing to do.
$\mathrm{K_{a2}= \dfrac{[asc^{2-}][H_3O^+]}{[Hasc^-]} = \dfrac{[asc^{2-}](2.52\times10^{-3})}{(2.52\times10^{-3})} = 1.51\times10^{-12}}$
$\mathrm{[asc^{2–}] = 1.51\times10^{-12}}$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/09_Solutions_of_Polyprotic_Acid_Base_Systems_Problem_A.txt |
2. Calculate the pH of a 0.089 M solution of sodium carbonate.
The carbonate ion is part of the carbonic acid system. If we look up carbonic acid in the table, we find that it is a diprotic acid. The two equilibria are as follows. Also, looking at the two pKa values (6.35 and 10.33) we notice that there is an appreciable difference between the two, suggesting that it may well be necessary to consider one of the two reactions in this problem.
$\ce{H2CO3 + H2O \leftrightarrow HCO3- + H3O+} \hspace{60px} \mathrm{K_{a1}}$
$\ce{HCO3- + H2O \leftrightarrow CO3^2- + H3O+} \hspace{60px} \mathrm{K_{a2}}$
Before continuing, we need to know whether sodium carbonate refers to NaHCO3 or Na2CO3. We will adopt a particular system in this course for naming these types of species, but in this case note that sodium carbonate refers to Na2CO3. The $\ce{HCO3-}$ ion is known as the bicarbonate ion and the species NaHCO3 is known as sodium bicarbonate (also known as baking soda). The name bicarbonate for $\ce{HCO3-}$ is not a systematic name, but a common name for this ion.
The naming system we will adopt can be demonstrated for the phosphoric acid (H3PO4) series of reactions. Phosphoric acid has three dissociable hydrogen ions, leading to the following possible species.
\begin{align} &\ce{H3PO4} &&\textrm{– phosphoric acid}\nonumber\ &\ce{NaH2PO4} &&\textrm{– sodium dihydrogen phosphate}\nonumber\ &\ce{Na2HPO4} &&\textrm{– disodium hydrogen phosphate}\nonumber\ &\ce{Na3PO4} &&\textrm{– sodium phosphate}\nonumber \end{align}\nonumber
Note that the species sodium phosphate refers to the one in which all the dissociable hydrogen ions have been replaced with sodium cations. The others contain a prefix that tells you how many hydrogen atoms or sodium ions are involved in the salt. You must be careful when using or purchasing species like the intermediate ones since the names given above may not be used. Sometimes these are referred to as sodium phosphate monobasic and sodium phosphate dibasic. Presumably the label actually gives the formula so that you can be certain which species you actually have.
If we go back to our solution of sodium carbonate, this means that we have the $\ce{CO3^2-}$ species in solution. Remember, if you add sodium carbonate to water, the ions will dissociate to produce sodium ions and carbonate ions. Since the sodium ion is the cation of a strong base (sodium hydroxide – NaOH), this species won’t form and the sodium ion is what we call a spectator ion (it effectively watches things but does not get involved in any important reactions). Since the carbonate ion is the anion of a weak acid, it's actually a base and we can write the following reactions to describe what will occur in this solution.
$\ce{CO3^2- + H2O \leftrightarrow HCO3- + OH-}\hspace{60px}\mathrm{K_b\: of\: K_{a2}}$
$\ce{HCO3- + H2O \leftrightarrow H2CO3 + OH-}\hspace{60px}\mathrm{K_b\: of\: K_{a1}}$
Because of the significant distinction between the two Kb values, we only need to consider the first reaction in the series above to calculate the pH. The amount of hydroxide produced by the second reaction will be insignificant.
We can then treat this as a monobasic base, a process we have seen before.
\begin{align} & &&\ce{CO3^2-}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{HCO3-} \hspace{25px} + &&\ce{OH-}\nonumber\ &\ce{Initial} &&0.089 &&0 &&0\nonumber\ &\ce{Equilibrium} &&0.089 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\ &\ce{Approximation} &&0.089 &&\ce{x} &&\ce{x}\nonumber \end{align}
Plugging this into the appropriate Kb expression gives:
$\mathrm{K_b = \dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]} = \dfrac{(x)(x)}{0.089} = 2.14\times10^{-4}}$
$\mathrm{x = [OH^-] = 4.36\times10^{-3}}$
$\mathrm{pOH = 2.37}$
$\mathrm{pH = 11.63}$
The pH value of 11.63 seems reasonable since this is a solution of a base. Checking the approximation shows that this was just under our 5% rule.
$\dfrac{4.36\times10^{-3}}{0.089} \times100=4.9\%$
11 Solutions of Polyprotic Acid Base Systems Problem C
3. Calculate the pH of a solution prepared by adding 30 ml of 0.1 M hydrochloric acid to 60 ml of 0.080 M potassium malonate.
The first key to solving this problem is to identify the nature of potassium malonate. If we look in the table we will find the species malonic acid (H2mal), a diprotic acid. Potassium malonate is therefore the species K2mal. When you add this to water, you would get two potassium cations and the malonate ion (mal2). Since potassium is the cation of a strong base (potassium hydroxide – KOH), it does not react in any way and is a spectator ion.
$\ce{H2mal + H2O \leftrightarrow Hmal- + H3O+} \qquad \mathrm{K_{a1} = 1.40 \times 10^{-3}}$
$\ce{Hmal- + H2O \leftrightarrow mal^2- + H3O+}\qquad \mathrm{K_{a2} = 2.01 \times 10^{-6}}$
So the malonate ion is a base. We have added hydrochloric acid, a strong acid to the solution. The hydrochloric acid will therefore undergo a neutralization reaction with the malonate ion to produce Hmal. If it turns out that all the mal2 gets used up in producing Hmal and there is still an excess of hydrochloric acid, the additional hydrochloric acid will then convert Hmal to H2mal. Remember, a strong acid will always lead to the neutralization of a base. Hmal can act as a base and accept another hydrogen ion to product H2mal.
Next we need to calculate the moles of mal2 and hydrochloric acid that we have in solution.
Moles of mal2: $\mathrm{\left(0.08\: \dfrac{mol}{L}\right)(0.060\: L) = 0.0048\: moles}$
Moles of HCl: $\mathrm{\left(0.10\: \dfrac{mol}{L}\right)(0.030\: L) = 0.0030\: moles}$
The reaction that describes what will occur is as follows:
\begin{align} & &&\ce{mal^2-}\hspace{25px} + &&\ce{H3O+} \hspace{25px}\leftrightarrow &&\ce{Hmal-} \nonumber\ &\ce{Initial} &&\mathrm{0.0048\: mol} &&\mathrm{0.0030\: mol} &&0\nonumber\ &\ce{Neutralization} &&\mathrm{0.0018\: mol} &&0 &&\mathrm{0.0030\: mol}\nonumber \end{align}
Note that there are appreciable amounts of both members of a conjugate pair, which constitutes a buffer. The only remaining question is whether we need to consider the other reaction that can occur for the Hmal.
$\ce{Hmal- + H2O \leftrightarrow H2mal + OH-}$
It turns out that just like the case of ascorbic acid or sodium carbonate, there is so much distinction between the K values for the two reactions that we can ignore the second one and only consider the mal2/Hmal reaction in determining the pH. Since we have a buffer, we can now use the appropriate Henderson-Hasselbalch expression for Ka2.
$\mathrm{pH=pK_{a2}+ \log\left( \dfrac{[mal^{2-}]}{[Hmal^-]}\right )=5.696+\log\left( \dfrac{0.0018}{0.0030}\right )=5.47}$
If you go ahead and substitute the amount of Hmal and $\ce{OH-}$ into the Ka1 expression, as shown below, you see that the amount of H2mal that forms is insignificant compared to the Hmal concentration and can be ignored.
$\mathrm{[Hmal^-] = (0.0030\: mol/0.090\: L) = 0.0333\: M}$
$\mathrm{[H_3O^+] = 3.39\times 10^{-6}}$
$\mathrm{K_{a1}=\dfrac{[Hmal^-][H_3O^+]}{[H_2mal]}=\dfrac{(0.0333)(3.39\times10^{-6})}{[H_2mal]}=1.40\times10^{-3}}$
$\mathrm{[H_2mal] = 8.06\times10^{-5}}$ | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/10_Solutions_of_Polyprotic_Acid_Base_Systems_Problem_B.txt |
4. Calculate the pH of a solution prepared by adding 55 ml of 0.098 M sodium phosphate to 65 ml of 0.136 M phosphoric acid.
\begin{align} \ce{&Phosphoric\: acid &&– H3PO4\ &Sodium\: phosphate &&– Na3PO4 \:(which\: dissociates\: to\: PO4^3- )} \end{align}
Both of these species are in the phosphoric acid system shown below. The two species in appreciable quantities are shown in boldface.
\begin{align} \ce{&\mathbf{H3PO4} + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\ \ce{& H2PO4- + H2O \leftrightarrow HPO4^2- + H3O+} &&\mathrm{K_{a2}}\ \ce{& HPO4^2- + H2O \leftrightarrow \mathbf{PO4^{3-}} + H3O+} &&\mathrm{K_{a3}} \end{align}
What we need to realize here is that H3PO4 is an acid, and $\ce{PO4^3-}$ is a base. As such they will react with each other according to the following neutralization reaction.
$\ce{H3PO4 + PO4^3- \leftrightarrow H2PO4- + HPO4^2-}$
If we use the Ka for H3PO4 (7.11×10-3) and the Kb for $\ce{PO4^3-}$ (2.4×10-2) and solve for Kn using our established equation (Kn = (Ka×Kb)/Kw) we get a value of 1.7×1010, a very large number. This neutralization will go to completion.
We can also take this a step further. In almost all cases for a multistep equilibria system such as this, we can anticipate that only one reaction of the series will be important in determining the pH. This does not happen in every situation (a little later in the course we will see an example where this does not rigorously work), but it does most of the time. One expectation is that eventually we will wind up with appreciable amounts of both members of a conjugate pair, which constitutes a buffer. The most straight forward way of working with this problem is to consider the moles of different chemicals that are present.
Moles of phosphate: $\mathrm{\left(0.098\: \dfrac{mol}{L}\right)\left(0.055\: L\right) = 0.00539\: moles}$
Moles of phosphoric acid: $\mathrm{\left(0.136\: \dfrac{mol}{L}\right)(0.065\: L) = 0.00884\: moles}$
\begin{align} & &&\ce{H3PO4}\hspace{25px} + &&\ce{PO4^3-} \hspace{25px}\leftrightarrow &&\ce{H2PO4-}\hspace{25px} + &&\ce{HPO4^2-}\nonumber\ &\ce{Initial} &&0.00884 &&0.00539 &&0 &&0\nonumber\ &\ce{Neutralization} &&0.00345 &&0 &&0.00539 &&0.00539\nonumber \end{align}
As seen by the data, all of the phosphate ion ($\ce{PO4^3-}$) has been used up in the solution. At this point, which is still an intermediate one, it can be helpful to put in boldface the three species that are present in appreciable quantities in the solution.
\begin{align} &\mathbf{H_3PO_4} + \ce{H2O} \leftrightarrow \mathbf{\ce{H2PO4-}} + \ce{H3O+} &&\mathrm{K_{a1}}\ &\mathbf{\ce{H2PO4-}} + \ce{H2O} \leftrightarrow \mathbf{\ce{HPO4^2-}} + \ce{H3O+} &&\mathrm{K_{a2}}\ &\ce{HPO4^2-} + \ce{H2O} \leftrightarrow \ce{PO4^3-} + \ce{H3O+} &&\mathrm{K_{a3}} \end{align}
What we now need to examine is the possibility for neutralization between H3PO4 and $\ce{HPO4^2-}$. Note that the product consists of two equivalents of the species $\ce{H2PO4-}$.
$\ce{H3PO4 + HPO4^2- \leftrightarrow 2H2PO4-}$
Evaluation of the Kn for this reaction involves the Ka value of H3PO4 (7.11×10-3) and the Kb value for $\ce{HPO4^2-}$ (Kb of Ka2 = 1.578×10-7). Using our established equation for evaluating Kn, we get a value of 1.12×105, which is a large number. This neutralization will essentially go to completion.
We can now assess the moles of material produced in the reaction.
\begin{align} & &&\ce{H3PO4}\hspace{25px} + &&\ce{HPO4^2-} \hspace{25px}\leftrightarrow &&\ce{2H2PO4-}\nonumber\ &\ce{Initial} &&0.00345 &&0.00539 &&0.00539 \nonumber\ &\ce{Neutralization} &&0 &&0.00194 &&0.01229\nonumber \end{align}
Be careful when calculating the amount of $\ce{H2PO4-}$ in the final solution. There is an amount already present (0.00539 moles) and we produce two equivalents in the reaction above (2×0.00345 moles), hence the total of 0.01229 moles of $\ce{H2PO4-}$.
Once again, it is helpful to boldface the species in the series of reactions that are present in appreciable quantities.
\begin{align} \ce{&H3PO4 + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\ &\mathbf{\ce{H2PO4-}} + \ce{H2O \leftrightarrow \mathbf{HPO4^{2-}} + H3O+} &&\mathrm{K_{a2}}\ \ce{&HPO4^2- + H2O \leftrightarrow PO4^3- + H3O+} &&\mathrm{K_{a3}} \end{align}
Note that we now have appreciable quantities of a conjugate pair. Since the distribution of the species in a conjugate pair will not change (these two cannot neutralize each other since they will simply reform each other), we can now calculate the pH using the appropriate Henderson-Hasselbalch equation.
$\mathrm{pH=pK_a+ \log\left( \dfrac{[HPO_4^{2-}]}{[H_2PO_4^-]}\right )=7.198+\log\left( \dfrac{0.00194}{0.01229}\right )=6.4}$
It might be useful to check and make sure that ignoring the other two reactions was a reasonable thing to do.
We can calculate the concentration of H3PO4 using the expression for Kn. We first need to convert the moles of the different species to molarity for the subsequent calculations.
$\mathrm{[H_2PO_4^-] =\dfrac{0.01229\: mol}{0.120\: L} = 0.102\: M}$
$\mathrm{[HPO_4^{2-}] =\dfrac{0.00194\: mol}{0.120\: L} = 0.016\: M}$
$\mathrm{K_n= \dfrac{[H_2PO_4^-]^2}{[H_3PO_4][HPO_4^{2-}]}=\dfrac{(0.102)^2}{(x)(0.016)}=1.12\times10^5}$
$\mathrm{x = [H_3PO_4] = 5.8\times 10^{-6}}$
This is a very small quantity of H3PO4, so ignoring the Ka1 reaction was justified. We can calculate the amount of $\ce{PO4^3-}$ using Ka3.
$\mathrm{K_{a3}= \dfrac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]} = \dfrac{(x)(3.98\times10^{-7})}{(0.016)} = 4.17\times10^{-13}}$
$\mathrm{x = [PO_4^{3-}] = 1.68\times 10^{-8}}$
Once again, we see that this is an exceptionally small quantity that can be ignored. As has been the pattern so far, in problems involving polyprotic acids or bases, we see that only one reaction is significant in determining the pH of the solution. This situation will occur in almost all cases for these reactions. The approach to these problems is to identify the important reaction and solve for the pH using only that one, then use the values that are obtained to check that we could ignore the other ones. Of course, if the pKa values for the reactions are appreciably different (value of 10 or greater), we would know ahead of time that only one reaction will be important. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/12_Solutions_of_Polyprotic_Acid_Base_Systems_Problem_D.txt |
5. Calculate the pH of a 0.240 M solution of sodium bicarbonate (NaHCO3).
This will dissociate to produce the $\ce{HCO3-}$ ion, which is in the carbonic acid system.
\begin{align} &\ce{H2CO3 + H2O \leftrightarrow HCO3- + H3O+} &&\mathrm{K_{a1}}\nonumber\ &\ce{HCO3- + H2O \leftrightarrow CO3^2- + H3O+} &&\mathrm{K_{a2}}\nonumber \end{align}\nonumber
What we observe in this case is that the $\ce{HCO3-}$ is an intermediate and we essentially have “all” of an intermediate. Some small amount of H2CO3 and $\ce{CO3^2-}$ will form, since the two equilibrium constants have finite values, but not enough of either one will form to constitute a buffer. The bicarbonate ion has the ability to react as an acid (reaction 2) or a base (reaction 1), and it might be tempting to determine whether it’s a stronger acid or base (by comparing the relative magnitude of Ka2 to the Kb of Ka1) and use that reaction to calculate the pH. The situation is complicated by the fact that any H2CO3 and $\ce{CO3^2-}$ that are formed can neutralize each other.
What we need to do in this case is write an expression for [H3O+] in terms of species in the carbonate system. Remember, there is some H3O+ around from dissociation of water, however, this value will be overcome by the carbonate system and can be ignored in this calculation.
We need to realize that we have a lot of $\ce{HCO3-}$ in solution and consider what each of the reactions above does to the concentration of H3O+ in solution. At the start, before the system equilibrates, there is no H2CO3 or $\ce{CO3^2-}$.
If we look at the second reaction (Ka2), we see that for every $\ce{CO3^2-}$ that is produced we need to produce one equivalent of H3O+. If this was the only reaction that occurred, it would allow us to write the following expression:
$\mathrm{[H_3O^+] = [CO_3^{2-}]}\nonumber$
If we look at the first reaction (Ka1), we see that in order to produce a molecule of H2CO3, we actually need to remove an H3O+ species. Therefore, every H2CO3 that we find in the final solution subtracts an H3O+. This would allow us to write:
$\mathrm{[H_3O^+] =\, –[H_2CO_3]}\nonumber$
Both $\ce{CO3^2-}$ and H2CO3 will be present in the final solution, so we can combine these two equations and come up with the following expression for the concentration of H3O+ in the final solution:
$\mathrm{[H_3O^+] = [CO_3^{2-}] - [H_2CO_3]}\nonumber$
Now we need to do some algebra. First, rearrange the above expression into the following:
$\mathrm{[H_3O^+] + [H_2CO_3] = [CO_3^{2-}]}\label{1}$
Now rearrange the Ka1 and Ka2 expressions to solve them for [H2CO3] and [$\ce{CO3^2-}$] respectively.
$\mathrm{K_{a1}= \dfrac{[HCO_3^-][H_3O^+]}{[H_2CO_3]} \hspace{60px} [H_2CO_3]= \dfrac{[HCO_3^-][H_3O^+]}{K_{a1}}}\label{2}$
$\mathrm{K_{a2}= \dfrac{[CO_3^{2-}][H_3O^+]}{[HCO_3^-]} \hspace{60px} [CO_3^{2-}]=\dfrac{K_{a2}[HCO_3^-]}{[H_3O^+]}}\label{3}$
Substitute the expressions for [H2CO3] and [$\ce{CO3^2-}$] into equation ($\ref{1}$).
$\mathrm{[H_3O^+] + \dfrac{[HCO_3^-][H_3O^+]}{K_{a1}}= \dfrac{K_{a2}[HCO_3^-]}{[H_3O^+]}}\label{4}$
Multiply each side through by Ka1[H3O+] to remove all terms from the denominator.
$\mathrm{K_{a1}[H_3O^+]^2 + [HCO_3^-][H_3O^+]^2 = K_{a1}K_{a2}[HCO_3^-]}\label{5}$
Pull out an [H3O+]2 term from the left-hand side of the equation.
$\mathrm{[H_3O^+]^2(K_{a1} + [HCO_3^-]) = K_{a1}K_{a2}[HCO_3^-]}\label{6}$
Divide both sides by (Ka1 + [$\ce{HCO3-}$]) to give the following:
$\mathrm{[H_3O^+]^2= \dfrac{K_{a1}K_{a2}[HCO_3^-]}{K_{a1}+[HCO_3^-]}}\label{7}$
Now comes a critical assessment of the terms in the denominator, as we want to compare the magnitude of Ka1 to [$\ce{HCO3-}$]. Usually, we would anticipate that Ka1 is a fairly small number since this is a weak acid. For example, in the current problem, Ka1 is 4.47×10-7. Usually the concentration of the intermediate ($\ce{HCO3-}$ in this case) is fairly high (0.240 M in this case). In many situations the concentration of the intermediate is a lot larger than the Ka value.
$\mathrm{[HCO_3^-] \gg K_{a1}}\nonumber$
In this case, we can ignore the Ka1 in the term (Ka1 + [$\ce{HCO3-}$]). That simplifies equation ($\ref{7}$) to the following:
$\mathrm{[H_3O^+]^2= \dfrac{K_{a1}K_{a2}[HCO_3^-]}{[HCO_3^-]}}\label{8}$
Notice how the [$\ce{HCO3-}$] terms now cancel out of the equation leaving:
$\mathrm{[H_3O^+]^2 = K_{a1} K_{a2}}\label{9}$
Using the properties of logs, this expression can be rewritten as follows:
$\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}}\label{10}$
There is a certain way in which this outcome seems to make sense. We stated at the onset that one of the problems was that the bicarbonate ion could act as an acid and a base. The pKa1 value represents bicarbonate acting as a base, and pKa2 represents bicarbonate acting as an acid. This equation essentially represents an average of these two values. That average will also reflect whether the intermediate is more likely to act as an acid or base, as the pH of the final solution will either be acidic or basic depending on the magnitudes of the two pKa values. One other wonderful aspect of the pH of this solution is that it is independent of the concentration of bicarbonate. Of course, we need to remember that we made an approximation to come up with this simple form to get the pH. At very dilute concentrations of intermediate, that approximation breaks down and then the calculation becomes more complicated. We would need to use equation ($\ref{7}$) in that case.
In this case, we can now substitute in the two pKa values for the carbonic acid system and determine the pH:
$\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}=\dfrac{6.35+10.33}{2}=8.34}\nonumber$
The value of 8.34 is slightly basic. Perhaps it is not surprising then that we could use a solution of sodium bicarbonate as an antacid if a night of pizzas, tacos, jalapeno poppers, and tequila sunrises (including the worm at the bottom of the bottle) left our stomach with excess acid.
It turns out that the generalized expression we derived in this case, in which the pH was equal to (pKa1 + pKa2)/2, can be applied to any intermediate in a polyprotic acid system. For example, consider the series of equations for phosphoric acid.
\begin{align} &\ce{H3PO4 + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\nonumber\ &\ce{H2PO4- + H2O \leftrightarrow HPO4^2- + H3O+} &&\mathrm{K_{a2}}\nonumber\ &\ce{HPO4^2- + H2O \leftrightarrow PO4^3- + H3O+} &&\mathrm{K_{a3}}\nonumber \end{align}\nonumber
Suppose we had a solution that to a first approximation was “all” sodium dihydrogen phosphate (NaH2PO4). This would dissolve to produce $\ce{H2PO4-}$, the first intermediate in this series of reactions. We can ignore the third reaction because it will be insignificant. We can perform a derivation analogous to what we did for bicarbonate and would come up with the following expression for the pH:
$\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}}\nonumber$
Suppose instead we had a solution that to a first approximation was “all” disodium hydrogen phosphate (Na2HPO4). This would dissolve to produce $\ce{HPO4^2-}$, the second intermediate in this series of reactions. In this case, we can ignore the first reaction and, doing a derivation analogous to what we did for bicarbonate, we would come up with the following expression for the pH:
$\mathrm{pH=\dfrac{pK_{a2}+ pK_{a3}}{2}}\nonumber$
Something to note in this case is that if we examined the comparable equation to equation ($\ref{7}$), it would look as follows:
$\mathrm{[H_3O^+]^2= \dfrac{K_{a2}K_{a3}[HPO_4^{2-}]}{K_{a2}+ [HPO_4^{2-}]}}\nonumber$
Note that in this case, we are comparing the magnitude of [$\ce{HPO4^2-}$] to Ka2. Since Ka2 is always smaller than Ka1 (and usually much smaller), the likelihood that we can ignore the magnitude of Ka2 relative to the concentration of [$\ce{HPO4^2-}$] is improved, allowing us to use this very straight forward way of calculating the pH. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/13_Solutions_of_Polyprotic_Acid_Base_Systems_Problem_E.txt |
Out of Class Assignment #4, Problem 2
Starting with 30 ml of 0.1 M citric acid, calculate the initial pH and the pH at each 5 ml increment of 0.1 M NaOH until you are 10 ml past the last equivalence point. Plot the data and determine whether 99.9% of the citric acid has been neutralized at the last equivalence point. Also calculate the concentration of all species in solution at the second equivalence point.
It is worth examining this problem in some detail since we have not done anything exactly like it in class. Essentially it consists of the titration of a polyprotic acid using a strong base. Citric acid is a common buffer but is an interesting example because the first two pKa values are fairly close to each other.
If we look in the table we find out that citric acid (H3cit) is a triprotic acid. The following three equilibrium reactions define the system.
\begin{align} &\ce{H3cit + H2O \leftrightarrow H2cit- + H3O+} &&\mathrm{K_{a1} = 7.45\times 10^{-4}}\ &\ce{H2cit^2- + H2O \leftrightarrow Hcit^2- + H3O+} &&\mathrm{K_{a2} = 1.73\times 10^{-5}}\ &\ce{Hcit- + H2O \leftrightarrow cit^3- + H3O+} &&\mathrm{K_{a3} = 4.02\times 10^{-7}} \end{align}
Even though the first two Ka values are fairly close to each other, we can still use only the Ka1 expression to solve for the initial pH.
\begin{align} & &&\ce{H3cit}\hspace{25px} + \hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{H2cit-} \hspace{25px} + &&\ce{H3O+} &&\mathrm{K_{a1}}\ &\ce{Initial} &&0.1 &&0 &&0 && \ &\ce{Equilibrium} &&0.1 - \ce{x} &&\ce{x} &&\ce{x} && \ &\ce{Approximation} &&0.1 &&\ce{x} &&\ce{x} && \end{align}
$\mathrm{K_{a1}=\dfrac{[H_2cit^-][H_3O^+]}{[H_3cit]}=\dfrac{(x)(x)}{0.1}=7.45\times10^{-4}}$
$\mathrm{x = [H_3O^+] = 0.00863 \hspace{60px} pH = 2.06}$
If we check the approximation, it actually turns out that the value is too high and that we should have used a quadratic if we wanted the exact answer. But the value of 2.06 will suffice for now.
The next step is to consider what happens when we start adding sodium hydroxide to the solution. This will convert H3cit into the other forms, and we can start the process by assuming it will occur in a stepwise manner (in other words, H3cit will be converted into H2cit by the base until all the H3cit is used up, then H2cit will be converted into Hcit2, etc.). This would allow us to construct the chart shown in Table 3 of the moles of each species that would occur over the course of the titration.
Table 3. Moles of each species in the titration of citric acid (0.1 M, 30 ml) with NaOH (0.1M).
ml NaOH H3cit H2cit Hcit2 cit3 pH
0 0.0030 mol 0 0 0 2.06
5 0.0025 0.0005 0 0
10 0.0020 0.0010 0 0
15 0.0015 0.0015 0 0
20 0.0010 0.0020 0 0
25 0.0005 0.0025 0 0
30 0 0.0030 0 0
35 0 0.0025 0.0005 0
40 0 0.0020 0.0010 0
45 0 0.0015 0.0015 0
50 0 0.0010 0.0020 0
55 0 0.0005 0.0025 0
60 0 0 0.0030 0
65 0 0 0.0025 0.0005
70 0 0 0.0020 0.0010
75 0 0 0.0015 0.0015
80 0 0 0.0010 0.0020
85 0 0 0.0005 0.0025
90 0 0 0 0.0030
95 0 0 0 0.0030
100 0 0 0 0.0030
If we examine the increments from 5 ml to 25 ml, we see that we have appreciable quantities of H3cit and H2cit, which are both members of a conjugate pair. This region is a buffer solution and the pH can be determined using Ka1.
$\textrm{5 ml:} \hspace{32px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\right )=2.43}$
$\textrm{10 ml:}\hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0010\: mol}{0.0020\: mol}\right )=2.83}$
$\textrm{15 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = 3.128 + \log\dfrac{0.0015\: mol}{0.0015\: mol} = 3.128}$
Note that the pH at this increment is pKa1.
$\textrm{20 ml:} \hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0020\: mol}{0.0010\: mol}\right )= 3.43}$
$\textrm{25 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = 3.128 + \log\left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 3.83}$
30 ml: This is the first equivalence point, since we have converted all of the H3cit to H2cit. At this point we have “all” of the first intermediate (H2cit) and can calculate the pH using the expression (pK1 + pK2)/2
$\mathrm{pH = \dfrac{pK_{a1}+ pK_{a2}}{2}=\dfrac{3.128+4.761}{2}=3.9}$
If we examine the region from 35 to 55 ml, we have appreciable quantities of H2cit and Hcit2–, a buffer solution based on Ka2.
$\textrm{35 ml:} \hspace{25px} \mathrm{pH = pK_{a2}+ \log\dfrac{[Hcit^{2-}]}{[H_2cit^-]} = 4.761 + \log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\right ) = 4.06}$
$\textrm{40 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right ) = 4.46}$
$\textrm{45 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 4.761}$
Note that at this point, the pH is equal to pKa2.
$\textrm{50 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right ) = 5.06}$
$\textrm{55 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 5.46}$
60 ml: This is the second equivalence point, since we have converted all of the H2cit- to Hcit2-. At this point we have “all” of the second intermediate (Hcit2-) and can calculate the pH using the expression (pK2 + pK3)/2
$\mathrm{pH = \dfrac{pK_{a2}+ pK_{a3}}{2}=\dfrac{4.761 + 6.396}{2}= 5.58}$
If we examine the region from 65 to 85 ml, we have appreciable quantities of Hcit2- and cit3-, a buffer solution based on Ka3.
$\textrm{65 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0005\: mol}{0.0025\: mol}\right ) = 5.70}$
$\textrm{70 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right ) = 6.09}$
$\textrm{75 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 6.396}$
Note that at this point, the pH is equal to pKa3.
$\textrm{80 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right ) = 6.70}$
$\textrm{85 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 7.09}$
90 ml: This is the third equivalence point, since we have converted all of the Hcit2– to cit3–. To a first approximation we only have cit3– in solution. This is a polybasic base, but as we have done before, we only need to consider the first reaction in the series to calculate the pH. The relevant reaction, which is the Kb value of Ka3, is shown below.
cit3– + H2O = Hcit2– + OH Kb of Ka3 = 2.5×10-8
We need to calculate the concentration of cit3– that is present in solution, recognizing that the titrant caused a dilution of the initial concentration of citric acid (30 ml of initial solution and 90 ml of additional titrant).
$\textrm{Molarity of citrate = (0.0030 mol/0.120 L) = 0.025 M}$
\begin{align} & &&\ce{cit^3-}\hspace{25px} + \hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{Hcit^2-} \hspace{25px} + &&\ce{OH-} \ &\ce{Initial} &&0.025 &&0 &&0 \ &\ce{Equilibrium} &&0.025 - \ce{x} &&\ce{x} &&\ce{x} \ &\ce{Approximation} &&0.025 &&\ce{x} &&\ce{x} \end{align}
$\mathrm{K_{b3} =\dfrac{[Hcit^{2-}][OH^-]}{[cit^{3-}]}=\dfrac{(x)(x)}{0.025} = 2.5\times10^{-8}}$
$\mathrm{x = [OH^-] = 2.5\times 10^{-5}}$
$\mathrm{pOH = 4.6 \hspace{60px} pH = 9.4}$
Checking the approximation shows that it was valid in this case. Note that the pH at this equivalence point is basic, which is not surprising since cit3– is a base.
$\dfrac{2.5×10^{-5}}{0.025}× 100 = 0.1\%$
95 ml: In this case we have a mixture of a strong base (NaOH) with a weaker base (citrate). The extra amount of strong base (5 ml or 0.0005 moles) will determine the pH.
$\mathrm{[OH^-] =\dfrac{0.0005\: mol}{0.125\: L} = 4.0\times10^{-3}}$
$\mathrm{pOH = 2.4 \hspace{60px} pH = 11.6}$
100 ml: Once again, the pH is determined by the amount of extra strong base present in the solution (10 ml or 0.0010 moles).
$\mathrm{[OH^-] =\dfrac{0.0010\: mol}{0.130\: L} = 7.7\times10^{-3}}$
$\mathrm{pOH = 2.1 \hspace{60px} pH = 11.9}$
We can now compile an entire chart (Table 4) of the changes that occur over this titration:
Table 4. Calculated pH values for the titration of citric acid (0.1 M, 30 mL) with NaOH (0.1 M).
ml NaOH H3cit H2cit Hcit2 cit3 pH
0 0.0030 m 0 0 0 2.06
5 0.0025 0.0005 0 0 2.43
10 0.0020 0.0010 0 0 2.83
15 0.0015 0.0015 0 0 3.128 (pKa1)
20 0.0010 0.0020 0 0 3.43
25 0.0005 0.0025 0 0 3.83
30 0 0.0030 0 0 3.94 (pKa1+pKa2)/2
35 0 0.0025 0.0005 0 4.06
40 0 0.0020 0.0010 0 4.46
45 0 0.0015 0.0015 0 4.761 (pKa2)
50 0 0.0010 0.0020 0 5.06
55 0 0.0005 0.0025 0 5.46
60 0 0 0.0030 0 5.58 (pKa2+pKa3)/2
65 0 0 0.0025 0.0005 5.70
70 0 0 0.0020 0.0010 6.09
75 0 0 0.0015 0.0015 6.396 (pKa3)
80 0 0 0.0010 0.0020 6.70
85 0 0 0.0005 0.0025 7.09
90 0 0 0 0.0030 9.40
95 0 0 0 0.0030 11.60
100 0 0 0 0.0030 11.90
It is especially helpful to plot these values versus the ml of titrant as shown in Figure 3.
Figure 3. pH versus ml titrant for the titration of citric acid (0.1 M, 30 mL) with NaOH (0.1 M).
There are several things worth noting in this plot. One is the way that the first two equivalence points blend together and there are no clear breaks in the plot. The only equivalence point in this titration that is readily observable is the third. The other is to note that citric acid has a significant buffer region that stretches from a pH of about 2.5 to 5.5. Citric acid is commonly used as a buffer for this pH region.
It is also worth examining what would be observed for a similar plot of a different triprotic acid. The data in Table 5 is for an identical titration of phosphoric acid.
Table 5. pH values for the titration of phosphoric acid (0.1 M, 30 ml) with NaOH (0.1 M).
ml NaOH H3PO4 $\ce{H2PO4-}$ $\ce{HPO4^2-}$ $\ce{PO4^3-}$ pH
0 0.0030 mol 0 0 0 -
5 0.0025 0.0005 0 0 1.45
10 0.0020 0.0010 0 0 1.85
15 0.0015 0.0015 0 0 2.148 (pKa1)
20 0.0010 0.0020 0 0 2.45
25 0.0005 0.0025 0 0 2.85
30 0 0.0030 0 0 4.673 (pKa1+pKa2)/2
35 0 0.0025 0.0005 0 6.50
40 0 0.0020 0.0010 0 6.90
45 0 0.0015 0.0015 0 7.198 (pKa2)
50 0 0.0010 0.0020 0 7.50
55 0 0.0005 0.0025 0 7.90
60 0 0 0.0030 0 9.789 (pKa2+pKa3)/2
65 0 0 0.0025 0.0005 11.68
70 0 0 0.0020 0.0010 12.08
75 0 0 0.0015 0.0015 12.38 (pKa3)
80 0 0 0.0010 0.0020 12.68
85 0 0 0.0005 0.0025 -
90 0 0 0 0.0030 -
95 0 0 0 0.0030 11.60
100 0 0 0 0.0030 11.90
It is worth realizing that a few data points have been omitted since there is a problem at the beginning and again at about 75 ml of titrant. In the early part, the acid is strong enough that a fairly significant proportion dissociates. At the latter part of the titration, the base is so strong that we really do not convert all of the $\ce{HPO4^2-}$ to $\ce{PO4^3-}$ as implied. Even with this problem, we can examine a generalized plot for the titration of phosphoric acid with sodium hydroxide (Figure 4).
Figure 4. pH versus ml titrant for the titration of phosphoric acid (0.1 M, 30 mL) with NaOH (0.1 M).
Note here that the first two equivalence points are obvious, whereas the third equivalence point will not be distinguishable because of the very high pKa3 value. The concentration of citric or phosphoric acid can be determined through a titration with sodium hydroxide, provided you realize which equivalence points can be successfully monitored during the titration.
Now we can examine the last two parts of the homework problem. The first is whether 99.9% of the species is in the form cit3– at the third equivalence point. Going back to the calculation at 90 ml of titrant, we determined that [cit3–] was 0.025 M and [Hcit2–] was 2.5×10-5 M.
$\dfrac{2.5 \times 10^{-5}}{0.025} \times 100 = 0.1\%$
If 0.1% is in the form Hcit2–, then 99.9% is in the form cit3– and it just makes it.
The other part of the problem was to calculate the concentration of all species in solution at the second equivalence point. The first thing we ought to do is compile a list of what all the species are so we know what we have to calculate. In doing this, we can ignore any spectator ions such as sodium. That means there are six species whose concentration we need to calculate.
H3cit H2cit Hcit2– cit3– H3O+ OH
Since we already calculated the pH of this solution (5.58), we can readily calculate the concentration of H3O+ and OH.
[H3O+] = 2.63×10-6
[OH] = 3.80×10-9
We also said that to a first approximation it was “all” Hcit2– (0.0030 mol). With 30 ml of initial solution and 60 ml of titrant, we have a total volume of 90 ml.
[Hcit2–] = 0.0030 mol/0.090 L = 0.033 M
Since we now have [H3O+] and [Hcit2–], we can use the appropriate Ka expressions to calculate the three other citrate species.
Use the Ka3 expression to calculate [cit3–]:
$\mathrm{K_{a3} =\dfrac{[cit^{3-}][H_3O^+]}{[Hcit^{2-}]}=\dfrac{[cit^{3-}](2.63\times10^{-6})}{0.033} = 4.02\times10^{-7}}$
$\mathrm{[cit^{3-}] = 0.005\: M}$
Use the Ka2 expression to calculate [H2cit]:
$\mathrm{K_{a2} =\dfrac{[Hcit^{2-}][H_3O^+]}{[H_2cit^{2-}]}=\dfrac{(0.033)(2.63\times10^{-6})}{[Hcit^{2-}]} = 1.733\times10^{-5}}$
$\mathrm{[H_2cit^–] = 0.005\: M}$
Using the value of H2cit that was just calculated, we can substitute this into the Ka1 expression and calculate the concentration of H3cit.
$\mathrm{K_{a1} =\dfrac{[H_2cit^{2-}][H_3O^+]}{[H_3cit]} =\dfrac{(0.005)(2.63\times10^{-6})}{[H_3cit]} = 7.45\times10^{-4}}$
$\mathrm{[H_3cit] = 1.77 \times 10^{-5}}$
One last set of things to examine are the calculated values for [H2cit], [Hcit2–] and [cit3–].
[Hcit2–] = 0.033 M
[H2cit] = 0.005 M
[cit3–] = 0.005 M
What we need to appreciate is that there is a problem with these numbers. We started this calculation by claiming that “all” of the material was in the form of Hcit2–. What these calculations show is that there are appreciable amounts of H2cit and cit3– in solution. If these two values were accurate, it would mean that the concentration of Hcit2– could only be 0.023 M.
Why does this happen? It has to do with how close the pKa values are for citric acid. The approximation that we could examine this as a stepwise manner, where we proceeded from one reaction to the other and that intermediates were overwhelmingly the predominant form at the equivalence points, breaks down in this case because of how close the pKa values are. The interesting part of this is that the pH of the solution would be 5.58, and that we will get exactly equivalent concentrations of H2cit and cit3–, although they will not be exactly 0.005 M. In reality, I do not think we would ever try to calculate the exact amount of each of these species at a pH like this, although later in the course we are going to come back to the citric acid situation and see a way to calculate the exact concentration of species present provided we know the pH.
One thing to keep in mind is that often times we do not use equilibrium calculations to arrive at exact values of substances. For one thing, concentrations are an approximation of activities and this may not always be a good one. For another, we often use equilibrium calculations to provide ballpark values to let us know whether a particular process we may be considering is even feasible. In this case, for example, these values show that we could never use the second equivalence point in a citric acid titration for measurement purposes. This does not mean that citric acid cannot be used as a buffer, because it frequently is. However, if we prepare a citric acid buffer (or any buffer), we do not rely on calculated amounts to ensure that the pH is where we want it, we use a pH meter to monitor the buffer and use small amounts of a strong acid or strong base to adjust the pH to the value we want. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/02_Acid-Base_Chemistry/14_Titration_of_a_Polyprotic_Weak_Acid_with_Sodium_Hydroxi.txt |
In-class Problem Set #4
1. Calculate the concentration of free calcium(II) ions in a solution prepared with initial concentrations of calcium of 0.020 M and EDTA4 of 0.10 M.
This is our first example involving a water-soluble metal complex. The general form of these equations is that a metal ion (M) reacts with a ligand (L) to form a water-soluble complex.
$\mathrm{M(aq) + L(aq) \leftrightarrow ML(aq) \hspace{60px} K_f}$
Usually we appreciate that these are all water-soluble species and omit the (aq) notation from the equations. The general form of the equation is referred to as a formation constant, hence the notation Kf for the equilibrium constant. If you examine the values of Kf in the table, you will see that many of these are reasonably large values. In fact, because they are large, it is common for tables to report the logKf value rather than the Kf value. It makes sense that many of these in the table would have large values because people would be interested in metal complexes that had relatively high formation constants if they wanted to use them for analytical purposes. Another thing to realize is that many ligands can form a series of stepwise complexes with a metal, as illustrated below.
\begin{align} &\ce{M + L \leftrightarrow ML} &&\mathrm{K_{f1}}\ &\ce{ML + L \leftrightarrow ML_2} &&\mathrm{K_{f2}}\ &\ce{ML_2 + L \leftrightarrow ML_3} &&\mathrm{K_{f3}}\ &\ce{ML_3 + L \leftrightarrow ML_4} &&\mathrm{K_{f4}} \end{align}
It is also worth examining what types of species function as ligands. First, it’s worth realizing that the metal in these reactions is almost always a cation. Therefore, anions are one group of compounds that have to be examined as possible ligands. The way to see if complexation occurs is to determine all of the anions and metal ions that exist in a particular solution, and then see if any possible combination of a metal ion with an anion has a value or values in the table of formation constants. If any do, then that process needs to be incorporated into any calculations of equilibrium concentrations. One thing to note is that every aqueous solution has some amount of hydroxide ion, and many metal ions form water-soluble complexes with hydroxide. The presence of hydroxide ion therefore provides an additional complication when assessing the distribution of most metal species in solution.
Another thing to realize is that anions are always the conjugate bases of acids. In other words, any anion has the potential to be protonated with a hydrogen ion. If the anion is the conjugate base of a strong acid (halide, nitrate, perchlorate), then its concentration will not vary as a function of pH because it will not be protonated in water. If the anion is the conjugate base of a weak acid, which is actually far more common, then the concentration of the anion in water is a function of the pH. This means that the complexation of the metal by the anion will depend on the pH as well. What you might begin to realize now is that metal complexation in water is a complicated process that is influenced by the pH and the availability of different ligands. We will see that there are a series of systematic ways to address metal complexation and handle all of the simultaneous equilibria that occur.
The other large group of ligands are the nitrogen bases. These are neutral ligands that have the ability to form donor-acceptor complexes with the metal ion. The ligand acts as an electron-pair donor, something we refer to as a Lewis base. The positive metal ion acts as an electron-pair acceptor, something we refer to as a Lewis acid. (Actually, the Lewis acid-Lewis base interaction also occurs for anions when they bond to metals). For example, ammonia is a ligand that forms water-soluble complexes with many metal ions. Ethylenediamine is another important nitrogen-containing ligand that forms water-soluble complexes with many metal ions. In this case, what is particularly interesting is that both nitrogen atoms bond to the metal ion at the same time, forming what is called a chelate complex.
Anions can bond to metal ions in a chelate manner as well. For example, the carboxylate ion actually bonds to metal ions through a chelate arrangement of the two oxygen atoms.
Another very important ligand is the species ethylenediaminetetraacetic acid (H4EDTA or H4E). EDTA is a tetraprotic acid that can act as a ligand and form very stable chelate complexes with metal ions. We have a table of Kf values for different EDTA complexes and note that these are very large numbers. An interesting thing to realize is that we always think of the fully deprotonated ligand (E4–) as the species that actually forms the complex. The other forms (H4E, H3E, H2E2–, HE3–) are not involved in the complexation. Note that this is a general observation. For example, in the phosphoric acid system, the species $\ce{PO4^3-}$ would be the ligand, not the partially protonated $\ce{HPO4^2-}$ and $\ce{H2PO4^2-}$ species.
Note that one E4– ligand has the ability to completely surround a metal ion and fill all of the coordination sites simultaneously. It might be tempting to think that this ability to fill all the sites at one time accounts for the exceptionally large formation constants. In actuality, the large formation constants are driven more by entropy changes. If we think about the nature of a metal cation in water, we realize that the positive metal ion is surrounded by negative ends of water molecules as shown below. In this picture, the oxygen atoms of the water molecules are said to be in the first coordination sphere of the metal ion. This is typically what occurs with many metal salts in water. The cation and anion separate from each other and are solvated by water.
The reaction that we can now write to represent the bonding of E4– to a metal ion (M+) is shown below:
$\ce{M(H2O)6+ + E^4- \leftrightarrow ME^3- + 6 H2O}$
Note that the reaction side of the equation has two species, the product side has seven species. There is considerably larger entropy associated with the seven species on the product side, and this huge gain in favorable entropy is what primarily accounts for the large formation constants of metal complexes with EDTA. We will rarely write the metal ion as $\ce{M(H2O)6+}$ and instead simplify it to M+. But it’s sometimes worth remembering that it is surrounded by some number of water molecules. Also note that the ME3– species has a net charge of –3. Many of these water-soluble complexes have a net charge, which in part is responsible for their water solubility.
One last thing about EDTA. If you look in the structures shown in our table, you will see that EDTA is actually a zwitterion in solution (See below a neutral and zwitterionic form of EDTA). What’s important to realize is that it really does not matter what form you write it in. Both forms below have four dissociable protons. Both forms have a net neutral charge. Both can be expressed as H4EDTA.
In theory, it might actually be possible to protonate both nitrogen atoms of EDTA, producing a species of the form H6EDTA2+.
H6EDTA2+
This species does not occur because the first two protons are so acidic that, even if we dissolved EDTA in a strong acid such as concentrated hydrochloric acid, it is doubtful that these sites would be protonated. In water, at pH values of 1 or higher, we will never be able to put more than four protons onto the EDTA.
Now we are ready to calculate the answer to the first problem.
02 Concentration of Unreacted Metal Ion
Calculate the concentration of free calcium(II) ion in a solution prepared with initial concentrations of calcium of 0.020 M and EDTA4- of 0.10 M.
In solving this problem, we will start under very naïve circumstances. We will not consider any other complexation of the calcium ion (for example, by something like hydroxide), and we have been given a concentration of E4– and do not need to worry about the pH of this solution, how we got a concentration of 0.10 M, or whether protonation occurs.
The first thing to do is to write the reaction and look up the relevant formation constant.
$\ce{Ca^2+ + E^4- \leftrightarrow CaE^2-} \hspace{60px} \mathrm{K_f = 5.0\times 10^{10}}$
What we see is that has a very large formation constant. That means that this reaction will go to completion. The approach to solving this problem is to allow it to go to completion, realizing that one of the reagents will limit the amount of product that forms. Then we need to allow a small amount of back reaction to occur. We can construct the following chart.
\begin{align} & &&\ce{Ca^2+}\hspace{25px} + &&\ce{E^4-} \hspace{25px}\leftrightarrow &&\ce{CaE^2-} \ &\ce{Initial} &&\mathrm{0.020\: M} &&\mathrm{0.10\: M} &&0 \ &\ce{Complete\: Reaction} &&0 &&0.08 &&0.02 \ &\ce{Back\: Reaction} &&\ce{x} &&0.08 +\ce{x} &&0.02 - \ce{x} \ &\ce{Assumption} && &&0.08 \gg\ce{x} &&0.02 \gg\ce{x} \ &\ce{Approximation} &&\ce{x} &&0.08 &&0.02 \end{align}
The approximation that 0.08 $\gg$ x and 0.02 $\gg$ x is reasonable since the Kf value is so large, therefore the amount of back reaction will be excessively small.
$\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E^{4-}]} =\dfrac{0.02}{(x)(0.08)} = 5.0\times10^{10}}$
$\mathrm{x = [Ca^{2+}] = 5\times 10^{-12}}$
The amount of unreacted calcium ion is incredibly small. Obviously the approximations we made were justified and we see how far these reactions can go toward completion. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/03_Water-Soluble_Complexes/01_Introduction_and_Nomenclature.txt |
2. Calculate the concentration of free calcium(II) ions in a solution prepared with initial concentrations of calcium of 0.020 M and total EDTA of 0.10 M. The solution is buffered at a pH of 2.
Now we have added a complicating factor. We will incorporate the fact that the EDTA has several protonated forms, and that these forms depend on the pH of the solution. Since only the E4– form bonds to the metal ion, protonated forms of EDTA reduce the concentration of E4– available for complexation. If very little E4– is available for complexation, very little of the calcium ion will be complexed.
It is also worth realizing that, in most cases, we do know the pH of a solution. One reason is because we want a particular pH so we have prepared the solution in a buffer. The other is that it is very easy to measure the pH of a solution using a pH meter, so if pH is a relevant issue, we simply measure it.
The set of reactions below show what we now know will occur in this solution (note, we are ignoring the possibility that the calcium can complex with hydroxide ion – more on that later).
\begin{align} \ce{Ca^2+ +\: &E^4- \leftrightarrow CaE^2- \ &\:\:\Updownarrow\nonumber\ &HE^3- \nonumber\ &\:\:\Updownarrow\nonumber\ &H2E^2- \nonumber\ &\:\:\Updownarrow\nonumber\ &H3E- \nonumber\ &\:\:\Updownarrow\nonumber\ &H4E\nonumber} \end{align}
The problem we face here is that the reaction we want to examine is the Kf reaction for CaE2–. If we want to use our established way of doing that calculation, we need to know the initial amounts of Ca2+ and E4– that we have in solution, but some of the E4– has been protonated and we do not know how much we have. Also, we have one other problem. Suppose, in a solution with a total amount of EDTA of 0.10 M, we could calculate how much of the EDTA was in the E4– form. We could conceivably allow this to complex with the Ca2+, but we have an additional source of EDTA (the HE3, H2E2, H3E, and H4E forms) that will redistribute to some extent and provide additional amounts of E4–.
There is a very interesting observation about acid-base systems. It turns out that if we know the pH of the solution, the fraction of the total that exists in any one form is fixed. In other words, the fraction of total EDTA that exists in the H4E, H3E, H2E2, HE3, and E4– forms is only a function of the pH of the solution. It does not depend on the total amount of EDTA in solution. We refer to these fractions as α-values. We are usually interested in the α-value for the fully deprotonated anion, since that is the form that complexes with the metal. It’s important to realize that we can calculate α-values for any of the species involved in a series of reactions for a polyprotic acid.
It is worth showing that the fraction of EDTA that exists in solution as E4– is only a function of pH. This will also demonstrate the general procedure that we can use to calculate α-values.
First, write an expression for the fraction of EDTA that exists in solution as E4–.
$\mathrm{α_{E^{4-}} =\dfrac{[E^{4-}]}{[H_4E] + [H_3E^-] + [H_2E^{2-}] + [HE^{3-}] + [E^{4-}]}}$
Note that this is just the concentration of E4– over the total EDTA in solution. The next step is to take the reciprocal of this expression. Doing so will allow you to divide the equation into a set of separate terms.
$\mathrm{\dfrac{1}{α_{E^{4-}}}=\dfrac{[H_4E]}{[E^{4-}]}+\dfrac{[H_3E^-]}{[E^{4-}]}+\dfrac{[H_2E^{2-}]}{[E^{4-}]}+\dfrac{[HE^{3-}]}{[E^{4-}]}+\dfrac{[E^{4-}]}{[E^{4-}]}}$
The next step is to use the Ka values for EDTA to substitute in for each of the ratio terms. The first one to examine is the ratio of [HE3]/[E4–], which we can obtain using only the Ka4 expression. Rearranging Ka4 as shown below gives the following term to substitute in.
$\mathrm{HE^{3-} + H_2O \leftrightarrow E^{4-} + H_3O^+ \hspace{60px} K_{a4}}$
$\mathrm{K_{a4} =\dfrac{[E^{4-}][H_3O^+]}{[HE^{3-}]} \hspace{60px} \dfrac{[HE^{3-}]}{[E^{4-}]} =\dfrac{[H_3O^+]}{K_{a4}}}$
Next we can evaluate an expression to substitute in for [H2E2]/[E4–]. This will involve using the Ka3 and Ka4 expression for EDTA. The easiest way to see this is to add up the Ka3 and Ka4 reactions. Remember, the equilibrium constant for the resulting reaction is the product of the equilibrium constants for those added together.
\begin{align} &\ce{H2E^2- + H2O \leftrightarrow HE^3- + H3O+} &&\mathrm{K_{a3}}\ &\underline{\ce{HE^3- + H2O \leftrightarrow E^4- + H3O+}\hspace{20px}} &&\mathrm{K_{a4}}\nonumber\ &\ce{H2E^2- + 2H2O \leftrightarrow E^4- + 2H3O+} &&\mathrm{K = K_{a3} K_{a4}}\nonumber \end{align}
$\mathrm{K_{a3} K_{a4} =\dfrac{[E^{4-}][H_3O^+]^2}{[H_2E^{2-}]} \hspace{60px} \dfrac{[H_2E^{2-}]}{[E^{4-}]} = \dfrac{[H_3O^+]^2}{K_{a3} K_{a4}}}$
Perhaps by this point we see a pattern developing. Evaluating the [H3E]/[E4–] term will require the use of Ka2, Ka3, and Ka4 and yield the following term:
$\mathrm{\dfrac{[H_3E^-]}{[E^{4-}]} = \dfrac{[H_3O^+]^3}{K_{a2} K_{a3} K_{a4}}}$
Similarly, evaluating the [H4E]/[E4–] term will, which requires using Ka1, Ka2, Ka3, and Ka4, will yield the following term:
$\mathrm{\dfrac{[H_4E]}{[E^{4-}]} = \dfrac{[H_3O^+]^4}{K_{a1} K_{a2} K_{a3} K_{a4}}}$
The final result is shown below.
$\mathrm{\dfrac{1}{α_{E^{4-}}} = \dfrac{[H_3O^+]^4}{K_{a1} K_{a2} K_{a3} K_{a4}} + \dfrac{[H_3O^+]^3}{K_{a2} K_{a3} K_{a4}} + \dfrac{[H_3O^+]^2}{K_{a3} K_{a4}} + \dfrac{[H_3O^+]}{K_{a4}} + 1}$
What we see is that the only variable in this expression is [H3O+], so the fraction of EDTA that exists in solution as E4– is only a function of the pH. There is no term for the total amount of EDTA in the equation, so that does not matter. It only depends on the pH.
Similarly, we can evaluate the fraction of the other species as a function of pH. For example, let’s begin the process of evaluating the α-value of H4E. The general procedure is the same as used for $\mathrm{α_{E^{4-}}}$. The first step is to write the relevant equation for H4E over the total.
$\mathrm{α_{H_4E} =\dfrac{[H_4E]}{[H_4E] + [H_3E^-] + [H_2E^{2-}] + [HE^{3-}] + [E^{4-}]}}$
The next step is to take the reciprocal and divide the equation into separate terms.
$\mathrm{\dfrac{1}{α_{H_4E}}=\dfrac{[H_4E]}{[H_4E]} + \dfrac{[H_3E^-]}{[H_4E]} + \dfrac{[H_2E^{2-}]}{[H_4E]} + \dfrac{[HE^{3-}]}{[H_4E]} + \dfrac{[E^{4-}]}{[H_4E]}}$
The next step is to use the Ka expressions to express each ratio in terms of [H3O+] and the Ka values. I will not show all these rearrangements here, but doing them yields the following result for 1/$\mathrm{α_{H_4E}}$. You ought to try this and convince yourself that this is correct.
$\mathrm{\dfrac{1}{α_{H_4E}}= 1 +\dfrac{K_{a1}}{[H_3O^+]} + \dfrac{K_{a1} K_{a2}}{[H_3O^+]^2} +\dfrac{K_{a1} K_{a2} K_{a3}}{[H_3O^+]^3} + \dfrac{K_{a1} K_{a2} K_{a3} K_{a4}}{[H_3O^+]^4}}$
Similarly, we could write expressions, take reciprocals, and evaluate the terms for α-values for H3E, H2E2–, and HE3–. The full 1/α-value expressions that would result for all of the species involved in the EDTA system are shown below. Note the characteristic patterns that result.
$\mathrm{\dfrac{1}{α_{E^{4-}}}= \dfrac{[H_3O^+]^4}{K_{a1}K_{a2}K_{a3}K_{a4}} + \dfrac{[H_3O^+]^3}{K_{a2}K_{a3}K_{a4}} +\dfrac{[H_3O^+]^2}{K_{a3}K_{a4}} + \dfrac{[H_3O^+]}{K_{a4}}+1 }$
$\mathrm{\dfrac{1}{α_{HE^{3-}}}= \dfrac{[H_3O^+]^3}{K_{a1}K_{a2}K_{a3}} + \dfrac{[H_3O^+]^2}{K_{a2}K_{a3}} +\dfrac{[H_3O^+]}{K_{a3}} +1 + \dfrac{K_{a4}}{[H_3O^+]} }$
$\mathrm{\dfrac{1}{α_{H_2E^{2-}}}= \dfrac{[H_3O^+]^2}{K_{a1}K_{a2}} + \dfrac{[H_3O^+]}{K_{a2}}+1 +\dfrac{K_{a3}}{[H_3O^+]} + \dfrac{K_{a3}K_{a4}}{[H_3O^+]^2} }$
$\mathrm{\dfrac{1}{α_{H_3E^-}}= \dfrac{[H_3O^+]}{K_{a1}} +1 + \dfrac{K_{a2}}{[H_3O^+]} +\dfrac{K_{a2}K_{a3}}{[H_3O^+]^2} + \dfrac{K_{a2}K_{a3}K_{a4}}{[H_3O^+]^3} }$
$\mathrm{\dfrac{1}{α_{H_4E}}= 1 + \dfrac{K_{a1}}{[H_3O^+]} +\dfrac{K_{a1}K_{a2}}{[H_3O^+]^2} + \dfrac{K_{a1}K_{a2}K_{a3}}{[H_3O^+]^3} + \dfrac{K_{a1}K_{a2}K_{a3}K_{a4}}{[H_3O^+]^4}}$
The other important thing to do with α-values is examine a plot of α-values versus pH for a series of compounds. Examples are shown in Figure 5 for carbonic acid, phosphoric acid, and citric acid.
Carbonic Acid (pKa values of 6.4 and 10.4)
Phosphoric Acid (pKa values of 2.12, 7.21, and 12.66)
Citric Acid (pKa values of 3.13, 4.76, and 5.41)
Figure 5. Plots of α-values as a function of pH for carbonic, phosphoric and citric acid.
There are several important items to note about these figures on the previous page. At very low pH, meaning highly acidic conditions, the fully protonated forms predominate. At very high pH, meaning highly basic conditions, the fully deprotonated forms predominate. Since the deprotonated forms are the ones that will bond to metal ions, complexation of metal ions by ligands is favored at more basic pH values. (Note, we could do a similar plot of α-values for the different forms of ethylenediamine, H2En2+, HEn+, and En. Remembering that the neutral form, En, is the one that complexes with metals, we would see that this form is favored at more basic pH values). At intermediate pH values, different intermediate forms increase and then diminish as the pH is raised from acidic to basic conditions. At the crossing points of two of the α-value plots, we typically have a 50:50 mixture of a conjugate pair. In other words, these crossing points are the excellent buffer regions for these reagents.
Observe that the forms of the carbonic and phosphoric acid systems are very regularized. There is some pH where only one species of the series predominates and the concentrations of all others are minimal. This is not the case with citric acid. We see a somewhat unusual situation in which at a pH of around 4 and 5, we see that the H2cit and Hcit2– forms respectively do not reach 99%. At these points we find appreciable amounts of the neighboring species. Also note that the amounts of H2cit and cit3– present at the point at which Hcit2– is maximized are exactly equal to each other. The same thing occurs for the two neighboring species at pH 4. This is a rather rare occasion of having appreciable quantities of three species from an acid-base system present at the same time. The reason for this unusual behavior is that the pKa values are very close to each other. (Note: pKa3 in this chart is reportedly 5.41, which is different than the value of 6.396 in our table of K values!) We have already examined citric acid when it was titrated with sodium hydroxide and introduced its somewhat unusual behavior. The plot of α-values really points out the effects of having close pKa values and how this influences the concentrations of species in solution.
We are now in a position to finally see how to incorporate an α-value into the calculation involving complexation of Ca2+ by E4– at a pH of 2. Write the formation constant expression and recognize that we can substitute in for the [E4–] term.
$\ce{Ca^2+ + E^4- \leftrightarrow CaE^2-}\hspace{60px} \mathrm{K_f}$
$\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E^{4-}]}}$
$\mathrm{[E^{4-}] =α_{E^{4-}}[E]_{TOT}}$
Substituting the [E4–] expression into the Kf equation gives the following:
$\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}]α_{E^{4-}}[E]_{TOT}}}$
The $\mathrm{α_{E^{4-}}}$ term is a constant, since the pH is known, and we can rearrange the expression into the following:
$\mathrm{(K_f)(α_{E^{4-}}) =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E]_{TOT}}}$
$\mathrm{(K_f)(α_{E^{4-}})}$ is something we call a conditional constant (KCOND). Why a conditional constant? It turns out that the pH is a “condition” in this solution that influences the concentration and availability of E4–. Incorporating the α-value into the conditional constant will allow us to assess whether it is likely that the Ca2+ will complex with the E4–. The conditional constants for the complexation of Ca2+ with E4– are shown in Table 6 as a function of pH. What we need to do is calculate the conditional constant, and then examine its magnitude. If the conditional constant is large, the reaction goes to completion. If the conditional constant is small, the reaction does not go to completion.
Table 6. Conditional constants for the complexation of Ca2+ with E4-.
pH $\mathrm{α_{E^{4-}}}$ $\mathrm{(K_f)(α_{E^{4-}})}$ Extent of reaction
1 3.66×10-18 1.83×10-7 Very small
2 2.00×10-14 1.00×10-3 Fairly small
3 1.61×10-11 0.805 Intermediate
4 2.48×10-9 1.24×102 Intermediate
5 2.47×10-7 1.24×104 Close to completion
6 1.67×10-5 8.35×106 Completion
7 3.89×10-4 1.95×107 Completion
8 4.47×10-3 2.24×108 Completion
9 4.36×10-2 2.18×109 Completion
10 0.314 1.57×1010 Completion
11 0.820 4.10×1010 Completion
12 0.979 4.90×1010 Completion
13 0.998 4.99×1010 Completion
Note how the reaction, based on the magnitude of the conditional constant, goes further to completion as the pH is made more basic. At very acidic pH values, very little reaction occurs. Also note that the conditional constant is large by a pH of 6, even though the α-value for the E4– is still fairly small (1.67×10-5). This shows how the very large formation constant (5×1010) leads to formation of the complex (as the E4– is used up, we have a source of additional E4– from the protonated E species HE3–, H2E2–, H3E, and H4E). But also note, the α-values for the E species do not change as long as the pH remains fixed. If E4– is removed by complexation, some new E4– will form to maintain the same distribution of α-values for all of the E species.
If we have the expression:
$\mathrm{(K_f)(α_{E^{4-}}) =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E]_{TOT}}}$
We could write this as belonging to the following reaction. Note that the procedure we use assumes that only a small amount of reaction occurs since the conditional constant is only 1×10-3.
\begin{align} & &&\ce{Ca^2+}\hspace{25px} + &&\ce{E_{TOT}} \hspace{25px}\leftrightarrow &&\ce{CaE^2-} &&\mathrm{K_{COND} = (K_f)(α_{E^{4-}}) = 1\times 10^{-3}} \ &\ce{Initial} &&0.02 &&0.10 &&0 && \nonumber\ &\ce{Equilibrium} &&0.02 - \ce{x} &&0.10 - \ce{x} &&\ce{x} && \nonumber\ &\ce{Approximation} &&0.02 &&0.10 &&\ce{x} &&\nonumber \end{align}
Substitute these in to calculate the value of [CaE2–]:
$\mathrm{K_{COND} = \dfrac{[CaE^{2-}]}{[Ca^{2+}][E]_{TOT}} =\dfrac{x}{(0.02)(0.1)} = 1\times10^{-3}}$
$\mathrm{x = [CaE^{2-}] = 2\times 10^{-6}}$
Checking the approximation shows that it was valid to assume that very little of the Ca2+ and E4– complexes.
$\dfrac{2×10^{-6}}{0.02} × 100 = 0.01\%$
Let’s also consider how we would handle this if we had a pH with a large conditional constant. For example, consider the situation at pH 6. In this case, we treat it assuming that the reaction goes to completion and that some back reaction then occurs.
\begin{align} & &&\ce{Ca^2+}\hspace{25px} + &&\ce{E_{TOT}} \hspace{25px}\leftrightarrow &&\ce{CaE^2-} &&\mathrm{K_{COND} =8.35\times 10^6} \ &\ce{Initial} &&0.02 &&0.10 &&0 && \nonumber\ &\ce{Completion} &&0 &&0.08 &&0.02 && \nonumber\ &\ce{Back\: reaction} &&\ce{x} &&0.08 + \ce{x} &&0.02 - \ce{x} && \nonumber\ &\ce{Approximation} &&\ce{x} &&0.08 &&0.02 &&\nonumber \end{align}
Substituting this in gives the following concentration of free calcium ion. This concentration is very low such that we know the approximations were valid.
$\mathrm{K_{COND} =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E]_{TOT}} =\dfrac{0.02}{(x)(0.08)} = 8.35×10^6}$
$\mathrm{x = [Ca^{2+}] = 2.99\times 10^{-8}}$
An important thing to notice is that the total concentration of EDTA drops in this process because some of it reacts with the Ca2+ to form the complex. Remember that [E]TOT refers only to those forms of EDTA that are not complexed with Ca2+ and does not include the complexed form (CaE2–). | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/03_Water-Soluble_Complexes/03_Utilization_of_-Values_for_the_Ligand.txt |
There is one other aspect to this problem we have not yet considered. That is whether calcium can complex with the hydroxide ion and whether this complexation is significant enough to alter any of the results we have seen before regarding complexation of Ca2+ by E4-. Looking in the table of formation constants indicates that calcium ions can complex with hydroxide according to the following equation. Since only one Kf value is listed, it is only a one-step process. We might also notice that it’s a fairly small association constant, so that we might anticipate that this reaction would never represent that much of an interference in the complexation of Ca2+ with E4-.
$\ce{Ca^2+ + OH- \leftrightarrow Ca(OH)+} \hspace{60px} \mathrm{K_f = 1.99×10}$
We can couple this process into the overall scheme as shown below:
\begin{align} &\ce{Ca^2+} \hspace{25px} + &&\ce{E^4-} \hspace{25px}\leftrightarrow \hspace{25px}\ce{CaE^2-} \ \ce{OH-} &\:\Updownarrow &&\:\Updownarrow\nonumber\ &\ce{Ca(OH)+} &&\ce{HE^3-} \nonumber\ & &&\:\Updownarrow\nonumber\ & &&\ce{H2E^2-} \nonumber\ & &&\:\Updownarrow\nonumber\ & &&\ce{H3E-}\nonumber\ & &&\:\Updownarrow\nonumber\ & &&\ce{H4E}\nonumber \end{align}
The approach we will use is analogous to that employed with the protonation of E4–. If we know the concentration of ligand, it is possible to calculate α-values for the uncomplexed metal ion and the metal-ligand species. In this case of hydroxide ion, the concentration is known and fixed provided the pH is known and fixed. With other ligands, we may need to assess whether the initial ligand concentration we are provided remains fixed. In some cases, the ligand will complex with the metal and this causes the concentration to drop from its initial value, changing the α-values that were calculated.
For the situation in this problem, we need to calculate $\mathrm{α_{Ca^{2+}}}$. We do this by setting up a ratio of Ca2+ to the total of other calcium species. There is one important thing to realize in setting up this ratio. We only want to look at the distribution of calcium species in the set of reactions involving complexation with hydroxide ion. We therefore do not include CaE2 as a term in the ratio.
$\mathrm{α_{Ca^{2+}} = \dfrac{[Ca^{2+}]}{[Ca(OH)^+] + [Ca^{2+}]}}$
The next step is to take the reciprocal, and divide the equation into a series of separate terms.
$\mathrm{\dfrac{1}{α_{Ca^{2+}}} = \dfrac{[Ca(OH)^+]}{[Ca^{2+}]}+\dfrac{[Ca^{2+}]}{[Ca^{2+}]}}$
We can now use the Kf expression to substitute in for the first term in this equation. Using Kf for complexation of Ca2+ with hydroxide, we get as follows:
$\ce{Ca^2+ + OH- \leftrightarrow Ca(OH)+} \hspace{60px} \mathrm{K_f = 1.99×10}$
$\mathrm{K_f =\dfrac{[Ca(OH)^+]}{[Ca^{2+}][OH^-]}}$
Rearrange the Kf expression as follows:
$\mathrm{\dfrac{[Ca(OH)^+]}{[Ca^{2+}]} = K_f[OH^-]}$
Substitute this into the 1/$\mathrm{α_{Ca^{2+}}}$ expression to get:
$\mathrm{\dfrac{1}{α_{Ca^{2+}}} = K_f[OH^-] + 1}$
What we see is that the fraction of calcium that exists as Ca2+ only depends on the Kf value and the concentration of ligand (hydroxide in this case). We could also write the following expression:
$\mathrm{[Ca^{2+}] = α_{Ca^{2+}}[Ca]_{TOT}}$
Remember, [Ca]TOT = [Ca2+] + [Ca(OH)+] in this expression.
We can substitute this into our original Kf expression for the complexation of Ca2+ with E4–, just as we did previously to account for the protonation of EDTA as a function of pH.
$\ce{Ca^2+ + E^4- \leftrightarrow CaE^2-}$
$\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E^{4-}]}}$
$\mathrm{K_f =\dfrac{[CaE^{2-}]}{(α_{Ca^{2+}}[Ca]_{TOT})(α_{E^{4-}}[E]_{TOT})}}$
$\mathrm{(K_f)(α_{Ca^{2+}})(α_{E^{4-}}) =\dfrac{[CaE^{2-}]}{[Ca]_{TOT}[E]_{TOT}}}$
This provides a conditional constant $\mathrm{(K_f)(α_{Ca^{2+}})(α_{E^{4-}})}$ that incorporates both conditions that are present: protonation of the E4– and complexation of Ca2+ by hydroxide ion. This conditional constant is essentially the equilibrium constant for the following reaction:
$\ce{[Ca]_{TOT} + [E]_{TOT} \leftrightarrow CaE^2-}$
What we then need to do is examine the magnitude of this conditional constant to assess whether the complexation of calcium with EDTA will occur.
Table 7 is a compilation of $\mathrm{α_{E^{4-}}}$, $\mathrm{α_{Ca^{2+}}}$, conditional constants, and extent of reaction for this entire process as a function of pH.
Table 7. Conditional constants for the complexation of Ca2+ with E4-.
pH $\mathrm{α_{E^{4-}}}$ $\mathrm{α_{Ca^{2+}}}$ $\mathrm{(K_f)(α_{E^{4-}})(α_{Ca^{2+}})}$ Extent of reaction
1 3.66×10-18 1 1.83×10-7 Very small
2 2.00×10-14 1 1.00×10-3 Fairly small
3 1.61×10-11 1 0.805 Intermediate
4 2.48×10-9 1 1.24×102 Intermediate
5 2.47×10-7 1 1.24×104 Close to completion
6 1.67×10-5 1 8.35×106 Completion
7 3.89×10-4 1 1.95×107 Completion
8 4.47×10-3 1 2.24×108 Completion
9 4.36×10-2 1 2.18×109 Completion
10 0.314 0.998 1.57×1010 Completion
11 0.820 0.980 4.02×1010 Completion
12 0.979 0.834 4.09×1010 Completion
13 0.998 0.334 1.67×1010 Completion
First, let’s consider the situation at pH 2. The $\mathrm{α_{Ca^{2+}}}$ value is 1, which means that essentially none of the Ca2+ is complexed with the hydroxide ion. This makes sense since there is a very low level of hydroxide ion at pH 2 ([OH-] = 10-12) and because the Kf value for calcium complexation with hydroxide is not that large. The complexation of calcium by hydroxide has no significant effect on the system at this pH. As the pH becomes more basic, notice how $\mathrm{α_{Ca^{2+}}}$ eventually falls below 1. This means that some of the calcium ion will complex with hydroxide. But if we examine the overall conditional constant, we also see that there is so much E4– available at the more basic pH values, that complexation with hydroxide is never sufficient enough to overcome the complexation of Ca2+ with the E4–. It would take a much larger Kf value for complexation of Ca2+ with hydroxide for this reaction to compete with the reaction with E4–.
Calcium ion in tap water forms an insoluble precipitate with soap molecules and prevents the formation of lots of suds. Because it’s hard to get suds when a high concentration of calcium ion is present, the water is referred to as “hard water”. The classic procedure for analyzing the calcium concentration in hard water is to perform a titration with EDTA. The solution is buffered at a pH of 10 to ensure that there is complete complexation of the calcium with the EDTA. The conditional constants in the table above show the reason why a pH of 10 is used.
One last thing we need to consider is how we would handle a metal complex in which there were multiple formation constants. For example, if we look up the complexation of Cd2+ with hydroxide, we see that there are four steps in the process and that the Kf values are larger than the one with Ca2+. If we had substituted Cd2+ for Ca2+ in the problem above, the competing complexation of Cd2+ with hydroxide might have had more of an influence on the complexation of Cd2+ with E4–. Of course, we also need to examine the complexation of Cd2+ with E4–, which has a Kf value of 3.16×1016 from the table.
The relevant equilibria for Cd2+ in this case are as follows:
\begin{align} &\ce{Cd^2+ + OH- \leftrightarrow Cd(OH)+} &&\mathrm{K_{f1}}\ &\ce{Cd(OH)+ + OH- \leftrightarrow Cd(OH)2} &&\mathrm{K_{f2}}\ &\ce{Cd(OH)2 + OH- \leftrightarrow Cd(OH)3-} &&\mathrm{K_{f3}}\ &\ce{Cd(OH)3- + OH- \leftrightarrow Cd(OH)4^2-} &&\mathrm{K_{f4}} \end{align}
The evaluation of $\mathrm{α_{Cd^{2+}}}$ would involve the initial equation shown below:
$\mathrm{α_{Cd^{2+}} =\dfrac{[Cd^{2+}]}{[Cd(OH)_4^{2-}] + [Cd(OH)_3^-] + [Cd(OH)_2] + [Cd(OH)^+] + [Cd^{2+}]}}$
Taking the reciprocal leads to the following terms:
$\mathrm{ \dfrac{1}{α_{Cd^{2+}}} =\dfrac{[Cd(OH)_4^{2-}]}{[Cd^{2+}]} + \dfrac{[Cd(OH)_3^-]}{[Cd^{2+}]} + \dfrac{[Cd(OH)_2]}{[Cd^{2+}]} + \dfrac{[Cd(OH)^+]}{[Cd^{2+}]} + \dfrac{[Cd^{2+}]}{[Cd^{2+}]} }$
Using the Kf expressions for the complexation of Cd2+ with hydroxide, each ratio can be evaluated in terms of Kf values and [OH-], leading to the following equation.
$\mathrm{\dfrac{1}{α_{Cd^{2+}}} =K_{f1} K_{f2} K_{f3} K_{f4}[OH^-]^4 + K_{f1} K_{f2} K_{f3}[OH^-]^3 + K_{f1} K_{f2}[OH^-]^2 + K_{f1}[OH^-] + 1}$
$\mathrm{(K_f)(α_{Cd^{2+}})(α_{E^{4-}}) =\dfrac{[CdE^{2-}]}{[Cd]_{TOT}[E]_{TOT}}}$
Evaluation of a similar set of conditional constants over the entire pH range for the complexation of Cd2+ with E4- leads to the set of data in Table 8.
Table 8. Conditional constants for the complexation of Cd2+ with E4-.
pH $\mathrm{α_{E^{4-}}}$ $\mathrm{α_{Cd^{2+}}}$ $\mathrm{(K_f)(α_{Cd^{2+}})(α_{E^{4-}})}$ Extent of reaction
1 3.66×10-18 1 1.20×10-1 Intermediate
2 2.00×10-14 1 6.32×102 Intermediate
3 1.61×10-11 1 5.09×105 Close to completion
4 2.48×10-9 1 7.84×107 Completion
5 2.47×10-7 1 7.81×109 Completion
6 1.67×10-5 1 5.28×1011 Completion
7 3.89×10-4 0.998 1.23×1013 Completion
8 4.47×10-3 0.980 1.38×1014 Completion
9 4.36×10-2 0.830 1.14×1015 Completion
10 0.314 0.284 2.81×1015 Completion
11 0.820 1.09×10-2 2.82×1014 Completion
12 0.979 2.84×10-5 8.79×1011 Completion
13 0.998 8.30×10-9 2.62×108 Completion
If we compare this data to that for Ca2+, we see that a much higher proportion of the Cd2+ is complexed with hydroxide ion at the more basic pH values. The complexation with hydroxide is sufficient enough at pH 12 and 13 to significantly lower the conditional constant compared to the maximum at pH 10. Nevertheless, the complexation of Cd2+ by the EDTA is still complete at pH 12 and 13 because of such a high formation constant.
What we see for a species like Cd2+ is some optimum pH for complexation with EDTA. At low pH, protonation of the EDTA reduces the extent of complexation. At high pH, complexation of the Cd2+ with hydroxide competes with the EDTA to some extent. If we wanted to perform an analysis of Cd2+ using EDTA, we would buffer the solution at the pH that produces the maximum conditional constant, which is at a pH of 10. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/03_Water-Soluble_Complexes/04_Utilization_of_-Values_for_the_Metal_Ion.txt |
Sometimes when performing a calculation concerning a process at chemical equilibrium there are no simplifying assumptions that can be made. If the system is a relatively simple one, like the dissolution of a weak acid or base in water, the problem can be answered by solving a quadratic formula. At other times, the problem can be considerably more complex with many more species found in the solution. In such an instance the problem usually must be answered by writing and solving a set of simultaneous equations. To determine the number of equations needed, one must first determine the number of unknowns in the solution. As an example, consider the first problem that we solved in this course, a solution of ammonia in water. It turns out that in this case, there are four unknowns in the solution.
\begin{align} \ce{&Ammonia &&NH3\ &Ammonium\: ion &&NH4+\ &Hydronium\: ion &&H3O+\ &Hydroxide\: ion &&OH-} \end{align}
Did we use four equations to solve this? We used the Kb for ammonia and the Kw for water (remember, using the Kb, we ended up calculating the pOH, which we then converted to pH using Kw). A third equation we used (probably without you realizing it) is what is known as a mass balance. In this case, if we were told that the initial concentration of ammonia was 0.10 M, we wrote an expression for the final concentration as (0.10 – x). Another way of saying this is:
$\mathrm{[NH_3]_{Final} + [NH_4^+]_{Final} = [NH_3]_{Initial} = 0.10\: M.}$
Before going on, convince yourself that the equation above is correct.
The fourth equation we used to solve the problem was to say that the concentration of ammonium ion in the final solution equaled the concentration of hydroxide ion (remember, we assumed that the initial amount of hydroxide ion was small compared to what was produced by the reaction of the ammonia).
$\mathrm{[NH_4^+]_{Final} = [OH^-]_{Final}}$
This equation is known as a charge balance. It is important to realize that all solutions must be electrically neutral; that is, for every substance of positive charge there must be an equivalent amount of negative charge to balance it out. If something dissolves in water and produces positive ions, then there must be negative ions around to balance them out.
It is also worth pointing out that the equation shown above is not really the entire charge balance for that solution, (we ignored some original hydroxide and hydronium ion in solution). The exact form would actually be:
$\mathrm{[NH_4^+]_{Final} + [H_3O^+]_{Final} = [OH^-]_{Final}}$
When faced with a problem requiring a set of simultaneous equations, in addition to all of the relevant equilibrium constant expressions, the mass and charge balances are usually needed to come up with as many equations as there are unknowns.
Consider another example, that of dissolving sodium acetate in water to make up a 0.10 M solution. We can write two mass balance expressions.
[Na+] = 0.10 M
Remember that the sodium acetate will dissociate into its component ions. The sodium ion does not undergo any reaction with water, but acetate does to produce acetic acid. The concentration of acetic acid in the final solution will drop below 0.10 M, but the total of the two species must equal 0.10 M, the initial amount that was put into solution.
[Acetic acid] + [acetate] = 0.10 M
The charge balance must account for all positively charged (sodium and hydronium ions) and negatively charged (acetate and hydroxide ions) species in solution. We can only write one complete charge balance for a solution.
[Na+] + [H3O+] = [acetate] + [OH-]
Charge balances get interesting when one of the ions has a charge greater than one. If you consider calcium(II)chloride (CaCl2), note that two chloride ions result for each calcium ion.
CaCl2 = Ca2+ + 2Cl
The charge balance for a solution of calcium chloride in water is written as follows (assuming that neither calcium nor chloride ions undergo any reactions with water, hydronium, or hydroxide).
2[Ca2+] + [H3O+] = [Cl] + [OH-]
You must convince yourself that the above equation is correct, especially that the concentration of calcium ion should be multiplied by two. Many people are initially troubled that the (2+) ion gets multiplied by two, since that seems counter-intuitive. What you must realize is that the equation actually equates concentrations of species in solution. Leave out the hydronium and hydroxide ions from the equation, and notice again in the reaction written above, that for every one calcium ion there are two chloride ions produced. If you plug in a 1 for calcium in the charge balance equation, you will see that the concentration of chloride calculates to be 2. Once you appreciate that the coefficient is in the right place, you may also appreciate that this can be generalized. The concentration of an ion with a charge of (3–) will be multiplied by 3, the concentration of an ion with a charge of (4+) will be multiplied by 4, etc. Knowing how to write mass and charge balances correctly is a critical skill to have when solving equilibrium problems. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/04_Mass_and_Charge_Balances.txt |
In-class Problem Set #5
Calculate the solubility of lead(II)phosphate under the following constraints.
SOLUBILITY: For our purposes, the solubility of a substance is defined as the moles of the solid that will dissolve in one liter of solution.
a) No other simultaneous equilibria occur.
The first step in a problem like this is to write the relevant reaction that describes the process. This involves the solubility of a sparingly soluble substance. Reactions of sparingly soluble substances are always written with the solid on the reactant side and the dissolved ions on the product side.
$\ce{Pb3(PO4)2(s) \leftrightarrow 3Pb^2+(aq) + 2PO4^3- (aq)} \hspace{60px} \mathrm{K_{sp} = 8.1\times 10^{-47}}$
The equilibrium expression for this reaction is written as follows:
$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2}$
and is known as the solubility product. Note that the solid does not appear in the equilibrium constant expression.
The way to solve this problem is to write two expressions for the solubility (S), one in terms of lead ion, the other in terms of phosphate ion. What we need to consider is that the only way we get lead or phosphate ions in solution is to have some of the lead phosphate dissolve. Remember, solubility refers to the moles of solid that dissolve in a liter of solution.
If we consider the equation, one thing we would see is that for every one molecule of solid lead phosphate that dissolves, we get three lead ions. This leads to the following expression for solubility:
$\mathrm{S =\dfrac{[Pb^{2+}]}{3} \hspace{30px} or \hspace{30px} [Pb^{2+}] = 3S}$
Before we continue, we need to make sure that this makes sense. Remember, S is a measure of the number of lead phosphate molecules that dissolve, and if we have three lead ions, only one lead phosphate has dissolved. If [Pb2+] = 3, S = 1 in the above equation.
We can write a similar equation for phosphate ion, keeping in mind that for every one molecule of solid lead phosphate that dissolves, we get two phosphate ions.
$\mathrm{S =\dfrac{[PO_4^{3-}]}{2} \hspace{30px} or \hspace{30px} [PO_4^{3-}] = 2S}$
We can now substitute these two solubility expressions into the Ksp expression:
$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = (3S)^3(2S)^2 = 108S^5}$
$\mathrm{S = 2.37\times 10^{-10}}$
So this is a sparingly soluble material and we have an exceptionally low solubility.
05 Solubility Equilibria
b) Calculate the solubility at pH 3.
Now you need to consider the protonation of phosphate that can occur. If we look back at the Ksp expression, we notice that it only contains lead ion and phosphate ion. Protonation of the phosphate will reduce the concentration of phosphate in solution, thereby causing more of the lead phosphate to dissolve based on Le Chatelier’s principle.
\begin{align} \ce{Pb3(PO4)2 \leftrightarrow 3Pb^2+ +\: &2PO4^3- \ &\:\:\:\Updownarrow\ &HPO4^2- \ &\:\:\:\Updownarrow\ &H2PO4^- \ &\:\:\:\Updownarrow\ &H3PO4} \end{align}
The problem with trying to solve this is that we do not know the concentration of phosphate ($\ce{PO4^3-}$) because it no longer relates directly to the amount of lead in solution. Once again, the way to approach solving this is to write two expressions for the solubility, one in terms of lead ion, the other in terms of phosphate species.
The situation for lead has not changed from part (a) of this problem, so we have the same expression for solubility for lead.
$\mathrm{S =\dfrac{[Pb^{2+}]}{3} \hspace{30px} or \hspace{30px} [Pb^{2+}] = 3S}$
For phosphate, we know that the only source of phosphate is by dissolution of lead phosphate. If there was a way for us to find the total amount of all phosphate species in solution, we could relate that back to the amount of lead phosphate that had to dissolve. This leads to the following expression relating the concentrations of phosphate species to solubility:
$\mathrm{S =\dfrac{[H_3PO_4] + [H_2PO_4^-] + [HPO_4^{2-}] + [PO_4^{3-}]}{2} =\dfrac{[PO_4]_{TOT}}{2}}$
$\mathrm{[PO_4]_{TOT} = 2S}$
But we also know the following:
$\mathrm{[PO_4^{3-}] = α_{PO_4^{3-}}[PO_4]_{TOT}}$
Since we were told the pH of this solution, we realize that we can evaluate the α-value and it’s a fixed number. We can then substitute in from the solubility expression above to get:
$\mathrm{[PO_4^{3-}] = α_{PO_4^{3-}}(2S)}$
If we now substitute the terms for [Pb2+] and [$\ce{PO4^3-}$] back into the Ksp expression, we get the following equation:
$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = (3S)^3[α_{PO_4^{3-}}(2S)]^2}$
$\mathrm{K_{sp} = 108S^5(α_{PO_4^{3-}})^2 = 8.1\times 10^{-47}}$
We now need to evaluate $\mathrm{α_{PO_4^{3-}}}$ at a pH of 3. The form of the 1/α-value expression is as follows:
$\mathrm{\dfrac{1}{α_{PO_4^{3-}}}=\dfrac{[H_3O^+]^3}{K_{a1} K_{a2} K_{a3}} + \dfrac{[H_3O^+]^2}{K_{a2} K_{a3}} + \dfrac{H_3O^+}{K_{a3}} + 1}$
Substituting in for [H3O+] and the Ka values gives an α-value of 2.315×10-14 at a pH of 3. Putting this value into the Ksp expression above gives a final solubility of:
S = 6.75×10-5
At this point, it would be worthwhile comparing the solubility in part (a) (no competing equilibria) to the solubility at pH 3.
(a) S = 2.37×10-10
(b) S = 6.75×10-5
Notice how the solubility is much higher at pH 3. This is reasonable since protonation of the phosphate ion was expected to increase the solubility. This trend points out an important aspect of the solubility of metal ions. Assuming that the anion of the solid is the anion of a weak acid, lowering the pH of the solution will cause a higher extent of protonation of the anion and increase the solubility of the solid.
In general, the solubility of sparingly soluble substances increases with the acidity of the water. It turns out that this is one of the principle concerns of acid rain. Acid rain into unbuffered natural waters raises the acidity (lowers the pH) of the water. The higher acidity causes solid metal salts and minerals in the lake or river bed to dissolve at higher levels. For example, there are lakes with poor buffering in which the impact of acid rain has increased the levels of dissolved aluminum ion (Al3+). Aluminum ion is known to form a highly insoluble complex with hydroxide ion [Al(OH)3, Ksp = 2.20×10-32]. Obviously the solubility of this complex is critically dependent on pH. At acidic pH values, it will dissolve because hydroxide is low. At neutral to basic pH, it will precipitate because the hydroxide level becomes high enough. Aluminum hydroxide is a very gelatinous solid that is sometimes used as a sticky flocculent in water treatment processes (undesirable impurities essentially stick to this material and slowly settle out with it). When the fish take the water through their gills (which are at a pH of 7.4) to remove the dissolved oxygen, the pH of the water increases and the aluminum ion now precipitates out as aluminum hydroxide. The gelatinous precipitate clogs up the gills of the fish and actually causes the fish to die of suffocation. The fish deaths that have occurred in some lakes heavily impacted by acid rain are attributable to this phenomenon. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/05_Solubility_Equilibria/01_Accounting_for_pH.txt |
c) Now you realize for the solution in part (b) that lead can form soluble hydroxide complexes. Incorporate these into the expression.
The scheme below shows the total set of reactions that occur in this solution.
\begin{align} \ce{Pb3(PO4)2}\hspace{25px} \leftrightarrow \hspace{25px} &\ce{3Pb^2+} \hspace{25px}+\: &&\ce{2PO4^3-} \ \ce{&\:\:\:\Updownarrow &&\:\:\:\Updownarrow\ &Pb(OH)+ &&HPO4^2- \ &\:\:\:\Updownarrow &&\:\:\:\Updownarrow\ &Pb(OH)2 &&H2PO4^- \ &\:\:\:\Updownarrow &&\:\:\:\Updownarrow\ &Pb(OH)3- &&H3PO4} \end{align}
The approach in this case is going to be analogous to what we just did for the protonation of the phosphate ion. We know the concentration of hydroxide because the pH is known. This enables us to calculate an $\mathrm{\alpha_{Pb^{2+}}}$ value and incorporate that into the Ksp expression.
The next step is to write two expressions for the solubility, one in terms of lead species, the other in terms of phosphate species.
The equation in terms of phosphate is identical to what was just done in part (b).
$\mathrm{S = \dfrac{[PO_4]_{TOT}}{2} \hspace{60px} [PO_4]_{TOT} = 2S}$
$\mathrm{[PO_4^{3-}] = α_{PO_4^{3-}}[PO_4]_{TOT} = α_{PO_4^{3-}}(2S)}$
The equation for lead is as follows:
$\mathrm{S =\dfrac{[Pb^{2+}] + [Pb(OH)^+] + [Pb(OH)_2] + [Pb(OH)_3^-]}{3} =\dfrac{[Pb]_{TOT}}{3}}$
$\mathrm{[Pb]_{TOT} = 3S}$
$\mathrm{[Pb^{2+}] = \alpha_{Pb^{2+}} [Pb]_{TOT} = \alpha_{Pb^{2+}}(3S)}$
Evaluation of $\mathrm{\alpha_{Pb^{2+}}}$ is done by writing the ratio of Pb2+ over the total, taking the reciprocal so that there is a set of individual terms, and then using the Kf expressions for lead complexation with hydroxide to substitute in for each of the terms. The final equation for 1/$\mathrm{\alpha_{Pb^{2+}}}$ is shown below.
$\mathrm{\dfrac{1}{α_{Pb^{2+}}} = K_{f1} K_{f2} K_{f3}[OH^-]^3 + K_{f1} K_{f2}[OH^-]^2 + K_{f1}[OH^-] + 1}$
Evaluation of $\mathrm{\alpha_{Pb^{2+}}}$ at a pH of 3 gives a value of 0.999984. So very little of the lead actually complexes with hydroxide, which should not be that surprising given the small amount of hydroxide ion in solution at pH 3.
Above we have expressions for [Pb2+] and [$\ce{PO4^3-}$] that are in terms of α-values and S. These can be substituted into the Ksp expression to give the following:
$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = [\alpha_{Pb^{2+}}(3S)]^3[\alpha_{PO_4^{3-}} (2S)]^2 = 8.1\times 10^{-47}}$
$\mathrm{8.1\times 10^{-47} = 108S^5(\alpha_{Pb^{2+}})^3(\alpha_{PO_4^{3-}})^2}$
$\mathrm{S = 6.75\times 10^{-5}}$
If we compare this to the answer in part (b), it turns out that the two are the same. This means that so little lead complexes with the hydroxide ion at pH 3 that it does not lead to any increase in the solubility. If we were to make the solution more basic, complexation of lead by hydroxide would become more important. But also note that protonation of the phosphate would become less important, so the overall solubility is a balance between two processes that influence the solubility in opposite ways as a function of pH. What we might well observe for lead phosphate is that its solubility is smallest at some intermediate pH. At low pH, protonation of the phosphate increases the solubility. At high pH, complexation of lead with hydroxide increases the solubility. If we wanted to use precipitation of lead phosphate as a way to analyze lead (say by collecting the precipitate by filtration and weighing) or remove lead from a solution, we would need to perform a calculation over the entire pH range to find the best value for precipitation of the most amount of material.
d) Revisit problem (a). What is the actual solubility of lead phosphate in unbuffered water given that other equilibria will simultaneously occur?
This is a difficult situation because we know that hydroxide complexes of lead can form and that protonation of phosphate can occur, but it does not seem like we can use α-values because we really do not know the pH. The best approach might be to try some simplifying treatments to see if anything will work.
One thing we could do is assume that the pH of the water is 7, and that dissolving of the lead phosphate does not change it. If that were the case, we should evaluate $\mathrm{α_{PO_4^{3-}}}$ at pH 7 to see what fraction of the phosphate stays in this form. Evaluation of $\mathrm{α_{PO_4^{3-}}}$ at pH 7 gives a value of 1.62×10-6. This means that only a small fraction of the phosphate species will exist as $\ce{PO4^3-}$ and more of it will be protonated. The protonation has the possibility of changing the pH enough from 7 to make a difference. Similarly, if we evaluate $\mathrm{α_{Pb^{2+}}}$ at a pH of 7 we get a value of 0.864, so some lead complexes as well. If we go ahead and plug in these values into the Ksp expression:
$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = [\alpha_{Pb^{2+}}(3S)]^3[\alpha_{PO_4^{3-}}(2S)]^2 = 8.1\times10^{-47}}$
$\mathrm{S = 7.49\times 10^{-6}}$
This is a small number, but the problem is that it’s an appreciable number compared to the concentration of H3O+ at a pH of 7. This means that the pH will probably change enough from 7 to make a difference in the solubility.
It turns out that we cannot make any simplifying assumptions in this case. In this event, we need to solve a series of simultaneous equations. If we write all the unknown species, you find that there are a total of ten for this solution.
\begin{align} \ce{&[H3O+] &&[Pb^2+] &&[PO4^3- ]\ &[OH- ] &&[Pb(OH)+] &&[HPO4^2- ]\ & &&[Pb(OH)2] &&[H2PO4- ]\ & &&[Pb(OH)3- ] &&[H3PO4]} \end{align}
We may be able to eliminate some of these as insignificant, since it might be unlikely that we would get any significant levels of hydroxide complexes or protonation of phosphate besides the first species (Pb(OH)+ and $\ce{HPO4^2-}$). Even if that is the case, we would still need to solve a set of simultaneous equations.
What would be the ten equations? Eight of them are equilibrium constant expressions needed to describe the reactions taking place.
Ksp Ka1 Ka2 Ka3 Kf1 Kf2 Kf3 Kw
One is the mass balance, which involves the relationship between the two solubility expressions we can write for this solution.
$\mathrm{S = \dfrac{[Pb]_{TOT}}{3} \hspace{60px} S = \dfrac{[PO_4]_{TOT}}{2}}$
$\mathrm{\dfrac{[Pb]_{TOT}}{3} = \dfrac{[PO_4]_{TOT}}{2}}$
The final equation is the charge balance:
$\ce{[H3O+] + 2[Pb^2+] + [Pb(OH)+]} = \ce{[OH- ] + [Pb(OH)3- ] + [H2PO4- ] + 2[HPO4^2- ] + 3[PO4^3- ] }$
Next step? HAVE FUN! | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/05_Solubility_Equilibria/02_Accounting_for_the_Presence_of_Complexing_Ligands.txt |
Suppose you are given a question that asks whether a precipitate of a particular compound will form?
This is actually a common question to ask. Many solutions have a complex mix of metal cations and anions. It is quite likely that some of these combinations have small Ksp values, and so are sparingly soluble. In this case, we might be interested to understand ahead of time whether it is likely that a precipitate will form in the solution. Another common example is that many metals form insoluble hydroxide complexes. We therefore may want to know whether a change in pH is going to cause a dissolved metal ion to precipitate out of solution.
The thing to keep in mind is that the solubility product can never exceed the value of Ksp. For example, suppose you were to consider the species silver carbonate (Ag2CO3). The solubility reaction and Ksp expression is shown below.
$\ce{Ag2CO3 \leftrightarrow 2Ag+ + CO3^2-}$
$\mathrm{K_{sp} = [Ag^+]^2[CO_3^{2-}] = 7.7\times10^{-12}}$
Suppose we had a process that would lead to a solution with silver and carbonate ions in it. Suppose that we were also able to calculate the starting value of each ion that we expected in the solution.
If we expected a concentration of silver of 5×10-4 M and a concentration of carbonate of 1×10-3 M, would a precipitate form? What we need to do is take these values and put them into the form of the Ksp expression. Since these are not likely to be equilibrium concentrations, instead of calling this expression K, we use the notation Q.
$\mathrm{Q = [Ag^+]^2[CO_3^{2-}] }$
$\mathrm{Q = [5×10^{-4}]^2[1×10^{-3}] = 2.5×10^{-10}}$
What we now need to do is compare the magnitude of Q (2.5×10-10) to the magnitude of Ksp (7.7×10-12). If Q is greater than Ksp, a precipitate will form since the solubility product term can never exceed Ksp. If Q is less than Ksp, no precipitate will form (this is not yet a saturated solution).
Since 2.5×10-10 > 7.7×10-12, a precipitate will form in this case. Not all of the silver and carbonate will precipitate out of solution. Instead, the concentrations will be lowered so that the concentrations exactly satisfy the Ksp expression.
Another common question is whether it is possible to quantitatively precipitate (99.9%) of one metal cation in the presence of another.
If we assume that the concentrations of the metal ions in the solution are known, we can calculate the concentration of the precipitating anion that is the highest possible value that will not cause any precipitation. We can also calculate the concentration of the precipitating anion that is needed to precipitate 99.9% of the metal ion.
For example, suppose we had a solution that was 1×10-3 M in Pb2+, and we wanted to try to precipitate 99.9% of the lead as its bromide salt. The relevant reaction and equilibrium expression is shown below.
$\ce{PbBr2 \leftrightarrow Pb^2+ + 2Br-}$
$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6}}$
We could calculate the concentration of bromide ion that is the highest one at which none of the lead ion will precipitate. This will be the value where the solubility product exactly equals the value of Ksp.
$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6} = (1\times 10^{-3})[Br^-]^2}$
$\mathrm{[Br^-]^2 = 6.2\times 10^{-3} \hspace{60px} [Br^-] = 7.87\times 10^{-2}}$
Any concentration of bromide higher than 7.87×10-2 M will cause some of the lead to precipitate as lead bromide. Suppose we had another metal ion in solution besides lead, and this other ion formed a bromide complex that was much less soluble than lead bromide. We could calculate the concentration of bromide needed to precipitate 99.9% of this other ion, and then compare that value to 7.87×10-2 M. If the value was less than 7.87×10-2 M, it is theoretically possible to precipitate this other ion in the presence of lead. If the value is greater than 7.87×10-2 M, lead bromide will start to precipitate and interfere with the separation.
If we want to precipitate 99.9% of the lead, that means that 0.1% remains. Since the lead concentration was initially 1×10-3 M, the final concentration of Pb2+after 99.9% precipitates will be 1×10-6 M. We can plug this into the Ksp expression to solve for the concentration of bromide that is needed to precipitate 99.9% of the lead.
$\mathrm{K_{sp} = [Pb^{2+}][Br^-]^2 = 6.2\times 10^{-6} = (1\times 10^{-6})[Br^-]^2}$
$\mathrm{[Br^-]^2 = 6.2 \hspace{60px} [Br^-] = 2.49}$
So a bromide concentration of 2.49 M would be needed to precipitate 99.9% of the lead ion as lead bromide in this solution. This is a reasonably high concentration of bromide ion. We could presumably get that high a level with a solution of hydrobromic acid. The solubility of sodium bromide might be as high as this, but it is getting to be a bit of a high concentration of bromide to precipitate out the lead. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/02_Text/05_Solubility_Equilibria/03_Additional_Questions.txt |
The problem sets on chemical equilibrium can be used in at least two different manners. The primary intent is to use these as a set of in-class, collaborative learning exercises. Groups of 3-4 students work together in discussing and working through the problems. When using the problem sets in this manner, the instructor must actively facilitate and guide students through the material. This manual will guide instructors through each of the problem sets, identifying possible student responses to the questions and the response and activities of the instructor during the progression of the problem.
An alternative to the use of the problems in class is to assign them as out-of-class activities, preferably done as a group activity among students or as a peer-led learning activity (REF). The accompanying text that goes with each problem provides a detailed discussion of each step of the thought process of solving it, such that students could work back and forth between the problem and text on an iterative basis to gain an understanding of the material.
There is no perfect way to assemble groups for such collaborative learning activities. I gather information on the first day of class (year in college, major, prior chemistry courses) and then use this to set groups of 3-4 students that start on the second day of class. I try to make the groups as heterogeneous as possible and they work together for the entire semester. Another strategy is to assign groups for a shorter period of time that might encompass completion of a specific topic or unit, and to then create new groups for the next unit. One other possibility is to have different groups every day of class. Since it is important for groups to work well together, having new groups every day may be less successful than allowing groups to work together for more extended periods of time. I would recommend that the instructor assign groups rather than allowing the students to pick their own. This avoids the potential problem of friends who want to be in the same group but who then do not work well together or stay focused on the assigned task. It also avoids the problem of the student who is left without a group at the end of the selection process, something that can be especially problematic if it is a member of a minority group. When using collaborative groups, it is also important for the instructor to monitor the functioning of the groups and to step in to address either dysfunctional groups or the recalcitrant individual within a group. (Ref articles on group learning). Peer-evaluation processes are often used by instructors who employ group activities as a way of assessing how well groups are working (REF).
I also expect the groups to meet outside of class for any homework assignments, something that is aided because I am at a residential college. An alternative to this is to schedule a room on the evening before a homework assignment is due and encourage them to come to this place and work in any arrangement they wish on the homework. I have run such sessions for several years now and attend them as a facilitator (one result is that it has cut down considerably the individual traffic to my office seeking help on the homework problems) and it has been an excellent way to promote collaboration among the students.
The instructor has an especially important role to fulfill during such group activities. I have observed that the more engaged that I am in the process in helping to guide the students through the material, the more effective the learning that occurs. In most instances, it seems that the students are initially stumped by the question, that they begin to explore things that they do know that might apply to answering the question, and that help from the instructor either by letting them know that they are on the right track or by suggesting another direction in which to take their thinking is necessary. As they begin a question, I roam around the room listening in on conversations and looking over their shoulders at what might be written in their notebook. If I hear something interesting, I indicate that to the group. If I see that someone has written something interesting and relevant in their notebook, I tell other group members that they ought to talk with this individual about what they have written, and that the individual should explain to the other group members why they wrote that down. If I hear a group going entirely in the wrong direction, I probe them on why they are heading in that way and then offer suggestions about things to consider that will set them off in the right direction. When all groups have realized an important point, I call time out and summarize the concept at the board. Then I send them back to continue with the next part of the problem. Most of the problems are handled in such an iterative manner where the students work through some important part of the problem, I summarize it at the board when they have developed the concept, and then they return to the next part of the problem. Occasionally a group will just not see something, whereas every other group has gotten the point, and it may require a direct intervention from the instructor with that group to explain the concept. Similarly, there are times when I call their attention to the board to summarize a point when one of the groups still has not gotten the concept but waiting would slow down the remainder of the class to an unacceptable level.
When using these materials, I want the students to discuss and discover the concepts inherent in the problems, so they do not have the text when working on the problems. After they have completed a particular problem, I then give them a copy of that portion of the text (everyone is instructed to have a three-ring loose-leaf binder of a certain minimum thickness that will accommodate the entire text that will be passed out in increments as the semester develops). The text thoroughly goes through the thought process for solving each problem and I encourage the students to read it over that evening to reinforce the concepts developed in class that day. I also give homework problems designed to reinforce the concepts developed in class.
04 Instructor's Manual
You are a chemist involved in developing a new product for a textile company. As part of the new process, a suspension of the compound lead phosphate will be used to treat the surface of the textile. The lead phosphate will end up in the waste effluent from your plant. This effluent will be discharged to the local municipal waste water treatment plant. Unfortunately, from your standpoint (fortunately, from the standpoint of an environmentalist) the waste water treatment plant faces strict requirements on the amount of lead that is permitted in their end products. (A waste water treatment plant ends up with "clean" water and a solid sludge. Most lead ends up in the sludge, and the Environmental Protection Agency has set a limit on how much lead is permitted in the sludge.) Most municipalities will require you to enter into a pre-treatment agreement, under which you will need to remove the lead before discharging to the plant. For example, the City of Lewiston will require you to discharge a material that contains no more than 0.50 mg of total lead per liter.
Lead phosphate is a sparingly soluble material so most of it will actually be a solid in your waste, thereby allowing you to filter it out before discharge to the treatment plant.
What is the concentration of total dissolved lead in the discharge?
I use this problem on the first day of class to show the types of problems we will be addressing by the end of the equilibrium unit and to set the stage for the different processes (acid-base reactions, formation of water-soluble metal complexes, solubility of a sparingly soluble salt) that we will be using over the term. It allows me to introduce the different tables of equilibrium constants that we will use over the term and to show how these different processes usually occur simultaneously in real systems (e.g., environment, living organisms) and that we need to develop the expertise to handle these real systems.
The in-class problems on chemical equilibrium are intended for a student who has taken general chemistry and had a previous introduction to the topics of chemical equilibrium, acid-base chemistry, solubility equilibria, and complex-formation equilibria. I usually find that, while students have been introduced to these topics, their understanding is still marginal and they need a refresher on the concepts that we had hoped they would learn in the general chemistry course.
Before passing out the first problem set to the class, I spend a relatively brief period of time discussing with them some background information on chemical equilibrium. Using the generalized equation shown below, I ask them what things we might say about the equilibrium state of this reaction.
aA + bB = cC + dD
Usually it does not take too long before students in the class indicate that one characteristic of the equilibrium state is that the concentrations of the chemicals remain fixed. I point out that viewing the system at the macroscopic level (i.e., concentration), we would define equilibrium as a static state. Students also offer up that another thing that characterizes the equilibrium state is that the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore, at the microscopic level of individual species, we find that the system is dynamic and species are actually changing their identities.
I then ask if someone would provide me with the equilibrium constant expression, and many of the students can usually give the answer I am looking for.
Equilibrium constant expression
I then ask the person who provided the answer (and also throw this out more generally to the class) whether they are so certain that this is the correct expression that they would be willing to stake their entire grade in the course on it – if it’s correct, they receive an A for the course and do not have to attend; if it’s incorrect, they receive an F for the course and also do not have to attend (one bright side (?) to taking the offer no matter whether the answer is correct or not is that they don’t need to attend the class). The students immediately sense that the answer is likely not correct (why the offer if it is correct?), but rarely do they come up with why (we either do not cover the concept of activity in general chemistry or if we do, since we then do all the calculations using terms for concentration, they forget this).
This allows me to introduce the concept of activity, and how equilibrium constant expressions are correctly written in terms of activity and not in terms of concentration. I write something to the effect of the following on the board, and ask them if they could identify an A species in the picture that might be regarded as “inactive”. They readily identify the one shown in bold as an “inactive” form of A.
A B B A A A B B A B
I then ask them why they think equilibrium constant expressions and calculations in general chemistry were always done using concentrations rather than activities. They are usually stumped by this and I do not let them spend too long thinking about it. I point out that chemists rarely shy away from difficult calculations, so it is likely something different than that, and indicate that in many instances we simply do not know how to accurately express the activity of a chemical, so that using concentration as a approximation for activity is the best we can do. I also point out that since any equilibrium calculation that uses concentrations is at best an approximation of the system, that it will reasonably allow us to make other approximations when doing equilibrium calculations that will not compromise the outcomes. Furthermore, I point that from the perspective of an analytical chemist, what we usually care about is whether a reaction goes to completion or not (how large is K) or whether a reaction hardly occurs at all (how small is K), and that many analytical procedures are predicated on using systems that have either exceptionally large or exceptionally small values of K so that we can be assured that we are measuring all of what we desire or that no other substance is interfering in the measurement.
Realizing that we will use concentration as an approximation of concentration, I then ask them to consider whether this approximation is more valid at high or low concentration. I encourage them to talk to their neighbors (they are not yet in assigned groups) and then take a poll of the class. Usually it seems that most students think the approximation will be better at high concentration. I go back to my example of As and Bs that is still on the board and ask them to put in more A species and tell me whether it will lead to more or fewer inactive forms. The students then realize that the concentration is a better approximation of activity at lower values. I also point that those methods we do have for rigorously determining activity also break down at higher concentrations, so that they rarely help us in the cases where we would most benefit by using activity instead of concentration. I also point out that while we could get more accurate answers for systems at low concentrations, the use of concentrations is valid enough that all the extra work to use activities is often not warranted.
With this background, I indicate that throughout the remainder of the course we will always use concentration as an approximation of activity, that with our use of concentration we must always keep in mind that we are obtaining “ballpark” figures for the amounts of species in a solution that is at equilibrium, and that because we are only obtaining ballpark values, it will allow us to use other approximations to simplify many of the calculations.
Occasionally throughout the unit I like to ask them to just generally assess the extent to which the utilization of concentrations as an approximation of activities is valid for the system described in a particular problem.
With this background the students are ready to divide into their groups and start on the first problem on the in-class set. When I have passed out the first set, and asked them to read the first question, I point out that the concentration provided in this (0.155 M) and all subsequent problems refers to the initial concentration of species in solution (rigorously referred to as the Formality of the chemical) rather than the equilibrium concentration – that the is prepared with a certain concentration but then proceeds to equilibrium such that the final concentration will be different than the initial concentration. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/01_Overview%3A_Significance_of_Chemical_Equilibrium.txt |
In-class Problem Set #1
1. Calculate the pH of a solution that is 0.155 M in ammonia.
As students begin to ponder this question, and as the instructor begins to circulate among the groups, some things to ask are:
What is ammonia? Is it an acid or a base? Is it strong or weak?
After about five minutes, everyone should have identified ammonia as a weak base and have the correct chemical formula. I write the correct chemical formula on the board and that it is a weak base. With this information, they can next be asked:
What does ammonia react with? Can you write the correct chemical equation representing this reaction?
Students may not recall that the solution contains water and that the water is necessary in the proper reaction. Once groups have written the correct reaction, I call timeout, write it on the board, and indicate how we can use this to describe the general reaction of a base – a base reacts with water to produce its protonated form and hydroxide ion. I then ask the following question:
What is the K expression for this equation?
Which invariably leads to the question:
Should [H2O] appear in this expression?
Using a timeout, we spend about five minutes as a class discussing what [H2O] is and why it shouldn’t appear in the equation. It may be worth mentioning other species that do not appear in the K expression such as other solvents and solids.
What subscript do we attach to this K?
Students usually realize that it is Kb. The equilibrium constant tables that we use in the course only have pKa values so there is only a value for the ammonium ion. This leads to the following questions:
What is the relationship between Ka, Kb, pKa, and pKb? What reaction is the Ka expression describing? What is the relationship between Kb, Kw and this reaction?
Allow the students several minutes to discuss this and to try to figure out the relationships. At some point in the discussion the concept of a conjugate pair will come up and this should be summarized by the instructor at the board. Also, the relationship between pK and K will come up in the discussion. Some students will remember that Ka times Kb for a conjugate pair equals Kw (or at least remember that there is some connection between these three equilibrium constants even if they do not remember the exact relationship). When groups get to the correct expression, I ask them to:
Prove that Ka times Kb for a conjugate pair equals Kw.
Some groups immediately write the two K expressions and multiply them together. Others need prompting to do this. It is then worth summarizing at the board the proof that Ka times Kb for a conjugate pair equals Kw. Students can then be asked:
What is the value of Kb for ammonia? What does the magnitude of Kb tell you about the strength of ammonia as a base?
It may take the groups a few minutes arrive at a Kb value that they all agree on. They should be able to recognize that a small Kb value implies that the base is weak and therefore not very reactive. This is a good time to talk about what the magnitude of any K value tells you about the reaction. At this point, students can now perform the actual calculation.
Instruct students to create a table of concentrations below each species in the reaction where the first row is the starting concentration and the second row is the concentration at equilibrium. Ask them just to fill in the first row.
Groups usually correctly write the concentrations of ammonia and ammonium, but will probably have a “0” as the initial concentration of OH-.
Is the initial concentration of OH- really 0?
Students will remember that water dissociates and it is worth spending a brief amount of time discussing the auto-protolysis of water and how the initial concentrations of both OH- and H3O+ must be 10-7 in order for Kw to have a value of 10-14.
Instruct them to fill in the next row on the table.
Students should very quickly identify that the concentration of NH3 will decrease by x while the concentration of NH4+ and OH- will increase by x.
What happens if these values were substituted directly into the Kb expression?
They will probably recognize that this would result in having to solve a quadratic.
Ask the students to think about the magnitude of x as it might compare to the initial concentration of ammonia and the initial concentration of hydroxide.?
Most realize it will be small compared to the initial amount of ammonia and large compared to the initial amount of hydroxide. It is helpful to then briefly explain why these two conclusions are valid.
If x is small relative to the initial concentration of ammonia and large relative to the initial concentration of hydroxide, are there any approximations that can be made?
Groups will usually realize right away that the value of x can be ignored relative to the amount of initial ammonia and that the initial amount of hydroxide can be ignored relative to x. It is then worth summarizing these conclusions and explaining that we will use a 5% or less criteria throughout the semester to determining whether simplifying approximations are valid. I usually remind them of our overall approximation in which we are using concentration instead of activity and ask them to consider whether the use of concentration is truly warranted for an initial ammonia concentration of 0.155 M.
What is the final concentration of OH-?
Are the two approximations valid?
Some groups may need to be reminded how to calculate a percent.
What is the relationship between OH- and pH?
What is the pH of the solution at equilibrium? Is it basic and does the value seem about right given the magnitude of the Kb value? What would we expect for a pH if the Kb was smaller and what would we expect if the Kb was larger? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/01_Solution_of_a_Weak_Base.txt |
2. Calculate the pH of a solution that is 0.147 M in pyridine and 0.189 M in pyridinium chloride.
Before beginning this problem, spend a few minutes talking about nomenclature of acids, bases, cations and anions, including the suffixes that are used to denote negatively (“ate”, “ide”, “ite”) and positively (“ium”) charged species.
Write the correct chemical equation that represents this solution at equilibrium.
Allow the students about ten minutes to get started on this problem. Many students may be tempted to have both pyridine and pyridinium appear as reactants in the same chemical equation. Talk about the possibility of having two expressions representing this system. Groups are then usually able to write both the Ka and Kb expressions that could be used to describe the system. Sometime during this discussion the students need to think about the following two questions:
What happens to the pyridinium chloride when dissolved in water? What is the role of Cl- in the solution?
Talk for about five minutes about spectator ions and their role in the system.
Instruct the students to pick one equation (either Ka or Kb) and use it to find the pH of the solution using the system established in problem 1.
It may take them a while to arrive at a pH that they are satisfied with. Some may need a reminder that there are initial amounts of both pyridine and pyridinium. If some groups finish before others instruct them to try finding the pH using the other equation.
Are they two pH values the same? Is this a surprise?
How does the magnitude of Ka compare to the magnitude of Kb? What does this tell us about which equation we should use?
Discuss how both reactions cannot be going towards product at the same time, so there must be one expression that more accurately represents this system.
What is the relationship between the two species present?
Talk for about fifteen minutes about conjugate pairs and buffers. Talk about what sorts of factors influence the buffer capacity of a system. Introduce the Henderson-Hasselbalch expression. Discuss how by examining the form of the Henderson-Hasselbalch equation, we can explain why diluting a buffer does not lead to a change in pH (unless the solution becomes excessively dilute – which then is no longer really a buffer).
Is the pH found using the Henderson-Hasselbalch expression the same as the one found before?
03 Solution of a Weak Acid
3. Calculate the pH of a solution that is 0.332 M in anilinium iodide.
Is anilinium an acid or a base?
After about ten minutes everyone should have the correct chemical equations representing this system. Some students may identify iodine as a spectator ion and some may write out the reaction of iodine with a proton. Some students may be tempted to say that the presence of HI will affect the acidity of anilinium.
What is the pH of this solution?
After about ten minutes the students should have all calculated a pH for this system. They should be able to identify this as a straightforward problem based on the procedure developed in problem 1. They may need to be reminded to use approximations. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/02_Solution_Containing_a_Conjugate_Pair_%28.txt |
4. Calculate the pH of a solution that is prepared by mixing 45 mL of 0.224 M chlorobenzoic acid (3-) with 30 mL of 0.187 M ethylamine.
Is ethylamine an acid or a base? What is its chemical formula?
After a few minutes the groups should have identified the species as an acid or base. It may be helpful at this point to introduce abbreviations for the species (e.g., Hcba = chlorobenzoic acid) present that take into account their acidic or basic properties that will simplify writing the reaction and equilibrium constant expression.
What happens when we add an acid to a base?
Students usually realize that neutralization occurs.
What is the correct reaction to express the neutralization that occurs? Write the Kn expression for this reaction?
It should take the students no more than five minutes to correctly answer both of these questions.
Do we have a table of Kn values?
Since we do not, is there a way we could construct the Kn reaction by adding together a series of reactions for which we do know the K values?
In general, if a reaction can be broken down into steps, how do we calculate the K value for a net equation based on the K values for the steps?
Most likely a few students will recall that K values are multiplied together to find a K value for the net equation.
Based on this relationship, write an expression for Kn in terms of values we know.
All of the groups realize that the Ka and Kb expressions are needed, but most ignore the H3O+, OH- and H2O in these reactions and initially believe that Kn = Ka x Kb. If so, ask them to evaluate the value for Kn using this expression. Finding that the value is small:
What does it mean if Kn is quite small and does this make sense based on your knowledge of neutralizations?
Some students may not be bothered by the small Kn because it came from the reaction of a weak acid with a weak base.
What about the reaction involving H2O that is also occurring in the solution?
After a few minutes students should identify the reaction as the opposite of the Kw reaction. They will probably need to be reminded that when the opposite of a reaction with a known K is occurring, you multiply by the inverse.
What is the expression for Kn and what is the actual value of Kn in the problem. In general, how can we predict the magnitude of Kn?
Spend about ten minutes talking about Kn values in general. When will Kn be big? What would it take for Kn to be small? Point out that Kn will always be large when either the acid or base is strong.
Groups can then be instructed to set up a table of values under the neutralization reaction that show the initial concentrations and to construct other appropriate rows under that (e.g., equilibrium concentration, etc.)
Most groups often subtract x from the initial amounts of both reactants. If so, it helps to ask:
What should x be when K is large?
Groups soon realize that x would be large and that a large K means that the reaction will go to completion. Indicate that the second line in the table should be the concentrations present in solution after the reaction has gone to zero.
What are the concentrations of the four species once the reaction has gone to completion?
One of the values in this line will be 0 and it is useful to ask whether this value could actually be zero. Groups will realize that it cannot and must be a finite number. The concept of a back reaction occurring for a reaction that goes to completion can be introduced.
Is there any substantial back reaction?
Most groups realize that the back reaction must be small for a reaction with a large K. They can now write the third line of the chart, which has terms for the concentrations of each species at equilibrium.
Groups are now asked to calculate the pH for the solution?
Most groups see that there is extra chlorobenzoic acid and use this extra amount with its Ka expression to calculate a pH. Rarely do the students see that appreciable amounts of chlorobenzoic acid and the chlorobenzoate anion exist such that this constitutes a buffer with appreciable quantities of a conjugate pair. At this point I usually like to go up to the board where I have written the reaction and say let’s circle all those species that we have appreciable amounts of. Prompting that there are appreciable amounts of chlorobenzoic acid and its conjugate anion usually eventually triggers the realization that it is a buffer. It is worth mentioning that neutralization reactions frequently result in buffer solutions.
At this point, the groups can calculate the pH using the Henderson-Hasselbach expression for the chlorobenzoic acid. They can also calculate the pH by substituting the amounts of aniline and anilinium obtained from the neutralization reaction into its corresponding Henderson-Hasselbach expression. We then discuss why both values must be identical if we have done the problem correctly. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/04_Solutions_that_are_Mixtures_of_Acids_and.txt |
5. Calculate the pH of a solution that is prepared by mixing 75 mL of 0.088 M aniline with 50 mL of 0.097 M nitrophenol (2-).
Since this is just another neutralization reaction, the groups can usually advance through the following steps with minimal prompting.
Which is the acid and which is the base?
What is the chemical equation representing this system at equilibrium?
What is Kn for this neutralization? Does it go to completion?
In this case, the value of Kn is small and most groups realize that they know how to do a set up a calculation for a reaction with a small K value.
What assumptions can we make? What is the value of x for this system?
What do we have in the solution when it has reached equilibrium?
Which conjugate pair should we use to calculate the pH? Does it matter?
It is useful to have every group or different groups calculate the pH using the two different sets of conjugate pairs to show that the pH is the same for each of the two.
06 Titration of a Weak Base
In-class Problem Set #2
1. Starting with 30 mL of a solution that is 0.1 M in butylamine, calculate the original pH, and then the pH as 5 mL increments of 0.1 M hydrochloric acid are added. Continue the series of calculations until 40 mL of acid have been added. Plot the data (pH on the y-axis, volume of added acid on the x-axis).
Has 99.9% of the butylamine been titrated at the equivalence point?
What is the initial pH of the solution?
It should take the groups a few minutes to calculate the initial pH of the system. Point out after that this is a solution of a weak base so the initial pH being basic is not a surprise.
What happens when we add 5 mL of 0.1 M HCl?
Allow the students a few minutes to think about what will happen then spend some time discussing the neutralization and conversion of butyl amine to butyl ammonium, leading to the formation of a buffer. Talk about how the concentration of butylamine will change with each addition of HCl, but if we use moles we can ignore the change in volume (consider the form of the Henderson-Hasselbalch and how each component of the buffer is in the same volume so a mole ratio is the same as the ratio of molarities).
Sometimes students ask about the amount of butyl amine that reacted to produce butyl ammonium in the initial solution (0.002 moles of the 0.030 moles in solution) and whether they should start with 0.030 or 0.028 moles of butyl amine. Or you may wish to raise this as a question for the class.
Discuss the fact that 0.002 moles of hydroxide ion (a strong base) is produced in the initial solution and that this reacts first with 0.002 moles of the HCl. Once the hydroxide is used up, the butyl amine begins to react.
What will happen after an additional 5 mL of 0.1 M HCl is added?
Have students make a table to keep track of the moles of each species present at each step up to 30 ml of added titrant. It may take the students about fifteen minutes to correctly set up the table and decide how to fill in each column.
What happens after the addition of 30 mL of 0.1 M HCl? How do you find the pH at that point?
Students should recognize that once all of the butylamine has been used up there is no longer a buffer. Talk about the equivalence point and various methods of indicating the progress of a titration. Students may be tempted to calculate the pH by evaluating how much of the excess strong acid has been added, but should recognize that only butylammonium is present and so the question simply becomes a weak acid problem. It may take them about ten minutes to realize this and calculate the pH.
What is the pH when 35 and 40 ml of titrant has been added?
Most groups readily see that there is now extra strong acid that will be most important in determining the pH.
What would a plot of pH vs. mL of acid added look like?
Have the students make a plot of the titration curve and point out distinguishing features such as the equivalence point, buffer region, and the inflection point. Discuss how the plot would look if the base had a pKa of 8. How would it look if the pKa was 6?
Has 99.9% of the butylamine been titrated at the equivalence point?
Students may have some trouble recognizing that calculating the x value for the Ka expression at the equivalence point can provide them with information regarding how much butylamine is in the system at the end of the titration. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/05_Solutions_that_are_Mixtures_of_Acids_and.txt |
In-class Problem Set #3
1. Calculate the pH of a 0.127 M solution of ascorbic acid.
Groups quickly see that ascorbic acid has two pKa values.
What does it mean if something has two pKa values? What two chemical expressions represent the two equilibria that are occurring?
Students may be tempted at first to ignore the presence of a second pKa value, but encourage them to investigate what it implies about the acid. Once they have realized that there are two dissociable protons, ask them to write the correct chemical expressions making sure to include the proper charges. This should take about ten minutes. Some groups then use Ka1 to solve for a concentration of H3O+ and Ka2 to solve for a second value of H3O+. If they do so, they realize that the second value is much smaller than the first. But they also apparently have a solution with two different concentrations of H3O+.
What do we know about [H3O+] in the two expressions?
Remind the students that since only one pH value can be measured, the concentration of H3O+ in the two equations must be equal.
Write an expression for [H3O+] in terms of the ascorbic acid species.
Students should have no trouble recognizing that for every one mole of ascorbic acid that is deprotonated, one mole of H3O+ is produced, but they may have some trouble understanding that for every one mole of the fully deprotonated ascorbate formed, two moles of H3O+ are formed.
What is x for the Ka1 expression?
What is x for the Ka2 expression? What does this tell us about the second reaction?
Students should realize very quickly that the second x value is equal to the Ka2. With an x value this small, they should recognize that the second reaction is insignificant and that solving for the pH using just the pKa1 is sufficient. Spend several minutes talking about the general differences between pKa values and when the second one can be ignored. Make sure students are clear that this was an example that began with the fully protonated species of a polyprotic acid.
What is the pH of the solution?
08 Solutions of Polyprotic Acid Base System
2. Calculate the pH of a 0.089 M solution of sodium carbonate.
Before starting this problem, talk about nomenclature for polyprotic acids. For example:
H3PO4 phosphoric acid
NaH2PO sodium dihydrogen phosphate
Na2HPO4 disodium hydrogen phosphate
Na3PO4 sodium phosphate
What are the two equilibria are occurring in this solution? What are the chemical expressions representing them?
Students may be tempted to initially say that the presence of sodium will result in the formation of sodium hydroxide which could shift the equilibrium. Remind them when ions can be assumed to be spectators. It should take the students about five minutes to write the correct expressions and recognize that they correspond to the Kb expressions. Make sure that the students determine the Kb1 and Kb2 values from the correct Ka values.
Can the second expression be ignored?
Students should recognize that a large difference in Kb values implies that the second reaction is negligible. Point out that this is just the opposite of the previous scenario; in this case we started with all of a deprotonated polyprotic acid.
What is the pH of the solution?
09 Solutions of Polyprotic Acid Base System
3. Calculate the pH of a solution prepared by adding 30 mL of 0.1 M hydrochloric acid to 60 mL of 0.080 M potassium malonate.
What sort of reaction is occurring?
After several minutes the groups should recognize that the malonate can be protonated as it is neutralized by the strong acid.
Since a neutralization reaction is occurring, what is Kn?
Students should immediately recognize that since HCl is a strong acid Kn will be large and the reaction can be assumed to go to completion.
What is in the solution after the neutralization occurs?
After several minutes, the students should recognize that once the HCl has converted some of the malonate, a buffer is formed. Make sure that they convert the amounts into moles since dilution will occur. They may be tempted to incorporate the fully protonated malonic acid into the equilibrium but remind them that the Ka values differ so much that formation of malonic acid is negligible.
Calculate the concentration of malonic acid at equilibrium to ensure that it is negligible.
Which Ka value should be used in the Henderson-Hasselbalch equation?
Students should recognize that since the K value that relates the two species forming the buffer is Ka2, they should use that value in the Henderson-Hasselbalch equation.
What is the pH of the solution? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/07_Solutions_of_Polyprotic_Acid_Base_System.txt |
4. Calculate the pH of a solution prepared by adding 55 mL of 0.098 M sodium phosphate to 65 mL of 0.136 M phosphoric acid.
If groups are having trouble getting started on this problem, encourage them to write all of the reactions corresponding to the phosphate species and identify (circle) the species that are initially present. When groups have written the proper reactions, write them on the board and circle the two species that are present in the initial solution.
What type of reaction is going to occur? What type of species is phosphoric acid? What type of species is phosphate?
Students should recognize very quickly that phosphoric acid is an acid and phosphate is a base (but not conjugate pairs) and that these two can neutralize each other leading to the formation of intermediate species.
Write the neutralization reaction that will occur.
Write this reaction on the board once the groups have written it.
What is the Kn for this reaction? Does it go to completion?
Students should not have trouble with this question at this point.
What is present after the initial neutralization?
Students should have no trouble calculating the concentration (or moles) of all of the phosphate species present after neutralization, and should recognize that one of the reactants is in excess. Encourage them to rewrite the equations corresponding to this system and circle all of the species present.
What will happen next?
Students may struggle with the next step of this problem. They may be tempted to assess the extent of back-reaction. They may need help seeing that another neutralization reaction can occur. If they are tempted to treat this system as a buffer, ask them which one they would use to calculate the pH in order to get them thinking more about species present. Spend about five to ten minutes discussing this.
Write the neutralization reaction that will occur.
Write this reaction on the board once the groups have written it. Remind groups to carry forward any amounts that were already produced in the first neutralization reaction as the initial concentrations for the second neutralization reaction.
What is the Kn for the neutralization reaction that will occur with the remaining species?
Students may have trouble knowing which Ka and which Kb to use when calculating the Kn. Encourage the students to take their time with this step.
What is present after the second neutralization reaction? Are there any further neutralization reactions that can occur with those species that are now in solution? What is the pH?
Remind students to keep the stoichiometric coefficients in mind when carrying out these calculations. It should take students about ten minutes to complete this problem. Spend some time at the board to summarize the entire calculation and describe why it makes sense for the system to eventually get to the state where only one reaction is needed to calculate the pH. It can also be helpful to have groups use the major species and calculated pH to calculate the concentration of the other species in solution. Writing the final concentrations of everything in solution further demonstrates how the other reactions and other species are insignificant compared to the two major species.
11 Solutions of Polyprotic Acid Base System
5. Calculate the pH of a 0.240 M solution of sodium bicarbonate.
Write the appropriate reactions that incorporate the bicarbonate species?
Give the students about five minutes to begin working on this problem. They will probably identify sodium bicarbonate as both a weak acid and a weak base.
Write an expression for H3O+ in terms of the other species?
Groups should eventually recognize that for every one mole of CO32- formed, one mole of H3O+ was formed, and for every mole of H2CO3 formed, one mole of H3O+ was used up. They may need some prompting to think about what will occur for the H2CO3. Once the groups have the correct expression, spend time at the board going through the algebraic manipulation that incorporates this expression into the Ka expressions to arrive at the formula that can be used to calculate the pH.
At the end of this problem, it is beneficial to most students to spend several minutes generalizing the methods used in the last five problems. Address how to approach problems of each of the following starting species: a fully protonated acid, appreciable amounts of neighboring intermediates, appreciable amounts of varying intermediates, one intermediate, a fully deprotonated base. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/02_Acid-Base_Chemistry/10_Solutions_of_Polyprotic_Acid_Base_System.txt |
In-class Problem Set #4
1. Calculate the concentration of free calcium(II) in a solution initially prepared with 0.020 M calcium and 0.10 M EDTA4-.
Before beginning this problem set, spend about twenty minutes discussing amino acids and zwitterions, hydrochloride salts, and common ligands such as EDTA and how they function.
Give the students about five minutes to think about what might be occurring and look up Kf values for the complex. Spend about thirty minutes talking about metal-ligand complexes and chelates. Talk about H2O as a ligand and how a hydrated metal ion compares entropically to a metal chelated with EDTA. Briefly discuss the geometry of metal-ligand complexes and how the complexation stoichiometry varies depending on the metal ion and ligand. With the realization that the Kf value is large, groups are able to set up a table with initial amounts, a line for a complete reaction, and a line for the back reaction.
What is the value of x for the back-reaction based on this Kf expression?
Examine the amount of uncomplexed calcium(II) to confirm that it is quite small.
02 Utilization of -Values
2. Calculate the concentration of free calcium(II) in a solution initially prepared with 0.020 M calcium and 0.10 M total EDTA. The solution is buffered at a pH of 2.
What affect will pH have on the ligand?
Students should realize that EDTA4- is a base and that in the presence of an acid the protonated species will form. After groups have discussed this, state this to the class as a whole.
How do we expect the formation of the calcium-EDTA complex to change based on this pH?
Students should recognize that with increased amounts of protonated EDTA, according to LeChatlier’s principle, the amount of calcium-EDTA that can form is reduced. After groups have discussed this, state this conclusion to the class as a whole.
Can we now solve for the amount of uncomplexed calcium ion?
I use a set of equilibrium constant tables that has an additional table that contains a listing of the fraction of EDTA that exists as EDTA4- (α-values) reported for each half a pH unit. This table is on the same page as the Ka values for EDTA and Kf values for metals with EDTA.
Students see this table and invariably take the α-value for pH 2 and multiply it the total concentration of EDTA to arrive at a concentration of EDTA4-. They then tend to think that, since Kf is large, they can use the initial amount of calcium(II), calculated amount of EDTA4-, and set up a table for a reaction that goes to completion. In the line of the table for the reaction at completion, they will list the complexed calcium as equivalent to the initial concentration of EDTA4- they just calculated, and show that EDTA4- goes to zero. At this time, it is important to point out to the entire class that the concentration of EDTA4- cannot go to zero, since there are large quantities of the protonated forms in solution that must redistribute to produce some finate amount of EDTA4-. With this being the case, the students can appreciate that the amount of complexed calcium must be larger than they calculated, but are stumped about how to actually do the calculation. Some are now tempted to think that the presence of a buffer with a pH of 2 is insignificant because the large Kf will shift the equilibrium towards the formation of the complex.
At this point the concept of an α-value should be introduced.
I spend about an hour introducing the concept of α-values. The first thing I point out is that having a table of α-values for EDTA4- implies that the fraction of EDTA that exists in solution as EDTA4- only depends on the pH and has no dependence on the total concentration of EDTA, and that we need to show that this is the case.
I ask the groups to write an expression for the fraction of EDTA that exists in solution as EDTA4-.
Most groups are able to write this expression without any difficulty. I write it on the board and then instruct them to take the reciprocal of this expression and examine how the equation now consists of a series of ratios of species. Write the reciprocal expression on the board.
I now ask them if they can evaluate terms for each ratio that will only depend on the pH (or concentration of hydronium ion).
Some groups realize that they will need to use the Ka expressions for EDTA to evaluate terms for the ratios. Others need to be prompted by asking them what else we know about the system or about EDTA. I also write the four dissociation reactions for EDTA on the board as a way of getting them to think about how we might evaluate these terms. Once they realize that the Ka expressions will need to be used, it can be pointed out that the ratio of HEDTA3- to EDTA4- can be evaluated using only Ka4. I ask them to use appropriate Ka expressions to evaluate that ratio as well as the ratio of H2EDTA2- to EDTA4- and to then see if a pattern is emerging such that they can guess the forms of the final two ratios. They usually can guess the final two terms after seeing the pattern for the earlier two. We can then examine how the only variable in the expression is the concentration of hydronium ion. I then point out how we could do a similar process for any other of the protonated EDTA species in solution, arriving at a set of α-value expressions for each species observed in the dissociation of EDTA. I then draw the axes of a graph on the board that has the value of α on the y-axis and pH from 0-14 on the x-axis.
I then ask the groups to draw a plot for the α-values for each species observed in a triprotic acid (H3A, H2A-, HA2-, A3-).
Remind them that the sum of the α-values for all four species must be 1 at every pH value. It also helps to have them think about the two extreme species (H3A and A3-) first and to think about what the plot of their α-values would look like as a function of pH. Most groups are able to reason out that there should be a lot of H3A and not much A3- at highly acidic pH and the reverse of this at highly basic pH. Once they see that, they can usually rationalize that the intermediate species must have an α-value that is low at highly acidic and highly basic pH with some maximum point at an intermediate pH. At this point, I show them examples of plots of α-values for different systems and we look at where the buffers are on the plots, how the plots vary with different species depending on the Ka values for the system, etc.
Now we can ask how the α-value can be used to calculate the concentration of EDTA4- at a pH of 2?
Groups can usually substitute the α-value expression into the Kf expression for the reaction. I then show them how the α-value, as a constant for the particular pH, can be brought up onto the side with Kf to create a conditional constant, and how the magnitude of the conditional constant is then used to determine whether or not the reaction goes to completion and what strategy we will use to actually calculate the equilibrium concentrations. We also write the reaction that is described by the conditional constant. We can now construct a table of initial and equilibrium values under the conditional reaction and calculate the amount of calcium complex that forms.
Are there any other calcium complexes that can form with the ions present? Are they significant?
The students will probably not think of hydroxide as a ligand, but remind them that since metals will exist in solution as cations, hydroxide can often form metal complexes. Have them look up the Kf for calcium hydroxide and assess whether or not this is significant. Remind them that the system is buffered at a pH of 2.
Evaluate a second conditional constant that incorporates in the complexation of calcium by hydroxide ion.
I indicate that this will probably be a process analogous to what we just did for the protonation of EDTA and ask whether they think it might be possible to calculate an α-value for the calcium ion. I also ask them what might be the variable in this expression and most realize it should likely be the concentration of competing ligand.
Derive an expression for the α-value for the calcium ion as a function of hydroxide.
Incorporate the α-value into Kf and calculate a second conditional constant that accounts for both protonation of the EDTA and complexation of calcium by hydroxide.
What is the concentration of calcium EDTA complex in the solution?
What would a plot of complexation of calcium with EDTA look like as a function of pH?
Students will often think that one of the extremes leads to maximized complexation, but challenge them to consider what would happen at an acidic pH and a basic pH. I show them a page in the accompanying text that has a compilation of the conditional constants for the calcium complex with EDTA to illustrate this trend.
What if we had a metal that forms multiple hydroxide complexes such as cadmium? How do we factor hydroxide complexation?
Students should realize that if they had an α-value for Cd2+, they could find a conditional constant. Allow them time to derive the expression for αCd2+ and find its value. Suggest also that they start with the terms that include Cd(OH)+ and Cd(OH)2 and see if they can see a pattern that would allow them to guess the final two terms. They may need help getting started but, as with the α-values for ligands, allow them to simplify the individual terms. I then show them a table from the text that shows the conditional constants for the cadmium complex with EDTA as a function of pH to illustrate the trend in complexation with pH.
What if there were more competing ligands?
Discuss how a total α-value could be written with separate sets of terms for all competing ligands, emphasizing how we must know the concentration of uncomplexes ligand in the final solution to be able to calculate an α-value. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/03_Water-Soluble_Complexes/01_Concentration_of_Unreacted_Metal_Ion.txt |
In-class Problem Set #5
Calculate the solubility of lead(II)phosphate under the following constraints.
a) No other simultaneous equilibria will occur.
What reaction is occurring and what is the equilibrium constant expression?
Students should recognize that lead(II)phosphate will dissociate in water, but may have trouble remembering what the stoichiometric coefficients are in front of each ion. Allow the students a few minutes to write the correct reaction and Ksp expression. Once everyone has the correct expression, spend a few minutes discussing sparingly soluble complexes, the solubility product, saturated solutions, and solubility.
What two expressions equate the individual dissociated ions to the solubility of the solid?
If students are struggling with this step, ask them something along the lines of:
If I measured the concentration of free lead(II), what would that tell me about how many moles of the solid dissolved?
Students may try to incorporate Ksp into their expression or try to account for both ions in the same expression. Remind them that there is one expression for each ion. It may take about ten minutes for every student to have and comprehend the right expressions.
What is the value of S?
Students will probably not be able to answer this question on their own. Remind them of the two expressions they just wrote for S and that they also have a Ksp expression for the solution. Groups will eventually realize that the two expressions for S can be rearranged and substituted into the Ksp expression leaving S as the only variable.
Using the value for S, what is the concentration of free lead(II) in this system?
Again, the students will not necessarily know how to approach this questions, but remind them that they have an expression which relates S to the concentration of each ion.
04 Solubility Equilibria
b) The solution is buffered at a pH of 3 and you will need to consider the protonation of phosphate that can occur.
What will happen to the solubility of lead(II)phosphate at this pH?
Allow the students a few minutes to consider this question. They should realize that as phosphate is protonated, the equilibrium will shift towards dissociated ions and thus the solubility is increased.
This is a chance to take some time to discuss the impact that acid rain has on metals in the environment. A good example is the increased solubility of aluminum leading to the precipitation of aluminum(III)hydroxide in fish gills.
What two expressions equate the species in solution to the solubility of the solid?
Groups usually realize that we will need an expression in terms of lead ion and phosphate species.
Where did every mole of phosphate (protonated and deprotonated) come from? What is the S expression for the phosphate ion?
Students should realize that every mole of phosphate must have come from the original lead(II)phosphate and that the S expression must be modified to include the total concentration of phosphate species present. They also realize that since we know the pH, and since we are considering total phosphate species, that an α-value is likely used somehow to solve the problem.
How would the α-value for PO43- help? What is the modified Ksp expression that includes the α-value?
Allow students a few minutes to consider how to factor the α-value into the S and Ksp expression. Summarize the final Ksp expression that includes the α-value at the board.
What is the value of αPO43-?
Have students write the correct expression for αPO43-. It may take them about five minutes to all have the correct expression. In order to save time, once they have written the correct expression give them the exact value of αPO43- instead of having them evaluate it.
What is the value of S?
Students should all be able to calculate S. Discuss how the magnitude of S compares to that of part (a). What is the concentration of free lead(II)? | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/04_Solubility_Equilibria/01_Accounting_for_pH.txt |
c) Now you realize that, for the solution in part (b), lead can form soluble hydroxide complexes. Incorporate these into the expression.
What complexes can form? What is the modified expression for S for the lead species?
By looking up Kf values, students should realize that lead can form three soluble hydroxide complexes in water. They should recall from part (b) that they need to incorporate the total lead into the S expression and that knowing value of αPb2+ will allow them to solve for S.
What is the value of αPb2+? What does the value of αPb2+ tell us about the formation of hydroxide complexes?
As with αPO43-, once they have written the correct expression, provide them with the exact numerical value. It should take them no more than five minutes to write the correct expression. The students should realize that with an α-value of approximately 1, there is essentially no complexation of lead with hydroxide.
What would happen at a high pH? What does a plot of the solubility of lead(II)phosphate versus pH look like? What would happen to the solubility at highly acidic, intermediate, and highly basic pHs.
Where is the solubility minimized? Spend about ten minutes discussing what can occur at extreme pH values and what factors determine the pH at which solubility is minimized. Also discuss methods that are used to remove metals as their insoluble hydroxide species (a common industrial strategy to remove toxic metals from waste streams) from solutions.
d) Revisit part (a). What is the actual solubility of lead phosphate in unbuffered water given that other equilibria will simultaneously occur?
Allow the students ten minutes to think about how to approach this problem. Students may initially be tempted to find α-values for the ions at a pH of 7 forgetting that α-values are only valid for a constant pH. Allow them to calculate αPO43- in order to demonstrate that the pH will change. Talk about how in order for the pH to remain constant, we would have to make unrealistic assumptions such as no H3O+ or OH- reacting or both sets of competing reactions consuming exactly the same amount of H3O+ and OH-. Discuss how this is an example where there are no simplifying assumptions that we can make and that the only way to solve this problem is to solve a set of simultaneous equations.
How many unknown variables are there in this system? What are the equations?
Students may have a surprising amount of trouble counting the unknowns. Once they have all agreed on a number, challenge them to come up with the same number of equations to relate those unknown variables. They may need some assistance figuring out what sorts of equations will be helpful. Remind them that something like Ksp is an equation that relates some of the unknown variables. Allow them about five minutes to come up with as many equations as they can. They may be tempted to list the S expressions as equations, but remind them that doing so would introduce S as a new variable. They will most likely come up with eight or nine on their own but not know about a mass and charge balance. Point out that equating the two S expressions is the mass balance for the system.
Ask them whether there is something else besides mass needs to be balanced in a solution?
Groups usually realize that the solution needs to be neutral so that the change must be balanced.
What is the expression for the charge balance for the system?
Instruct them to write all of the cations on one side and all of the anions on the other. Make sure that they correctly account for ions with charges other than one. Many groups put the coefficient in the wrong place in their initial attempt at writing a charge balance – they want to take a 3- ion and divide its concentration by three rather than multiplying it by three. A good example is to ask them to consider Na3PO4 dissolving in water and to write a charge balance for only that species, keeping in mind the relationship between the concentrations of the ion. | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Chemical_Equilibrium/04_Instructor's_Manual/04_Solubility_Equilibria/02_Accounting_for_the_Presence_of_Complex.txt |
• Goal 1: This module will frame the roles of signal and noise in chemical measurements.
• Objective 1: Define analytical signals and estimate signal parameters that correlate to analyte concentrations
• Objective 2: Define noise, estimate the magnitude of noise, and investigate how the presence of noise interferes with the measurement of analytical signals
• Objective 3: Define signal-to-noise ratio (S/N) as it relates to method performance and investigate how S/N is used to determine the detection limit of an analytical method
• Goal 2: This module will describe how to improve the signal-to-noise ratio of analytical signals
• Objective 1: Provide an introduction to the behavior of passive electronic circuits and show how they are used to improve the S/N of an analytical measurement
• Objective 2: Provide an introduction to software-based methods and show how they are used to improve the S/N of an analytical measurement
01 Figure of Merit
Defining Signal and Noise
All analytical data sets contain two components: signal and noise
Signal
1. This is the part of the data that contains information about the chemical species of interest (i.e. analyte).
2. Signals are often proportional to the analyte mass or analyte concentration
1. Beer-Lambert Law in spectroscopy where the absorbance, A, is proportional to concentration, C.
$\mathrm{A = εbC}$
There are other significant relationships between signal and analyte concentration:
2. The Nernst equation where a measured potential (E) is logarithmically related to the activity of an analyte (ax)
$\mathrm{E = E^\circ ± \dfrac{RT}{nF} \ln a_x}$
3. Competitive immunoassays (e.g. ELISA) where labeled (analyte spike) and unlabeled analyte molecules (unknown analyte) compete for antibody binding sites
$\mathrm{A = kN_{binding\: sites} = \dfrac{C_{labeled}} {C_{(labeled+unlabeled)}}}$
Noise
1. This is the part of the data that contains extraneous information.
2. Noise originates from various sources in an analytical measurement system, such as:
1. Detectors
2. Photon Sources
3. Environmental Factors
Therefore, characterizing the magnitude of the noise (N) is often a difficult task and may or may not be independent of signal strength (S).
3. A more detailed discussion on specific relationships between signal and noise may be obtained by clicking here and reading Section 3.
02 Signals and Noise
Why is Noise Unwanted?
Noise degrades the accuracy and precision of a signal, and therefore our knowledge about how much analyte is present.
Signal-to-Noise Ratio (S/N): A Figure of Merit
The quality of a signal may be expressed by its signal-to-noise ratio
$\mathrm{\dfrac{S}{N} = \dfrac{mean}{standard\: deviation} = \dfrac{\bar{x}}{σ}}$
02 Measuring Signals
If the signal is at steady-state, as in the case of flame atomic absorption spectroscopy (FAAS), S is best estimated as the average signal magnitude, shown below by the solid line.
If the signal is transient, as in the case of chromatographic peaks, S is best estimated as the peak height or peak area. In the figure below, the peak height is measured from the midpoint of the baseline fluctuations (bottom horizontal line) to the top of the peak.
The peak area of a transient signal is the integrated response, which in this case has units of (μV*min). The peak area of this response is roughly equivalent to the area of the shaded triangle superimposed on the chromatographic peak.
03 Quantifying Noise
All data contains some level of uncertainty due to random fluctuations in the measurement process. We will focus on describing random fluctuations that may be described mathematically using a Gaussian distribution shown below.
In this relationship:
• y is the frequency that a value x will occur
• μ is the population mean
• σ is the standard deviation of the population
$\mathrm{y = \dfrac{exp\left\{\dfrac{-(x - μ)^2}{2σ^2}\right\}}{σ \sqrt{2π}}}$
Of course, there are such a myriad of samples and measurement methods that each case yields a unique distribution with a unique mean and standard deviation.
In order to generally describe the Gaussian distribution, one must represent the Gaussian distribution in a standardized format. This can be done in two steps:
1. Mean-Centering
subtracting the population mean from all the members of the data set so that μ = 0
2. Normalization
dividing each member of the data set by the distribution standard deviation so that σ = 1
The x-axis is now represented by a unitless quantity, z
$\mathrm{z = (x-μ)/σ}$
04 Normal Error Curve
If we look at a standardized Gaussian distribution — the so-called Normal Error Curve shown below — you can see that the probability of any one measurement being a member of this particular distribution increases as the magnitude of z increases.
The area underneath the curve represented by "z" multiples of the standard deviation are shown in the table below:
±z Area Represented Under Normal Error Curve
(Confidence Level)
1.00 68.3%
1.64 90.0%
1.96 95.0%
2.58 99.0%
3.00 99.7%
05 Calculating S N
Calculating the signal to noise ratio based on our brief discussion of Gaussian statistics can be achieved as follows:
1. Find a section of the data that contains a representative baseline. Notice that on the chart, the representative baseline does not contain any signal.
Estimate peak-to-peak noise (VN)
If the data is on a piece of paper, draw two lines that are parallel with the baseline and tangential to the edges of the baseline. See the example above.
If the data is digitized (e.g. in a spreadsheet or text file), locate the maximum and minimum values in a representative section of the dataset that only represents the noise level.
Estimate root mean square noise
1. Calculate the standard deviation of the noise voltage (expressed as VRMS, the square root of the mean squared voltage for a given frequency range). At the 99% confidence level: VN = ±2.58σ
$\mathrm{σ = V_{RMS} = \dfrac{VN}{±2.58} = \dfrac{(V_{max} - V_{min})}{5.16} = \dfrac{(2.50 - (- 2.50))}{5.16} = 0.97\: V}$
2. Estimate the S/N. The signal is 16.0 μV.
Estimating S/N
First, calculate the standard deviation (VRMS) of the noise. At the 99% confidence level: VN = ±2.58σ.Therefore:
$\mathrm{σ = \dfrac{V_N}{±2.58} = \dfrac{(V_{max} - V_{min})}{5.16} = \dfrac{(2.50 - (-2.50))}{5.16} = 0.97\: V}$
Second, calculate the S/N. The signal is 16.0 μV.
$\mathrm{\dfrac{S}{N} = \dfrac{16.0\: μV}{0.97\: μV} = 16.5}$
03 Sources of Noise
Johnson Noise
Also called thermal noise, this source of noise results in random voltage fluctuations produced by the thermal agitation of electrons as they pass through resistive elements in the electronics of an instrument. The relationship between Johnson Noise and experimental parameters is as follows:
$\mathrm{V_{RMS} =\sqrt{4kTRΔf}}$
where
• VRMS: Root-mean-square noise voltage with a frequency bandwidth of Δf (in Hertz).
• k: Boltzmann’s constant (1.38 x 10-23 J/K)
• T: Temperature (K)
• R: Resistance of resistive element (Ω)
Reduction of Johnson Noise is accomplished most easily by:
• Cooling the detector (reducing T)
• Decreasing the frequency bandwidth of the signal (reducing Δf)
• Actual measurements of Johnson Noise may be found by clicking here
Shot Noise
This source of noise results in current fluctuations produced by electrons crossing a junction in a random fashion, which highlights the quantized nature of electron flow. The relationship between Shot Noise and experimental parameters is as follows:
$\mathrm{i_{RMS} = \sqrt{2IeΔf}}$
with
• iRMS: Root-mean-square current fluctuation (in Amperes)
• I: Average direct current (A)
• e: electronic charge (1.60 x 10-19 C)
• Δf: frequency bandwidth (Hz)
Reduction of Shot Noise is accomplished most easily by decreasing the frequency bandwidth of the signal (reducing Δf). A good discussion of Shot Noise may be found by clicking here
Flicker Noise
Flicker noise is also called 1/f noise because the magnitude of flicker noise is inversely proportional to frequency. The source of flicker noise is uncertain and it seems to be significant only at low frequencies (<100 Hz). A good summary of flicker noise (and Johnson noise) may be found by clicking here
Environmental Noise
These are sources of noise that interfere with analytical measurements. Examples of such sources include:
• electrical power lines (e.g. 50 or 60 Hz line noise)
• electrical equipment (e.g. motors, fluorescent lights, etc.)
• RF sources (e.g. cell phones)
• environmental factors (drift in temperature, aging of electronic components) | textbooks/chem/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Courseware/Introduction_to_Signals_and_Noise/01_Goals_and_Objectives.txt |
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