D. Harvey - Modern Analytical Chemistry (794078), страница 64
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For example, when acetone is the solvent, a Soxhlet extraction is limited to 56 °C. With a microwave-assisted extraction, however, a temperature of over 150 °C can be obtained when using acetone as the solvent.Two other examples of a continuous extraction deserve mention. Volatile organic compounds (VOCs) can be quantitatively removed from liquid samples by aliquid–gas extraction.
As shown in Figure 7.19, the VOCs are removed by passingan inert purging gas, such as He, through the sample. The He removes the VOCs,which are then carried by the He to a tube where they are collected on a solid adsorbent. When the extraction is complete, the VOCs can then be removed from thetrap for analysis by rapidly heating the tube while flushing with He. This techniqueis known as a purge and trap. Recoveries for analytes using a purge and trap maynot be reproducible, requiring the use of internal standards for quantitative work.Primaryadsorbent trapSampleSecondaryadsorbent trapFigure 7.19Schematic diagram of a purge-and-trapsystem.
Analyte is collected in the primaryadsorption trap. The secondary adsorptiontrap is monitored for evidence ofbreakthrough.Purge gas1400-CH07 9/8/99 4:04 PM Page 215Chapter 7 Obtaining and Preparing Samples for AnalysisContinuous extractions also can be accomplished with supercritical fluids.19When a substance is heated above its critical temperature and pressure, it forms asupercritical fluid whose properties are between those of a gas and a liquid. Supercritical fluids are better solvents than gases, making them a better reagent for extractions.
In addition, the viscosity of a supercritical fluid is significantly less thanthat of a liquid solvent, allowing it to pass more readily through particulate samples.One example of a supercritical extraction is the determination of total petroleumhydrocarbons (TPHs) in soils, sediments, and sludges with supercritical CO2. Approximately 3 g of sample is placed in a 10-mL stainless steel cartridge, and supercritical CO2, at a pressure of 340 atm and a temperature of 80 °C, is passed throughthe cartridge for 30 min at flow rate of 1–2 mL/min. The petroleum hydrocarbonsare collected by passing the effluent from the cartridge through 3 mL of tetrachloroethylene at room temperature. At this temperature the CO2 reverts to the gasphase and is released to the atmosphere.20Chromatographic Separations In an extraction, the sample is initially present inone phase, and the component of interest is extracted into a second phase.
Separations can also be accomplished by continuously passing one sample-free phase,called the mobile phase, over a second sample-free phase that remains fixed or stationary. The sample is then injected or placed into the mobile phase. As the sample’s components move with the mobile phase, they partition themselves betweenthe mobile and stationary phases. Those components having the largest partitioncoefficients are more likely to move into the stationary phase, taking longer to passthrough the system.
This is the basis of all chromatographic separation techniques.As currently practiced, modern chromatography provides a means both of separating analytes and interferents and of performing a qualitative or quantitative analysisof the analyte. For this reason a more thorough treatment of chromatography isfound in Chapter 12.7G Liquid–Liquid ExtractionsA liquid–liquid extraction is one of the most important separation techniques usedin environmental, clinical, and industrial laboratories. Two examples from environmental analysis serve to illustrate its importance. Public drinking water suppliesare routinely monitored for trihalomethanes (CHCl3, CHBrCl2, CHBr2Cl, andCHBr3) because of their known or suspected carcinogeneity.
Before their analysisby gas chromatography, trihalomethanes are separated from their aqueous matrixby a liquid–liquid extraction using pentane.21 A liquid–liquid extraction is alsoused in screening orange juice for the presence of organophosphorous pesticides. Asample of orange juice is mixed with acetonitrite and filtered. Any organophosphorous pesticides that might be present in the filtrate are extracted with petroleumether before a gas chromatographic analysis.22In a simple liquid–liquid extraction the solute is partitioned between two immiscible phases. In most cases one of the phases is aqueous, and the other phase isan organic solvent such as diethyl ether or chloroform.
Because the phases are immiscible, they form two layers, with the denser phase on the bottom. The solute isinitially present in one phase, but after extraction it is present in both phases. Theefficiency of a liquid–liquid extraction is determined by the equilibrium constantfor the solute’s partitioning between the two phases.
Extraction efficiency is also influenced by any secondary reactions involving the solute. Examples of secondary reactions include acid–base and complexation equilibria.215supercritical fluidA state of matter where a substance isheld at a temperature and pressure thatexceeds its critical temperature andpressure.1400-CH07 9/8/99 4:04 PM Page 216216Modern Analytical Chemistry7G.1 Partition Coefficients and Distribution RatiosEarlier we learned that the partitioning of a solute between two phases is describedby a partition coefficient. If the solute is initially in an aqueous phase and is extracted into an organic phase*Saqt Sorgthe partition coefficient isKD =distribution ratioA ratio expressing the totalconcentration of solute in one phaserelative to a second phase; all forms ofthe solute are considered in defining thedistribution ratio (D).[Sorg ][Saq ]A large value for KD indicates that the extraction of the solute into the organic phaseis favorable.In evaluating the efficiency of an extraction, however, we must consider thesolute’s total concentration in each phase.
We define the distribution ratio, D, tobe the ratio of the solute’s total concentration in each phase.D =[Sorg ]tot[Saq ]totWhen the solute exists in only one form in each phase, then the partition coefficientand the distribution ratio are identical. If, however, the solute exists in more thanone form in either phase, then KD and D usually have different values. For example,if the solute exists in two forms in the aqueous phase, A and B, only one of which,A, partitions itself between the two phases, thenD =[Sorg ]A[Sorg ]A≤ KD =[Saq ]A + [Saq ]B[Saq ]AThis distinction between KD and D is important. The partition coefficient is anequilibrium constant and has a fixed value for the solute’s partitioning between thetwo phases. The value of the distribution ratio, however, changes with solution conditions if the relative amounts of forms A and B change.
If we know the equilibriumreactions taking place within each phase and between the phases, we can derive analgebraic relationship between KD and D.7G.2 Liquid–Liquid Extraction with No Secondary ReactionsSorgOrganicIn the simplest form of liquid–liquid extraction, the only reaction affecting extraction efficiency, is the partitioning of the solute between the two phases (Figure 7.20).In this case the distribution ratio and the partition coefficient are equal.AqueousSaqFigure 7.20Scheme for a simple liquid–liquid extractionwithout any secondary reactions.D =[Sorg ]tot[Sorg ]=[Saq ]tot[Saq ]7.19Conservation of mass requires that the moles of solute initially present in onephase equal the combined moles of solute in the aqueous and organic phases afterthe extraction; thus(Moles aq)0 = (moles aq)1 + (moles org)17.20*Although the following treatment assumes that the solute is initially present in the aqueous phase, the resultingequations for the distribution of the solute between the two phases are independent of which phase originally containsthe solute.1400-CH07 9/8/99 4:04 PM Page 217Chapter 7 Obtaining and Preparing Samples for Analysiswhere the subscript indicates the extraction number.
The concentration of S in theaqueous phase after the extraction is[Saq ] =(moles aq)1Vaq7.21whereas the solute’s concentration in the organic phase is[Sorg ] =(moles org)1Vorg7.22where Vaq and Vorg are the volumes of the aqueous and organic phases.
Solvingequation 7.20 for (moles org)1 and substituting into equation 7.22 leave us with[Sorg ] =(moles aq)0 − (moles aq)1Vorg7.23Substituting equations 7.21 and 7.23 into equation 7.19, we obtainD =[(moles aq)0 − (moles aq)1]/ Vorg(moles aq)0 Vaq − (moles aq)1Vaq=(moles aq)1 / Vaq(moles aq)1VorgRearranging and solving for the fraction of solute remaining in the aqueous phaseafter one extraction, (qaq)1, gives(qaq )1 =Vaq(moles aq)1=(moles aq)0DVorg + Vaq7.24The fraction present in the organic phase after one extraction, (qorg)1, is(qorg )1 =DVorg(moles org)1= 1 − (qaq )1 =(moles org)0DVorg + VaqExample 7.14 shows how equation 7.24 is used to calculate the efficiency of a simpleliquid–liquid extraction.EXAMPLE 7.14A solute, S, has a KD between water and chloroform of 5.00.
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