Effect of Sample Volume on Quantitative Analysis by Solid-phase Microextraction Part 1. Theoretical Considerations Tadeusz G�orecki† and Janusz Pawliszyn* Department of Chemistry and Waterloo Centre for Groundwater Research, University of Waterloo, Waterloo, Ontario, Canada N2L 3G1 This paper discusses the effect of sample volume on the amount of analyte extracted from a sample by solid-phase microextraction (SPME) in two-phase (sample–fiber coating) and three-phase (sample–headspace–fiber coating) systems.Up-to-date knowledge is summarized, and new concepts are introduced. The effect of sample volume on quantification and precision of results can be neglected only in rare cases. The minimum sample volume which ensures that the amount extracted, n, is lower than 1% of the initial amount of the analyte present in the sample, as well as the volume for which exactly half of the initial amount of the analyte is extracted, have been calculated for both two- and three-phase systems.It is critical that the volumes of samples and standards are the same during analysis by SPME. Extraction kinetics in headspace analysis is dependent on the headspace capacity. If it is sufficiently large, the analyte is extracted almost exclusively from the gaseous phase, and equilibration can be very fast. On the other hand, this causes a significant loss of sensitivity. The effect of sample volume on the determination of the value of the partition coefficient, K, is also discussed. If the change in concentration of the analyte in the sample at equilibrium is not taken into account, erroneous results are obtained.Even when a proper procedure is used, there are practical limitations to the accuracy of the K value determination. Large sample volumes should always be used for K value determination, as they enable broader ranges of K values to be covered with good accuracy. Keywords: Solid-phase microextraction; sample volume; partition coefficient Solid-phase microextraction (SPME) was introduced in 1990.1 Since then, the interest in this method continues to grow very rapidly.SPME utilizes a small, fused-silica fiber coated with a polymeric stationary phase for analyte extraction from the matrix. The fiber is mounted for protection in a syringe-like device. The stationary phase can be a very viscous liquid (e.g., polydimethylsiloxane, PDMS) or a porous solid. For liquid phases, analytes are absorbed until an equilibrium is reached in the system.The amount extracted under these conditions is dependent on the partition coefficient between the sample and the coating. Sampling in SPME can be carried out directly from gaseous or liquid samples, or from the headspace over liquid or solid samples.2 SPME has been used for many applications, including the determination of substituted benzene compounds, 3–5 caffeine in beverages,6 volatile organic compounds in water,7–9 polyaromatic hydrocarbons and polychlorinated biphenyls,10 chlorinated hydrocarbons,11 phenols,12–14 pesticides, 15–22 and fatty acids,23 as well as lead and tetraethyllead.24 Although, initially, SPME was used in conjunction with GC only, its applicability has been broadened to LC25,26 and supercritical fluid chromatography (SFC).27 Theoretical aspects of SPME analysis have been studied in detail.Louch et al.28 developed the theory for two-phase systems (sample and fiber coating), and Zhang and Pawliszyn2 for three-phase systems (sample–headspace–coating). Practical aspects have been discussed in numerous papers, including those of Arthur et al.29 and Motlagh and Pawliszyn.30 Although the concepts behind SPME are relatively simple, analysis of the contents of some papers, as well as conversations with numerous users, indicate that certain aspects of SPME remain unclear or are misunderstood by many people, which can easily lead to confusion and frustration with the technique.A notorious example is the often large discrepancies between the partition coefficient (distribution constant) values reported for the same coatings and compounds by different groups, or even different researchers in the same group. These discrepancies have led Langenfeld et al.31 to propose the term ‘effective distribution constant’ to describe the partition coefficients observed for given sample–coating systems. The aim of this series of papers is to discuss several aspects of SPME analysis that might be directly responsible for the problems observed.In particular, the effect of sample volume on the amount of analyte extracted in two- and three-phase systems, as well as on determination of partition coefficients, K, will be discussed. Part 1 briefly summarizes the current knowledge and introduces several new concepts. Part 2 will present experimental verification of the ideas presented. Sample Volume versus Amount Extracted in Two-phase Systems In two-phase systems (gaseous sample–coating or liquid sample–coating) at equilibrium, the initial amount of the analyte present in the sample is distributed between the sample and the SPME fiber coating.The mass balance in such systems can be described in the following way:28 C0Vs = CHs Vs + CHf Vf (1) where C0 is the initial concentration of the analyte in the sample, Vs the sample volume, CHs the concentration of the analyte in the sample at equilibrium, CHf the concentration of the analyte in the fiber coating at equilibrium and Vf the volume of the coating.It should be emphasized that for liquid samples the above mass balance applies only to systems with no headspace, or for analytes whose Henry’s constants are so low that their amount in the sample headspace can be neglected. Partitioning between the sample and fiber coating is governed by the partition coefficient, K, also called the distribution constant: K C C = ¥ ¥ f s (2) † On leave from the Faculty of Chemistry, Technical University of Gda�nsk, Poland.Analyst, October 1997, Vol. 122 (1079–1086) 1079Combination of eqns. (1) and (2) and a few simple rearrangements yield the final expression describing the amount extracted by the SPME fiber at equilibrium, n:28 n KC V V V KV = + 0 s f s f (3) It is a common notion that the term KVf in the denominator of eqn. (3) is usually so small that it can be neglected, in which case the amount of analyte extracted by the fiber coating at equilibrium would be independent of sample volume and simply described by n = KC0Vf.While this may be true for analytes characterized by very low coating–sample partition coefficients or for samples of very large volumes, in most cases the above assumption is incorrect, which can lead to significant errors, as will be discussed in the following sections. The concentration of the analyte in the sample at equilibrium can be determined from a simple mass balance: C C V n V s s s ¥ = - 0 (4) Dividing both sides of eqn.(4) by C0 yields: C C C V n C V n C V KV V KV V V KV s s s s f s f s s f ¥ = - = - = - + = + 0 0 0 0 1 1 (5) Also, by definition C C n n s ¥ = 0 0 (6) where n is the amount of analyte extracted from a sample of volume Vs and n0 is the amount that would be extracted by the fiber from a sample of infinite volume (in which case no significant decrease in the concentration of the analyte in the sample would be observed at equilibrium).Fig. 1 presents the dependence of n/n0 on Vs for (a) a 100 mm fiber (Vf = 0.65 ml) and (b) a 30 mm fiber (Vf = 0.14 ml), for sample volumes of up to 1 l and K values ranging from 1000 to 1 000 000. This situation would be characteristic of air sampling from typical glass bulbs, and hydrocarbons from the gasoline range. According to Martos et al.,32 K Å 1000 corresponds to C7, while K Å 1 000 000 corresponds to C15.It is obvious from Fig. 1 that except for compounds with the lowest K values, sample volume has a very significant effect on the amount extracted by the fiber. Practical consequences of this are very important. Calibration of the fiber is usually performed by exposing it to a standard gaseous mixture generated in a glass bulb of known volume by the static me.33 The results of such calibrations are valid only for gaseous samples of exactly the same volume. If the fiber is exposed to ambient air (an unlimited supply of sample), the amount extracted by the fiber can be significantly larger compared with limited volume samples.On the other hand, if the containers used for grab sampling of air are smaller than the bulb used for calibration, the amount extracted will be smaller than expected from the calibration graph. In both cases, erroneous quantitative results will be obtained. It is obvious, therefore, that the same volume should be used for both the sample and the standard.Alternatively, a different calibration procedure can be used. Dynamic generation of standard gas mixtures provides unlimited volumes of gas standards, and thus can be used to calibrate the fiber for direct ambient air sampling. Alternatively, calibration based on retention parameters as described by Martos and Pawliszyn34 can be used for PDMS fibres. In this procedure no standards are necessary. Direct analysis of aqueous samples is usually performed by exposing the fiber to a small volume of sample contained in a vial.Fig. 2 presents similar dependences determined for sample volumes of up to 40 ml, and three coating thicknesses, viz., 100, 30 and 7 mm (Vf = 0.028 ml). It should be emphasized again that no headspace can be present in the vial if the two-phase system model is to be applied. The thin coating (7 mm) fibers are often used for direct sampling of semi-volatile compounds, since thicker coatings cause equilibration times to be excessively long.It is clear from Fig. 2 that the sample volume is an Fig. 1 Dependence of n/n0 on Vs for (a) 100 mm fiber (Vf = 0.65 ml), and (b) 30 mm fiber (Vf = 0.14 ml), for direct sampling from large volume samples (two-phase system). Fig. 2 Dependence of n/n0 on Vs for (a) 100 mm fiber (Vf = 0.65 ml), (b) 30 mm fiber (Vf = 0.14 ml), and (c) 7 mm fiber (Vf = 0.028 ml), for direct sampling from small volume samples (two-phase system). 1080 Analyst, October 1997, Vol. 122important parameter even for compounds with relatively small K values, especially when very small sample volumes (1 ml) are used. Moreover, headspace sampling is normaly used for volatile compounds with small K values and direct sampling is used for semi-volatile compounds with large K values. Hence, sample volume cannot usually be neglected even when the 7 mm coating is used. The course of the curves indicates that in some cases (steep part of the curve, e.g., K = 1000 for a 100 mm coating, Vs between 1 and 5 ml) small differences in sample volume can result in relatively large differences in the amount extracted, which adversely affects the precision of the determination.Care should be taken, therefore, to ensure that the sample volume during direct sampling is always the same. In view of the above discussion, it is interesting to establish what the sample volume should be in order for the amount extracted by the fiber to be insignificant compared with the amount remaining in the sample after extraction. Eqn.(3) can be used for this purpose. Let us assume that the amount extracted, n = CfVf, fulfils the above condition when it is smaller than, or equal to, 1% of the initial analyte amount in the sample, i.e., CHf Vf @ 0.01 C0Vs (7) The equation can be rearranged to yield: C C V V KV V f f s f s ¥ @ 0 0 01 0 01 @ @ . . (8) since, according to the initial assumption, C0 � CHs . Under the above conditions, eqn.(3) can be simplifid to n = KC0Vf (9) Simple rearrangement of eqn. (8) yields the final condition: VS ! 100KVf (10) For a 100 mm fiber, whose volume is approximately 0.65 ml, eqn. (10) can be written as: VS ! 0.065 K [ml] (11) It follows from eqn. (11) that even for analytes whose partition coefficient values are as low as 100, the minimum sample volume when using a 100 mm thick coating should be 6.5 ml if the effect of sample volume on the amount extracted is to be neglected.Similar reasoning can be applied to calculate the sample volume from which exactly 50% of the analyte is extracted, i.e., n = CfVf = 0.5 C0Vs. In this case 0 5 0 0 . C V KC V V V KV s s f s f = + (12) After a few rearrangements, the final condition is obtained: Vs 50% = KVf (13) or, for a 100 mm fiber, Vs = K 3 0.65 3 1023 [ml] (14) It follows from eqn. (14) that when 1 ml samples are analyzed with 100 mm fibers, 50% of the initial amount of the analyte will be extracted when the K value is approximately 1540.Typical partition coefficients of semi-volatile compounds are usually higher; it is obvious, therefore, that sample volume has a very significant effect on the amount extracted when small sample volumes are used. Sample Volume versus Amount Extracted in Three-phase Systems In most cases liquid samples are placed in vials with some headspace remaining inside. At equilibrium, by definition the chemical potentials of the analyte in all three phases (liquid sample–headspace–fiber coating) must be the same.2 A very important consequence of this fact is that the amount of the analyte extracted by the fiber at equilibrium in a three-phase system is the same independently of where the fiber is located, be it the headspace or the liquid.Consequently, in systems where a headspace is present, a different dependence should be used to calculate the amount of analyte extracted by the fiber regardless of where the fiber is located, viz:2 n KC V V KV K V V = + + 0 s f f hs hs s (15) where Khs is the headspace–liquid partition coefficient Khs = CH hs/CHs (16) (CH hs is the concentration of the analyte in the headspace at equilibrium), and Vhs is the headspace volume.Compared with a two-phase system described by eqn. (3), the difference is the additional term KhsVhs in the denominator of eqn. (4). For volatile compounds, Khs is usually close to 1, which means that headspace volume can be neglected only when it is close to zero (a two-phase system). Semi-volatile compounds have much lower values of Khs; therefore, the KhsVhs term may be negligibly small; however, such an assumption should always be verified. For obvious reasons the dependence of the amount of analyte extracted by the fiber on sample volume is much more complex in three-phase systems.However, it can be still dealt with relatively easily if the total volume of the system (sample plus headspace) remains constant.In practice, this will be the usual case, as the vial volume remains constant, and the headspace volume is the difference between the total volume and the sample volume. In such systems the concentration of an analyte in the sample at equilibrium is determined by the following mass balance: C C V C V n V s s hs hs s ¥ ¥ = - - 0 (17) Dividing both sides of eqn. (17) by C0 yields: C C C V C V n C V s hs hs s s ¥ ¥ = - - 0 0 0 1 (18) From the definition of Khs [eqn.(16)], CH hs can be replaced by KhsCHs , while n is described by eqn. (15), which yields C C K C V C V KV KV K V V s ks s hs s f f hs hs s ¥ ¥ = - - + + 0 0 1 (19) Simple rearrangements yield the final dependence: C C V KV K V V s s f hs hs s ¥ = + + 0 (20) Similarly to a two-phase system, Cs H/C0 = n/n0 when the headspace volume is constant. Fig. 3 illustrates the effect of headspace volume on the amount of analyte extracted in a system of constant volume (4, 15 and 40 ml—typical vial sizes) by a 100 mm fiber, typically used for the analysis of volatile compounds in sample headspace.To allow for better comparison of the results, the xaxis has been defined as the ratio of headspace volume to sample volume. The following values of partition coefficients have been used to calculate the course of the curves: K = 400, 2500 and 1000, and Khs = 0.15, 0.7 and 1.24, respectively. These values correspond approximately to those for chloroform, 1,1,1-trichloroethane and carbon tetrachloride.It is interesting that n/n0 is always the largest for the dependence corresponding Analyst, October 1997, Vol. 122 1081to chloroform, which has the lowest Khs value, even though its K value is also the lowest. This is because, with such a combination of Khs and K, only araction of the analyte is present in the sample headspace; therefore, the equilibrium concentration of the analyte in the sample remains relatively high.The course of the two other curves is also interesting. For small headspace volumes, n/n0 is higher for the curve corresponding to carbon tetrachloride than for that corresponding to 1,1,1-trichloroethane, which is consistent with the values of their partition coefficients, K. However, as carbon tetrachloride has the largest Khs, n/n0 drops faster for this compound than for 1,1,1-trichloroethane, as a result of which the two lines cross each other.This illustrates the significance of headspace volume on the analytical results. It is also interesting that the largest relative increase in n/n0 ( Å 44%) when moving from small to large vials is observed for 1,1,1-trichloroethane, with the highest K value. For compounds with lower K values the relative increase in the amount extracted is much lower ( Å 16 and Å 12% for chloroform and carbon tetrachloride, respectively). It is, therefore, the combination of K and Khs for a given compound that determines the magnitude of the effect of the sample volume on the amount extracted in three-phase systems with headspace.Headspace volume can be the critical factor determining the precision of the results in three-phase systems. It is relatively easy to measure sample volume accurately. However, vials are not manufactured to have exactly the same volume. Wall thicknesses and bottom shapes may differ from vial to vial. Also, the shape of the septum in a closed vial can vary from concave to convex.All these factors will affect the total volume of the system, and, therefore, the headspace volume, which is the difference between the total volume and sample volume. As illustrated in Fig. 3, for compounds with large Khs, changing the headspace volume can significantly affect the amount extracted, especially in the range of headspace volumes usually applied (Vhs/Vs @ 1). Such differences (especially related to septum shape) are usually more pronounced in small vials; therefore, worse precision can usually be expected when using such vials for headspace sampling. Similarly to two-phase systems, it is possible to calculate the minimum sample volume necessary for the amount extracted by the fiber to be insignificant compared with the amount remaining in the sample after extraction.It is again assumed that the condition is fulfilled when less than 1% of the initial amount present in the sample is extracted by the fiber, i.e., n = CHf Vf @ 0.01 C0Vs.However, in this case it cannot be assumed that CHs � Co, because even if the amount extracted by the fiber constitutes only 1% of the initial amount of the analyte, a significant fraction of the analyte might be present in the headspace. Eqn. (15) will, therefore, be used to calculate the criterion: 001 0 0 . C V KC V V KV K V V s s f f hs hs s ! + + (21) The headspace volume, Vhs, can be expressed as a fraction of the sample volume, Vhs = aVs.Simple rearrangements yield the final criterion: V KV aK s f hs ! 99 1+ (22) Compared with the criterion for a two-phase system [eqn. (10)], there is an additional term in the denominator of eqn. (22), viz., aKhs. For non-zero headspace volumes this term is always greater than 0, which means that the sample volume fulfilling the criterion is smaller in three-phase systems than in two-phase systems. This is because, ultimately, the amount of analyte extracted by the fiber is determined by the concentration of the analyte in the liquid sample at equilibrium.This concentration will be lower in three-phase systems than in two-phase systems, since a significant fraction of the analyte may be present in the headspace of the sample. As a consequence, the sensitivity of headspace SPME can only be lower than, or equal to, that of direct SPME if the system is allowed to reach equilibrium. The higher sensitivity for headspace SPME reported sometimes (especially for compounds with relatively large K and Khs values) is clearly due to non-equilibrium conditions when sampling directly from liquid samples.This is not unusual in the light of the fact that equilibration times for direct sampling from water can be as long as a few hours (especially with inefficient agitation), and the increases in the amount of analyte extracted might not be noticeable if the experiment is not carried out for sufficiently long times.In addition, during prolonged sampling, analyte losses via adsorption onto the sample vial walls, absorption by the exposed parts of the silicone rubber septum (the Teflon lining of the septum has to be pierced to introduce the fiber; since the needle of the SPME device is blunt, the lining usually breaks open exposing a small area of silicone substrate), microbial decomposition, etc., can more than outweigh the expected increase in the amount of analyte extracted, as a result of which a smaller extracted amount can actually be observed after a longer time.This and other factors, including elimination of matrix effects related to the presence of high-boiling compounds in the sample and significantly faster equilibration times, usually make headspace SPME the preferred method for the analysis of mostly volatile compounds. It is also easy to determine the sample volume from which 50% of the analyte will be extracted by the SPME fiber placed in the sample headspace.The procedure is similar to that Fig. 3 Effect of headspace volume on the amount of analyte extracted in a system of a constant volume of (a) 4, (b) 15 and (c) 40 ml by a 100 mm fiber for chloroform, 1,1,1-trichloroethane and carbon tetrachloride (see text for the values of partition coefficients used for calculations). 1082 Analyst, October 1997, Vol. 122presented for two-phase systems. The final criterion in this case is: V KV aK s f hs 50 1 % = + (23) The difference between eqns.(23) and (13) is again the additional term in the denominator, aKhs. The higher the headspace–sample partition coefficient, Khs, or the headspaceto- sample volume ratio, a, the smaller the sample volume from which 50% of the initial amount of the analyte is extracted by the SPME fiber. Effect of Headspace Capacity on Extraction Kinetics Headspace volume can have a significant effect on equilibration times (extraction kinetics). If the headspace capacity is low (small Khs), and K is large, the equilibration process is very slow.A rapid initial increase in the amount of analyte extracted is usually observed, followed by a much slower increase that lasts for a long time. The first stage corresponds to analyte extraction from the gaseous phase only. As soon as the headspace concentration of the analyte falls below the equilibrium level with respect to the aqueous phase, analyte molecules start to move from the liquid sample to the headspace.At any given moment there can only be so many molecules in the headspace, depending on the Khs value, which causes the equilibration process to be very slow. The headspace acts in this case as a bottleneck for analyte transport to the fiber. On the other hand, if the amount of the analyte extracted by the fiber at equilibrium is negligible compared with the amount present in the headspace equilibrated with the sample, only a very small amount of the analyte has actually to be transported from the liquid sample through the headspace to the fiber coating, i.e., the analyte is extracted almost exclusively from the gaseous phase, and the process is much faster than for the case described above.Assuming this situation occurs when 95% of the analyte extracted by the fiber at equilibrium comes exclusively from the headspace, the criterion that must be fulfilled can be described by eqn. (24). The assumption is reasonable, as a 5% difference usually falls within the limits of experimental error for trace SPME–GC analysis: R KV K V KV K aV = = f hs hs f hs s @ 0.05 (24) The above criterion means that the capacity of the headspace (KhsVhs) needs to be at least 20 times larger than the capacity of the fiber (KVf) to achieve rapid extraction.For a given sample volume, Vs, this c be achieved by using a sufficiently large headspace volume (corresponding to a large headspace-tosample volume ratio, a), or by increasing Khs.The latter can be accomplished by increasing the temperature or by salting the analyte out of the liquid phase. When the criterion described by eqn. (24) is fulfilled, equilibration can take as little as a few minutes, and is almost independent of the agitation conditions (provided that the analyte is equilibrated between the liquid phase and its headspace before the extraction begins). It should be emphasized, however, that great care must be exercised when moving the method from one vial size to another, as illustrated in Fig. 4. When the K value is large and/or Khs is small [Fig. 4(a)] a significant portion of the analyte extracted by the fiber has to be transported from the liquid sample through the headspace independently of the headspace volume, as evidenced by the values of R larger than 0.05 for all three values of a illustrated except for a = 2 and sample volumes larger than 28 ml (hence headspace volumes larger than 56 ml).The equilibration times will be relatively long for all vial sizes up to 40 ml. On the other hand, when the K value is smaller and/or Khs is larger, the R value can be lower than 0.05 for sample and headspace volumes often found in practice. Whether the criterion is fulfilled or not depends on the actual headspace volume [Fig. 4(b)]. For the same headspace-to-sample volume ratio, a, the criterion is fulfilled for large headspace volumes, but not for small ones. For example, when using a headspace-tosample volume ratio a = 0.5, the criterion is fulfilled for headspace volumes greater than 7 ml ( = sample volumes greater than 14 ml); therefore, under such conditions equilibration is very fast. However, when a 2 ml sample is used and the headspace volume is 1 ml, R is equal to 0.345, which indicates that a significant portion of the analyte has to be transported to the fiber from the liquid sample.As a consequence, the equilibration time will be longer even though the overall sample volume is smaller. If this is not accounted for and extraction times determined for large vials are used with small vials, the accuracy and precision of the determination suffers.It should be remembered, therefore, that the equilibration time has to be examined whenever the volume of the sample and/or headspace, or the extraction conditions (temperature, salt addition) are changed. It should also be emphasized that the increase in headspace capacity causes a significant loss of sensitivity.On the other hand, increasing the headspace capacity (which can be accomplished, for example, by increasing the extraction temperature) might dramatically shorten the equilibration times for some semi-volatile analytes while maintaining sufficient sensitivity. Both these effects need to be taken into account when developing a method involving headspace extraction. Effect of Sample Volume on the Determination of K in Two-phase Systems The fiber coating–sample partition coefficient [eqn.(2)] can be determined in several ways. The best way is to use eqn. (3). After a few rearrangements, the following dependence is obtained: K nV V C V n = - s f s ( ) 0 (25) Fig. 4 Dependence of R = KVf/KhsaVs on sample volume for three headspace-to-sample volume ratios, a = 0.5, 1 and 2, and a 100 mm PDMS fiber (Vf = 0.65 ml); (a) K = 1000, Khs = 0.25; (b) K = 250, Khs = 0.5. Analyst, October 1997, Vol. 122 1083If the amount extracted by the fiber, n, is negligible compared with the initial amount of the analyte in the sample, C0Vs, this equation is simplified to: K n C V = 0 f (26) which is a simple rearrangement of eqn. (9). Often, K is calculated directly from its definition [eqn. (2)]. In this case, if n is negligible compared with the initial amount of the analyte in the sample, C0Vs, then CHs can be substituted by C0, and the calculation is straightforward. However, it is our belief that often this substitution is made even when it is not legitimate. The consequences of this can be very significant, as will be illustrated in the following paragraphs.In both cases, the fiber volume necessary to calculate the concentration of the analyte in the fiber coating at equilibrium must be known with sufficiently good accuracy to obtain valid results. Let us introduce a new term, K apparent (Kap), this being the ratio of the concentration of the analyte in the fiber at equilibrium to the initial concentration C0: K C C ap f = ¥ 0 (27) Kap is what is calculated when depletion of the initial concentration of the analyte in the sample is not taken into account (i.e., when C0 is used instead of CHs ).The ratio of K to Kap is equal to: K K C C ap s = ¥ 0 (28) Therefore, K C C K ap s = � ¥ 0 (29) Since (5), s s s f C C V V KV ¥ = + 0 K KV V KV ap s s f = + (30) The true value of K can be calculated using the following equation derived from eqn.(29): K K V K V V = - - ap s ap f s (31) However, in many cases it will not be possible to determine K accurately by this method owing to the phenomena illustrated in Fig. 5. Fig. 5 presents the relationship between Kap and K for a wide range of partition coefficients, for two different coatings (100 and 7 mm) and three sample volumes (2, 10 and 30 ml). The dependences level off in all cases. It is obvious that the relationship reaches a certain limit, above which the same value of Kap is determined independently of the true value of K.The limit can be easily determined mathematically: lim lim K K K KV V KV V V ®¥ ®¥ = + æ è ç ö ø ÷ = ap s s f s f (32) It follows from eqn. (31) that Kap is determined only by the phase ratio Vs/Vf. The smaller the sample size (or the thicker the coating), the lower the value of Kap. It is possible that this phenomenon is behind certain very low K values reported in the literature for compounds that should have very large partition coefficients (e.g., polyaromatic hydrocarbons), especially when small sample volumes are used.Also, it is clear that when Kap is calculated instead of K, the values for thinner fiber coatings are much higher than those for thicker coatings (the difference in volume between a 100 mm fiber coating and a 7 mm coating is more than an order of magnitude). Such apparent discrepancies have also been reported in the literature.Even if proper methodology is used, owing to practical limitations it is not always possible to determine K accurately, as will be illustrated in the following paragraphs. Table 1 presents the amount of analyte extracted from a 2 ml sample and the corresponding per cent. increase in the amount extracted when going from a smaller K to a larger K for a 100 mm thick fiber coating and 1 mg l21 initial concentration, for different K values. For small K values the increase in the amount extracted when K increases is significant, although not linear as would be the case for infinite sample volumes.For large K values the increase in the amount extracted when K increases becomes insignificant and falls within the limits of experimental error. Table 2 illustrates the significance of this phenomenon. It presents the ranges of K values that are obtained when the amount of analyte extracted by the fiber, n, determined experimentally falls within ±5% (relative) of the true value, for two sample volumes (2 and 35 ml) and two fiber coating thicknesses (100 and 7 mm).A ±5% error can be assumed typical for trace analysis by SPME–GC. Fig. 5 Relationship between Kap and K for two different coatings (100 and 7 mm) and three sample volumes (2, 10 and 30 ml). Table 1 Amount of analyte extracted from a 2 ml sample and % increase in the amount extracted for a 100 mm thick fiber coating and 1 mg l21 initial concentration, for different K values K n/ng Increase (%) 100 0.0629540 — 1 000 0.4905660 717 10 000 1.5294118 212 100 000 1.9402985 28 1 000 000 1.9938650 2.8 10 000 000 1.9993848 0.28 100 000 000 1.9999385 0.3 1084 Analyst, October 1997, Vol. 122It is clear from Table 2 that practical limitations to K determinationan be very severe, especially when thick fiber coatings and low sample volumes are used. For a 2 ml sample and a 100 mm thick fiber, the error of K determination for K values around 10 000 is around ±20% for a ±5% relative error of n determination.If the true K values are higher by an order of magnitude, the values determined experimentally can range from Å 36 000 to infinity, which is of no use. In practice, owing to experimental errors, for large K values the value of n found experimentally can be larger than the initial amount of analyte in the sample. In such cases, negative K values would be obtained, which, of course, have no physical meaning. It is necessary in such cases to assume that the amount extracted, n, is equal to the initial amount of analyte in the sample (quantitative extraction), in which case infinite K values are obtained. On the other hand, K values of Å 1 000 000 can be determined with reasonable accuracy when large sample volumes (35 ml) and thin fiber coatings (7 mm) are used.It is our belief that the above-described phenomena account for most of the discrepancies between K values reported in the literature for the same substances and coatings.Other possible sources of errors include limited solubility of some analytes in water, as well as losses of analytes due to volatilization (when headspace is present in the system), absorption by exposed parts of silicone rubber septa, adsorption on glass, or biodegradation. The first two phenomena are more probable for volatile analytes, while the remaining phenomena are more significant for semi-volatile compounds, since when K values are very high, equilibration times become very long.From a practical point of view, partition coefficients, K, should always be determined using large sample volumes (nothing can be gained by using small volumes). For semi-volatile compounds characterized by large K values, only thin coatings (7 mm for PDMS) are practical for the experimental determination of K. Above a certain level, the accurate determination of K becomes virtually impossible. Conclusions Sample volume plays an extremely important role in analysis by SPME.In order to avoid errors or poor precision, care should be taken to ensure that the volumes of samples and standard solutions used for calibration are the same. In headspace SPME, the combination of K and Khs determines the magnitude of the effect of sample volume on the amount extracted by the fiber, n. With the same headspace-to-sample volume ratio, the magnitude of the effect is greatest for compounds with low Khs. The minimum sample volume which ensures that n is lower than 1% of the initial amount of analyte present in the sample (which means that the change in concentration of the analyte in the sample at equilibrium is negligible) can easily be calculated for both two- and three-phase systems, and so is the volume for which exactly half of the initial analyte amount is extracted. Variable vial volumes might cause poorer precision in headspace sampling, as only the volume of the liquid sample can be accurately controlled.Extraction kinetics in headspace sampling is determined by the headspace capacity, KhsVhs. If it is sufficiently large compared with the fiber capacity, KVf, then the analyte is extracted almost exclusively from the headspace, and equilibration can be very fast. The headspace capacity is determined by the actual headspace volume rather than by the headspace-tosample volume ratio a; therefore, dramatically different equilibration times might be observed when scaling down the method from large to small vials, even though a remains constant.On the other hand, a large headspace capacity causes a significant loss of method sensitivity. Sample volume should always be taken into account when determining K values. If the change in concentration of the analyte in the sample at equilibrium is not taken into account, erroneous results are obtained. Even when a proper procedure is used, there are practical limitations to the accuracy of the K value determination.Large sample volumes should always be used for K value determination, as they enable broader ranges of K values to be covered with good accuracy. Improper procedure is, in our opinion, the main cause of discrepancies between K values reported in the literature for the same compound by different sources, or for different coating thicknesses. Financial support of Supelco, Varian and National Sciences and Engineering Research Council of Canada is acknowledged.References 1 Arthur, C. L., and Pawliszyn, J., Anal. Chem., 1990, 62, 2145. 2 Zhang, Z., and Pawliszyn, J., Anal. Chem., 1993, 65, 1843. 3 Arthur, C. L., Killam, L. M., Motlagh, S., Lim, M., Potter, D. W., and Pawliszyn, J., Environ. Sci. Technol., 1992, 26, 979. 4 Potter, D. W., and Pawliszyn, J., J. Chromatogr., 1992, 625, 247. 5 Wittkamp, B. L., and Tilotta, D. C., Anal. Chem., 1995, 67, 600. 6 Hawthorne, S. B., Miller, D. J., Pawliszyn, J., and Arthur, C.L., J. Chromatogr., 1992, 603, 185. 7 Arthur, C. L., Pratt, K., Motlagh, S., Pawliszyn, J., and Belardi, R. P., J. High Resolut. Chromatogr., 1992, 15, 741. 8 Langenfeld, J. J., Hawthorne, S. B., and Miller, D. J., Anal. Chem., 1996, 68, 144. 9 Nilsson, T., Ferrari, F., and Facchetti, S., in Proceedings of the 18th International Symposium on Capillary Chromatography, Riva del Garda (Italy), May 20–24, 1996, ed. Sandra, P., and Davos, G., IOPMS, Kortrijk, 1996, p. 618. 10 Potter, D., and Pawliszyn, J., Environ.Sci. Technol., 1994, 28, 298. 11 Chai, M., Arthur, C. L., Pawliszyn, J., Belardi, R. P., and Pratt, K. F., Analyst, 1993, 118, 1501. 12 Buchholz, K., and Pawliszyn, J., Environ. Sci. Technol., 1993, 27, 2844. 13 Buchholz, K., and Pawliszyn, J., Anal. Chem., 1994, 66, 160. 14 Schaefer, B., and Engewald, W., Fresenius’ J. Anal. Chem., 1995, 352, 535. 15 Boyd-Boland, A. A., and Pawliszyn, J., J. Chromatogr., 1995, 704, 163. 16 Eisert, R., Levsen, K., and Wuensch, G., J.Chromatogr., 1994, 683, 175. Table 2 Ranges of K values obtained when the amount of analyte extracted by the fiber, n, determined experimentally falls within ±5% (relative) of the true value, for two different fiber coating thicknesses and two sample volumes 100 mm— 2 ml sample 35 ml sample K Range of K K Range of K 100 105–95 100 105–95 1 000 1067–935 1 000 1051–949 10 000 12 537–8172 10 000 10 598–9413 100 000 H–36 190 100 000 115 748–86 928 1 000 000 H–55 072 1 000 000 4 700 000–492 593 10 000 000 H–58 104 10 000 000 H–923 611 100 000 000 H–58 426 100 000 000 H–1 012 177 7 mm— 2 ml sample 35 ml sample K Range of K K Range of K 100 105–95 100 105–95 1 000 1051–949 1 000 1050–950 10 000 10 574–9434 10 000 10 504–9496 100 000 112 903–88 785 100 000 105 422–94 622 1 000 000 3 500 000–558 824 1 000 000 1 093 750–913 462 10 000 000 H–1 187 500 10 000 000 17 500 000–6 785 714 100 000 000 H–1 338 028 100 000 000 H19 000 000 Analyst, October 1997, Vol. 122 108517 Popp, P., Kalbitz, K., and Oppermann, G., J. Chromatogr., 1994, 687, 133. 18 Lee, X., Kumazawa, T., Taguchi, T., Sato, K., and Suzuki, O., Hochudoku, 1995, 13, 122. 19 Eisert, R., and Levsen, K., Fresenius’ J. Anal. Chem., 1995, 351, 555. 20 Levsen, K., and Eisert, R., J. Am. Soc. Mass Spectrom., 1995, 6, 1119. 21 Graham, K. N., Sarna, L. P., Webster, G. R. B., Gaynor, J. D., and Ng, H. Y. F., J. Chromatogr., 1996, 725, 129. 22 Magdic, S., and Pawliszyn, J., J.Chromatogr., 1996, 723, 111. 23 Pan, L., Adams, M., and Pawliszyn, J., Anal. Chem., 1995, 67, 4396. 24 G�orecki, T., and Pawliszyn, J., Anal. Chem., 1996, 68, 3008. 25 Chen, J., and Pawliszyn, J., Anal. Chem., 1995, 67, 2530. 26 Boyd-Boland, A. A., and Pawliszyn, J., Anal. Chem., 1996, 68, 1521. 27 Hirata, Y., and Pawliszyn, J., J. Microcol. Sep., 1994, 6, 443. 28 Louch, D., Motlagh, S., and Pawliszyn, J., Anal. Chem., 1992, 64, 1187. 29 Arthur, C. L., Killam, L.M., Buchholz, K. D., and Pawliszyn, J., Anal. Chem., 1992, 64, 1960. 30 Motlagh, S., a Pawliszyn, J., Anal. Chim. Acta, 1993, 284, 265. 31 Langenfeld, J. J., Hawthorne, S. B., and Miller, D. J., Anal. Chem., 1996, 68, 144. 32 Martos, P., Saraullo, A., and Pawliszyn, J., Anal. Chem., 1997, 69, 402. 33 Namiesnik, J., J. Chromatogr., 1984, 300, 79. 34 Martos, P., and Pawliszyn, J., Anal. Chem., 1997, 69, 206. Paper 7/01303E Received February 25, 1997 Accepted June 4, 1997 1086 Analyst, October 1997, Vol. 122 Effect of Sample Volume on Quantitative Analysis by Solid-phase Microextraction Part 1. Theoretical Considerations Tadeusz G�orecki† and Janusz Pawliszyn* Department of Chemistry and Waterloo Centre for Groundwater Research, University of Waterloo, Waterloo, Ontario, Canada N2L 3G1 This paper discusses the effect of sample volume on the amount of analyte extracted from a sample by solid-phase microextraction (SPME) in two-phase (sample–fiber coating) and three-phase (sample–headspace–fiber coating) systems.Up-to-date knowledge is summarized, and new concepts are introduced. The effect of sample volume on quantification and precision of results can be neglected only in rare cases. The minimum sample volume which ensures that the amount extracted, n, is lower than 1% of the initial amount of the analyte present in the sample, as well as the volume for which exactly half of the initial amount of the analyte is extracted, have been calculated for both two- and three-phase systems.It is critical that the volumes of samples and standards are the same during analysis by SPME. Extraction kinetics in headspace analysis is dependent on the headspace capacity. If it is sufficiently large, the analyte is extracted almost exclusively from the gaseous phase, and equilibration can be very fast. On the other hand, this causes a significant loss of sensitivity. The effect of sample volume on the determination of the value of the partition coefficient, K, is also discussed. If the change in concentration of the analyte in the sample at equilibrium is not taken into account, erroneous results are obtained.Even when a proper procedure is used, there are practical limitations to the accuracy of the K value determination. Large sample volumes should always be used for K value determination, as they enable broader ranges of K values to be covered with good accuracy.Keywords: Solid-phase microextraction; sample volume; partition coefficient Solid-phase microextraction (SPME) was introduced in 1990.1 Since then, the interest in this method continues to grow very rapidly. SPME utilizes a small, fused-silica fiber coated with a polymeric stationary phase for analyte extraction from the matrix. The fiber is mounted for protection in a syringe-like device. The stationary phase can be a very viscous liquid (e.g., polydimethylsiloxane, PDMS) or a porous solid.For liquid phases, analytes are absorbed until an equilibrium is reached in the system. The amount extracted under these conditions is dependent on the partition coefficient between the sample and the coating. Sampling in SPME can be carried out directly from gaseous or liquid samples, or from the headspace over liquid or solid samples.2 SPME has been used for many applications, including the determination of substituted benzene compounds, 3–5 caffeine in beverages,6 volatile organic compounds in water,7–9 polyaromatic hydrocarbons and polychlorinated biphenyls,10 chlorinated hydrocarbons,11 phenols,12–14 pesticides, 15–22 and fatty acids,23 as well as lead and tetraethyllead.24 Although, initially, SPME was used in conjunction with GC only, its applicability has been broadened to LC25,26 and supercritical fluid chromatography (SFC).27 Theoretical aspects of SPME analysis have been studied in detail.Louch et al.28 developed the theory for two-phase systems (sample and fiber coating), and Zhang and Pawliszyn2 for three-phase systems (sample–headspace–coating).Practical aspects have been discussed in numerous papers, including those of Arthur et al.29 and Motlagh and Pawliszyn.30 Although the concepts behind SPME are relatively simple, analysis of the contents of some papers, as well as conversations with numerous users, indicate that certain aspects of SPME remain unclear or are misunderstood by many people, which can easily lead to confusion and frustration with the technique.A notorious example is the often large discrepancies between the partition coefficient (distribution constant) values reported for the same coatings and compounds by different groups, or even different researchers in the same group. These discrepancies have led Langenfeld et al.31 to propose the term ‘effective distribution constant’ to describe the partition coefficients observed for given sample–coating systems.The aim of this series of papers is to discuss several aspects of SPME analysis that might be directly responsible for the problems observed. In particular, the effect of sample volume on the amount of analyte extracted in two- and three-phase systems, as well as on determination of partition coefficients, K, will be discussed. Part 1 briefly summarizes the current knowledge and introduces several new concepts. Part 2 will present experimental verification of the ideas presented.Sample Volume versus Amount Extracted in Two-phase Systems In two-phase systems (gaseous sample–coating or liquid sample–coating) at equilibrium, the initial amount of the analyte present in the sample is distributed between the sample and the SPME fiber coating. The mass balance in such systems can be described in the following way:28 C0Vs = CHs Vs + CHf Vf (1) where C0 is the initial concentration of the analyte in the sample, Vs the sample volume, CHs the concentration of the analyte in the sample at equilibrium, CHf the concentration of the analyte in the fiber coating at equilibrium and Vf the volume of the coating.It should be emphasized that for liquid samples the above mass balance applies only to systems with no headspace, or for analytes whose Henry’s constants are so low that their amount in the sample headspace can be neglected. Partitioning between the sample and fiber coating is governed by the partition coefficient, K, also called the distribution constant: K C C = ¥ ¥ f s (2) † On leave from the Faculty of Chemistry, Technical University of Gda�nsk, Poland.Analyst, October 1997, Vol. 122 (1079–1086) 1079Combination of eqns. (1) and (2) and a few simple rearrangements yield the final expression describing the amount extracted by the SPME fiber at equilibrium, n:28 n KC V V V KV = + 0 s f s f (3) It is a common notion that the term KVf in the denominator of eqn. (3) is usually so small that it can be neglected, in which case the amount of analyte extracted by the fiber coating at equilibrium would be independent of sample volume and simply described by n = KC0Vf.While this may be true for analytes characterized by very low coating–sample partition coefficients or for samples of very large volumes, in most cases the above assumption is incorrect, which can lead to significant errors, as will be discussed in the following sections. The concentration of the analyte in the sample at equilibrium can be determined from a simple mass balance: C C V n V s s s ¥ = - 0 (4) Dividing both sides of eqn.(4) by C0 yields: C C C V n C V n C V KV V KV V V KV s s s s f s f s s f ¥ = - = - = - + = + 0 0 0 0 1 1 (5) Also, by definition C C n n s ¥ = 0 0 (6) where n is the amount of analyte extracted from a sample of volume Vs and n0 is the amount that would be extracted by the fiber from a sample of infinite volume (in which case no significant decrease in the concentration of the analyte in the sample would be observed at equilibrium).Fig. 1 presents the dependence of n/n0 on Vs for (a) a 100 mm fiber (Vf = 0.65 ml) and (b) a 30 mm fiber (Vf = 0.14 ml), for sample volumes of up to 1 l and K values ranging from 1000 to 1 000 000. This situation would be characteristic of air sampling from typical glass bulbs, and hydrocarbons from the gasoline range. According to Martos et al.,32 K Å 10orresponds to C7, while K Å 1 000 000 corresponds to C15.It is obvious from Fig. 1 that except for compounds with the lowest K values, sample volume has a very significant effect on the amount extracted by the fiber. Practical consequences of this are very important. Calibration of the fiber is usually performed by exposing it to a standard gaseous mixture generated in a glass bulb of known volume by the static method.33 The results of such calibrations are valid only for gaseous samples of exactly the same volume.If the fiber is exposed to ambient air (an unlimited supply of sample), the amount extracted by the fiber can be significantly larger compared with limited volume samples. On the other hand, if the containers used for grab sampling of air are smaller than the bulb used for calibration, the amount extracted will be smaller than expected from the calibration graph. In both cases, erroneous quantitative results will be obtained. It is obvious, therefore, that the same volume should be used for both the sample and the standard.Alternatively, a different calibration procedure can be used. Dynamic generation of standard gas mixtures provides unlimited volumes of gas standards, and thus can be used to calibrate the fiber for direct ambient air sampling. Alternatively, calibration based on retention parameters as described by Martos and Pawliszyn34 can be used for PDMS fibres. In this procedure no standards are necessary.Direct analysis of aqueous samples is usually performed by exposing the fiber to a small volume of sample contained in a vial. Fig. 2 presents similar dependences determined for sample volumes of up to 40 ml, and three coating thicknesses, viz., 100, 30 and 7 mm (Vf = 0.028 ml). It should be emphasized again that no headspace can be present in the vial if the two-phase system model is to be applied. The thin coating (7 mm) fibers are often used for direct sampling of semi-volatile compounds, since thicker coatings cause equilibration times to be excessively long. It is clear from Fig. 2 that the sample volume is an Fig. 1 Dependence of n/n0 on Vs for (a) 100 mm fiber (Vf = 0.65 ml), and (b) 30 mm fiber (Vf = 0.14 ml), for direct sampling from large volume samples (two-phase system). Fig. 2 Dependence of n/n0 on Vs for (a) 100 mm fiber (Vf = 0.65 ml), (b) 30 mm fiber (Vf = 0.14 ml), and (c) 7 mm fiber (Vf = 0.028 ml), for direct sampling from small volume samples (two-phase system). 1080 Analyst, October 1997, Vol. 122important parameter even for compounds with relatively small K values, especially when very small sample volumes (1 ml) are used. Moreover, headspace sampling is normaly used for volatile compounds with small K values and direct sampling is used for semi-volatile compounds with large K values. Hence, sample volume cannot usually be neglected even when the 7 mm coating is used. The course of the curves indicates that in some cases (steep part of the curve, e.g., K = 1000 for a 100 mm coating, Vs between 1 and 5 ml) small differences in sample volume can result in relatively large differences in the amount extracted, which adversely affects the precision of the determination.Care should be taken, therefore, to ensure that the sample volume during direct sampling is always the same. In view of the above discussion, it is interesting to establish what the sample volume should be in order for the amount extracted by the fiber to be insignificant compared with the amount remaining in the sample after extraction. Eqn.(3) can be used for this purpose. Let us assume that the amount extracted, n = CfVf, fulfils the above condition when it is smaller than, or equal to, 1% of the initial analyte amount in the sample, i.e., CHf Vf @ 0.01 C0Vs (7) The equation can be rearranged to yield: C C V V KV V f f s f s ¥ @ 0 0 01 0 01 @ @ . . (8) since, according to the initial assumption, C0 � CHs .Under the above conditions, eqn. (3) can be simplifid to n = KC0Vf (9) Simple rearrangement of eqn. (8) yields the final condition: VS ! 100KVf (10) For a 100 mm fiber, whose volume is approximately 0.65 ml, eqn. (10) can be written as: VS ! 0.065 K [ml] (11) It follows from eqn. (11) that even for analytes whose partition coefficient values are as low as 100, the minimum sample volume when using a 100 mm thick coating should be 6.5 ml if the effect of sample volume on the amount extracted is to be neglected.Similar reasoning can be applied to calculate the sample volume from which exactly 50% of the analyte is extracted, i.e., n = CfVf = 0.5 C0Vs. In this case 0 5 0 0 . C V KC V V V KV s s f s f = + (12) After a few rearrangements, the final condition is obtained: Vs 50% = KVf (13) or, for a 100 mm fiber, Vs = K 3 0.65 3 1023 [ml] (14) It follows from eqn. (14) that when 1 ml samples are analyzed with 100 mm fibers, 50% of the initial amount of the analyte will be extracted when the K value is approximately 1540.Typical partition coefficients of semi-volatile compounds are usually higher; it is obvious, therefore, that sample volume has a very significant effect on the amount extracted when small sample volumes are used. Sample Volume versus Amount Extracted in Three-phase Systems In most cases liquid samples are placed in vials with some headspace remaining inside. At equilibrium, by definition the chemical potentials of the analyte in all three phases (liquid sample–headspace–fiber coating) must be the same.2 A very important consequence of this fact is that the amount of the analyte extracted by the fiber at equilibrium in a three-phase system is the same independently of where the fiber is located, be it the headspace or the liquid.Consequently, in systems where a headspace is present, a different dependence should be used to calculate the amount of analyte extracted by the fiber regardless of where the fiber is located, viz:2 n KC V V KV K V V = + + 0 s f f hs hs s (15) where Khs is the headspace–liquid partition coefficient Khs = CH hs/CHs (16) (CH hs is the concentration of the analyte in the headspace at equilibrium), and Vhs is the headspace volume.Compared with a two-phase system described by eqn. (3), the difference is the additional term KhsVhs in the denominator of eqn. (4). For volatile compounds, Khs is usually close to 1, which means that headspace volume can be neglected only when it is close to zero (a two-phase system). Semi-volatile compounds have much lower values of Khs; therefore, the KhsVhs term may be negligibly small; however, such an assumption should always be verified.For obvious reasons the dependence of the amount of analyte extracted by the fiber on sample volume is much more complex in three-phase systems. However, it can be still dealt with relatively easily if the total volume of the system (sample plus headspace) remains constant.In practice, this will be the usual case, as the vial volume remains constant, and the headspace volume is the difference between the total volume and the sample volume. In such systems the concentration of an analyte in the sample at equilibrium is determined by the following mass balance: C C V C V n V s s hs hs s ¥ ¥ = - - 0 (17) Dividing both sides of eqn. (17) by C0 yields: C C C V C V n C V s hs hs s s ¥ ¥ = - - 0 0 0 1 (18) From the definition of Khs [eqn.(16)], CH hs can be replaced by KhsCHs , while n is described by eqn. (15), which yields C C K C V C V KV KV K V V s ks s hs s f f hs hs s ¥ ¥ = - - + + 0 0 1 (19) Simple rearrangements yield the final dependence: C C V KV K V V s s f hs hs s ¥ = + + 0 (20) Similarly to a two-phase system, Cs H/C0 = n/n0 when the headspace volume is constant. Fig. 3 illustrates the effect of headspace volume on the amount of analyte extracted in a system of constant volume (4, 15 and 40 ml—typical vial sizes) by a 100 mm fiber, typically used for the analysis of volatile compounds in sample headspace.To allow for better comparison of the results, the xaxis has been defined as the ratio of headspace volume to sample volume. The following values of partition coefficients have been used to calculate the course of the curves: K = 400, 2500 and 10, and Khs = 0.15, 0.7 and 1.24, respectively. These values correspond approximately to those for chloroform, 1,1,1-trichloroethane and carbon tetrachloride.It is interesting that n/n0 is always the largest for the dependence corresponding Analyst, October 1997, Vol. 122 1081to chloroform, which has the lowest Khs value, even though its K value is also the lowest. This is because, with such a combination of Khs and K, only a fraction of the analyte is present in the sample headspace; therefore, the equilibrium concentration of the analyte in the sample remains relatively high.The course of the two other curves is also interesting. For small headspace volumes, n/n0 is higher for the curve corresponding to carbon tetrachloride than for that corresponding to 1,1,1-trichloroethane, which is consistent with the values of their partition coefficients, K. However, as carbon tetrachloride has the largest Khs, n/n0 drops faster for this compound than for 1,1,1-trichloroethane, as a result of which the two lines cross each other.This illustrates the significance of headspace volume on the analytical results. It is also interesting that the largest relative increase in n/n0 ( Å 44%) when moving from small to large vials is observed for 1,1,1-trichloroethane, with the highest K value. For compounds with lower K values the relative increase in the amount extracted is much lower ( Å 16 and Å 12% for chloroform and carbon tetrachloride, respectively). It is, therefore, the combination of K and Khs for a given compound that determines the magnitude of the effect of the sample volume on the amount extracted in three-phase systems with headspace.Headspace volume can be the critical factor determining the precision of the results in three-phase systems. It is relatively easy to measure sample volume accurately. However, vials are not manufactured to have exactly the same volume. Wall thicknesses and bottom shapes may differ from vial to vial.Also, the shape of the septum in a closed vial can vary from concave to convex. All these factors will affect the total volume of the system, and, therefore, the headspace volume, which is the difference between the total volume and sample volume. As illustrated in Fig. 3, for compounds with large Khs, changing the headspace volume can significantly affect the amount extracted, especially in the range of headspace volumes usually applied (Vhs/Vs @ 1). Such differences (especially related to septum shape) are usually more pronounced in small vials; therefore, worse precision can usually be expected when using such vials for headspace sampling.Similarly to two-phase systems, it is possible to calculate the minimum sample volume necessary for the amount extracted by the fiber to be insignificant compared with the amount remaining in the sample after extraction. It is again assumed that the condition is fulfilled when less than 1% of the initial amount present in the sample is extracted by the fiber, i.e., n = CHf Vf @ 0.01 C0Vs.However, in this case it cannot be assumed that CHs � Co, because even if the amount extracted by the fiber constitutes only 1% of the initial amount of the analyte, a significant fraction of the analyte might be present in the headspace. Eqn. (15) will, therefore, be used to calculate the criterion: 001 0 0 . C V KC V V KV K V V s s f f hs hs s ! + + (21) The headspace volume, Vhs, can be expressed as a fraction of the sample volume, Vhs = aVs. Simple rearrangements yield the final criterion: V KV aK s f hs ! 99 1+ (22) Compared with the criterion for a two-phase system [eqn.(10)], there is an additional term in the denominator of eqn. (22), viz., aKhs. For non-zero headspace volumes this term is always greater than 0, which means that the sample volume fulfilling the criterion is smaller in three-phase systems than in two-phase systems. This is because, ultimately, the amount of analyte extracted by the fiber is determined by the concentration of the analyte in the liquid sample at equilibrium.This concentration will be lower in three-phase systems than in two-phase systems, since a significant fraction of the analyte may be present in the headspace of the sample. As a consequence, the sensitivity of headspace SPME can only be lower than, or equal to, that of direct SPME if the system is allowed to reach equilibrium.The higher sensitivity for headspace SPME reported sometimes (especially for compounds with relatively large K and Khs values) is clearly due to non-equilibrium conditions when sampling directly from liquid samples. This is not unusual in the light of the fact that equilibration times for direct sampling from water can be as long as a few hours (especially with inefficient agitation), and the increases in the amount of analyte extracted might not be noticeable if the experiment is not carried out for sufficiently long times.In addition, during prolonged sampling, analyte losses via adsorption onto the sample vial walls, absorption by the exposed parts of the silicone rubber septum (the Teflon lining of the septum has to be pierced to introduce the fiber; since the needle of the SPME device is blunt, the lining usually breaks open exposing a small area of silicone substrate), microbial decomposition, etc., can more than outweigh the expected increase in the amount of analyte extracted, as a result of which a smaller extracted amount can actually be observed after a longer time.This and other factors, including elimination of matrix effects related to the presence of high-boiling compounds in the sample and significantly faster equilibration times, usually make headspace SPME the preferred method for the analysis of mostly volatile compounds. It is also easy to determine the sample volume from which 50% of the analyte will be extracted by the SPME fiber placed in the sample headspace.The procedure is similar to that Fig. 3 Effect of headspace volume on the amount of analyte extracted in a system of a constant volume of (a) 4, (b) 15 and (c) 40 ml by a 100 mm fiber for chloroform, 1,1,1-trichloroethane and carbon tetrachloride (see text for the values of partition coefficients used for calculations). 1082 Analyst, October 1997, Vol. 122presented for two-phase systems. The final criterion in this case is: V KV aK s f hs 50 1 % = + (23) The difference between eqns.(23) and (13) is again the additional term in the denominator, aKhs. The higher the headspace–sample partition coefficient, Khs, or the headspaceto- sample volume ratio, a, the smaller the sample volume from which 50% of the initial amount of the analyte is extracted by the SPME fiber. Effect of Headspace Capacity on Extraction Kinetics Headspace volume can have a significant effect on equilibration times (extraction kinetics).If the headspace capacity is low (small Khs), and K is large, the equilibration process is very slow. A rapid initial increase in the amount of analyte extracted is usually observed, followed by a much slower increase that lasts for a long time. The first stage corresponds to analyte extraction from the gaseous phase only. As soon as the headspace concentration of the analyte falls below the equilibrium level with respect to the aqueous phase, analyte molecules start to move from the liquid sample to the headspace.At any given moment there can only be so many molecules in the headspace, depending on the Khs value, which causes the equilibration process to be very slow. The headspace acts in this case as a bottleneck for analyte transport to the fiber. On the other hand, if the amount of the analyte extracted by the fiber at equilibrium is negligible compared with the amount present in the headspace equilibrated with the sample, only a very small amount of the analyte has actually to be transported from the liquid sample through the headspace to the fiber coating, i.e., the analyte is extracted almost exclusively from the gaseous phase, and the process is much faster than for the case described above.Assuming this situation occurs when 95% of the analyte extracted by the fiber at equilibrium comes exclusively from the headspace, the criterion that must be fulfilled cabe described by eqn.(24). The assumption is reasonable, as a 5% difference usually falls within the limits of experimental error for trace SPME–GC analysis: R KV K V KV K aV = = f hs hs f hs s @ 0.05 (24) The above criterion means that the capacity of the headspace (KhsVhs) needs to be at least 20 times larger than the capacity of the fiber (KVf) to achieve rapid extraction. For a given sample volume, Vs, this can be achieved by using a sufficiently large headspace volume (corresponding to a large headspace-tosample volume ratio, a), or by increasing Khs.The latter can be accomplished by increasing the temperature or by salting the analyte out of the liquid phase. When the criterion described by eqn. (24) is fulfilled, equilibration can take as little as a few minutes, and is almost independent of the agitation conditions (provided that the analyte is equilibrated between the liquid phase and its headspace before the extraction begins). It should be emphasized, however, that great care must be exercised when moving the method from one vial size to another, as illustrated in Fig. 4. When the K value is large and/or Khs is small [Fig. 4(a)] a significant portion of the analyte extracted by the fiber has to be transported from the liquid sample through the headspace independently of the headspace volume, as evidenced by the values of R larger than 0.05 for all three values of a illustrated except for a = 2 and sample volumes larger than 28 ml (hence headspace volumes larger than 56 ml).The equilibration times will be relatively long for all vial sizes up to 40 ml. On the other hand, when the K value is smaller and/or Khs is larger, the R value can be lower than 0.05 for sample and headspace volumes often found in practice. Whether the criterion is fulfilled or not depends on the actual headspace volume [Fig. 4(b)]. For the same headspace-to-sample volume ratio, a, the criterion is fulfilled for large headspace volumes, but not for small ones.For example, when using a headspace-tosample volume ratio a = 0.5, the criterion is fulfilled for headspace volumes greater than 7 ml ( = sample volumes greater than 14 ml); therefore, under such conditions equilibration is very fast. However, when a 2 ml sample is used and the headspace volume is 1 ml, R is equal to 0.345, which indicates that a significant portion of the analyte has to be transported to the fiber from the liquid sample.As a consequence, the equilibration time will be longer even though the overall sample volume is smaller. If this is not accounted for and extraction times determined for large vials are used with small vials, the accuracy and precision of the determination suffers. It should be remembered, therefore, that the equilibration time has to be examined whenever the volume of the sample and/or headspace, or the extraction conditions (temperature, salt addition) are changed. It should also be emphasized that the increase in headspace capacity causes a significant loss of sensitivity.On the other hand, increasing the headspace capacity (which can be accomplished, for example, by increasing the extraction temperature) might dramatically shorten the equilibration times for some semi-volatile analytes while maintaining sufficient sensitivity. Both these effects need to be taken into account when developing a method involving headspace extraction. Effect of Sample Volume on the Determination of K in Two-phase Systems The fiber coating–sample partition coefficient [eqn.(2)] can be determined in several ways. The best way is to use eqn. (3). After a few rearrangements, the following dependence is obtained: K nV V C V n = - s f s ( ) 0 (25) Fig. 4 Dependence of R = KVf/KhsaVs on sample volume for three headspace-to-sample volume ratios, a = 0.5, 1 and 2, and a 100 mm PDMS fiber (Vf = 0.65 ml); (a) K = 1000, Khs = 0.25; (b) K = 250, Khs = 0.5.Analyst, October 1997, Vol. 122 1083If the amount extracted by the fiber, n, is negligible compared with the initial amount of the analyte in the sample, C0Vs, this equation is simplified to: K n C V = 0 f (26) which is a simple rearrangement of eqn. (9). Often, K is calculated directly from its definition [eqn. (2)]. In this case, if n is negligible compared with the initial amount of the analyte in the sample, C0Vs, then CHs can be substituted by C0, and the calculation is straightforward.However, it is our belief that often this substitution is made even when it is not legitimate. The consequences of this can be very significant, as will be illustrated in the following paragraphs. In both cases, the fiber volume necessary to calculate the concentration of the analyte in the fiber coating at equilibrium must be known with sufficiently good accuracy to obtain valid results. Let us introduce a new term, K apparent (Kap), this being the ratio of the concentration of the analyte in the fiber at equilibrium to the initial concentration C0: K C C ap f = ¥ 0 (27) Kap is what is calculated when depletion of the initial concentration of the analyte in the sample is not taken into account (i.e., when C0 is used instead of CHs ).The ratio of K to Kap is equal to: K K C C ap s = ¥ 0 (28) Therefore, K C C K ap s = � ¥ 0 (29) Since (5), s s s f C C V V KV ¥ = + 0 K KV V KV ap s s f = + (30) The true value of K can be calculated using the following equation derived from eqn.(29): K K V K V V = - - ap s ap f s (31) However, in many cases it will not be possible to determine K accurately by this method owing to the phenomena illustrated in Fig. 5. Fig. 5 presents the relationship between Kap and K for a wide range of partition coefficients, for two different coatings (100 and 7 mm) and three sample volumes (2, 10 and 30 ml). The dependences level off in all cases.It is obvious that the relationship reaches a certain limit, above which the same value of Kap is determined independently of the true value of K. The limit can be easily determined mathematically: lim lim K K K KV V KV V V ®¥ ®¥ = + æ è ç ö ø ÷ = ap s s f s f (32) It follows from eqn. (31) that Kap is determined only by the phase ratio Vs/Vf. The smaller the sample size (or the thicker the coating), the lower the value of Kap. It is possible that this phenomenon is behind certain very low K values reported in the literature for compounds that should have very large partition coefficients (e.g., polyaromatic hydrocarbons), especially when small sample volumes are used.Also, it is clear that when Kap is calculated instead of K, the values for thinner fiber coatings are much higher than those for thicker coatings (the difference in volume between a 100 mm fiber coating and a 7 mm coating is more than an order of magnitude). Such apparent discrepancies have also been reported in the literature.Even if proper methodology is used, owing to practical limitations it is not always possible to determine K accurately, as will be illustrated in the following paragraphs. Table 1 presents the amount of analyte extracted from a 2 ml sample and the corresponding per cent. increase in the amount extracted when going from a smaller K to a larger K for a 100 mm thick fiber coating and 1 mg l21 initial concentration, for different K values.For small K values the increase in the amount extracted when K increases is significant, although not linear as would be the case for infinite sample volumes. For large K values the increase in the amount extracted when K increases becomes insignificant and falls within the limits of experimental error. Table 2 illustrates the significance of this phenomenon. It presents the ranges of K values that are obtained when the amount of analyte extracted by the fiber, n, determined experimentally falls within ±5% (relative) of the true value, for two sample volumes (2 and 35 ml) and two fiber coating thicknesses (100 and 7 mm).A ±5% error can be assumed typical for trace analysis by SPME–GC. Fig. 5 Relationship between Kap and K for two different coatings (100 and 7 mm) and three sample volumes (2, 10 and 30 ml). Table 1 Amount of analyte extracted from a 2l sample and % increase in the amount extracted for a 100 mm thick fiber coating and 1 mg l21 initial concentration, for different K values K n/ng Increase (%) 100 0.0629540 — 1 000 0.4905660 717 10 000 1.5294118 212 100 000 1.9402985 28 1 000 000 1.9938650 2.8 10 000 000 1.9993848 0.28 100 000 000 1.9999385 0.3 1084 Analyst, October 1997, Vol. 122It is clear from Table 2 that practical limitations to K determination can be very severe, especially when thick fiber coatings and low sample volumes are used. For a 2 ml sample and a 100 mm thick fiber, the error of K determination for K values around 10 000 is around ±20% for a ±5% relative error of n determination.If the true K values are higher by an order of magnitude, the values determined experimentally can range from Å 36 000 to infinity, which is of no use. In practice, owing to experimental errors, for large K values the value of n found experimentally can be larger than the initial amount of analyte in the sample. In such cases, negative K values would be obtained, which, of course, have no physical meaning.It is necessary in such cases to assume that the amount extracted, n, is equal to the initial amount of analyte in the sample (quantitative extraction), in which case infinite K values are obtained. On the other hand, K values of Å 1 000 000 can be determined with reasonable accuracy when large sample volumes (35 ml) and thin fiber coatings (7 mm) are used. It is our belief that the above-described phenomena account for most of the discrepancies between K values reported in the literature for the same substances and coatings.Other possible sources of errors include limited solubility of some analytes in water, as well as losses of analytes due to volatilization (when headspace is present in the system), absorption by exposed parts of silicone rubber septa, adsorption on glass, or biodegradation. The first two phenomena are more probable for volatile analytes, while the remaining phenomena are more significant for semi-volatile compounds, since when K values are very high, equilibration times become very long.From a practical point of view, partition coefficients, K, should always be determined using large sample volumes (nothing can be gained by using small volumes). For semi-volatile compounds characterized by large K values, only thin coatings (7 mm for PDMS) are practical for the experimental determination of K. Above a certain level, the accurate determination of K becomes virtually impossible.Conclusions Sample volume plays an extremely important role in analysis by SPME. In order to avoid errors or poor precision, care should be taken to ensure that the volumes of samples and standard solutions used for calibration are the same. In headspace SPME, the combination of K and Khs determines the magnitude of the effect of sample volume on the amount extracted by the fiber, n. With the same headspace-to-sample volume ratio, the magnitude of the effect is greatest for compounds with low Khs.The minimum sample volume which ensures that n is lower than 1% of the initial amount of analyte present in the sample (which means that the change in concentration of the analyte in the sample at equilibrium is negligible) can easily be calculated for both two- and three-phase systems, and so is the volume for which exactly half of the initial analyte amount is extracted. Variable vial volumes might cause poorer precision in headspace sampling, as only the volume of the liquid sample can be accurately controlled.Extraction kinetics in headspace sampling is determined by the headspace capacity, KhsVhs. If it is sufficiently large compared with the fiber capacity, KVf, then the analyte is extracted almost exclusively from the headspace, and equilibration can be very fast. The headspace capacity is determined by the actual headspace volume rather than by the headspace-tosample volume ratio a; therefore, dramatically different equilibration times might be observed when scaling down the method from large to small vials, even though a remains constant.On the other hand, a large headspace capacity causes a significant loss of method sensitivity. Sample volume should always be taken into account when determining K values. If the change in concentration of the analyte in the sample at equilibrium is not taken into account, erroneous results are obtained. Even when a proper procedure is used, there are practical limitations to the accuracy of the K value determination.Large sample volumes should always be used for K value determination, as they enable broader ranges of K values to be covered with good accuracy. Improper procedure is, in our opinion, the main cause of discrepancies between K values reported in the literature for the same compound by different sources, or for different coating thicknesses. Financial support of Supelco, Varian and National Sciences and Engineering Research Council of Canada is acknowledged.References 1 Arthur, C. L., and Pawliszyn, J., Anal. Chem., 1990, 62, 2145. 2 Zhang, Z., and Pawliszyn, J., Anal. Chem., 1993, 65, 1843. 3 Arthur, C. L., Killam, L. M., Motlagh, S., Lim, M., Potter, D. W., and Pawliszyn, J., Environ. Sci. Technol., 1992, 26, 979. 4 Potter, D. W., and Pawliszyn, J., J. Chromatogr., 1992, 625, 247. 5 Wittkamp, B. L., and Tilotta, D. C., Anal. Chem., 1995, 67, 600. 6 Hawthorne, S. B., Miller, D. J., Pawliszyn, J., and Arthur, C. L., J. Chromatogr., 1992, 603, 185. 7 Arthur, C. L., Pratt, K., Motlagh, S., Pawliszyn, J., and Belardi, R. P., J. High Resolut. Chromatogr., 1992, 15, 741. 8 Langenfeld, J. J., Hawthorne, S. B., and Miller, D. J., Anal. Chem., 1996, 68, 144. 9 Nilsson, T., Ferrari, F., and Facchetti, S., in Proceedings of the 18th International Symposium on Capillary Chromatography, Riva del Garda (Italy), May 20–24, 1996, ed. Sandra, P., and Davos, G., IOPMS, Kortrijk, 1996, p. 618. 10 Potter, D., and Pawliszyn, J., Environ. Sci. Technol., 1994, 28, 298. 11 Chai, M., Arthur, C. L., Pawliszyn, J., Belardi, R. P., and Pratt, K. F., Analyst, 1993, 118, 1501. 12 Buchholz, K., and Pawliszyn, J., Environ. Sci. Technol., 1993, 27, 2844. 13 Buchholz, K., and Pawliszyn, J., Anal. Chem., 1994, 66, 160. 14 Schaefer, B., and Engewald, W., Fresenius’ J. Anal. Chem., 1995, 352, 535. 15 Boyd-Boland, A. A., and Pawliszyn, J., J. Chromatogr., 1995, 704, 163. 16 Eisert, R., Levsen, K., and Wuensch, G., J. Chromatogr., 1994, 683, 175. Table 2 Ranges of K values obtained when the amount of analyte extracted by the fiber, n, determined experimentally falls within ±5% (relative) of the true value, for two different fiber coating thicknesses and two sample volumes 100 mm— 2 ml sample 35 ml sample K Range of K K Range of K 100 105–95 100 105–95 1 000 1067–935 1 000 1051–949 10 000 12 537–8172 10 000 10 598–9413 100 000 H–36 190 100 000 115 748–86 928 1 000 000 H–55 072 1 000 000 4 700 000–492 593 10 000 000 H–58 104 10 000 000 H–923 611 100 000 000 H–58 426 100 000 000 H–1 012 177 7 mm— 2 ml sample 35 ml sample K Range of K K Range of K 100 105–95 100 105–95 1 000 1051–949 1 000 1050–950 10 000 10 574–9434 10 000 10 504–9496 100 000 112 903–88 785 100 000 105 422–94 622 1 000 000 3 500 000–558 824 1 000 000 1 093 750–913 462 10 000 000 H–1 187 500 10 000 000 17 500 000–6 785 714 100 000 000 H–1 338 028 100 000 000 H19 000 000 Analyst, October 1997, Vol. 122 108517 Popp, P., Kalbitz, K., and Oppermann, G., J. Chromatogr., 1994, 687, 133. 18 Lee, X., Kumazawa, T., Taguchi, T., Sato, K., and Suzuki, O., Hochudoku, 1995, 13, 122. 19 Eisert, R., and Levsen, K., Fresenius’ J. Anal. Chem., 1995, 351, 555. 20 Levsen, K., and Eisert, R., J. Am. Soc. Mass Spectrom., 1995, 6, 1119. 21 Graham, K. N., Sarna, L. P., Webster, G. R. B., Gaynor, J. D., and Ng, H. Y. F., J. Chromatogr., 1996, 725, 129. 22 Magdic, S., and Pawliszyn, J., J. Chromatogr., 1996, 723, 111. 23 Pan, L., Adams, M., and Pawliszyn, J., Anal. Chem., 1995, 67, 4396. 24 G�orecki, T., and Pliszyn, J., Anal. Chem., 1996, 68, 3008. 25 Chen, J., and Pawliszyn, J., Anal. Chem., 1995, 67, 2530. 26 Boyd-Boland, A. A., and Pawliszyn, J., Anal. Chem., 1996, 68, 1521. 27 Hirata, Y., and Pawliszyn, J., J. Microcol. Sep., 1994, 6, 443. 28 Louch, D., Motlagh, S., and Pawliszyn, J., Anal. Chem., 1992, 64, 1187. 29 Arthur, C. L., Killam, L. M., Buchholz, K. D., and Pawliszyn, J., Anal. Chem., 1992, 64, 1960. 30 Motlagh, S., and Pawliszyn, J., Anal. Chim. Acta, 1993, 284, 265. 31 Langenfeld, J. J., Hawthorne, S. B., and Miller, D. J., Anal. Chem., 1996, 68, 144. 32 Martos, P., Saraullo, A., and Pawliszyn, J., Anal. Chem., 1997, 69, 402. 33 Namiesnik, J., J. Chromatogr., 1984, 300, 79. 34 Martos, P., and Pawliszyn, J., Anal. Chem., 1997, 69, 206. Paper 7/01303E Received February 25, 1997 Accepted June 4, 1997 1086 Analyst, October 1997, Vol. 122