D. Harvey - Modern Analytical Chemistry (794078), страница 60
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In addition, the closed container prevents theloss of volatile gases. Disadvantages include the inability to add reagents during digestion, limitations on the amount of sample that can be used (typically 1 g or less),and safety concerns due to the use of high pressures and corrosive reagents. Applications include environmental and biological samples.Inorganic samples that resist decomposition by digestion with acids orbases often can be brought into solution by fusing with a large excess of an alkali metal salt, called a flux.
The sample and flux are mixed together in a crucible and heated till the substances fuse together in a molten state. The resultingmelt is allowed to cool slowly to room temperature. Typically the melt dissolvesreadily in distilled water or dilute acid. Several common fluxes and their usesare listed in Table 7.3. Fusion works when other methods of decomposition donot because of the higher temperatures obtained and the high concentration ofthe reactive flux in the molten liquid. Disadvantages include a greater risk ofcontamination from the large quantity of flux and the crucible and the loss ofvolatile materials.Finally, organic materials may be decomposed by dry ashing. In this methodthe sample is placed in a suitable crucible and heated over a flame or in a furnace.Any carbon present in the sample is oxidized to CO2, and hydrogen, sulfur, and nitrogen are removed as H2O, SO2 and N2.
These gases can be trapped and weighed todetermine their content in the organic material. Often the goal of dry ashing is theremoval of organic material, leaving behind an inorganic residue, or ash, that can befurther analyzed.7D Separating the Analyte from InterferentsWhen a method shows a high degree of selectivity for the analyte, the task of performing a quantitative, qualitative, or characterization analysis is simplified.
For example, a quantitative analysis for glucose in honey is easier to accomplish if themethod is selective for glucose, even in the presence of other reducing sugars, suchas fructose. Unfortunately, analytical methods are rarely selective toward a singlespecies.In the absence of interferents, the relationship between the sample’s signal,Ssamp, and the concentration of analyte, CA, isSsamp = kACA7.92011400-CH07 9/8/99 4:03 PM Page 202202Modern Analytical Chemistrywhere kA is the analyte’s sensitivity.* In the presence of an interferent, equation 7.9becomesSsamp = kACA + kICI7.10where kI and CI are the interferent’s sensitivity and concentration, respectively.
Amethod’s selectivity is determined by the relative difference in its sensitivity towardthe analyte and interferent. If kA is greater than kI, then the method is more selective for the analyte. The method is more selective for the interferent if kI is greaterthan kA.Even if a method is more selective for an interferent, it can be used to determine an analyte’s concentration if the interferent’s contribution to Ssamp is insignificant. The selectivity coefficient, KA,I, was introduced in Chapter 3 as a means ofcharacterizing a method’s selectivity.K A,I =kIkA7.11Solving equation 7.11 for kI and substituting into equation 7.10 gives, after simplifyingSsamp = kA(CA + KA,I × CI)7.12An interferent, therefore, will not pose a problem as long as the product of its concentration and the selectivity coefficient is significantly smaller than the analyte’sconcentration.KA,I × CI << CAWhen an interferent cannot be ignored, an accurate analysis must begin by separating the analyte and interferent.7E General Theory of Separation EfficiencyrecoveryThe fraction of analyte or interferentremaining after a separation (R).The goal of an analytical separation is to remove either the analyte or the interferentfrom the sample matrix.
To achieve a separation there must be at least one significant difference between the chemical or physical properties of the analyte and interferent. Relying on chemical or physical properties, however, presents a fundamentalproblem—a separation also requires selectivity. A separation that completely removes an interferent may result in the partial loss of analyte. Altering the separationto minimize the loss of analyte, however, may leave behind some of the interferent.A separation’s efficiency is influenced both by the failure to recover all the analyte and the failure to remove all the interferent.
We define the analyte’s recovery,RA, asRA =CA(C A )owhere CA is the concentration of analyte remaining after the separation, and (CA)ois the analyte’s initial concentration. A recovery of 1.00 means that none of the analyte is lost during the separation. The recovery of the interferent, RI, is defined inthe same mannerCIRI =7.13(CI )o*In equation 7.9, and the equations that follow, the concentration of analyte, CA, can be replaced by the moles ofanalyte, nA, when considering a total analysis technique.1400-CH07 9/8/99 4:03 PM Page 203Chapter 7 Obtaining and Preparing Samples for Analysiswhere CI is the concentration of interferent remaining after the separation, and(CI)o is the interferent’s initial concentration. The degree of separation is given by aseparation factor, SI,A, which is the change in the ratio of interferent to analytecaused by the separation.11SI, A =CI / C AR= I(CI )o /(C A )oRAEXAMPLE 7.10An analysis to determine the concentration of Cu in an industrial plating bathuses a procedure for which Zn is an interferent.
When a sample containing128.6 ppm Cu is carried through a separation to remove Zn, the concentrationof Cu remaining is 127.2 ppm. When a 134.9-ppm solution of Zn is carriedthrough the separation, a concentration of 4.3 ppm remains. Calculate therecoveries for Cu and Zn and the separation factor.SOLUTIONThe recoveries for the analyte and interferent areRCu =127.2 ppm= 0.9891, or 98.91%128.6 ppmandRZn =4.3 ppm= 0.032, or 3.2%134.9 ppmThe separation factor isS Zn,Cu =RZn0.032== 0.032RCu0.9891In an ideal separation RA = 1, RI = 0, and SI,A = 0.
In general, the separation factorshould be approximately 10–7 for the quantitative analysis of a trace analyte in thepresence of a macro interferent, and 10–3 when the analyte and interferent are present in approximately equal amounts.Recoveries and separation factors are useful ways to evaluate the effectivenessof a separation. They do not, however, give a direct indication of the relative errorintroduced by failing to remove all interferents or failing to recover all the analyte.The relative error introduced by the separation, E, is defined asE =*Ssamp − Ssamp*Ssamp7.14*where Ssampis the expected signal for an ideal separation when all the analyte is recovered.*Ssamp= kA(CA)oSubstituting equations 7.12 and 7.15 into 7.14 givesE =kA (C A + K A,I × CI ) − kA (C A )okA (C A )o7.15203separation factorA measure of the effectiveness of aseparation at separating an analyte froman interferent (SI,A).1400-CH07 9/8/99 4:03 PM Page 204204Modern Analytical Chemistrywhich simplifies toE ==C A + K A,I × CI − (C A )o(C A )o(C )CAK × CI− A o + A,I(C A )o (C A )o(C A )o= (RA − 1) +K A,I × CI(C A )o7.16A more useful equation for the relative error is obtained by solving equation 7.13for CI and substituting back into equation 7.16 K × (CI )oE = (RA − 1) + A,I× RI (C A )o7.17The first term of equation 7.17 accounts for the incomplete recovery of analyte, andthe second term accounts for the failure to remove all the interferent.EXAMPLE 7.11Following the separation outlined in Example 7.10, an analysis is to be carriedout for the concentration of Cu in an industrial plating bath.
The concentrationratio of Cu to Zn in the plating bath is 7:1. Analysis of standard solutionscontaining only Cu or Zn give the following standardization equationsSCu = 1250 × (ppm Cu)SZn = 2310 × (ppm Zn)(a) What error is expected if no attempt is made to remove Zn before analyzingfor Cu? (b) What is the error if the separation is carried out? (c) What is themaximum acceptable recovery for Zn if Cu is completely recovered and theerror due to the separation must be no greater than 0.10%?SOLUTION(a) If the analysis is carried out without a separation, then RCu and RZn areequal to 1.000, and equation 7.17 simplifies toE =KCu, Zn × (ppm Zn)o(ppm Cu)oFrom equation 7.11 the selectivity coefficient isKCu, Zn =kZn2310== 1.85kCu1250Although we do not know the actual concentrations of Zn or Cu in the sample,we do know that the concentration ratio (ppm Zn)o/(ppm Cu)o is 1/7. Thus(1.85)(1)E == 0.264, or 26.4%(7)(b) To calculate the error, we substitute the recoveries calculated in Example7.10 into equation 7.171400-CH07 9/8/99 4:03 PM Page 205Chapter 7 Obtaining and Preparing Samples for Analysis205 (1.85)(1)E = (0.9891 − 1) + × 0.032 (7)= (−0.0109) + (0.0085)= −0.0024, or − 0.24%Note that a negative determinate error introduced by failing to recover allthe analyte is partially offset by a positive determinate error due to a failureto remove all the interferent.(c) To determine the maximum allowed recovery for Zn, we make appropriatesubstitutions into equation 7.17 (1.85)(1)0.0010 = (1.000 − 1) + × RZn (7)and solve for RZn, obtaining a recovery of 0.0038, or 0.38%.
Thus, at least99.62% of the Zn must be removed by the separation.7F Classifying Separation TechniquesAn analyte and an interferent can be separated if there is a significant difference inat least one of their chemical or physical properties. Table 7.4 provides a partial listof several separation techniques, classified by the chemical or physical property thatis exploited.7F.1 Separations Based on SizeThe simplest physical property that can be exploited in aseparation is size.
The separation is accomplished using aporous medium through which only the analyte or interferent can pass. Filtration, in which gravity, suction, orpressure is used to pass a sample through a porous filter isthe most commonly encountered separation techniquebased on size.Particulate interferents can be separated from dissolved analytes by filtration, using a filter whose poresize retains the interferent. This separation technique isimportant in the analysis of many natural waters, forwhich the presence of suspended solids may interfere inthe analysis.
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