D. Harvey - Modern Analytical Chemistry (794078), страница 65
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A 50.00-mLsample of a 0.050 M aqueous solution of the solute is extracted with 15.00 mLof chloroform. (a) What is the extraction efficiency for this separation?(b) What is the solute’s final concentration in each phase? (c) What volume ofchloroform is needed to extract 99.9% of the solute?SOLUTIONFor a simple liquid–liquid extraction, the distribution ratio, D, and thepartition coefficient, KD, are identical.(a) The fraction of solute remaining in the aqueous phase after the extractionis given by equation 7.2450.00 mL(qaq )1 == 0.400(5.00)(15.00 mL) + 50.00 mLThe fraction of solute present in the organic phase is, therefore, 0.600.Extraction efficiency is the percentage of solute successfully transferredfrom its initial phase to the extracting phase.
The extraction efficiency is,therefore, 60.0%.2171400-CH07 9/8/99 4:04 PM Page 218218Modern Analytical Chemistry(b) The moles of solute present in the aqueous phase before the extraction is(Moles aq)0 = [Saq ]0 × Vaq =0.050 mol× 0.05000 L = 0.0025 molLSince 40.0% of the solute remains in the aqueous phase, and 60.0% hasbeen extracted into the organic phase, the moles of solute in the twophases after extraction are(Moles aq)1 = (moles aq)0 × (qaq)1 = 0.0025 mol × (0.400) = 0.0010 mol(Moles org)1 = (moles aq)0 – (moles aq)1 = 0.0025 mol – 0.0010 mol = 0.0015 molThe solute’s concentration in each phase is[Saq ]1 =(moles aq)10.0010 mol== 0.020 MVaq0.05000 L[Sorg ]1 =(moles org)10.0015 mol== 0.10 MVorg0.01500 L(c) To extract 99.9% of the solute (qaq)1 must be 0.001.
Solving equation 7.24for Vorg, and making appropriate substitutions for (qaq)1 and Vaq givesVorg =Vaq − (qaq )1Vaq50.00 mL − (0.001)(50.00 mL)== 9990 mL(qaq )1 D(0.001)(5.00)Clearly, a single extraction is not reasonable under these conditions.In Example 7.14 a single extraction results in an extraction efficiency of only60%. If a second extraction is carried out, the fraction of solute remaining in theaqueous phase, (qaq)2, is given by(qaq )2 =Colorplate 4 shows an example of aliquid–liquid extraction.Vaq(moles aq)2=(moles aq)1DVorg + VaqIf the volumes of the aqueous and organic layers are the same for both extractions,then the cumulative fraction of solute remaining in the aqueous layer after two extractions, (Qaq)2, is(Qaq )2 =Vaq(moles aq)2= (qaq )1 (qaq )2 = (moles aq)0DV+Vorgaq 2In general, for a series of n identical extractions, the fraction of analyte remaining inthe aqueous phase after the last extraction isVaq(Qaq )n = DVorg + Vaq n7.25EXAMPLE 7.15For the extraction described in Example 7.14, determine (a) the extraction efficiency for two extractions and for three extractions; and(b) the number of extractions required to ensure that 99.9% of the solute isextracted.1400-CH07 9/8/99 4:04 PM Page 219219Chapter 7 Obtaining and Preparing Samples for AnalysisSOLUTION(a) The fraction of solute remaining in the aqueous phase after two and threeextractions is250.00 mL(Qaq )2 = = 0.160 (5.00)(15.00 mL) + 50.00 mL 350.00 mL(Qaq )3 = = 0.064 (5.00)(15.00 mL) + 50.00 mL Thus, the extraction efficiencies are 84.0% with two extractions and 93.6%with three extractions.(b) To determine the minimum number of extractions for an efficiency of99.9%, we set (Qaq)n to 0.001 and solve for n in equation 7.2550.00 mL0.001 = (5.00)(15.00 mL) + 50.00 mL n= (0.400)nTaking the log of both sideslog(0.001) = nlog(0.400)and solving for n givesThus, a minimum of eight extractions is necessary.An important observation from Examples 7.14 and 7.15 is that an extractionefficiency of 99.9% can be obtained with less solvent when using multiple extractions.
Obtaining this extraction efficiency with one extraction requires 9990 mLof the organic solvent. Eight extractions using separate 15-mL portions of the organic solvent, however, requires only 120 mL. Although extraction efficiency increases dramatically with the first few multiple extractions, the effect quickly diminishes as the number of extractions is increased (Figure 7.21). In most casesthere is little gain in extraction efficiency after five or six extractions. In Example7.15 five extractions are needed to reach an extraction efficiency of 99%, and anadditional three extractions are required to obtain the extra 0.9% increase in extraction efficiency.Extraction efficiencyn = 7.5410090807060504030201005010Number of extractionsFigure 7.21Plot of extraction efficiency versus numberof extractions for the liquid–liquid extractionscheme in Figure 7.20.7G.3 Liquid–Liquid Extractions Involving Acid–Base EquilibriaIn a simple liquid–liquid extraction the distribution ratio and the partition coefficient are identical.
As a result, the distribution ratio is unaffected by any change inthe composition of the aqueous or organic phase. If the solute also participates in a single-phase equilibrium reaction, then the distribution ratio andthe partition coefficient may not be the same. For example, Figure 7.22shows the equilibria occurring when extracting an aqueous solution containing a molecular weak acid, HA, with an organic phase in which ionic speciesare not soluble. In this case the partition coefficient and the distributionratio are[HA org ]KD =7.26[HA aq ]HAOrganicKDAqueousKaHA + H2OH3O+ + A–Figure 7.22Scheme for the liquid–liquid extraction of amolecular weak acid.1400-CH07 9/8/99 4:04 PM Page 220220Modern Analytical Chemistry[HA org ]tot[HA org ]=− ][HA aq ]tot[HA aq ] + [A aqD =7.27Since the position of an acid–base equilibrium depends on the pH, the distributionratio must also be pH-dependent.
To derive an equation for D showing this dependency, we begin with the acid dissociation constant for HA.+− ][H 3Oaq][A aq[HA aq ]Ka =7.28–]Solving equation 7.28 for [A aq−] =[ AaqKa[HA aq ]+ ][H 3Oaqand substituting into equation 7.27 givesD =[HA org ]+[HA aq ] + (Ka[HA aq ]/[H 3Oaq])Factoring [HAaq] from the denominator[HA org ]D =[HA aq ]{1 + (Ka /[H 3O+aq ])}and substituting equation 7.26D =KD+1 + (Ka /[H 3Oaq])gives, after simplifying, the sought-after relationship between the distribution ratioand the pH of the aqueous solutionD =+]KD[H 3Oaq+[H 3Oaq] + Ka7.29The value for D given by equation 7.29 can be used in equation 7.25 to determineextraction efficiency.EXAMPLE 7.16An acidic solute, HA, has an acid dissociation constant of 1.00 × 10–5, and apartition coefficient between water and benzene of 3.00.
Calculate theextraction efficiency when 50.00 mL of a 0.025 M aqueous solution of HAbuffered to a pH of 3.00, is extracted with 50.00 mL of benzene. Repeat forcases in which the pH of the aqueous solution is buffered to 5.00 and 7.00.SOLUTION+ ] is 1.00 × 10–3, and the distribution ratio forWhen the pH is 3.00, the [H3Oaqthe extraction isD =(3.00)(1.00 × 10 −3 )= 2.971.00 × 10 −3 + 1.00 × 10 −51400-CH07 9/8/99 4:04 PM Page 221221Chapter 7 Obtaining and Preparing Samples for AnalysisThe fraction of solute remaining in the aqueous phase is70.0050.00 mL= 0.252(2.97)(50.00 mL) + 50.00 mLExtraction efficiency(Qaq )1 =80.00The extraction efficiency, therefore, is almost 75%.
When the same calculationis carried out at a pH of 5.00, the extraction efficiency is 60%, but theextraction efficiency is only 3% at a pH of 7.00. As expected, extractionefficiency is better at more acidic pHs when HA is the predominate species inthe aqueous phase. A graph of extraction efficiency versus pH for this system isshown in Figure 7.23.
Note that the extraction efficiency is greatest for pHsmore acidic than the weak acid’s pKa and decreases substantially at pHs morebasic than the pKA. A ladder diagram for HA is superimposed on the graph tohelp illustrate this effect.60.0050.0040.0030.0020.0010.000.001357911 13pHFigure 7.23Plot of extraction efficiency versus pH of theaqueous phase for the liquid–liquid extractionof the molecular weak acid in Example 7.16.The same approach can be used to derive an equation for the distribution ratio when the solute is a molecular weak base, B, (Figure 7.24).The resulting distribution ratio isKD KaD =+Ka + [H 3Oaq]Bwhere Ka is the acid dissociation constant for the weak base’s conjugateweak acid.OrganicKDAqueousKbB + H2O7G.4 Liquid–Liquid Extractions Involving Metal ChelatorsOne of the most common applications of a liquid–liquid extraction isthe selective extraction of metal ions using a chelating agent.
Unfortunately, many chelating agents have a limited solubility in water orare subject to hydrolysis or air oxidation in aqueous solutions. Forthese reasons the chelating agent is added to the organic solvent instead of the aqueous phase. The chelating agent is extracted into theaqueous phase, where it reacts to form a stable metal–ligand complexwith the metal ion. The metal–ligand complex is then extracted intothe organic phase. A summary of the relevant equilibria is shown inFigure 7.25.If the ligand’s concentration is much greater than the metal ion’sconcentration, the distribution ratio is given as*D =βKD,c Kan CLn+ n + βK n C nnKD,L[H 3Oaq]a LScheme for the liquid–liquid extraction of amolecular weak base.MLnHLOrganicKD,HLAqueousHL+KD,cH2OKa*The derivation of equation 7.31 is considered in problem 33 in the end-of-chapter problem set.+ HB+Figure 7.247.30where CL is the initial concentration of ligand in the organic phase before the extraction.
The distribution ratio calculated using equation 7.30can be substituted back into equation 7.25 to determine the extractionefficiency. As shown in Example 7.17, the extraction efficiency for metalions shows a marked pH dependency.OH–nL–β+Mn +MLn+H3O+Figure 7.25Scheme for the liquid–liquid extraction of ametal ion by a metal chelator.1400-CH07 9/8/99 4:04 PM Page 222Modern Analytical ChemistryEXAMPLE 7.17A divalent metal ion, M2+, is to be extracted from an aqueous solution into anorganic solvent using a chelating agent, HL, dissolved in the organic solvent.The partition coefficients for the chelating agent, KD,L, and the metal–ligandcomplex, KD,c, are 1.0 × 104 and 7.0 × 104, respectively.
The acid dissociationconstant for the chelating agent, Ka, is 5.0 × 10–5, and the formation constantfor the metal–ligand complex, β, is 2.5 × 10 16 . Calculate the extractionefficiency when 100.0 mL of a 1.0 × 10–6 M aqueous solution of M2+, bufferedto a pH of 1.00, is extracted with 10.00 mL of an organic solvent that is 0.1 mMin the chelating agent.
Repeat the calculation at a pH of 3.00.SOLUTION+] = 0.10), the distribution ratio for the extraction isAt a pH of 1.00 ([H3OaqD =(2.5 × 1016 )(7.0 × 104 )(5.0 × 10 −5 )2 (1.0 × 10 −4 )2= 0.0438(1.0 × 104 )2 (0.10)2 + (2.5 × 1016 )(5.0 × 10 −5 )2 (1.0 × 10 −4 )2and the fraction of metal ion remaining in the aqueous phase is(Qaq )1 =100.0 mL= 0.996(0.0438)(10.00 mL) + 100.0 mLThus, at a pH of 1.00, only 0.40% of the metal is extracted. Changing the pH to3.00, however, gives an extraction efficiency of 97.8%. A plot of extractionefficiency versus the pH of the aqueous phase is shown in Figure 7.26.100.0090.00Extraction efficiency22280.0070.0060.0050.0040.0030.0020.0010.00Figure 7.260.001234pH567Typical plot of extraction efficiency versuspH for the liquid–liquid extraction of ametal ion by metal chelator.One of the advantages of using a chelating agent is the high degree of selectivity that it brings to the extraction of metal ions.
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