D. Harvey - Modern Analytical Chemistry (794078), страница 96
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Once the amount of Ni is known, theamount of Fe can be determined from the results for titration 2. Finally,titration 3 can be solved for the amount of Cr.Titration 1g Ni= M EDTA1 × VEDTA1AW Nig Ni = M EDTA1 × VEDTA1 × AW Ni0.05831 M × 0.02614 L × 58.69 g / mol = 0.08946 g NiTitration 2Moles EDTA1 + moles Fe = moles EDTA2g Fe= moles EDTA2 − moles EDTA1AW Feg Fe = (M EDTA2 × VEDTA2 − M EDTA1 × VEDTA1 ) × AW Fe(0.05831 M × 0.03543 L − 0.05831 M × 0.02614 L)× 55.847 g/mol = 0.03025 g FeTitration 3Moles EDTA2 + moles Cr + moles Cu = moles EDTA3g Cr= moles EDTA3 − moles EDTA2 − moles CuAW Crg Cr = (M EDTA3 × VEDTA3 − M EDTA2 × VEDTA2 − M Cu × VCu ) × AW Cr(0.05831 M × 0.05000 L − 0.05831 M × 0.03543 L– 0.06316 M × 0.00621 L) × 51.996 g/mol = 0.02378 g CrEach of these titrations was conducted on a 50.00-mL aliquot of the original250.0-mL sample.
The mass of each analyte, therefore, must be corrected bymultiplying by a factor of 5. Thus, the grams of Ni, Fe, and Cr in the originalsample are1400-CH09 9/9/99 2:13 PM Page 331Chapter 9 Titrimetric Methods of Analysis3310.08946 g × 5 = 0.4473 g Ni0.03025 g × 5 = 0.1513 g Fe0.02378 g × 5 = 0.1189 g Crand the %w/w for each metal is0.4473 g× 100 = 62.33% w/w Ni0.7176 g(a)1098765432109C.6 Evaluation of Complexation TitrimetryThe scale of operations, accuracy, precision, sensitivity, time, and cost of methodsinvolving complexation titrations are similar to those described earlier for acid–basetitrimetric methods.
Compared with acid–base titrations, however, complexationtitrations are more selective. Despite the ability of EDTA to form strong complexeswith virtually all metal ions, carefully controlling the pH at which the titration iscarried out makes it possible to analyze samples containing two or more analytes(see Example 9.10). The reason that pH can be used to provide selectivity is easilyappreciated by examining Figure 9.32. A titration of Ca2+ at a pH of 9 gives a distinct break in the titration curve because the conditional formation constant (K ´f= 2.6 × 109) is large enough to ensure that the reaction of Ca2+ and EDTA goes tocompletion. At a pH of 3, however, the conditional formation constant (K ´f = 1.23)is so small that very little Ca2+ reacts with the EDTA.Spectrophotometric titrations are particularly useful for the analysis of mixturesif a suitable difference in absorbance exists between the analytes and products, ortitrant.
For example, the analysis of a two-component mixture can be accomplishedif there is a difference between the absorbance of the two metal–ligand complexes(Figure 9.33).9D Titrations Based on Redox ReactionsRedox titrations were introduced shortly after the development of acid–basetitrimetry. The earliest methods took advantage of the oxidizing power of chlorine.In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl2, HCl, and HOCl) based on its ability to oxidize solutionsof the dye indigo (indigo is colorless in its oxidized state). In 1814, Joseph Louis GayLussac (1778–1850), developed a similar method for chlorine in bleaching powder.In both methods the end point was signaled visually.
Before the equivalence point,the solution remains clear due to the oxidation of indigo. After the equivalencepoint, however, unreacted indigo imparts a permanent color to the solution.(b)0.080.020.040.060.0Volume of EDTA (mL)100.0Figure 9.32Titration curve for 10–2 M Ca2+ with 10–2 MEDTA at (a) pH = 9 and (b) pH = 3.Corrected absorbance0.1189 g× 100 = 16.57% w/w Cr0.7176 gpCa0.1513 g× 100 = 21.08% w/w Fe0.7176 gSecondequivalence pointFirstequivalence pointVolume of titrantFigure 9.33Spectrophotometric titration curve for thecomplexation titration of a mixture.redox titrationA titration in which the reaction betweenthe analyte and titrant is anoxidation/reduction reaction.1400-CH09 9/9/99 2:13 PM Page 332332Modern Analytical ChemistryThe number of redox titrimetric methods increased in the mid-1800s with theintroduction of MnO4–, Cr2O72– and I2 as oxidizing titrants, and Fe2+ and S2O32– asreducing titrants.
Even with the availability of these new titrants, however, the routine application of redox titrimetry to a wide range of samples was limited by thelack of suitable indicators. Titrants whose oxidized and reduced forms differ significantly in color could be used as their own indicator. For example, the intensely purple MnO4– ion serves as its own indicator since its reduced form, Mn2+, is almostcolorless. The utility of other titrants, however, required a visual indicator thatcould be added to the solution.
The first such indicator was diphenylamine, whichwas introduced in the 1920s. Other redox indicators soon followed, increasing theapplicability of redox titrimetry.9D.1 Redox Titration CurvesTo evaluate a redox titration we must know the shape of its titration curve. In anacid–base titration or a complexation titration, a titration curve shows the changein concentration of H3O+ (as pH) or Mn+ (as pM) as a function of the volume oftitrant. For a redox titration, it is convenient to monitor electrochemical potential.You will recall from Chapter 6 that the Nernst equation relates the electrochemical potential to the concentrations of reactants and products participating in aredox reaction.
Consider, for example, a titration in which the analyte in a reducedstate, Ared, is titrated with a titrant in an oxidized state, Tox. The titration reaction isAred + Toxt Tred + AoxThe electrochemical potential for the reaction is the difference between the reduction potentials for the reduction and oxidation half-reactions; thus,Erxn = ETox /Tred – EAox /AredAfter each addition of titrant, the reaction between the analyte and titrant reaches astate of equilibrium. The reaction’s electrochemical potential, Erxn, therefore, iszero, andETox /Tred = EAox /AredConsequently, the potential for either half-reaction may be used to monitor thetitration’s progress.Before the equivalence point the titration mixture consists of appreciable quantities of both the oxidized and reduced forms of the analyte, but very little unreactedtitrant.
The potential, therefore, is best calculated using the Nernst equation for theanalyte’s half-reactionE Aox / Ared = EA°ox Ared −formal potentialThe potential of a redox reaction for aspecific set of solution conditions, suchas pH and ionic composition.RT [ Ared ]lnnF[ Aox ]Although EÅox /Ared is the standard-state potential for the analyte’s half-reaction,a matrix-dependent formal potential is used in its place.
After the equivalencepoint, the potential is easiest to calculate using the Nernst equation for the titrant’shalf-reaction, since significant quantities of its oxidized and reduced forms arepresent.ETox /Tred = ET°ox Tred ––RT [Tred ]lnnF[Tox ]1400-CH09 9/9/99 2:13 PM Page 333Chapter 9 Titrimetric Methods of AnalysisCalculating the Titration Curve As an example, let’s calculate the titration curvefor the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 MHClO4. The reaction in this case isFe2+(aq) + Ce4+(aq)t Ce3+(aq) + Fe3+(aq)9.16The equilibrium constant for this reaction is quite large (it is approximately6 × 1015), so we may assume that the analyte and titrant react completely.The first task is to calculate the volume of Ce4+ needed to reach the equivalencepoint.
From the stoichiometry of the reaction we knowMoles Fe2+ = moles Ce4+orMFeVFe = MCeVCeSolving for the volume of Ce4+VCe =(0.100 M)(50.0 mL)M FeVFe== 50.0 mL(0.100 M)M Cegives the equivalence point volume as 50.0 mL.Before the equivalence point the concentration of unreacted Fe2+ and the concentration of Fe3+ produced by reaction 9.16 are easy to calculate. For this reasonwe find the potential using the Nernst equation for the analyte’s half-reaction° 3+E = E Fe/ Fe 2 + − 0.05916 ln[ Fe 2 + ][ Fe 3 + ]9.17The concentrations of Fe2+ and Fe3+ after adding 5.0 mL of titrant are[Fe2 + ] ==moles unreacted Fe2 +M V − M CeVCe= Fe Fetotal volumeVFe + VCe(0.100 M)(50.0 mL) − (0.100 M)(5.0 mL)= 8.18 × 10 −2 M50.0 mL + 5.0 mL[Fe3 + ] ==moles Ce4 + addedM CeVCe=total volumeVFe + VCe(0.100 M)(5.0 mL)= 9.09 × 10 −3 M50.0 ml + 5.0 mLSubstituting these concentrations into equation 9.17 along with the formal potentialfor the Fe3+/Fe2+ half-reaction from Appendix 3D, we find that the potential is 8.18 × 10 −2 E = +0.767 V − 0.05916 log = +0.711 V 9.09 × 10 −3 At the equivalence point, the moles of Fe2+ initially present and the moles ofadded are equal.
Because the equilibrium constant for reaction 9.16 is large,the concentrations of Fe2+ and Ce4+ are exceedingly small and difficult to calculatewithout resorting to a complex equilibrium problem. Consequently, we cannot calculate the potential at the equivalence point, Eeq, using just the Nernst equation forthe analyte’s half-reaction or the titrant’s half-reaction.
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