D. Harvey - Modern Analytical Chemistry (794078), страница 93
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Asshown in Table 9.13 for CdY 2– , the conditional formation constant becomessmaller, and the complex becomes less stable at lower pH levels.EDTA Must Compete with Other Ligands To maintain a constant pH, we must adda buffering agent. If one of the buffer’s components forms a metal–ligand complexwith Cd2+, then EDTA must compete with the ligand for Cd2+. For example, anNH4+/NH3 buffer includes the ligand NH3, which forms several stable Cd2+–NH3complexes. EDTA forms a stronger complex with Cd2+ and will displace NH3.
Thepresence of NH3, however, decreases the stability of the Cd2+–EDTA complex.We can account for the effect of an auxiliary complexing agent, such as NH3,in the same way we accounted for the effect of pH. Before adding EDTA, a massbalance on Cd2+ requires that the total concentration of Cd2+, CCd, beCCd = [Cd2+] + [Cd(NH3)2+] + [Cd(NH3)22+] + [Cd(NH3)32+] + [Cd(NH3)42+]1400-CH09 9/9/99 2:13 PM Page 317Chapter 9 Titrimetric Methods of AnalysisTable 9.14317Values of αM for Selected Concentrations of Ammonian+[NH3] (M)αAg+αCa2+αCd2+αCo2+αCu2+αMg2+αNi2+αZn2+10.50.10.050.010.0050.0011.00 × 10–74.00 × 10–79.98 × 10–63.99 × 10–59.83 × 10–43.86 × 10–37.95 × 10–25.50 × 10–17.36 × 10–19.39 × 10–19.69 × 10–19.94 × 10–19.97 × 10–19.99 × 10–16.09 × 10–81.05 × 10–63.51 × 10–42.72 × 10–38.81 × 10–22.27 × 10–16.90 × 10–11.00 × 10–62.22 × 10–56.64 × 10–33.54 × 10–23.55 × 10–15.68 × 10–18.84 × 10–13.79 × 10–146.86 × 10–134.63 × 10–107.17 × 10–93.22 × 10–63.62 × 10–54.15 × 10–31.76 × 10–14.13 × 10–18.48 × 10–19.22 × 10–19.84 × 10–19.92 × 10–19.98 × 10–19.20 × 10–103.44 × 10–85.12 × 10–56.37 × 10–44.32 × 10–21.36 × 10–15.76 × 10–13.95 × 10–106.27 × 10–93.68 × 10–65.45 × 10–51.82 × 10–21.27 × 10–17.48 × 10–1The fraction, αCd2+, present as uncomplexed Cd2+ isα Cd 2 + =[Cd2 + ]CCd9.14Solving equation 9.14 for [Cd2+] and substituting into equation 9.13 givesK f′ = α Y 4 − × K f =[CdY 2 − ]α Cd 2 + CCd CEDTAIf the concentration of NH3 is held constant, as it usually is when using a buffer,then we can rewrite this equation asK f′′ = α Cd 2 + × α Y 4 − × K f =[CdY 2 − ]CCd CEDTA9.15where Kf˝ is a new conditional formation constant accounting for both pH and thepresence of an auxiliary complexing agent.
Values of αMn+ for several metal ions areprovided in Table 9.14.9C.2 Complexometric EDTA Titration CurvesNow that we know something about EDTA’s chemical properties, we are ready toevaluate its utility as a titrant for the analysis of metal ions. To do so we need toknow the shape of a complexometric EDTA titration curve. In Section 9B we sawthat an acid–base titration curve shows the change in pH following the addition oftitrant.
The analogous result for a titration with EDTA shows the change in pM,where M is the metal ion, as a function of the volume of EDTA. In this section welearn how to calculate the titration curve. We then show how to quickly sketch thetitration curve using a minimum number of calculations.Calculating the Titration Curve As an example, let’s calculate the titration curvefor 50.0 mL of 5.00 × 10–3 M Cd2+ with 0.0100 M EDTA at a pH of 10 and in thepresence of 0.0100 M NH3. The formation constant for Cd2+–EDTA is 2.9 × 1016.Since the titration is carried out at a pH of 10, some of the EDTA is presentin forms other than Y4–. In addition, the presence of NH3 means that the EDTAmust compete for the Cd2+.
To evaluate the titration curve, therefore, we mustuse the appropriate conditional formation constant. From Tables 9.12 and9.14 we find that αY4– is 0.35 at a pH of 10, and that αCd2+ is 0.0881 when the1400-CH09 9/9/99 2:13 PM Page 318318Modern Analytical Chemistryconcentration of NH3 is 0.0100 M. Using these values, we calculate that the conditional formation constant isKf˝ = αY4– × αCd2+ × Kf = (0.35)(0.0881)(2.9 × 1016) = 8.9 × 1014Because Kf˝ is so large, we treat the titration reaction as though it proceeds tocompletion.The first task in calculating the titration curve is to determine the volume ofEDTA needed to reach the equivalence point. At the equivalence point we knowthatMoles EDTA = moles Cd2+orMEDTAVEDTA = MCdVCdSolving for the volume of EDTAVEDTA =(5.00 × 10 −3 M)(50.0 mL)M CdVCd== 25.0 mL0.0100 MM EDTAshows us that 25.0 mL of EDTA is needed to reach the equivalence point.Before the equivalence point, Cd2+ is in excess, and pCd is determined by theconcentration of free Cd2+ remaining in solution.
Not all the untitrated Cd2+ is free(some is complexed with NH3), so we will have to account for the presence of NH3.For example, after adding 5.0 mL of EDTA, the total concentration of Cd2+ ismoles excess Cd 2 +M V − M EDTAVEDTACCd == Cd Cdtotal volumeVCd + VEDTA=(5.00 × 10 −3 M)(50.0 mL) − (0.0100 M)(5.0 mL)= 3.64 × 10 −3 M50.0 mL + 5.0 mLTo calculate the concentration of free Cd2+ we use equation 9.14.[Cd2+] = αCd2+ × CCd = (0.0881)(3.64 × 10–3 M) = 3.21 × 10–4 MThus, pCd ispCd = –log[Cd2+] = –log(3.21 × 10–4) = 3.49At the equivalence point, all the Cd2+ initially present is now present as CdY2–.The concentration of Cd2+, therefore, is determined by the dissociation of the CdY2–complex. To find pCd we must first calculate the concentration of the complex.M CdVCdinitial moles Cd2+[CdY 2 − ] ==VCd + VEDTAtotal volume=(5.00 × 10 −3 M)(50.0 mL)= 3.33 × 10 −3 M50.0 mL + 25.0 mLLetting the variable x represent the concentration of Cd2+ due to the dissociation ofthe CdY2– complex, we haveK f′′ =[CdY 2 − ]3.33 × 10 −3 − x== 8.94 × 1014CCdCEDTA(x)(x)x = CCd = 1.93 × 10 −9 M1400-CH09 9/9/99 2:13 PM Page 319Chapter 9 Titrimetric Methods of Analysis319Once again, to find the [Cd2+] we must account for the presence of NH3; thus[Cd2+] = αCd2+ × CCd = (0.0881)(1.93 × 10–9 M) = 1.70 × 10–10 Mgiving pCd as 9.77.After the equivalence point, EDTA is in excess, and the concentration of Cd2+ isdetermined by the dissociation of the CdY2– complex.
Examining the equation forthe complex’s conditional formation constant (equation 9.15), we see that to calculate CCd we must first calculate [CdY2–] and CEDTA. After adding 30.0 mL of EDTA,these concentrations are[CdY 2 − ] ==CEDTA ==initial moles Cd2 +M CdVCd=total volumeVCd + VEDTA(5.00 × 10 −3 M)(50.0 mL)= 3.13 × 10 −3 M50.0 mL + 30.0 mLmoles excess EDTA− M CdVCdMV= EDTA EDTAtotal volumeVCd + VEDTA(0.0100 M)(30.0 mL) − (5.00 × 10 −3 M)(50.0 mL)= 6.25 × 10 −4 M50.0 mL + 30.0 mLSubstituting these concentrations into equation 9.15 and solving for CCd gives[CdY 2 − ]3.13 × 10 −3 M= 8.94 × 1014=CCdCEDTACCd (6.25 × 10 −4 )CCd = 5.60 × 10 −15 MThus,[Cd2+] = αCd2+ × CCd = (0.0881)(5.60 × 10–15 M) = 4.93 × 10–16 Mand pCd is 15.31. Figure 9.27 and Table 9.15 show additional results for thistitration.18.0016.0014.00pCd12.0010.008.006.004.00Figure 9.272.000.000102030Volume of EDTA (mL)4050Complexometric titration curve for 50.0 mLof 5.00 × 10–3 M Cd2+ with 0.0100 M EDTAat a pH of 10.0 in the presence of0.0100 M NH3.1400-CH09 9/9/99 2:13 PM Page 320320Modern Analytical ChemistryTable 9.15Data for Titration of 5.00 × 10–3 MCd2+ with 0.0100 M EDTAat a pH of 10.0 and in thePresence of 0.0100 M NH3Volume of EDTA(mL)pCd0.005.0010.0015.0020.0023.0025.0027.0030.0035.0040.0045.0050.003.363.493.663.874.204.629.7714.9115.3115.6115.7815.9116.01Sketching an EDTA Titration Curve Our strategy for sketching an EDTA titrationcurve is similar to that for sketching an acid–base titration curve.
We begin bydrawing axes, placing pM on the y-axis and volume of EDTA on the x-axis. Aftercalculating the volume of EDTA needed to reach the equivalence point, we add avertical line intersecting the x-axis at this volume. Next we calculate and plot twovalues of pM for volumes of EDTA before the equivalence point and two values ofpM for volumes after the equivalence point. Straight lines are drawn through eachpair of points. Finally, a smooth curve is drawn connecting the three straight-linesegments.EXAMPLE 9.7Sketch the titration curve for 50.0 mL of 5.00 × 10–3 M Cd2+ with 0.010 MEDTA at a pH of 10, and in the presence of an ammonia concentration that isheld constant throughout the titration at 0.010 M.
This is the same titration forwhich we previously calculated the titration curve (Table 9.15 and Figure 9.27).SOLUTIONWe begin by drawing axes for the titration curve (Figure 9.28a). We havealready shown that the equivalence point is at 25.0 mL, so we draw a verticalline intersecting the x-axis at this volume (Figure 9.28b).Before the equivalence point, pCd is determined by the excessconcentration of free Cd2+. Using values from Table 9.15, we plot pCd for 5.0mL and 10.0 mL of EDTA (Figure 9.28c).After the equivalence point, pCd is determined by the dissociation of theCd2+–EDTA complex. Using values from Table 9.15, we plot pCd for 30.0 mLand 40.0 mL of EDTA (Figure 9.28d).1400-CH09 9/9/99 2:13 PM Page 321Chapter 9 Titrimetric Methods of AnalysisPercent titrated050100Percent titrated15020005016.016.014.014.012.012.010.010.08.01502008.06.06.04.04.02.02.00.00.0010.0020.0030.0040.00Volume of titrant (mL)0.00.0050.00(a)10.0020.0030.0040.00Volume of titrant (mL)50.00(b)Percent titrated050100Percent titrated150200018.05010015020018.016.016.014.014.012.012.010.010.0pCdpCd10018.0pCdpCd18.08.08.06.06.04.04.02.02.00.00.000.00.0010.0020.0030.0040.00Volume of titrant (mL)50.00(c)10.0020.0030.0040.00Volume of titrant (mL)50.00(d)Percent titrated050100Percent titrated150200018.05010015020018.016.016.014.014.012.012.010.010.0pCdpCd3218.08.06.06.04.04.02.02.00.00.000.00.0010.0020.0030.0040.00Volume of titrant (mL)50.00(e)Figure 9.28How to sketch an EDTA complexometric titration curve; see text for explanation.(f)10.0020.0030.0040.00Volume of titrant (mL)50.001400-CH09 9/9/99 2:13 PM Page 322Modern Analytical ChemistryAn approximate sketch of the titration curve is completed by drawingseparate straight lines through the two points before and after the equivalencepoint (Figure 9.28e).
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