D. Harvey - Modern Analytical Chemistry (794078), страница 87
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For thetitration data in Table 9.5, the initial point in the second derivative titration curve is∆2 pH 0.270 − 0.815== −0.2733.00 − 1.00∆V 2and is plotted as the average of the two volumes (2.00 mL). The remainder of thedata for the second derivative titration curve are shown in Table 9.5 and plotted inFigure 9.13c.Derivative methods are particularly well suited for locating end points in multiprotic and multicomponent systems, in which the use of separate visual indicatorsfor each end point is impractical. The precision with which the end point may belocated also makes derivative methods attractive for the analysis of samples withpoorly defined normal titration curves.Derivative methods work well only when sufficient data are recorded duringthe sharp rise in pH occurring near the equivalence point.
This is usually not aproblem when the titration is conducted with an automatic titrator, particularlywhen operated under computer control. Manual titrations, however, often containonly a few data points in the equivalence point region, due to the limited range ofvolumes over which the transition in pH occurs. Manual titrations are, however,information-rich during the more gently rising portions of the titration curve before and after the equivalence point.Consider again the titration of a monoprotic weak acid, HA, with a strong base.At any point during the titration the weak acid is in equilibrium with H3O+ and A–HA(aq) + H2O(l)t H3O+(aq) + A–1400-CH09 9/9/99 2:12 PM Page 293Chapter 9 Titrimetric Methods of Analysisfor whichKa =[H 3O + ][A − ][HA]Before the equivalence point, and for volumes of titrant in the titration curve’sbuffer region, the concentrations of HA and A– are given by the following equations.[HA] =moles HA − moles OH − addedM V − M bVb= a atotal volumeVa + Vb[A − ] =moles OH − addedM bVb=total volumeVa + VbSubstituting these equations into the Ka expression for HA, and rearranging givesKa ==[H3O + ](M bVb )/(Va + Vb )(M aVa − M bVb )/(Va + Vb )[H3O + ]M bVbM aVa − M bVbKa M aVa − Ka M bVb = [H 3O + ]M bVbKa M aVa− KaVb = [H 3O + ]VbMbFinally, recognizing that the equivalence point volume isVeq =M aVaMbleaves us with the following equation.Vb × [H3O+] = Ka × Veq – Ka × VbFor volumes of titrant before the equivalence point, a plot of Vb × [H3O+] versus Vbis a straight line with an x-intercept equal to the volume of titrant at the end pointand a slope equal to –Ka.* Results for the data in Table 9.5 are shown in Table 9.6and plotted in Figure 9.13d.
Plots such as this, which convert a portion of a titrationcurve into a straight line, are called Gran plots.Finding the End Point by Monitoring Temperature The reaction between an acidand a base is exothermic. Heat generated by the reaction increases the temperatureof the titration mixture. The progress of the titration, therefore, can be followed bymonitoring the change in temperature.An idealized thermometric titration curve (Figure 9.14a) consists of threedistinct linear regions.
Before adding titrant, any change in temperature is due tothe cooling or warming of the solution containing the analyte. Adding titrant initiates the exothermic acid–base reaction, resulting in an increase in temperature.This part of a thermometric titration curve is called the titration branch.
The*Values of Ka determined by this method may be in substantial error if the effect of activity is not considered.Gran plotA linearized form of a titration curve.2931400-CH09 9/9/99 2:12 PM Page 294Modern Analytical ChemistryTemperatureTable 9.6Titration branchExcesstitrationbranch0VtitrTemperature(a)0Vtitr(b)Figure 9.14Figure 9.15Titration curves for 50.00 mL of 0.0100 MH3BO3 with 0.100 M NaOH determined bymonitoring (a) pH, and (b) temperature.Volume(mL)Vb × [H3O+] × 1072.004.0010.0012.0014.0015.0015.5515.6015.7015.8060434812985.033.812.52.760.6210.08050.0135temperature continues to rise with each addition of titrant until the equivalencepoint is reached. After the equivalence point, any change in temperature is due tothe difference between the temperatures of the analytical solution and the titrant,and the enthalpy of dilution for the excess titrant. Actual thermometric titrationcurves (Figure 9.14b) frequently show curvature at the intersection of the titrationbranch and the excess titrant branch due to the incompleteness of the neutralization reaction, or excessive dilution of the analyte during the titration.
The latterproblem is minimized by using a titrant that is 10–100 times more concentratedthan the analyte, although this results in a very small end point volume and alarger relative error.The end point is indicated by the intersection of the titration branch andthe excess titrant branch. In the idealized thermometric titration curve (seeFigure 9.14a) the end point is easily located.
When the intersection betweenthe two branches is curved, the end point can be found by extrapolation(Figure 9.14b).Although not commonly used, thermometric titrations have one distinct advantage over methods based on the direct or indirect monitoring of pH. As discussed earlier, visual indicators and potentiometric titration curves are limitedby the magnitude of the relevant equilibrium constants. For example, the titration of boric acid, H3BO3, for which Ka is 5.8 × 10–10, yields a poorly definedequivalence point (Figure 9.15a). The enthalpy of neutralization for boric acidwith NaOH, however, is only 23% less than that for a strong acid (–42.7 kJ/molpHThermometric titration curves—(a) ideal;(b) showing curvature at the intersection ofthe titration and excess titrant branches.Equivalence points are indicated by thedots (•).Gran Plot Treatment of the Datain Table 9.5(a)14.012.010.08.06.04.02.00.00.00 1.00 2.00 3.00 4.00 5.00 6.00Volume of titrantTemperature29425.10025.06025.02024.980–1(b)023451Volume of titrant61400-CH09 9/9/99 2:12 PM Page 295Chapter 9 Titrimetric Methods of Analysisfor H3BO3 versus –55.6 kJ/mol for HCl), resulting in a favorable thermometrictitration curve (Figure 9.15b).9B.3 Titrations in Nonaqueous SolventsThus far we have assumed that the acid and base are in an aqueous solution.
Indeed, water is the most common solvent in acid–base titrimetry. When consideringthe utility of a titration, however, the solvent’s influence cannot be ignored.The dissociation, or autoprotolysis constant for a solvent, SH, relates the concentration of the protonated solvent, SH2+, to that of the deprotonated solvent, S–.For amphoteric solvents, which can act as both proton donors and proton acceptors, the autoprotolysis reaction is2SHt SH2+ + S–with an equilibrium constant ofKs = [SH2+][S–]You should recognize that Kw is just the specific form of Ks for water. The pH of asolution is now seen to be a general statement about the relative abundance of protonated solventpH = –log[SH2+]where the pH of a neutral solvent is given as1pKs2Perhaps the most obvious limitation imposed by Ks is the change in pH duringa titration.
To see why this is so, let’s consider the titration of a 50 mL solution of10–4 M strong acid with equimolar strong base. Before the equivalence point, thepH is determined by the untitrated strong acid, whereas after the equivalence pointthe concentration of excess strong base determines the pH.
In an aqueous solutionthe concentration of H3O+ when the titration is 90% complete ispH neut =[ H 3 O+ ] ==M aVa − M bVbVa + Vb(1 × 10 −4 M)(50 mL) − (1 × 10 −4 M)(45 mL)= 5.3 × 10 −6 M50 + 45corresponding to a pH of 5.3. When the titration is 110% complete, the concentration of OH– is[OH − ] ==M bVb − M aVaVa + Vb(1 × 10 −4 M)(55 mL) − (1 × 10 −4 M)(50 mL)= 4.8 × 10 −6 M50 + 55or a pOH of 5.3. The pH, therefore, ispH = pKw – pOH = 14.0 – 5.3 = 8.7The change in pH when the titration passes from 90% to 110% completion is∆pH = 8.7 – 5.3 = 3.42951400-CH09 9/9/99 2:12 PM Page 296296Modern Analytical ChemistrypH20.015.0(b)10.0(a)pH = pKs – pOH = 20.0 – 5.3 = 14.75.00.00.00In this case the change in pH of20.0040.00 60.00 80.00 100.00Volume of titrantFigure 9.1610–4Titration curves for 50.00 mL ofM HClwith 10–4 M NaOH in (a) water,Kw = l × 10–14, and (b) nonaqueous solvent,Ks = 1 × 10–20.20.0(b)pH15.0(a)10.05.00.00.00If the same titration is carried out in a nonaqueous solvent with a Ksof 1.0 × 10–20, the pH when the titration is 90% complete is still 5.3.However, the pH when the titration is 110% complete is now20.0040.00 60.00 80.00 100.00Volume of titrantFigure 9.17Titration curves for 50.00 mL of 0.100 Mweak acid (pKa = 11) with 0.100 M NaOH in(a) water, Kw = 1 × 10–14; and(b) nonaqueous solvent, Ks = 1 × 10–20.
Thetitration curve in (b) assumes that thechange in solvent has no effect on the aciddissociation constant of the weak acid.levelingAcids that are better proton donors thanthe solvent are leveled to the acidstrength of the protonated solvent; basesthat are better proton acceptors than thesolvent are leveled to the base strength ofthe deprotonated solvent.∆pH = 14.7 – 5.3 = 9.4is significantly greater than that obtained when the titration is carriedout in water. Figure 9.16 shows the titration curves in both the aqueous and nonaqueous solvents.
Nonaqueous solvents also may be usedto increase the change in pH when titrating weak acids or bases (Figure 9.17).Another parameter affecting the feasibility of a titration is the dissociation constant of the acid or base being titrated. Again, the solventplays an important role. In the Brønsted–Lowry view of acid–base behavior, the strength of an acid or base is a relative measure of the easewith which a proton is transferred from the acid to the solvent, orfrom the solvent to the base.
For example, the strongest acid that canexist in water is H3 O + . The acids HCl and HNO3 are consideredstrong because they are better proton donors than H3O+. Strong acidsessentially donate all their protons to H 2 O, “leveling” their acidstrength to that of H3O+. In a different solvent HCl and HNO3 maynot behave as strong acids.When acetic acid, which is a weak acid, is placed in water, the dissociation reactionCH3COOH(aq) + H2O(l)t H3O+(aq) + CH3COO–(aq)does not proceed to a significant extent because acetate is a stronger base than waterand the hydronium ion is a stronger acid than acetic acid.
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