D. Harvey - Modern Analytical Chemistry (794078), страница 84
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Sørenson’s establishment of the pH scale in 1909provided a rigorous means for comparing visual indicators. The determination ofacid–base dissociation constants made the calculation of theoretical titration curvespossible, as outlined by Bjerrum in 1914. For the first time a rational method existed for selecting visual indicators, establishing acid–base titrimetry as a useful alternative to gravimetry.9B.1 Acid–Base Titration CurvesIn the overview to this chapter we noted that the experimentally determined endpoint should coincide with the titration’s equivalence point.
For an acid–base titration, the equivalence point is characterized by a pH level that is a function of theacid–base strengths and concentrations of the analyte and titrant. The pH at the endpoint, however, may or may not correspond to the pH at the equivalence point. Tounderstand the relationship between end points and equivalence points we mustknow how the pH changes during a titration. In this section we will learn how toconstruct titration curves for several important types of acid–base titrations. Our1400-CH09 9/9/99 2:12 PM Page 280280Modern Analytical Chemistryapproach will make use of the equilibrium calculations described in Chapter 6. Wealso will learn how to sketch a good approximation to any titration curve using onlya limited number of simple calculations.Titrating Strong Acids and Strong Bases For our first titration curve let’s considerthe titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH.
For the reaction of astrong base with a strong acid the only equilibrium reaction of importance isH3O+(aq) + OH–(aq)t 2H2O(l)9.1The first task in constructing the titration curve is to calculate the volume of NaOHneeded to reach the equivalence point. At the equivalence point we know from reaction 9.1 thatMoles HCl = moles NaOHorMaVa = MbVbwhere the subscript ‘a’ indicates the acid, HCl, and the subscript ‘b’ indicates thebase, NaOH. The volume of NaOH needed to reach the equivalence point, therefore, isVeq = Vb =(0.100 M)(50.0 mL)M aVa== 25.0 mL(0.200 M)MbBefore the equivalence point, HCl is present in excess and the pH is determined bythe concentration of excess HCl.
Initially the solution is 0.100 M in HCl, which,since HCl is a strong acid, means that the pH ispH = –log[H3O+] = –log[HCl] = –log(0.100) = 1.00The equilibrium constant for reaction 9.1 is (Kw)–1, or 1.00 × 1014. Since this is sucha large value we can treat reaction 9.1 as though it goes to completion. After adding10.0 mL of NaOH, therefore, the concentration of excess HCl is[HCl] ==moles excess HClM V − M bVb= a atotal volumeVa + Vb(0.100 M)(50.0 mL) − (0.200 M)(10.0 mL)= 0.050 M50.0 mL + 10.0 mLgiving a pH of 1.30.At the equivalence point the moles of HCl and the moles of NaOH are equal.Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water.Kw = 1.00 × 10–14 = [H3O+][OH–] = [H3O+]2[H3O+] = 1.00 × 10–7 MThus, the pH at the equivalence point is 7.00.Finally, for volumes of NaOH greater than the equivalence point volume, thepH is determined by the concentration of excess OH–.
For example, after adding30.0 mL of titrant the concentration of OH– is1400-CH09 9/9/99 2:12 PM Page 281Chapter 9 Titrimetric Methods of AnalysisTable 9.2Data for Titration of 50.00 mL of0.100 M HCl with 0.0500 M NaOHVolume (mL) of Titrant0.005.0010.0015.0020.0022.0024.0025.0026.0028.0030.0035.0040.0045.0050.00[OH − ] ==pH1.001.141.301.511.852.082.577.0011.4211.8912.5012.3712.5212.6212.70moles excess NaOHM bVb − M aVa=total volumeVa + Vb(0.200 M)(30.0 mL) − (0.100 M)(50.0 mL)= 0.0125 M50.0 mL + 30.0 mLTo find the concentration of H3O+, we use the Kw expression[H 3 O + ] =Kw1.00 × 10 −14== 8.00 × 10 −13−0.0125[OH ]giving a pH of 12.10.
Table 9.2 and Figure 9.1 show additional results for this titration curve. Calculating the titration curve for the titration of a strong base with astrong acid is handled in the same manner, except that the strong base is in excessbefore the equivalence point and the strong acid is in excess after the equivalencepoint.Titrating a Weak Acid with a Strong Base For this example let’s consider the titration of 50.0 mL of 0.100 M acetic acid, CH3COOH, with 0.100 M NaOH.
Again, westart by calculating the volume of NaOH needed to reach the equivalence point;thusMoles CH3COOH = moles NaOHMaVa = MbVbVeq = Vb =(0.100 M)(50.0 mL)M aVa== 50.0 mL(0.100 M)Mb2811400-CH09 9/9/99 2:12 PM Page 282282Modern Analytical ChemistryBefore adding any NaOH the pH is that for a solution of 0.100 M acetic acid.Since acetic acid is a weak acid, we calculate the pH using the method outlined inChapter 6.CH3COOH(aq) + H2O(l)Ka =t H3O+(aq) + CH3COO–(aq)[H 3O+ ][CH 3COO − ](x )(x )== 1.75 × 10 −5[CH 3COOH]0.100 − xx = [H3O+] = 1.32 × 10–3At the beginning of the titration the pH is 2.88.Adding NaOH converts a portion of the acetic acid to its conjugate base.CH3COOH(aq) + OH–(aq)t H2O(l) + CH3COO–(aq)9.2Any solution containing comparable amounts of a weak acid, HA, and its conjugateweak base, A–, is a buffer. As we learned in Chapter 6, we can calculate the pH of abuffer using the Henderson–Hasselbalch equation.pH = pKa + log[A − ][HA]The equilibrium constant for reaction 9.2 is large (K = Ka/Kw = 1.75 × 109), so wecan treat the reaction as one that goes to completion.
Before the equivalence point,the concentration of unreacted acetic acid is[CH 3COOH] =moles unreacted CH3COOHM V − M bVb= a atotal volumeVa + Vband the concentration of acetate is[CH 3COO − ] =moles NaOH addedM bVb=total volumeVa + VbFor example, after adding 10.0 mL of NaOH the concentrations of CH3COOH andCH3COO– are[CH 3COOH] =(0.100 M)(50.0 mL) − (0.100 M)(10.0 mL)= 0.0667 M50.0 mL + 10.0 mL[CH 3COO − ] =(0.100 M)(10.0 mL)= 0.0167 M50.0 mL + 10.0 mLgiving a pH ofpH = 4.76 + log0.0167= 4.160.0667A similar calculation shows that the pH after adding 20.0 mL of NaOH is 4.58.At the equivalence point, the moles of acetic acid initially present and the molesof NaOH added are identical.
Since their reaction effectively proceeds to completion, the predominate ion in solution is CH3COO–, which is a weak base. To calculate the pH we first determine the concentration of CH3COO–.[CH 3COO − ] =moles CH3COOH(0.100 M)(50.0 mL)== 0.0500 Mtotal volume50.0 mL + 50.0 mLThe pH is then calculated as shown in Chapter 6 for a weak base.1400-CH09 9/9/99 2:12 PM Page 283Chapter 9 Titrimetric Methods of AnalysisCH3COO–(aq) + H2O(l)Kb =283t OH–(aq) + CH3COOH(aq)[OH − ][CH 3COOH](x )(x )== 5.71 × 10 −10−0.0500 − x[CH3COO ]x = [OH–] = 5.34 × 10–6 MThe concentration of H3O+, therefore, is 1.87 × 10–9, or a pH of 8.73.After the equivalence point NaOH is present in excess, and the pH is determined in the same manner as in the titration of a strong acid with a strong base. Forexample, after adding 60.0 mL of NaOH, the concentration of OH– is[OH − ] =(0.100 M)(60.0 mL) − (0.100 M)(50.0 mL)= 0.00909 M50.0 mL + 60.0 mLgiving a pH of 11.96.
Table 9.3 and Figure 9.6 show additional results for this titration. The calculations for the titration of a weak base with a strong acid are handledin a similar manner except that the initial pH is determined by the weak base, thepH at the equivalence point by its conjugate weak acid, and the pH after the equivalence point by the concentration of excess strong acid.14.00Data for Titration of 50.0 mL of0.100 M Acetic Acid with 0.100 MNaOHVolume of NaOH(mL)pH0.005.0010.0015.0020.0025.0030.0035.0040.0045.0048.0050.0052.0055.0060.0065.0070.0075.0080.0085.0090.0095.00100.002.883.814.164.394.584.764.945.135.365.716.148.7311.2911.6811.9612.1212.2212.3012.3612.4112.4612.4912.5212.0010.008.00pHTable 9.36.004.002.000.00020406080100Volume NaOH (mL)Figure 9.6Titration curve for 50.0 mL of 0.100 M acetic acid (pKa = 4.76) with0.100 M NaOH.1400-CH09 9/9/99 2:12 PM Page 284284Modern Analytical ChemistryThe approach that we have worked out for the titration of a monoprotic weakacid with a strong base can be extended to reactions involving multiprotic acids orbases and mixtures of acids or bases.
As the complexity of the titration increases,however, the necessary calculations become more time-consuming. Not surprisingly, a variety of algebraic1 and computer spreadsheet2 approaches have been described to aid in constructing titration curves.Sketching an Acid–Base Titration Curve To evaluate the relationship between anequivalence point and an end point, we only need to construct a reasonable approximation to the titration curve. In this section we demonstrate a simple method forsketching any acid–base titration curve. Our goal is to sketch the titration curvequickly, using as few calculations as possible.To quickly sketch a titration curve we take advantage of the following observation. Except for the initial pH and the pH at the equivalence point, the pH at anypoint of a titration curve is determined by either an excess of strong acid or strongbase, or by a buffer consisting of a weak acid and its conjugate weak base.
As wehave seen in the preceding sections, calculating the pH of a solution containing excess strong acid or strong base is straightforward.We can easily calculate the pH of a buffer using the Henderson–Hasselbalch equation. We can avoid this calculation, however, if we make the following assumption.You may recall that in Chapter 6 we stated that a buffer operates over a pH range extending approximately ±1 pH units on either side of the buffer’s pKa. The pH is at thelower end of this range, pH = pKa – 1, when the weak acid’s concentration is approximately ten times greater than that of its conjugate weak base.
Conversely, the buffer’spH is at its upper limit, pH = pKa + 1, when the concentration of weak acid is tentimes less than that of its conjugate weak base. When titrating a weak acid or weakbase, therefore, the buffer region spans a range of volumes from approximately 10% ofthe equivalence point volume to approximately 90% of the equivalence point volume.*Our strategy for quickly sketching a titration curve is simple. We begin by drawing our axes, placing pH on the y-axis and volume of titrant on the x-axis. After calculating the volume of titrant needed to reach the equivalence point, we draw a verticalline that intersects the x-axis at this volume. Next, we determine the pH for two volumes before the equivalence point and for two volumes after the equivalence point.
Tosave time we only calculate pH values when the pH is determined by excess strong acidor strong base. For weak acids or bases we use the limits of their buffer region to estimate the two points. Straight lines are drawn through each pair of points, with eachline intersecting the vertical line representing the equivalence point volume. Finally, asmooth curve is drawn connecting the three straight-line segments. Example 9.1 illustrates this approach for the titration of a weak acid with a strong base.EXAMPLE 9.1Sketch the titration curve for the titration of 50.0 mL of 0.100 M acetic acidwith 0.100 M NaOH.
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