chernicd principle/ revi~i ted
Edited by MURIEL BOYDBISHOP
Clemson Universiw Clemson. SC 29631
Chemical Equilibrium VI. Buffer Solutions Adon A. Gordus The University of Michigan, Ann Arbor, MI 48109
In this article in the series' on chemical equilibrium we consider the conditions that permit use of the HendenonHasselhalch equation to descrihe the titration mixture in the buffer region of a weak acid-strong base titration. We also discuss the region following the last equivalence point in the titration. The Buffer Reglon
Most introductory texts (especially biochemistry texts) use the Henderson-Hasselbalch equation to descrihe the mixture existing in the intermediate buffer region of the titration of a weak acid with a strong base. This expression is strictly valid only for a limited range of concentrations and K, values. We can show this by starting with the thermodynamic equilibrium constant for a weak acid. We define the solution as consisting of C. M HA and Ccb M A- (where the subscript cb refers to the conjugate base). These are the concentrations prior to any shifts that may he required to establish equilihrium. The equilibrium mixture has concentrations of [HA] and [A-I. The thermodynamic equilihrium constant is:
where a denotes the activity of the species. If we assume that the solution is ideal, then:
-a
If negative logs are taken and the expression rearranged, we have the usual form of the Henderson-Hasselhalch equation: CCh + log -
c,
(4)
Shown in Figure 1is a plot of K, versus the fraction of HA titrated with a strong hase of equal concentration. Two sets of regions are depicted: one for the case where C. + Ccb =
856
Journal of Chemical Education
10
20
30
50
40
mL NaOH
Figure 1. Regions over which various simplified equations are valid for the tibation mixture mnslsting of C, M HA CCbM NaA. Solid lines correspond to C, + Ccb = 0.100 M: dashed lines correspond to C, + C& = 0.001 M. Applicable equations are given in Tables 1 and 2: separate equations that include me water equilibrium are required in the shaded regions which apply only to the C, + CCb= 0.001 M curves. The Hendersan-Hasselbalch equation is valid in region Ill.
+
+
If we further assume that there is no shift required to establish equilibrium (for instance, that less than 1% HA dissociates and that i t contributes less than 1% to the Ccb concentration or that less than 1%A- undergoes hydrolysis and that i t contributes less than than 1%t o the C, concentration), then [HA] = C, and [A-] = C,b so that:
pH = pK,
0
0.100 M (solid lines) and another for the case where C. C,b = 0.001 M (dashed lines). As in the Henderson-Hasselbalch equation, C, and C,b represent the concentrations of the acid HA and the conjugate hase A- after any acid hase reaction (if it is a titration) hut prior t o any shifts required to establish equilibrium. These two extremes of C. Ccb include the
+
+
'The previous articles are: Gwdus. A. A. "I. Thermodynamic Equilibrium Constant", J. Chem. Educ. 1991, 66, 138-140; "11. Derlving an Exact Equilibrium Equation", J. Chem.-Educ. 1991, 68, 215217; "111. A Few Math Tricks", J. Chem. Educ. 1991, 68, 291-293; "IV. Weak Acidsand Bases", J. Chem. Educ. 1991,68,397-399; "V. Seeing an Endpoint in Acid-Base Titrations", J. Chem. Educ. 1991, 645666567. The method of inequalities used to derive the equations for these regions is similar lo that used in article Ill in this series (seefootnote 1) to- establish OH ranae over which the water eauilibrium can be ~the ~ neglecteo as well as to construct Flgure 1 of article i ~ A. supplement describ ng tne derivations (and those in artic e V in lhis ser es') is available from the author ~
~~7
~-~~
Table 2. Equations for Cunoa In Figure 1
Table 1. Slmpllfled Equalions Valid l a HA Tllrallan wlth NaOH Prlor to Equivalence Polnl Assumptions'
Simplitled Equationb
[H+]= C.
a+d a
[H+I2
.
.
+ (& + Ka)[Ht] - C,K, = 0 ~
Regions Separatedab
~egionin Figure 3
*
< < 0.5
a a = 0.5
~
a+e
< a < 1.0 IV from V
0.5
(C,
+ CdK. = lo-''
+
a = hamion titrated = CCb/(C. C..): a = 0.5 cmeoponds to ~ a a where e Henderso-Hasselbabh equation reduces 10 pH = pK, 0 ~ ~ u s t i o far n r curves se~aratlngI1 from Ill and Ill from IV do not sppv to very low aM very h/gh a valves fa solutions &re C, is less than a b m 0.01 M alnce =sparate equations wet inclde me water squilibirum are required, m a e regions are h w n as C, = 0.001M. shaded areas in ~i~~~~ t for me specific ceae = c. is HA concentration and C*..is A- (canjugbte base) ooncemration after add-bare reaction but plor to any shins requiredto establish equilibrium. 600, concemrstiona are mnefted b r diiution mat occurs during me titration. See fwmote 2.
+ era
+ cC.
~
concentration ranges of typical laboratory titrations. The regions on the graph (based on ideal solutions) show where various assumptions are in error by less than 1%2.The equations that apply in each of the five regions of Figure l are riven inTahle 1.The eauations that separate the regians are given in Tahle 2. The Henderson-Hasselhalch refion (111) is in the center of -~~~ the graph. It is here that, as a resuit of the shifts to establish eauilihrium. there is less than 1%dissociation of HA and less tcan 1%additional A- or less than 1% hydrolysis of A- and less than 1%additional HA and where eqs 3 and 4 apply. This is a fairly narrow range of K, values, especially for C, CCb= 0.001 M. Immediately above this range, region 11,there is more than 1% HA dissociation or additional A-; immediately below the Henderson-Hasselhalch range, region IV, there is more than 1% hydrolysis of A- or additional HA. Region I at the top of the graph corresponds to strong acids such as HCI where the degree of ionization of HA is greater than 99%. Similarly, the regionat the hottomof the graph, V, corresnonds to verv. - . verv- weak acids that, therefore, have very strong conjugate bases and where the degree of hydrolysis of the coniurate base is greater than 996. " " From a practical standpoint, when performing titrations, onlv the reeion of the Henderson-Hasselhalch equation and ah&e is oiimportance in Figure 1. The reason is that the lower regions of the graph correspond to K, values for which a sharp change in pH is not seen in a titration (because C,K. is less than 10-9 as noted in the article V in this series1) and thus of no interest from the standpoint of titration analysis. However, the lower portion of Figure 1still can be used when considering buffer solutions consisting of mixtures of C. M HA Ccb M NaA, even if a sharp pH change would not he seen a t the equivalence point in the titration of HA.
-
+
~~~~
+
= 0.16 (pK. = 0.79). If a 0.100 M solution of HI03 is titrated with 0.100 M NaOH, then halfway to the equivalence point3 the (ideal solution) pH = 1.61, which, if eq 5 is used, corresponds toK, = 0.025; this is in error by afactor of O.16lO.025 = 6.4 or 540%. In fact, pH is never equal to pK. because the initial DHof a 0.100 M HIOa (ideal) solution3 is 1.16 and the pH ineleases during the tigation. The Henderson-Hasselhalch equation is meaningless for the titration of a strong acid with a strpng base since C, is essentially zero. In this case the equation for region I applies. For instance, if 25.00 mL of 0.100 M HCI is titrated with 0.120 M NaOH, then when Vb = 10.00 mL, the residual [Ht], according to the equation for region I is C. = [Ht] = I(25.00)(0.100) - (10.00)(0.120)]/(25.00 10.00) = 0.0371
+
Buffer Solutions
The Henderson-Hasselhalch equation is often used t o illustrate the properties of weak acidlconjugate base huffer solutions by showing that the ratio: C&. changes only slightly when a small amount of acid or base is added to the mixture or if the mixture is diluted. This huffer effect c h also he illustrated from the titration graphs because the addition of a small amount of OH- results in a small displacement t o the right on the graphs whereas the addition of a small amount of H+ results in a small displacement to the left on these same graphs. In the huffer region (typically corresponding to the region between 15% and 85% of the equivalence point volume) the titration-pH curves are relativelv flat as is seen in Fieure 2 where data for the titration of acetic acid are given; very little change in pH occurs when a small amount of OH- or Hf is added. The effect of dilution is also obvious in this figure. Titration curves for various initial concentrationsranplinr from C , = 1.0 M to LO-? M, where the titrant strong-hase~c¢ra 2.0 X 10-12/(Ca + Ccb) and also that K. < 5.0 X 10-3(C, C d . Outside this region considerable error could result. In many cases the pH may not differ much from pK., hut i t is important to emphasize that a difference in log x of only 0.3 corresponds to a factor of 2 and, therefore, represents a 100%error. For instance, consider the more extreme case of iodic acid, HI03, which has&
+
~
The calculation Is performed as follows: If 12.50 mL of 0.100 M NaOH (1.25 mmol) is added to 25.00 mL of 0.100 M H103 (2.50 mmol). then the solution contains 1.25 mmol of HiO3and 1.25 mmol of Na103 all in 37.50 mL so that C. = Ccb = 1.25137.50 = 0.0333 M. The coordinates for this mixture with Ca C* = 0.0666 and pK. = 0.790 place it in region II of Figure 1 so that the quadratic equation given in Table 1 Is required. The solution of this equation is [Hf] = 0.0245 M or pH = 1.610. The pH at the start of the titration also requires the use of a quadratic equation since the cwrdlnates of K. and the original concentration of C. = 0.100 place the solution in region 2 of Figure 1 of the second article In this series: see footnote 1 for reference.
+
Volume 68 Number 8 August 1991
657
10-1 6
V ,
,
,
I
16 ,
,
,
,
0.2
0.4 0.6 0.8 1.0 Fraction Titrated Figure 2. Titration ol an ideal solution of 25.0 mL of C . Macetic acid (K, = 1.00 X lo-') with NaoH of H x same concentcation. C, values are: = 1.0M; o = 0.1 M; t = 0.01 M; 0 = 0.001 M; A = 0.0001 M: A = 0.00001 M. 0
acid, are shown in Figure 2. As is seen for the higher concentrations of 1.0 to 10-3 M, the huffer-region portions of the titration curves are superimposed indicating that the pH is independent of concentration. After the Equlvalence Polnt Most texts indicate that the pH in the region following the equivalence point can be calculated based solely on the excess strong base added (corrected for dilution). This is avery
good approximation for most titrations and, when ahout 25% past the equivalence point, isvalid4 (within 1%)for titrations involving all acids with KJK. = K,b less than 10-4 (i.e., K, greater than 10-lo) for a titration involving C, = 0.1 M. Only for much weaker acids, for which Kcb is greater than about is it necessary to consider the additional contribution of OH- from the hydrolysis of the conjugate base. This is the reason that the portion of the titration curves for different acids (all of the same initial concentration) in the regions past the (last) equivalence point are superimposed. The pH of the solution, for practical purposes, is determined solely by the amount of excess strong hase. For example, when Vb = 40.00 mL of Cb = 0.100 M strong base is added to 25.00 mL of a C. = 0.100 M typical strong or weak acid, this corresponds to 15.00 mL excess NaOH. There is negligible OHarising from the hydrolysis of the conjugate base compared with the OH- from the excess strong base. Therefore, [OH-] = (15.00)(0.100)/(40.00 25.00) = 0.0231, and the pH = 12.363.
+
Tltration of Bases Although the various equations described in this article weredevelo~edforthe titrationof an acid witha strone base. the opposit'e titration of a base with a strong acid Ean bd easily considered. The equations and figures will be identical. Only the notation will change. T o convert any of the equations (or Fig. 1) in this paper for use in the titration of a hase simply substitute [OH-] for [H+], [H+]for [OH-], pOH for pH, and change subscript b to a and a to b.
'See Figure 6-6. In Gordus, A. A. Schaum3 Outline of Analyiical Chemistw McGraw-HIII: New York, 1985.
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