chemical principles revisted
edited by
MURIELBISHOP Ciernson University Clernson, SC 29634
Chemical Equilibrium VII. pH Approximations in Acid-Base Titrations Adon A. Gordus The University of Michigan, Ann Arbor, MI 48109 I n this article in the series' on chemical equilibrium we use the Henderson-Hasselbalch equation to describe a titration mixture i n the buffer region and other simplified equations to approximate the pH at various other stagcs in the titrationofa tvoical . weakacid. Althouch therauations" developed apply to the titration ofweak acids with a strong base, they can be easily converted for use when weak bases are titrated with a strong acid. Only a few substitutions are
.
'The previous articles are: Gordus, A. A. "I. Thermodynamic Equilibrium Constant", J. Chem. Educ. 1991, 68, 13&140 "11. Deriving an Exact Eauilibrium Eouation". J. Chem. Educ. 1991. 68. 215-217: "111. A Few ~ i t hricks", J. Chem Educ. 1991,68,291-293: "IV. weak Acids and Bases", J. Chem. Educ. 1991,68,397-399; "V. Seeing an Endpoint in Acid-BaseTitrationsn,J. Chem. Educ.1991,68,566568; VI. Buffer Solutions",J. Chem. Educ. 1991 68, 656458. 'To avoid clutter in the equations, we will designate equilibrium constants by the symbol K with concentrations in rnolarities. If the solution is assumed to be ideal, K will equal the thermodynamic constatn K,. If the solution is nonideal, Kwill equal K, 1 &where K, is the activity coefficient"constant' as described in article I (footnote1). ~
~
needed: [OH-] for [Ht1, tH+1for [OH-], and subscripts b for a, and a for b. We illustrate these approximations for the titration of a weak polyprotic acid H& with a strong base.
The Titration The Beginning ofthe Titration. At the start of the titration we can use the simple approximationderived in article W' for the first hydrogen dissociation. [H+l= This is valid only if the water equilibrium can be neglected and if there is less than 1%dissociation of H3X. The equation can be converted to:
Halfuray to the 1st Equivalence Point. T h e HendersonHasselbalch equation (eq 5 of article VI)' can be used for a typical weak acid. Using the approximation C. = Ccb the result is
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Number 9
September 1991
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PH = PKI (2) The First Equivalence Point. If we assume that we can neglect the third step i n the acid ionization, HX2- + H+ + X3-, then the result is3 rH+12 =
KI(Kz[HZI + K J [ H Z 1 + KI
1
PH = -(PKI + P&) 2
Halfway to the Second EquivalencePoint. Again, u s i n g the Henderson-Hasselbalch equation, we have PH = PKZ (6) The Second Equivalence Point. An expression similar to eq 5 results
(3)
If we further assume that W-Iis much larger than Kl we can neglect Kl i n the denominator. I n addition, if we assume that K2[H2X-I is much larger than K, we can neglect K, in the numerator. Equation 3 reduces to: [H'I' = K,K,
(5)
1
PH =-(PIG + P K ~ ) (7) 2 Halfway to the Third Equivalence Point. T h e Henderson-Hasselbalch equation is again used: PH = P& (8) The Third (Last) Equivalence Point. This would also correspond to the second equivalence point i n the titration of a diprotic acid or to the single equivalence point i n the titrationofamonoproticacid(H.4). In thiscase, we have
(4,
Taking negative logs gives %ee Gordus, A. A. Schaum's Outline of Analytical Chemktfy, McGraw-Hill: New York, 1985, Chapter 7.
Table 1. Calculated pH in Titration of 25.00 mL of 0.100 M Nitrilotriacetic Acidewith 0.100 M NaOH
Stage in titration
Vb,
Start
mL
0.00
Principal species
Cof Species
Approximate equation
Calculated pH Based on: Approx. Eq. Idealb NonidealC
H3X
Ca = 0.1 00
1
1.45
1.48
1.47
H3X
Ca= 0.0333 Ccbi = 0.0333
2
1.89
1.91
1.83
NaHzX
1I2 way to 1st eq. pt.
12.5
1st eq. pt.
25.0
NaH2X
Ccbi = 0.0500
5
2.19
2.28
2.15
112 way to 2nd eq. pt.
37.5
NaHzX Na2HX
Ccbi = 0.0200
6
2.49
2.72
2.43
Ccm = 0.0200
2nd eq. pt.
50.0
NazHX
CCW= 0.0333
7
6.11
6.10
5.78
112 way to 3rd eq. pt.
62.5
NazHX Na3X
Ccm = 0.0143 C m = 0.0143
8
9.73
9.73
9.16
3rd eq. pt.
75.0
Na3X
Ccm = 0.0250
10
11.06
11.05
10.69
50% past last eq. pt.
87.5
Na3X NaOH
CCW= 0.0222
12
12.05
12.05
11.91
Csb = 0.01 II
f i = 1.29 X,'ol fi = 3.24 X lo3, ffi = 1.86 X 10~''. Titration curve is shown in article V in this series; see footnote 1. pH calculated using exact equation for ideal salutlan: see article 11, footnote 1. pH calculated using exact equation with activity wefficients calculated using the extended Debye-Huckel equation; see article 11, footnote 1.
Table 2. Calculated pH in Titration of 25.00 mL of 0.100 M HA Acids with 0.100 M NaOH
Stage in titration
1R way
Vb,
mL
Principal C of species species
HA
Approx. eq.
H103,
ffi = 0.16
HAc, Ka = 1.8 x 1o5
HIO, 16 = 5.0 x loA3
0.79
1.61
1.58
4.75
4.75
4.62
12.30
12.02
11.88
NaA
Ca= 0.033 Ccb= 0.033
2
to eq. pt.
50%past37.5
NaA Ccb=0.040 NaOH Csh= 0.020
12
12.30
12.30
12.35
12.30
12.30
12.35
12.30
12.54
12.37
ea. ot. a
12.5
pH calculated using approximate equation listed in column 5. pH calculated using exact equation for ideal solution; see article 11, footnote 1. pH calculated using exact equation with activity wefficients calculated using the extended Debye-HGckeiequation: see article 11, footnote 1.
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Journal of Chemical Education
an expression analogous to eq 1, but the solution now contains the conjugate base so that
or substituting pOH = pK,
- pH and pKGb,= pK,-
pK,
If the acid being titrated is H a , then n = 3 in eqs 9 and 10; for a diprotic acid, n = 2; for a mouoprotic acid, n = 1. Following the Last Equivalence Point. When slightly past the last equivalence point the hydroxide concentration of the solution can be approximated as due solely to the excess strong base (sb)that was added (corrected for dilution), as discussed in article VI' so that [OH-] = C,b or POH = PC&
(11)
Because pH = &-pOH, this is the same as PH = PK, - P C . ~
(12)
These are reasonable approximations for many weak acids as shown by the data in Table 1 where the pH is calculated using eqs 1-12. Values are also calculated separately using exact-ideal-solutionequations and exact-nonideal-solution equations4 for the titration of 25.00 mL of 0.100 M nitrilotriacetic acid with 0.100 M NaOH. Then the three K values are between 1.3 x lo4 and 1.9 x 10-lo. (Similar data for the titration of H3P04and citric acid are given elsewhereJ3 In this example it is only toward the end of the titration that the approximate or even ideal-equation pH values differ significantly from the pH calculated using nonideal corrections. This is expected since multiplycharged ions predominate near the third equivalence point, and the activity coefficients for these ions show the greatest deviations from ideality. Thus, the pH a t the third equivalence point is calculated as 11.09usingthe approximateequation, but as 10.69using the exact equation with nonideality corrections. This is a difference of 0.40 pH units corresponding to an antilog of 2.51, for a 151%difference in the hydrogen ion activity The approximations given by eqs 1, 2, and S 1 2 would also be used for the titration of weak monoprotic acid (HA) with a strong base. Table 2 shows data for various monoprotic 0.100 M acids that range from moderately strong (HI03;K. = 0.16) to weak (acetic acid; Ka = 1.8 x lo6) to the very weak hypoiodons acid (HIO; K, = 5.0 x 10-13). The initial and buffer-regionvalues for the moderately strong HI03 calculated according to the approximate equations have considerable error and give an interesting pattern of pH values when using the approximate equations. The approximated pH increases and then decreases. The correct increasing progression in pH is seen only when using the exact equation for either the ideal or nonideal 4Theexact equation isderived in anicle II in this series (seefootnote
1 for reference.
calculation. This is typical of both moderately strong and strong acids. Using the Henderson-Hasselbalch equation, for example, causes large error, even though it is often used for such moderately strong acids. The data for acetic acid using the approximate equations agree with the values calculated from the exact (nonideal) equation throughout the titration range. This is one reason that HAc is usually chosen as a "typical" weak acid in calculations in texts. Corrections for nonideality, however, do show differences for HAc, especially at the equivalence point where the nonideal exact pH is 8.57 compared with the approximate (or ideal-exact) value of 8.73, a difference of 0.16 pH units. This is an error in hydrogen ion wncentration or activity of 45%. The very, very weak HI0 shows yet another pattern. HI0 is such a weak acid that its conjugate base I O i s a relatively strong base with KA = 0.020. Thus. there is moderate OHconthhution from ;he conjugate base even when 50% past the eauivalence mint as e\idc:nced bv the (ideal,calcula~ed pH =-12.54 cokpared with the ap&ox&ate pH = 12.30 calculation which assumes no OH- contribution from the conjugate base. For even weaker acids, eq 1and eq 2 (the HendersonHasselbalch equation) will begin to give error. As the acid becomes weaker, the solution begins to take on the properties of water, and the pH begins to approach 7.00. Other Buffer-Region Approximations When 50%to an equivalence point (and thus in the center of the buffer region) an equation of the form pH = pK, (eq 2, 6, or 8) is a suitable approximation for a typical weak acid. It is also possible to estimate the pH a t other positions in the buffer region based on the obsemation that the plot of pH vs. mL of titrant shows a fairly linear increase at 15585% of the equivalence point. The HendersonHasselbalch equation (eq 4 of article VI') can be used to calculate the change for every 10%increment. It amounts to about 0.2 pH units and is calculated as follows. When.%%to the equivalence point Cc&. = xI(100-1)
so that pH = pK.
+ log [xl(100- d l
Whenx = 60%. the logterm = 0.18 = 0.2. Whenx = 70%,the log term = 0.37 = 0.4. Whenx = 80%,the log term = 0.60. Values of x corresponding to incremental decreases of 10%beginning at 50%give the negatives of the values just calculated for the logs. Thus, on the average, there is a change of about 0.2 pH units for every incremental change of 10%from the midpoint 50% pH value which is pK, For instance, a solutionthat is 0.070MHAand also 0.030 M NaA is the same as that which would result when 0.030/(0.030 + 0.070) = 0.30 = 30% to the equivalence point in the titration of the acid. This is two 10%increments less than the 50% point in the titration. If K. = 3.5 x 1P(pK. = 5.46), then the predicted pH of the solution would be about 0.4 pH units less than pK, or 5.5 - 0.4 = 5.1. (The calculated pH = 5.09.)
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