Corrections for Simple Equations for
W. P. Cortelyou
Towson State College Baltimore, Maryland 21204
Titration Curves of Monoprotic Acids
The [H+]a t any point in the titration of a weak monoprotic acid up to and including the equivalence point may be calculated quite precisely by the Charlot equation,' first introduced to American readers by DeFord2
where C, is the stoichiometric molarity of the acid remaining a t that point and Cb is the stoichiometric molarity of the conjugate base (or salt) generated up to that point. The [H+] in the weak acid, before titration begins, is usually approximated by [H +I
=
(K,. CJ/.
(2)
After the titration starts the [H+] may be approximated by [H+l = K-(1
- f)/f
where f is the fraction titrated. At the equivalence point, where f but precise equation is IHc] = (K,)'/a[K.I(Cb
(3) =
1.000, a simple
+ Kc)]'/>
(4)
which follows from eqn. (1) when C, = 0 and Cbis much greater than the rest of the denominator. Beyond the equivalence point, where fexceeds 1.000,
Presented May 5, 1967, at the Joint Meeting of the Maryland and Washington Sections of the American Chemical Society. CHARLOT,G., Anal. Chirn. Ada, 1, 59, (1947). 27, 554, (1950). DEFORD,D. D., J. CHEM.EDUC.,
eqn. (1) no longer applies. The precise equation that does apply is
The derivation of this equation is given a t the end of this article. A good approximation results if the final ratio, which corrects for hydrolysis, is omitted. Then
Assuming K , = 1.00 X 10-14, eqns. (1) and (5) were used to calculate precise values for [H+]and pH a t nine f values on 48 titration curves. Then the simplified equations were used to calculate approximate TJHvalues. The corrections that must be added to these approximate values to give precise values are shown in Tables 1 and 2, which refer to 0.50 N and 0.01 N, the maximum and minimum concentrations used in this study. (Corrections are available a t 0.20 N, 0.10 N, 0.05 N, and 0.02 N.) Note the following principles, illustrated by these tables. (1) The largest corrections are needed for dilute solutions and for extreme pK, values. (2) For the higher half of the pK, values the corrections before the equivalence point are the same as those after the equivalence point, but they carry the opposite sign. The significance of this will be discussed later. (3) The maximum correction for low pK, values is a t half titration (f = 0.5). (4) No correction is required for eqn. (4),a t the equivalence point. (5) When pK. is 3.00 or less no correction is needed when thc simple formula is used to find [H+] after the equivalence point. The significance of these equations and the tables
Volume 45, Number 10, October 1968
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677
Table 1.
Corrections for pH when N. = Na =
0.500
Table 2.
Corrections for pH when N, = Na =
0.010
Eon.
f
1.00
2.00
9% 3.00 4.00 5.00
6.00 Corre~tiona
7.00
-
8.00
Correct DH near the Equivalence Point, when
Table 3.
ever a crude titration (within 10%) is feasible, since the p H jumps from 7.95 to 10.68 between if = 0.9) and (f = 1.1) If the complete set of tables is used, with interpolation, we can answer most of the questions on feasibility of titration that arise in practical situations. Also, the feasibility of titration can be determined, conservatively, from the simple equations. Because of Principle 2, demonstrated in Tables 1 and 2, when pK. exceeds 4.00, if the p H jump across the equivalence point is calculated by the simple equations i t will always be less than or equal t o the pH jump calculated by the precise equations. If a titration is feasible by the simple equations we can he sure it is feasible by the precise equations (when pK. exceeds 4.00). This is not true when pK. is less than 3.00 but that fact causes no difficulty. Such acids are so highly ionized that, typically, they are easy to titrate. Furthermore i t is easy to calculate very nearly the correct pH in this region. We use eqn. (6), which requires no correction is this region, to find the appropriate p H after the equivalence point. Then we find the pH a t a n equal distance before the equivalence point, by symmetry. For instance, (pH st 0.99) - (pH at eq.pt.) = (pH at eq.pt.) - (pH at 1.01)
5 6.95 0 8.00 0 8.97 0 9.70 0 10.42 o 11.40 12.38
7.95 8.99 9.89 10.20 10.51 11.40 12.38
8.95 9.98 10.59 10.70 10.81 11.41 12.38
Correct pH near the Equivalence Point, when N. = Na= 0.010
Table 4.
1 . 0 0 2.00
f
PE
0,900 0.980 0.999 1.000 1.001 1.010 1.100
1.00 2.00 3.00
...
3.00 2.00 1.00
3.30 4.32 5.32 7.01 8.70 9.70 10.68
3.46 4.48 5.48 7.09 8.70 9.70 10.68
PK-
3.00
4.00 5.00 6.00 Correct PH
7.00
8.00
4.04 5.08 6.08 7.39 8.70 9.70 10.68
4.96 5.95 6.00 7.00 7.00 7.93 7.85 8.35 8.71 8.77 9.70 9.70 10.68 10.68
7.95 8.93 9.30 9.35 9.40 9.77 10.68
8.95 8.88 9.83 9.85 9.87 10.00 10.69
6.95 8.00 8.70 8.85 9.00 9.71 10.68
SO
(pH at 0.99) = 2(pH st eq.pt.1 - (pH at 1.01)
and the pH jump across the equivalence point is (pH a t 1.01) - (pH a t 0.99) or 2[(pH a t 1.01) - (pH a t eq.pt)I. Of course all of this discussion assumes that there is no need to make activity corrections. The magnitude of such corrections is the next step in this study. Derivation of Equation (5)
At the equivalence point [OH-] exceeds lo-' because of hydrolysis of the conjugate base. Beyond the equivalence point this effect decreases but [OH-] continues to increase because of the excess NaOH added, [OH1.. = Cdf
may be clearer if the correct p H of the solution is shown as a function of the titration error before and after the equivalence point. So the situations covered in Tables 1 and 2 are treated again in Tables 3 and 4 from this point of view. I n these tables, E, the error of under-titration or overtitration is (1 - f) before the equivalence point and (f - 1) after the equivalence point. pE is, of course, -log(l - f) or -log(f - 1). These tables can quite easily he used to determine the feasibility of a titration. For instance: Can we titrate a 0.01 N solution of an acid for which pK. = 7.00, using 0.01 N NaOH? Table 4 provides the answer. Between if = 0.999) and if = 1.001) the pH changes only from 9.30 to 9.40, so we cannot do thc titration if we wish to do it within one part per thousand. How-
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Journol of Chemical Education
- l M 1 + ( f - 1)/2)
(7)
where Ch is numerically equal to the molarity of the NaOH added up to the equivalence point, (f - 1) is the over-titration fraction, and the denominator corrects for dilution. [OH-lhld [OH-]
=
[HA]hyd = (Kw/Ko)(lA-I/[OH-I)
=
(lH+I/K.)(Ca/(l
=
[OH-],,
" l + ( f - 1)/2
= -
Cdf - 1 (f
+
+ (f - 11/21)
+ [OH-Iw [(f
(8) (9)
- 1) + IH+I/Kal
(10)
- 1) ,Kdf - 1) + lH+l - 1)/2 K d f - 1)
[H+]t,t = &/[OH-lm =
K, -. 1 Ca
+ (f - 1)/2, (f - 1)
K J f - 1) Ks(f - 1) lH+l
+
(5)
