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Chapter 9 Titrimetric Methods of Analysis 295

for H 3 BO 3 versus –55.6 kJ/mol for HCl), resulting in a favorable thermometric
titration curve (Figure 9.15b).

9B.3 Titrations in Nonaqueous Solvents

Thus far we have assumed that the acid and base are in an aqueous solution. In-
deed, water is the most common solvent in acid–base titrimetry. When considering
the utility of a titration, however, the solvent’s influence cannot be ignored.
The dissociation, or autoprotolysis constant for a solvent, SH, relates the con-
–
+
centration of the protonated solvent, SH 2 , to that of the deprotonated solvent, S .
For amphoteric solvents, which can act as both proton donors and proton accep-
tors, the autoprotolysis reaction is
+
2SH t SH 2 +S –
with an equilibrium constant of
–
+
K s = [SH 2 ][S ]
You should recognize that K w is just the specific form of K s for water. The pH of a
solution is now seen to be a general statement about the relative abundance of pro-
tonated solvent
+
pH = –log[SH 2 ]
where the pH of a neutral solvent is given as
1
pH neut = p s K
2
Perhaps the most obvious limitation imposed by K s is the change in pH during
a titration. To see why this is so, let’s consider the titration of a 50 mL solution of
10 –4 M strong acid with equimolar strong base. Before the equivalence point, the
pH is determined by the untitrated strong acid, whereas after the equivalence point
the concentration of excess strong base determines the pH. In an aqueous solution
+
the concentration of H 3 O when the titration is 90% complete is
MV b b
aa -M V
[HO + ] =
3
V a +V b
( ´10 -4 M)(50 mL) - (1 ´10 -4 M)(45 mL) -6
1
= = 53 M
. ´10
50 +45
corresponding to a pH of 5.3. When the titration is 110% complete, the concentra-
–
tion of OH is
bb -M V
MV a a
[OH - ] =
V a +V b
1
( ´10 - 4 M)(55 mL) - (1 ´10 - 4 M)(50 mL)
. ´10
= = 48 - 6 M
50 +55
or a pOH of 5.3. The pH, therefore, is
pH=pK w – pOH = 14.0 – 5.3 = 8.7
The change in pH when the titration passes from 90% to 110% completion is
DpH = 8.7 – 5.3 = 3.4