Page 326 - Modern Analytical Chemistry
P. 326
1400-CH09 9/9/99 2:13 PM Page 309
Chapter 9 Titrimetric Methods of Analysis 309
9B.7 Characterization Applications
Two useful characterization applications involving acid–base titrimetry are the de-
termination of equivalent weight, and the determination of acid–base dissociation
constants.
Equivalent Weights Acid–base titrations can be used to characterize the chemical
and physical properties of matter. One simple example is the determination of the
equivalent weight* of acids and bases. In this method, an accurately weighed sample
of a pure acid or base is titrated to a well-defined equivalence point using a mono-
protic strong acid or strong base. If we assume that the titration involves the trans-
fer of n protons, then the moles of titrant needed to reach the equivalence point is
given as
Moles titrant = n ´moles analyte
and the formula weight is
g analyte g analyte
FW = =n ´
moles analyte moles titrant
Since the actual number of protons transferred between the analyte and titrant is
uncertain, we define the analyte’s equivalent weight (EW) as the apparent formula
weight when n = 1. The true formula weight, therefore, is an integer multiple of the
calculated equivalent weight.
FW = n ´EW
Thus, if we titrate a monoprotic weak acid with a strong base, the EW and FW are
identical. If the weak acid is diprotic, however, and we titrate to its second equiva-
lence point, the FW will be twice as large as the EW.
9
EXAMPLE .6
A 0.2521-g sample of an unknown weak acid is titrated with a 0.1005 M
solution of NaOH, requiring 42.68 mL to reach the phenolphthalein end point.
Determine the compound’s equivalent weight. Which of the following
compounds is most likely to be the unknown weak acid?
ascorbic acid C 6 H 8 O 6 FW = 176.1 monoprotic
malonic acid C 3 H 4 O 4 FW = 104.1 diprotic
succinic acid C 4 H 6 O 4 FW = 118.1 diprotic
citric acid C 6 H 8 O 7 FW = 192.1 triprotic
SOLUTION
The moles of NaOH needed to reach the end point is
–3
M b ´V b = 0.1005 M ´0.04268 L = 4.289 ´10 mol NaOH
giving an equivalent weight of
.
g analyte 0 2521 g
EW = = = 58 78 g/mol
.
.
moles titrant 4 289 ´10 -3 mol
*See Section 2B.2 for a review of chemical equivalents and equivalent weights.