Page 324 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:12 PM Page 307

Chapter 9 Titrimetric Methods of Analysis 307

The mass of protein, therefore, is
.
638 g protein
.
.
´ 0 03549 g N = 0 2264 g protein
g N
and the %w/w protein is
.
g protein 0 2264 g
´100 = ´100 =23 1.% w/w protein
g sample 0.9814 g

Earlier we noted that an acid–base titration may be used to analyze a mixture of
acids or bases by titrating to more than one equivalence point. The concentration of
each analyte is determined by accounting for its contribution to the volume of
titrant needed to reach the equivalence points.

EXAMPLE .5
9
–
2–
The alkalinity of natural waters is usually controlled by OH , CO 3 , and
–
HCO 3 , which may be present singularly or in combination. Titrating a
100.0-mL sample to a pH of 8.3 requires 18.67 mL of a 0.02812 M solution of
HCl. A second 100.0-mL aliquot requires 48.12 mL of the same titrant to
reach a pH of 4.5. Identify the sources of alkalinity and their concentrations
in parts per million.
SOLUTION
Since the volume of titrant needed to reach a pH of 4.5 is more than twice that
needed to reach a pH of 8.3, we know, from Table 9.8, that the alkalinity of the
2–
2–
sample is controlled by CO 3 and HCO 3 .
2–
–
Titrating to a pH of 8.3 neutralizes CO 3 to HCO 3 , but does not lead to a
–
reaction of the titrant with HCO 3 (see Figure 9.14b). Thus
2–
Moles HCl to pH 8.3 = moles CO 3
or
g CO 3 2-
M a ´ V a,pH 8.3 = 2-
FW CO 3
Solving for the grams of carbonate gives
2– 2–
g CO 3 = M a ´V a, pH 8.3 ´FW CO 3
2–
0.02812 M ´0.01867 L ´60.01 g/mol = 0.03151 g CO 3
2–
The concentration of CO 3 , therefore, is
mg CO 3 2- 31 51 mg 2-
.
.
= =315 1 ppm CO 3
.
liter 0 1000 L
2–
Titrating to the second end point at pH 4.5 neutralizes CO 3 to H 2 CO 3 , and
–
HCO 3 to H 2 CO 3 (see Figures 9.18b,c). The conservation of protons, therefore,
requires that
2– –
Moles HCl to pH 4.5 = 2 ´moles CO 3 + moles HCO 3
or