Page 325 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:12 PM Page 308

308 Modern Analytical Chemistry

2 ´ g CO 2- g HCO 3 -
3
M a ´ V a, pH 4.5 = +
FW CO 2- FW HCO -
3 3
Solving for the grams of bicarbonate gives
æ 2 ´ g CO 2- ö
3
g HCO - = M a ´ V a, pH 4.5 - 2- ÷ ´ FW HCO 3 -
ç
3
è FW CO 3 ø
æ 2 ´ 0 03151 gö 2-
.
.
.
.
.
ç 0 02812 M ´ 0 04812 L - ÷ ´ 61 02 g/mol = 0 01849 g HCO 3
è 60.01 g/mol ø
–
The concentration of HCO 3 , therefore, is
.
mg HCO 3 - = 18 49 mg =184 9 ppm HCO -
.
liter 0.1000 L 3
9B.6 Qualitative Applications
We have already come across one example of the qualitative application of
acid–base titrimetry in assigning the forms of alkalinity in waters (see Example 9.5).
This approach is easily extended to other systems. For example, the composition of
solutions containing one or two of the following species
– 2– 3–
H 3 PO 4 H 2 PO 4 HPO 4 PO 4 NaOH HCl
can be determined by titrating with either a strong acid or a strong base to the
methyl orange and phenolphthalein end points. As outlined in Table 9.11, each
species or mixture of species has a unique relationship between the volumes of
titrant needed to reach these two end points.

9
Table .11 Relationship Between End Point Volumes for Solutions of Phosphate Species
with HCl and NaOH

Solution Relationship Between End Point Relationship Between End Point
Composition Volumes with Strong Base Titrant a Volumes with Strong Acid Titrant a

H 3 PO 4 V PH =2 ´V MO — b
– V PH > 0; V MO =0 —
H 2 PO 4
2– — V MO > 0; V PH =0
HPO 4
3– —
PO 4 V MO =2 ´V PH
HCl V PH = V MO —
NaOH — V MO = V PH
HCl and H 3 PO 4 V PH < 2 ´V MO —
– —
H 3 PO 4 and H 2 PO 4 V PH > 2 ´V MO
– 2– V PH > 0; V MO =0 V MO > 0; V PH =0
H 2 PO 4 and HPO 4
2– 3– — V
HPO 4 and PO 4 V MO > 2 ´ PH
3– and NaOH —
PO 4 V MO < 2 ´V PH
a V MO and V PH are, respectively, the volume of titrant needed to reach the methyl orange and phenolphthalein end points.
b When no information is given, the volume of titrant needed to reach either end point is zero.