Page 372 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:14 PM Page 355

Chapter 9 Titrimetric Methods of Analysis 355

9
Table .22 Representative Examples
of Precipitation Titrations
Analyte Titrant a End Point b

3- AgNO 3 , KSCN Volhard
AsO 4
Br – AgNO 3 Mohr or Fajans
AgNO 3 , KSCN Volhard
Cl – AgNO 3 Mohr or Fajans
AgNO 3 , KSCN Volhard *
2- AgNO 3 , KSCN Volhard *
CO 3
2- AgNO 3 , KSCN Volhard *
C 2 O 4
2- AgNO 3 , KSCN Volhard *
CrO 4
I – AgNO 3 Fajans
AgNO 3 , KSCN Volhard
3- *
PO 4 AgNO 3 , KSCN Volhard
S 2- AgNO 3 , KSCN Volhard *
SCN – AgNO 3 , KSCN Volhard
a When two reagents are listed, the analysis is by a back titration. The first
reagent is added in excess, and the second reagent is used to back titrate
the excess.
b For Volhard methods identified by an asterisk (*) the precipitated silver
salt must be removed before carrying out the back titration.

+
titrant. Those titrations in which Ag is the titrant are called argentometric titra- argentometric titration
+
tions. Table 9.22 provides a list of several typical precipitation titrations. A precipitation titration in which Ag is
the titrant.
Quantitative Calculations The stoichiometry of a precipitation reaction is given by
the conservation of charge between the titrant and analyte (see Section 2C); thus

Moles of charge moles of charge
´ moles titrant = ´ moles analyte
mole titrant mole analyte
Example 9.17 shows how this equation is applied to an analysis based on a direct
titration.

9 7
EXAMPLE .1
A mixture containing only KCl and NaBr is analyzed by the Mohr method. A
0.3172-g sample is dissolved in 50 mL of water and titrated to the Ag 2 CrO 4
end point, requiring 36.85 mL of 0.1120 M AgNO 3 . A blank titration requires
0.71 mL of titrant to reach the same end point. Report the %w/w KCl and
NaBr in the sample.
SOLUTION

The volume of titrant reacting with the analytes is
V Ag = 36.85 mL – 0.71 mL = 36.14 mL
Conservation of charge for the titration requires that
+
Moles Ag = moles KCl + moles NaBr