Page 368 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 351

Chapter 9 Titrimetric Methods of Analysis 351

By now you are familiar with our approach to calculating titration curves. The
+
first task is to calculate the volume of Ag needed to reach the equivalence point.
The stoichiometry of the reaction requires that
+
Moles Ag = moles Cl –
or
M Ag V Ag = M Cl V Cl
Solving for the volume of Ag +
0
Cl
MV Cl (.0500 M)(50.0 mL)
V Ag = = = .0 mL
25
M Ag (0.100 M)
+
shows that we need 25.0 mL of Ag to reach the equivalence point.
–
Before the equivalence point Cl is in excess. The concentration of unreacted
–
+
Cl after adding 10.0 mL of Ag , for example, is
Cl Cl -M V
moles excess Cl - MV Ag Ag
[Cl - ] = =
total volume V Cl +V Ag
(.0500 M)(50.0 mL) - (.100 M)(10.0 mL)
0
0
=
50.0 mL + 10.0 mL
= . 250 ´10 - 2 M
–
If the titration curve follows the change in concentration for Cl , then we calculate
pCl as
–2
–
pCl = –log[Cl ] = –log(2.50 ´10 ) = 1.60
+
However, if we wish to follow the change in concentration for Ag then we must
first calculate its concentration. To do so we use the K sp expression for AgCl
+
–
K sp = [Ag ][Cl ] = 1.8 ´10 –10
Solving for the concentration of Ag +
K sp 18 - 10
. ´10
. ´ 10
[Ag + ] = - = - 2 = 72 - 9 M
[Cl ] . 250 ´10
gives a pAg of 8.14.
+
–
At the equivalence point, we know that the concentrations of Ag and Cl are
equal. Using the solubility product expression
+ 2
+
–
K sp = [Ag ][Cl ] = [Ag ] = 1.8 ´10 –10
gives
+
–5
–
[Ag ] = [Cl ] = 1.3 ´10 M
At the equivalence point, therefore, pAg and pCl are both 4.89.
+
After the equivalence point, the titration mixture contains excess Ag . The con-
+
centration of Ag after adding 35.0 mL of titrant is
Ag Ag -M V
moles excess Ag + MV Cl Cl
[Ag + ] = =
total volume V Cl +V Ag
0
(.100 M)(35.0 mL) - (.0500 M)(50.0 mL)
0
=
50.0 mL + 35.0 mL
= . 118 ´10 -2 M