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153Chapter 9 Titrimetric Methods
10 mol S O
2 mol S O
1 mol I
mol I
mol I
3 mol I
1 mol
2 mol MnO
1 mol O 1. 10 mol O
7.74 3
MnO 94
5
2 3
2
2 3
2
3
3
2
2 5
2
2
# # #
# # #=
- -
-
-
-
-
-
-
1.94 10 mol O
mol O
31.998 g O
6.21 10 g O5
2
2
2 4
2# # #=
- -
 he concentration of O2 in the original sample, therefore, is
100.0 mL
6.21 10 g O g
10 µg
6.21 ppm O
4
2
6
2
# #
=
-
55. he titration of KI with AgNO3 is an example of a precipitation 
titration. he titration’s equivalence point is reached when
n M V M V nI I Ag AgI Ag= = =
 where n is the moles of I– or of Ag+; thus 
V V M
M V
05
025 50
25
(0. 00 M)
(0. 0 M)( .0 mL)
.0 mL. .eq pt
Ag
I
Ag
I
= = = =
 Before the equivalence point, the concentration of I– is determined by 
the amount of excess I–, and the concentration of Ag+ is determined 
by the solubility of AgI in the presence of excess I–. For example, after 
adding 10.0 mL of AgNO3, we ind that
] V V
M V M V
[I
I Ag
I I Ag Ag
=
+
--
[I ] 50.0 mL 10.0 mL
(0.0250 M)(50.0 mL) (0.0500 M)(10.0 ml)
=
+
--
[I ] 0.0125 M=
-
] .
. .
K
0 0125
8 32 10 6 66 10[Ag ]
[I
M
17
15sp,AgI #
#= = =
+
-
-
-
 which gives pI as 1.90 and pAg as 14.18. After the equivalence point, 
the concentration of Ag+ is determined by the amount of excess Ag+, 
and the concentration of I– is determined by the solubility of AgI in 
the presence of excess Ag+. For example, after adding 35.0 mL of 
AgNO3, we ind that
] V V
M V M V
[Ag
Ag I
Ag Ag I I
=
+
-+
[ ] 35.0 mL 50.0 mL
(0.0500 M)(35.0 mL) (0.0250 M)(50.0 ml)
Ag =
+
-+
[ ] MAg 5.88 10 3
#=
+ -
] . .
.
K 8 32 10 1 41 10
5 88 10
[I
[Ag ]
M
17
14
3
sp,AgI #
#
#
= = =
-
+
-
-
-
For the titration curves in this problem 
and in the next problem, we will calcu-
late pAnalyte or pTitrant for one volume 
before each equivalence point and for one 
volume after the inal equivalence point.