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146 Solutions Manual for Analytical Chemistry 2.1
n M V M V n2Sn Sn Sn Ce CeCe#= = =
 where n is the moles of Sn2+ or of Ce4+; thus 
( )
V V M
M V2
0
2 0 25
50
(0. 100 M)
(0. 100 M)( .0 mL)
.0 mL
. .eq pt Ce
Ce
Sn Sn#
= = =
=
 Before the equivalence point, the potential is easiest to calculate by 
using the Nernst equation for the analyte’s half-reaction
( ) ( )eaq aq2Sn Sn42
?+
+ - +
.
. .
log
log
E E
E
0 05916
0 154 0 05916
2
2
[ ]
[ ]
[Sn ]
[Sn ]
Sn
Sn
/Sn
o
4
2
4
2
Sn4 2= -
=+ -
+
+
+
+
+ +
 For example, after adding 10.0 mL of titrant, the concentrations of 
Sn2+ and of Sn4+ are
]
.
V V
M V M V0 5
[Sn2
Sn Ce
Sn Sn Ce Ce#
=
+
-+
( . )0 5
[Sn ] 25.0 mL 10.0 mL
(0.0100 M)(25.0 mL) (0.0100 M)(10.0 mL)2
=
+
-+
.5 71[Sn ] 10 M2 3
#=
+ -
.
( . )
.
V V
M V0 5
0 5
101 43
[Sn ]
25.0 mL 10.0 mL
(0.0100 M)(10.0 mL)
M
4
3
Ce
Ce
Sn
Ce#
#
=
+
=
+
=
+
-
 which gives us a potential of
. .
.
. .logE 0 154 2
0 05916
1 43 10
5 71 10 0 136 V3
3
#
#
= - =-
-
 After the equivalence point, the potential is easiest to calculate by 
using the Nernst equation for the titrant’s half-reaction
( ) ( )eaq aqCeCe4 3
?+
+ - +
.
. .
log
log
E E
E
0 05916
1 72 0 05916
[Ce ]
[Ce ]
[Ce ]
[Ce ]
Ce /Ce
o
4
3
4
3
4 3= -
=+ -
+
+
+
+
+ +
 For example, after adding 60.0 mL of titrant, the concentrations of 
Ce3+ and of Ce4+ are
] V V
M V M V2
[Ce4
Ce Sn
Ce Ce SnSn#
=
+
-+
( )
25
60 2 25
60
[Ce ]
.0 mL .0 mL
(0.0100 M)( .0 mL) (0.0100 M)( .0 mL)4
=
+
-+