Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
Chapter 19
19-1
The electrode potential of a system that contains two or more redox couples is the
electrode potential of all half-cell processes at equilibrium in the system.
19-2
(a) Equilibrium is the state that a system assumes after each addition of reagent.
Equivalence refers to a particular equilibrium state when a stoichiometric amount of
titrant has been added.
(b) A true oxidation/reduction indicator owes its color change to changes in the electrode
potential of the system. A specific indicator exhibits its color change as a result of
reactions with a particular solute species.
19-3
The electrode potentials for all half-cell processes in an oxidation/reduction system have
the same numerical value when the system is at equilibrium.
19-4
For points before equivalence, potential data are computed from the analyte standard
potential and the analytical concentrations of the analyte and its reaction product. Postequivalence point data are based upon the standard potential for the titrant and its
analytical concentrations. The equivalence point potential is computed from the two
standard potentials and the stoichiometric relation between the analyte and titrant.
19-5
In contrast to all other points on the titration curve, the concentrations of all of the
participants in one of the half-reactions or the other cannot be derived from
stoichiometric calculations.
19-6
An asymmetric titration curve will be encountered whenever the titrant and the analyte in
a ratio that is not 1:1.
Fundamentals of Analytical Chemistry: 8th ed.
19-7
Chapter 19
(a)
E right  0.403 
0.0592  1 
log 
  0.441 V
2
 0.0511 
0.0592  1 
log 
  0.151 V
2
 0.1393 
 E right  Eleft  0.441  ( 0.151)   0.290 V
Eleft  0.126 
Ecell
Because Ecell is negative, the reaction will not proceed spontaneously in the direction
considered and an external voltage source is needed to force this reaction to occur.
(b)
E right  1.25 
0.0592  0.0620 
log 
 1.23 V
3 
2
 9.06  10 
0.0592  1 
log 
  0.806 V
2
 0.0364 
 E right  Eleft  1.23  ( 0.806)  2.04 V
Eleft  0.763 
Ecell
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered.
(c)
E right  0.250 
Eleft  0.000 
0.0592  1 
log 
  0.299 V
2
 0.0214 
0.0592  765 / 760 
log
 0.237 V
 1.00  10 4 2 
2


Ecell  E right  Eleft  0.299  ( 0.237)   0.0620 V
Because Ecell is negative, the reaction will not proceed spontaneously in the direction
considered and an external voltage source is needed to force this reaction to occur.
(d)
E right  0.854 
0.0592 
1

log 
 0.785 V
3 
2
 4.59  10 
Fundamentals of Analytical Chemistry: 8th ed.
[ Pb2 ][I  ]2  7.9  10 9 and [ Pb2 ] 
Eleft  0.126 
Chapter 19
7.9  10 9
[I  ]2
2
0.0592  0.0120 
  0.252 V
log 
9 
2
7
.
9

10


Ecell  E right  Eleft  0.785  ( 0.252)  1.04 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered.
(e)
[ H 3O  ][ NH 3 ]
[ H 3O  ]0.438
10
 5.70  10 

0.379
[ NH 4 ]
[ H 3O  ] 
5.70  10 0.379  4.93  10
10
10
0.438
E right  0.000 V
Eleft  0.000 

0.0592 
1.00
  0.551 V
log
2
 4.93  10 10  
2


Ecell  E right  Eleft  0.000  ( 0.551)  0.551 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered.
(f)


0.0784
  0.223 V
E right  0.359  0.0592 log 
2 
 0.13400.0538 


0.00918
  0.063 V
Eleft  0.099  0.0592 log 
2
 0.07901.47  10 2  


Ecell  E right  Eleft  0.223  ( 0.063)  0.286 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered.
Fundamentals of Analytical Chemistry: 8th ed.
19-8
Chapter 19
(a)
E right  0.277 
0.0592 
1

log 
 0.341 V
3 
2
 6.78  10 
0.0592  1 
log 
  0.793 V
2
 0.0955 
 E right  Eleft  0.341  ( 0.793)  0.452 V
Eleft  0.763 
Ecell
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered
(oxidation on the left, reduction on the right).
(b)
E right  0.854 
Eleft  0.771 
0.0592  1 
log 
  0.819 V
2
 0.0671 
0.0592  0.0681 
log 
  0.788 V
2
 0.1310 
Ecell  Eright  Eleft  0.819  0.788  0.031 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered
(oxidation on the left, reduction on the right).
(c)
E right  1.229 

0.0592 
1
  1.165 V
log 
4 
4
 1.120.0794  
 1 
Eleft  0.799  0.0592 log 
  0.751 V
 0.1544 
Ecell  E right  Eleft  1.165  0.751  0.414 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered
(oxidation on the left, reduction on the right).
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
(d)
E right  0.151  0.0592 log 0.1350  0.100 V
0.0592  1 
log 
  0.301 V
2
 0.0601 
 E right  Eleft  0.100  0.301   0.401 V
Eleft  0.337 
Ecell
Because Ecell is negative, the reaction does not proceed spontaneously in the direction
considered (reduction on the left, oxidation on the right).
(e)
[ H 3O  ][ HCOO  ]
[ H 3O  ]0.0764
4
 1.80  10 
[ HCOOH ]
0.1302
[ H 3O

1.80  10 0.1302  3.07  10
]
E right  0.000 
4
4
0.0764

0.0592 
1.00
  0.208 V
log
 3.07  10 4 2 
2


Eleft  0.000 V
Ecell  Eright  Eleft  0.208  0.000   0.208 V
Because Ecell is negative, the reaction does not proceed spontaneously in the direction
considered (reduction on the left, oxidation on the right).
(f)
 0.1134 
E right  0.771  0.0592 log 
  0.684 V
 0.003876 
Eleft

0.0592 
6.37  10 2
  0.040 V
 0.334 
log
 7.93  10 3 1.16  10 3 4 
2


Ecell  E right  Eleft  0.684  ( 0.040)  0.724 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered
(oxidation on the left, reduction on the right).
Fundamentals of Analytical Chemistry: 8th ed.
19-9
Chapter 19
(a)
E P b2  0.126 
0.0592  1 
log 
  0.158 V
2
 0.0848 
0.0592  1 
log 
  0.789 V
2
 0.1364 
 E right  Eleft  0.158  ( 0.789)  0.631 V
E Zn2  0.763 
Ecell
(b)
 0.0760 
E Fe3  0.771  0.0592 log 
  0.747 V
 0.0301 
 0.00309 
E Fe( CN ) 3  0.36  0.0592 log 
  0.461 V
6
 0.1564 
Ecell  E right  Eleft  0.747  0.461  0.286 V
(c)
ESHE  0.000 V


0.02723
  0.331 V
ET iO2  0.099  0.0592 log 
2
 1.46  10 3 10 3  


Ecell  E right  Eleft  0.000  ( 0.331)  0.331 V
19-10 (a) ZnZn2+(0.1364 M)Pb2+(0.0848 M)Pb
(b) PtFe(CN)64-(0.00309 M), Fe(CN)63-(0.1564 M)Fe3+(0.0301 M), Fe2+(0.0760 M)Pt
(c) PtTiO+(1.4610-3M), Ti3+(0.02723 M), H+(1.0010-3M)SHE
19-11 Note that in these calculations, it is necessary to round the answers to either one or two
significant figures because the final step involves taking the antilogarithm of a large
number.
(a)
Fe 3  V 2


Fe 2  V 3
o
E Fe
E Vo 3  0.256
3  0.771
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
 [ Fe 2 ] 
 [V 2 ] 

 3 
0.771  0.0592 log 


0
.
256

0
.
0592
log
3 
 [ Fe ] 
 [V ] 
 [ Fe 2 ][ V 3 ] 
0.771   0.256
  log K eq  17.348
 log 
3
2 
0.0592
 [ Fe ][ V ] 
[ Fe 2 ][ V 3 ]
 K eq  2.23  1017  2.2  1017
[ Fe 3 ][ V 2 ]
(b)
3
Fe(CN ) 6 Cr 2


4
Fe(CN ) 6 Cr 3
o
E Fe
 0.36 ECro 3  0.408
( CN ) 3
6
 [ Fe(CN ) 6 4 ] 
 [Cr 2 ] 



0.36  0.0592 log 


0
.
408

0
.
0592
log
3 
3 
 [Cr ] 
 [ Fe(CN ) 6 ] 
 [ Fe(CN ) 6 ][Cr 3 ] 
0.36   0.408
  log K eq  12.973
 log 
3
2 
0.0592
 [ Fe(CN ) 6 ][Cr ] 
4
4
[ Fe(CN ) 6 ][Cr 3 ]
 K eq  9.4  1012  9  1012
3
[ Fe(CN ) 6 ][Cr 2 ]
(c)

2V(OH ) 4  U 4
1.00 


2VO 2  UO 2
2
 4H 2O
o
E Vo ( OH )   1.00 E UO
2   0.334
4
2


0.0592 
[VO 2 ]2
0.0592 
[ U 4 ]



log 

0
.
334

log
 2
2

 4 

4
 [ UO ][ H ] 
2
2
[
V
(
OH
)
]
[
H
]
4
2




1.00  0.334 2  log  [VO 2 ]2 [ UO 2 2 ]   log K
 [V(OH )  ]2 [ U 4 ] 
4


0.0592
eq
 22.50
2
[VO 2 ]2 [ UO 2 ]
 K eq  3.2  10 22  3  10 22
 2
4
[V(OH ) 4 ] [ U ]
(d)
Tl3  2 Fe 2


Tl  2Fe 3
o
E Fe
ETo l  1.25
3  0.771
Fundamentals of Analytical Chemistry: 8th ed.
1.25 
Chapter 19
0.0592  [Tl ] 
0.0592  [ Fe 2 ]2 

log  3   0.771 
log 
3 2 
2
2
 [Tl ] 
 [ Fe ] 
1.25  0.771 2  log  [Tl ][ Fe3 ]2

 [Tl3 ][ Fe 2 ]2   log K eq  16.18


0.0592
[Tl ][ Fe 3 ]2
 K eq  1.5  1016  2  1016
[Tl3 ][ Fe 2 ]2
(e)
2Ce 4  H 3 AsO 3  H 2 O 
2Ce 3  H 3 AsO 4  2 H 

o
ECe
4  ( in 1 M HClO 4 )  1.70
1.70 
E Ho 3AsO4  0.577
0.0592  [Ce 3 ]2 
0.0592  [ H 3 AsO 4 ] 

  0.577 
log 
log 
4 2 
 2 
2
2
[
H
AsO
][
H
]
 [Ce ] 
3
3


1.70  0.577 2  log  [Ce 3 ]2 [ H 3AsO 3 ][ H  ]2   log K
 [Ce 4 ]2 [ H AsO ] 
3
4


3 2
 2
[Ce ] [ H 3 AsO 3 ][ H ]
 K eq  8.9  1037  9  1037
4 2
[Ce ] [ H 3 AsO 4 ]
0.0592
eq
 37.94
(f)

2V(OH ) 4  H 2SO3
1.00 


2VO 2  SO 4
2
 5H 2 O
o
E Vo ( OH )   1.00 ESO
2   0.172
4
4

0.0592 
[VO 2 ]2
0.0592  [ H 2SO 3 ] 
  0.172 

log 
log 
 2
2
 4 
 4 
2
2
 [V(OH ) 4 ] [ H ] 
 [SO 4 ][ H ] 
1.00  0.172 2  log 
2
[VO 2 ]2 [SO 4 ] 

 [V(OH )  ]2 [ H SO ]   log K eq  27.97
4
2
3 

0.0592
2
[VO 2 ]2 [SO 4 ]
 K eq  9.4  10 27  9  10 27
 2
[V(OH ) 4 ] [ H 2SO 3 ]
(g)
VO 2  V 2  2H 


2V 3  H 2 O
o
E VO
@   0.359
E Vo 3  0.256
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19


 [ V 2 ] 
[V 3 ]

 3 
0.359  0.0592 log 


0
.
256

0
.
0592
log
2
 2 
 [VO ][ H ] 
 [V ] 


0.359   0.256
[V 3 ]2
  log K eq  10.389
 log 
2
 2
2 
0.0592
 [VO ][ H ] [V ] 
[V 3 ]2
 K eq  2.4  1010
[VO 2 ][ H  ]2 [V 2 ]
(h)
TiO 2  Ti2  2H 


2Ti3  H 2 O
ETo iO2  0.099
ETo3  0.369


 [Ti2 ] 
[Ti3 ]

 3 
0.099  0.0592 log 


0
.
369

0
.
0592
log
2
 2 
 [TiO ][ H ] 
 [Ti ] 


0.099   0.369 
[Ti3 ]2
  log K eq  7.9054
 log 
2
 2
2 
0.0592
 [TiO ][ H ] [Ti ] 
[Ti3 ]2
 K eq  8.0  107
2
 2
2
[TiO ][ H ] [Ti ]
19-12 (a)
At equivalence, [Fe2+] = [V3+] and [Fe3+] = [V2+]
 [ Fe 2 ] 

Eeq  0.771  0.0592 log 
3 
 [ Fe ] 
 [V 2 ] 
Eeq  0.256  0.0592 log  3 
 [V ] 
 [ Fe 2 ][ V 2 ] 

2 Eeq  0.771   0.256  0.0592 log 
3
3 
 [ Fe ][ V ] 
0.771  0.256
Eeq 
 0.258 V
2
(b)
At equivalence, [Fe(CN)63-] = [Cr2+] and [Fe(CN)64-] = [Cr3+]
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
 [ Fe(CN ) 6 4 ] 

Eeq  0.36  0.0592 log 
3 
 [ Fe(CN ) 6 ] 
 [Cr 2 ] 

Eeq  0.408  0.0592 log 
3 
 [Cr ] 
 [ Fe(CN ) 6 4 ][Cr 2 ] 

2 Eeq  0.36   0.408  0.0592 log 
3
3 
 [ Fe(CN ) 6 ][Cr ] 
0.36  0.408
Eeq 
  0.024 V
2
(c)
At equivalence, [VO2+] = 2[UO22+] and [V(OH)4+] = 2[U4+]


[ VO 2 ]

Eeq  1.00  0.0592 log 

 2 
 [V(OH ) 4 ][ H ] 


[ U 4 ]

2 Eeq  20.344   0.0592 log 
2
 4 
 [ UO 2 ][ H ] 


[VO 2 ][ U 4 ]

3Eeq  1.00  20.344   0.0592 log 

2
 6 
 [ V(OH ) 4 ][ UO 2 ][ H ] 


1
  1.688  0.355  1.333 V
3Eeq  1.00  20.344   0.0592 log 
6 
 0.100 
1.333
Eeq 
 0.444 V
3
(d)
At equivalence, [Fe2+] = 2[Tl3+] and [Fe3+] = 2[Tl+]
 [ Fe 2 ] 

Eeq  0.771  0.0592 log 
3 
 [ Fe ] 
 [Tl ] 
2 Eeq  21.25  0.0592 log  3 
 [Tl ] 
 [ Fe 2 ][Tl ] 
 2[Tl3 ][Tl ] 



3Eeq  0.771  21.25  0.0592 log 

3
.
27

0
.
0592
log
3
3 

3 
 [ Fe ][Tl ] 
 2[Tl ][Tl ] 
3.27
Eeq 
 1.09 V
3
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
(e)
At equivalence, [Ce3+] = 2[H3AsO4], [Ce4+] = 2[H3AsO3] and [H+] = 1.00
 [Ce 3 ] 

Eeq  1.70  0.0592 log 
4 
 [Ce ] 
 [ H 3 AsO 3 ] 

2 Eeq  20.577   0.0592 log 
 2 
 [ H 3 AsO 4 ][ H ] 
 [Ce 3 ][ H 3 AsO 3 ] 

3Eeq  1.70  20.577   0.0592 log 
4
 2 
 [Ce ][ H 3 AsO 4 ][ H ] 
 1 
 2[ H 3 AsO 4 ][ H 3 AsO 3 ] 

  2.854  0.0592 log 
 2.854  0.0592 log 
2 
 2 
 2[ H 3 AsO 3 ][ H 3 AsO 4 ][ H ] 
 1.00 
2.854
Eeq 
 0.951 V
3
(f)
At equivalence, [V(OH)4+] = 2[H2SO3] and [VO2+] = 2[SO42-]


[VO 2 ]


Eeq  1.00  0.0592 log 

 2 
[
V
(
OH
)
][
H
]
4


 [ H 2SO 3 ] 

2 Eeq  20.172   0.0592 log 
2
 4 
[
SO
][
H
]
4




[ VO 2 ][ H 2SO 3 ]

3Eeq  1.00  20.172   0.0592 log 

2
 6 
 [ V(OH ) 4 ][SO 4 ][ H ] 


1
  1.344  0.355  9.89  10 1 V
3Eeq  1.00  20.172   0.0592 log 
6 
 0.100 
9.89  10 1
Eeq 
 0.330 V
3
(g)
At equivalence, [VO+] = [V2+]
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19


[V 3 ]

Eeq  0.359  0.0592 log 
2
 2 
 [VO ][ H ] 
 [ V 2 ] 
Eeq  0.256  0.0592 log  3 
 [V ] 


 [ V 2 ] 
[ V 2 ]

 2

2 Eeq  0.359   0.256  0.0592 log 

0
.
103

0
.
0592
log

 2 
 2 
 [ VO ][ H ] 
 [V ][ H ] 


1
  0.103  0.118  0.154 V
2 Eeq  0.103  0.0592 log 
2 


0
.
100


 0.154
Eeq 
  0.008 V
2
(h)
At equivalence, [Ti2+] = [TiO2+]


[Ti3 ]

Eeq  0.099  0.0592 log 
2
 2 
 [TiO ][ H ] 
 [Ti2 ] 
Eeq  0.369  0.0592 log  3 
 [Ti ] 


 [Ti2 ] 
[Ti2 ]

 2

2 Eeq  0.099   0.369   0.0592 log 

0
.
103

0
.
0592
log

 2 
 2 
[
TiO
][
H
]
[
Ti
][
H
]






1
  0.270  0.118  0.388 V
2 Eeq  0.270  0.0592 log 
2 
 0.100 
 0.388
Eeq 
  0.194 V
2
19-13 (a) In the solution to Problem 19-11(a) we find
K eq 
[V 3 ][ Fe 2 ]
 2.23  1017
[V 2 ][ Fe 3 ]
At the equivalence point,
[V 2 ]  [ Fe 3 ]  x
0.1000
[V 3 ]  [ Fe 2 ] 
 0.0500
2
Substituting into the first equation we find
Fundamentals of Analytical Chemistry: 8th ed.
2.23  10
17
Chapter 19
2

0.0500

x2
0.00250
 1.06  10 10
17
2.23  10
x
Thus,
[V 2 ]  [ Fe 3 ]  1.06  10 10 M
and
[V 3 ]  [ Fe 2 ]  0.0500 M
(b) Proceeding in the same way, we find
4
[Fe(CN) 6 ]  [Cr 3 ]  0.0500 M
3
[Fe(CN)6 ]  [Cr 2 ]  1.7  108 M
(c) At equivalence

[ V(OH ) 4 ]  2[ U 4 ]  x
2
2
[ VO 2 ]  2[ UO 2 ] 
2
[ UO 2 ] 
2 0.1000
0.2000
 2x 
 0.0667 M
3
3
0.1000
 0.0333 M
3
From the solution for Problem 19-11(c)
0.0667 0.0333
[VO 2 ]2 [ UO 2 ]
 3.2  10 22 
 2
4
[V(OH ) 4 ] [ U ]
x2 x
2
2
3
0.0667 0.0333
x

2
3.2  10 22
2
2
2
 
x  3 2  4.62  10 27   2.10  10 9 M

[V(OH ) 4 ]  2.1  10 9 M and [ U 4 ] 
2.1  10 9
 1.0  10 9 M
2
(d)
[ Fe 2 ]  2[Tl3 ]  x
[ Fe 3 ]  0.0667 M and [Tl ]  0.0333 M
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
From the solution for Problem 19-11(d)
0.03330.0667
[Tl ][ Fe 3 ]2
 1.5  1016 
3
2 2
x x2
[Tl ][ Fe ]
2
2
3
0.0667 0.0333
x

2
1.5  1016
2
 
x  3 2  9.88  10 21   2.70  10 7 M  [ Fe 2 ]
x 2.70  10 7

 1.4  10 7 M  [Tl3 ]
2
2
(e) Proceeding as in part (c), we find
[Ce 3 ]2 [ H 3 AsO 4 ]
[Ce 3 ]2 [ H 3AsO 4 ]
0.0667 0.0333
37

8
.
9

10


2
4 2
 2
4 2
x3
[Ce ] [ H 3 AsO 3 ][ H ]
[Ce ] [ H 3 AsO 3 ]1.00
2
2
When x  [Ce4 ]  2[H 3AsO 3 ]
[Ce 4 ]  6.9  10 14 M
[ H 3 AsO 3 ]  3.5  10 14 M
[Ce 3 ]  0.067 M
[ H 3 AsO 4 ]  0.033 M
(f) Proceeding as in part (c)
[ VO  ]  0.067 M
2
[SO 4 ]  0.033 M

[ V(OH ) 4 ]  3.2  10 11 M
[ H 2SO 3 ]  1.6  10 11 M
(g)
x  [V 2 ]  [VO 2 ]
0.200
[V 3 ] 
 0.100
2
Assume [H+] = 0.1000. From the solution to Problem 19-11(g)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
0.100
[V 3 ]2
 2.4  1010 
2
2
2
 2
2
[VO ][ V ][ H ]
[VO ][ V 2 ]0.100
1.00
2.4  1010  2
x
2
x  [VO ]  [V 2 ]  6.5  10 6 M
2
[V 3 ]  0.100 M
(h) Proceeding as in part (g), we find
[Ti3 ]  0.100 M
[ H  ]  0.100 M

[TiO 2 ]  [Ti2 ]  1.1  10 5 M
19-14
Eeq, V
Indicator
(a)
0.258
Phenosafranine
(b)
-0.024
None
(c)
0.444
Indigo tetrasulfonate or Methylene blue
(d)
1.09
1,10-Phenanthroline
(e)
0.951
Erioglaucin A
(f)
0.330
Indigo tetrasulfonate
(g)
-0.008
None
(h)
-0.194
None
19-15 (a)
2V 2  Sn 4


At 10.00 mL,
2V 3  Sn 2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
 0.0500 mmol Sn 4
2 mmol V 3

 10.00 mL 
mL
mmol Sn 4

3
[V ] 
60.00 mL


  0.0167 M
 0.1000 mmol V 2


 50.00 mL 
mL
  1.67  10 2 M  0.0667 M
[ V 2 ]  
60.00 mL
 [ V 2 ] 
 0.0667 
E  0.256  0.0592 log  3   0.256  0.0592 log 
  0.292 V
 0.0167 
 [V ] 
The remaining pre-equivalence point data are treated in the same way. The results appear
in the spreadsheet that follows.
At 50.00 mL,
Proceeding as in Problem 19-12, we write
 [ V 2 ] 
E  0.256  0.0592 log  3 
 [V ] 
 [Sn 2 ] 

2 E  2  0.154  0.0592 log 
4 
 [Sn ] 
 [V 2 ][Sn 2 ] 

3E  0.256  2  0.154   0.0592 log  3
4 
 [V ][Sn ] 
At equivalence, [V2+] = 2[Sn4+] and [V3+] = 2[Sn2+]. Thus,
Eeq 
 0.256  2  0.154 0.0592 log 1.00

 0.017 V
3
3
At 50.10 mL,
cSn 2
 0.1000 mmol V 2
1 mmol Sn 2

 50.00 mL 
mL
2 mmol V 2


100.10 mL


  0.025 M  [Sn 2 ]
cSn 4
 0.0500 mmol Sn 4


 50.10 mL 
mL
  0.025 M  5.0  10 5 M  [Sn 4 ]

100.10 mL
0.0592  [Sn 2 ] 
0.0592  2.5  10 2 
  0.154 
  0.074 V
E  0.154 
log 
log 
4 
5 
2
2
 [Sn ] 
 5.0  10 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
The remaining post-equivalence points are obtained in the same way and are found in the
table that follows.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
A
B
C
D
E
F
2+
4+
19-15 (a) Titration of 50.00 mL of 0.1000 M V with 0.0500 M Sn
2+
4+
3+
2+
Reaction: 2V + Sn  2V + Sn
3+ 2+
0
For V /V , E
-0.256
4+
2+
0
For Sn /Sn , E
0.154
2+
Initial conc. V
0.1000
4+
Conc. Sn
0.0500
Volume solution, mL
50.00
4+
3+
2+
4+
2+
Vol. Sn , mL
[V ]
[V ]
[Sn ]
[Sn ]
E, V
10.00
0.0167
0.0667
-0.292
25.00
0.0333
0.0333
-0.256
49.00
0.0495
0.0010
-0.156
49.90
0.0499
0.0001
-0.096
50.00
0.017
50.10
5.00E-05 0.0250 0.074
51.00
4.95E-04 0.0248 0.104
60.00
4.55E-03 0.0227 0.133
Spreadsheet Documentation
B9=$B$6*A9*2/($B$7+A9)
C9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)
F9=$B$3-0.0592*LOG10(C9/B9)
F13=($B$3+2*$B$4)/3
D14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)
E14=$B$7*$B$5/(2*($B$7+A14))
F14=$B$4-(0.0592/2)*LOG10(E14/D14)
G
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
(b)
3
Fe(CN ) 6  Cr 2


Fe(CN ) 6
4
 Cr 3
The pre-equivalence point data are obtained by substituting concentrations into the
equation
 [ Fe(CN ) 6 4 ] 

E  0.36  0.0592 log 
3 
[
Fe
(
CN
)
]
6


The post-equivalence point data are obtained with the Nernst expression for the Cr2+/ Cr3+
system. That is,
 [Cr 2 ] 

E  0.0408  0.0592 log 
3 
 [Cr ] 
The equivalence point potential is found following the procedure in Problem 19-12. The
results for all data points are found in the spreadsheet that follows.
Fundamentals of Analytical Chemistry: 8th ed.
1
Chapter 19
A
B
C
D
E
32+
19-15 (b) Titration of 50.00 mL of 0.1000 M Fe(CN) 6 with 0.1000 M Cr
3-
2
Reaction: Fe(CN)6 + Cr
40
Fe(CN)6 , E
3+
2+
0
2+
4-
 Fe(CN)6 + Cr
3+
3
4
For
For Cr /Cr , E
5
6
7
Initial conc. Fe(CN)6
0.1000
2+
Conc. Cr
0.1000
Volume solution, mL
50.00
2+
433+
2+
Vol. Cr , mL
[Fe(CN)6 ]
[Fe(CN)6 ]
[Cr ]
[Cr ]
10.00
0.0167
0.0667
25.00
0.0333
0.0333
49.00
0.0495
0.0010
49.90
0.0499
0.0001
50.00
50.10
0.0500 9.99E-05
51.00
0.0495
0.0010
60.00
0.0455
0.0091
Spreadsheet Documentation
B9=$B$6*A9/($B$7+A9)
C9=($B$5*$B$7-$B$6*A9)/($B$7+A9)
F9=$B$3-0.0592*LOG10(B9/C9)
F13=($B$3+$B$4)/2
D14=($B$5*$B$7/($B$7+A14)
E14=($B$6*A14-$B$5*$B$7)/($B$7+A14)
F14=$B$4-0.0592*LOG10(E14/D14)
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
F
0.36
-0.408
3-
E, V
0.40
0.36
0.26
0.20
-0.02
-0.248
-0.307
-0.367
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
(c) The data points for this titration, which are found in the spreadsheet that follows, are
obtained in the same way as those for parts (a) and (b).
1
A
B
C
D
E
43+
19-15 (c) Titration of 50.00 mL of 0.1000 M Fe(CN)6 with 0.0500 M Tl
4-
2
3+
Reaction: 2Fe(CN)6 + Tl
40
Fe(CN)6 , E
3+
+
0
3
4
For
For Tl /Tl , E
5
6
7
Initial conc. Fe(CN)6
0.1000
3+
Conc. Tl
0.0500
Volume solution, mL
50.00
3+
43Vol. Tl , mL
[Fe(CN)6 ] [Fe(CN)6 ]
10.00
0.0667
0.0167
25.00
0.0333
0.0333
49.00
0.0010
0.0495
49.90
0.0001
0.0499
50.00
50.10
51.00
60.00
Spreadsheet Documentation
B9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)
C9=$B$6*A9*2/($B$7+A9)
F9=$B$3-0.0592*LOG10(B9/C9)
F13=($B$3+2*$B$4)/3
D14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)
E14=($B$5*$B$7/2)/($B$7+A14)
F14=$B$4-(0.0592/2)*LOG10(E14/D14)
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
3-
F
+
 2Fe(CN)6 + Tl
0.36
1.25
4-
3+
[Tl ]
5.00E-05
4.95E-04
4.55E-03
+
[Tl ]
0.0250
0.0248
0.0227
E, V
0.32
0.36
0.46
0.52
0.95
1.17
1.20
1.23
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
(d) The data points for this titration, which are found in the spreadsheet that follows, are
obtained in the same way as those for parts (a) and (b).
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
A
B
C
D
E
3+
2+
19-15 (d) Titration of 0.1000 M Fe with Sn
3+
2+
2+
4+
Reaction: 2Fe + Sn  2Fe + Sn
3+
2+ 0
For Fe /Fe E
0.771
4+
2+
0
For Sn /Sn , E
0.154
3+
Initial conc. Fe
0.1000
2+
Conc. Sn
0.0500
Volume solution, mL
50.00
2+
3+
2+
4+
2+
Vol. Sn , mL
[Fe ]
[Fe ]
[Sn ]
[Sn ]
10.00
0.0667
0.0167
25.00
0.0333
0.0333
49.00
0.0010
0.0495
49.90
0.0001
0.0499
50.00
50.10
0.0250
5.00E-05
51.00
0.0248
4.95E-04
60.00
0.0227
4.55E-03
Spreadsheet Documentation
B9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)
C9=$B$6*A9*2/($B$7+A9)
F9=$B$3-0.0592*LOG10(B9/C9)
F13=($B$3+2*$B$4)/3
D14=(($B$5*$B$7/2)/($B$7+A14)
E14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)
F14=$B$4-(0.0592/2)*LOG10(E14/D14)
F
E, V
0.807
0.771
0.671
0.611
0.360
0.234
0.204
0.175
Fundamentals of Analytical Chemistry: 8th ed.

Chapter 19
(e) 2MnO 4  5U 4  2H 2 O 
2Mn 2  5UO 2

2
 4H 
At 10.00 mL,

0.02000 mmol MnO 4


 10.00 mL MnO 4  0.2000 mmol MnO 4
mL
0.05000 mmol U 4
 50.00 mL U 4  2.500 mmol U 4
mL
2

5 mmol UO 2 
 0.2000 mmol MnO 4  

 

2
mmol
MnO
2
2
4 
cUO 2  [ UO 2 ]  
 8.33  10 3 M UO 2
2
60.00 mL solution
c U 4   [ U 4 ] 
E  0.334 
2.5000 mmol U   8.33  10
4
3
60.00 mL
 0.0333 M U 4


0.0592 
[ U 4 ]
0.0592 
3.33  10 2


  0.316 V
log 

0
.
334

log
2
4 
 4 
3

2
2
 [8.33  10 1.00 
 [ UO 2 ][ H ] 
Additional pre-equivalence point data, obtained in the same way, are given in the
spreadsheet that follows.
At 50.00 mL,


[ Mn 2 ]

5Eeq  5  1.51  0.0592 log 

 8 
 [ MnO 4 ][ H ] 


[ U 4 ]

2 Eeq  2  0.334  0.0592 log 
2
 4 
 [ UO 2 ][ H ] 
Adding the two equations gives


[ Mn 2 ][ U 4 ]


7Eeq  5  1.51  2  0.334   0.0592 log 

2
 12 
[
MnO
][
UO
][
H
]
4
2


At equivalence, [MnO4-] = 2/5[U4+] and [Mn2+] = 2/5[UO22+]
Substituting these equalities and [H+] = 1.00 into the equation above gives
Eeq 
8.218 0.0592 log 1.00 8.218


 1.17 V
7
7
7
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 19
At 50.10 mL,

0.02000 MnO 4


mmol MnO 4 added 
 50.10 mL MnO 4  1.0020 MnO 4
mL
0.05000 mmol U 4
2 mmol Mn 2
mmol Mn 2 formed 
 50.00 mL U 4 
 1.000 mmol Mn 2
mL
5 mmol U 4


mmol MnO 4 remaining  1.0020  1.000  2.0  10 3 mmol MnO 4
E  1.51 


1.000 / 100.10   1.48 V
0.0592 
[ Mn 2 ]
0.0592 
  1.51 
log 
log 

8 
 8 
5
5
 1.0020 / 100.101.00 
 [ MnO 4 ][ H ] 
The remaining post equivalence point data are derived in the same way and are given in
the spreadsheet that follows.
Fundamentals of Analytical Chemistry: 8th ed.
1
A
B
C
D
E
F
4+
19-15 (e) Titration of 50.00 mL 0.05000 M U with 0.02000 M MnO4
-
2
3
Chapter 19
Reaction: 2MnO4 + 5U
4+
2+
0
For U /UO2 , E
MnO4 ,
0
4+
2+
+2H2O  2Mn
2+
+ 4H
+
0.334
4
5
For
E
4+
Initial conc. U
6
7
Conc. MnO4
0.0200
Volume solution, mL
50.00
4+
2+
2+
Vol. MnO4 , mL
[U ]
[UO2 ] [MnO4 ]
[Mn ]
10.00
0.0333 0.0083
25.00
0.0167 0.0167
49.00
0.0005 0.0247
49.90
0.0001 0.0250
50.00
50.10
2.00E-05
0.0100
51.00
0.0002
0.0099
60.00
0.0018
0.0091
Spreadsheet Documentation
B9=$B$5*$B$7-$B$6*A9*5/2)/($B$7+A9)
C9=($B$6*A9*5/2)/($B$7+A9)
F9=1.00 (entry)
G9=$B$3-(0.0592/2)*LOG10(B9/(C9*F9^4))
G13=((5*$B$4+2*$B$3)/7)-(0.0592/7)*LOG10(F13)
D14=($B$6*A14-$B$5*$B$7*2/5)/($B$7+A14)
E14=($B$5*$B$7*2/5)/($B$7+A14)
G14=$B$4-(0.0592/5)*LOG10(E14/(D14*F14^8))
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
+ 5UO2
G
1.51
0.0500
-
+
[H ]
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
E, V
0.316
0.334
0.384
0.414
1.17
1.48
1.49
1.50