Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
Chapter 20
20-1
(a) 2Mn 2  5S2O8
2
 8H 2O  10SO4
2

 2MnO 4  16H 
(b) NaBiO 3 ( s )  2Ce3  4 H   BiO   2Ce4  2 H 2 O  Na 
(c) H 2 O 2  U 4  UO 2
2
 2H 

(d) V(OH) 4  Ag ( s )  Cl   2 H   VO 2  AgCl ( s )  3H 2 O

(e) 2 MnO 4  5H 2 O 2  6H   5O 2  2 Mn 2  8H 2 O

(f) ClO 3  6I   6H   3I 2  Cl  3H 2O
20-2
(a) 2 Fe 3  SO 2 ( g )  2 H 2 O  2 Fe 2  SO 4
2
 4H 
(b) 2H 2 MoO 4  3Zn ( s)  12H   3Zn 2  2MoO3  8H 2 O


(c) 2MnO 4  5HNO2  H   2Mn 2  5NO3  3H 2O

(d) BrO 3  5Br   C6 H 5 NH2  3H   Br3C6 H 5 NH2  3Br   3H 2O
2
(e) 2HAsO 3  O2 ( g )  2HAsO 4
2
(f) 2I   2HNO 2  2H   I 2  2 NO  2H 2 O
20-3
Only in the presence of Cl- ion is Ag a sufficiently good reducing agent to be very useful
for prereductions. In the presence of Cl-, the half-reaction occurring in the Walden
reductor is
Ag ( s)  Cl   AgCl ( s)  e 
The excess HCl increases the tendency of this reaction to occur by the common ion
effect.
Fundamentals of Analytical Chemistry: 8th ed.
20-4
Chapter 20
Amalgamation of the zinc prevents loss of reagent by reaction of the zinc with hydronium
ions.
2
 2Ag ( s )  4 H   2Cl 


U 4  2AgCl ( s )  H 2 O
20-5
UO 2
20-6
2TiO 2  Zn ( s )  4 H 
20-7
Standard solutions of reductants find somewhat limited use because of their susceptibility


2Ti 3  Zn 2  2 H 2 O
to air oxidation.
20-8
Standard KMnO4 solutions are seldom used to titrate solutions containing HCl because of
the tendency of MnO4- to oxidize Cl- to Cl2, thus causing overconsumption of MnO4-.
20-9
Cerium (IV) precipitates as a basic oxide in alkaline solution.

20-10 2 MnO 4  3Mn 2  2 H 2 O  5MnO 2 ( s )  4 H 
20-11 Freshly prepared solutions of permanganate are inevitably contaminated with small
amounts of solid manganese dioxide, which catalyzes the further decompositions of
permanganate ion. By removing the dioxide at the outset, a much more stable standard
reagent is produced.
20-12 Standard permanganate and thiosulfate solutions are generally stored in the dark because
their decomposition reactions are catalyzed by light.

20-13 4MnO 4  2H 2 O  4MnO 2 ( s ) 3O 2 ( g )  4OH 
brown
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-14 Solutions of K2Cr2O7 are used extensively for back-titrating solutions of Fe2+ when the
latter is being used as a standard reductant for the determination of oxidizing agents.
20-15 Iodine is not sufficiently soluble in water to produce a useful standard reagent. It is quite
soluble in solutions containing excess I- because of formation of triiodide.
20-16 The solution concentration of I3- becomes stronger because of air oxidation of the excess
I-. The reaction is

6I   O2 ( g )  4H   2I 3  2H 2O
2

20-17 S2O3  H   HSO3  S( s)
20-18 When a measured volume of a standard solution of KIO3 is introduced into an acidic
solution containing an excess of iodide ion, a known amount of iodine is produced as a
consequence of a reaction.

IO3  5I   6H   3I 2  3H 2O
20-19

BrO 3  6I   6H   Br   3I 2  3H 2 O
excess
I 2  2S2 O 3
2
 2I   S 4 O 6
2
20-20
Cr2 O7
2
 6I   14H   2Cr 3  3I 2  7H 2 O
excess
I 2  2S2 O 3
2
 2I   S 4 O 6
2
20-21 2I 2  N 2 H 4  N 2 ( g )  4H   4I 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-22 Starch is decomposed in the presence of high concentrations of iodine to give products
that do not behave satisfactorily as indicators. This reaction is prevented by delaying the
addition of the starch until the iodine concentration is very small.
20-23
0.2256 g sample 
(a)
1000 mmol Fe 2
 4.03961 mmol Fe 2
55.847 g
4.03961 mmol Fe 2 1 mmol Ce4

 0.1142 M Ce4
2
35.37 mL
mmol Fe
2
4.03961 mmol Fe 2 1 mmol Cr2 O 7

(b)
35.37 mL
6 mmol Fe 2
 0.01904 M Cr2 O 7
2

(c)
4.03961 mmol Fe 2 1 mmol MnO 4


 0.02284 M MnO 4
35.37 mL
5 mmol Fe 2

4.03961 mmol Fe 2 1 mmol V(OH) 4


 0.1142 M V(OH) 4
(d)
2
35.37 mL
mmol Fe

4.03961 mmol Fe 2 1 mmol IO 3


 0.02855 M IO 3
(e)
2
35.37 mL
4 mmol Fe
20-24
0.02500 mmol K 2 Cr2 O 7
294.185 g K 2 Cr2 O 7
 500.0 mL 
 3.677 g K 2 Cr2 O 7
mL
1000 mmol
Dissolve 3.677 g K2Cr2O7 in water and dilute to 500.0 mL.
20-25
0.02500 mmol KBrO 3
1000 mL 167.001 g KBrO 3
 2.000 L 

 8.350 g KBrO 3
mL
L
1000 mmol
Dissolve 8.350 g KBrO3 in water and dilute to 2.000 L.
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-26
0.0500 mmol KMnO 4
1000 mL 158.034 g KMnO 4
 2.0 L 

 15.80 g KMnO 4
mL
L
1000 mmol
Dissolve about 16 g KMnO4 in water and dilute to 2.0 L.
20-27

0.0500 mmol I 3 1 mmol I 2
1000 mL 253.809 g I 2

 2.0 L 

 25.38 g I 2

mL
L
1000 mmol
mmol I 3
Dissolve between 25 and 26 g I2 in a concentrated solution of KI and dilute to 2.0 L.

20-28 2 MnO 4  5H 2 C 2 O 4  6H   2 Mn 2  10CO2 ( g )  8H 2 O
0.1756 g Na 2 C 2 O 4 1000 mmol Na 2 C 2 O 4 2 mmol KMnO 4


 0.01636 M KMnO 4
32.04 mL KMnO 4
133.999 g
5 mmol Na 2 C 2 O 4
20-29 Ce4  Fe2  Ce3  Fe3
0.1809 g Fe
1000 mmol Fe 1 mmol Fe 2 1 mmol Ce4



 0.1034 M Ce4
4
2
31.33 mL Ce
55.847 g
mmol Fe
mmol Fe
20-30
Cr2 O 7
2
 6I   14 H   2Cr 3  3I 2  7 H 2 O
I 2  2S 2 O 3
2
 2I   S 4 O 6
2
1 mmol Cr2O72-  3 mmol I2  6 mmol S2O322
0.1259 g K 2 Cr2 O 7 1000 mmol K 2 Cr2 O 7 6 mmol S 2 O 3


 0.06223 M Na 2S 2 O 3
41.26 mL Na 2S 2 O 3
294.185 g
mmol K 2 Cr2 O 7
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-31

BrO 3  6I   6H   Br   3I 2  3H 2 O
I 2  2S 2 O 3
2
 2I   S 4 O 6
2
1 mmol BrO3-  3 mmol I2  6 mmol S2O322
0.1017 g KBrO 3
1000 mmol KBrO 3 6 mmol S 2 O 3


 0.09192 M Na 2S 2 O 3
39.75 mL Na 2S 2 O 3
167.001 g
mmol KBrO 3
20-32 1 mmol I2  1 mmol Sb3+  2 mmol Sb2S3
(a)
0.02870 mmol I 2
1 mmol Sb3 121.760 g Sb3
 44.87 mL I 2 

mL
mmol I 2
1000 mmol
 100%  16.03% Sb3
0.978 g sample
(b)
 0.02870 mmol I 2
1 mmol Sb 3 1 mmol Sb 2S3 339.71 g Sb 2S3 


 44.87 mL I 2 


mL
mmol I 2
2 mmol Sb 3
1000 mmol 

 100%
0.978 g sample
 22.37% Sb 2S3
20-33
MnO 2  2I   4 H   Mn 2  I 2  2 H 2 O
I 2  2S 2 O 3
2
 2I   S 4 O 6
2
1 mmol MnO2  1 mmol I2  2 mmol S2O32 0.07220 mmol Na 2S 2 O 3
1 mmol MnO 2
86.937 g MnO 2 


 32.30 mL Na 2S 2 O 3 

mL
2 mmol Na 2S 2 O 3
1000 mmol 

 100%
0.1344 g sample
 75.42% MnO 2
Fundamentals of Analytical Chemistry: 8th ed.

20-34 3CS( NH2 ) 2  4BrO 3  3H 2O  3CO( NH2 ) 2  3SO4
Chapter 20
2
 4Br   6H 
4 mmol BrO3-  3 mmol CS(NH2)2
 0.00833 mmol KBrO 3
3 mmol CS( NH 2 ) 2 76.122 g CS( NH 2 ) 2 


 14.1 mL KBrO 3 

mL
4 mmol KBrO 3
1000 mmol

  100%
0.0715 g sample
 9.38% CS( NH 2 ) 2

20-35 MnO 4  5Fe 2  8H   Mn 2  5Fe 3  4 H 2 O
1 mmol MnO4-  5 mmol Fe2+  5/2 mmol Fe2O3
mmol KMnO 4 
0.02086 KMnO 4
 39.21 mL KMnO 4  0.8179 mmol KMnO 4
mL
(a)

5 mmol Fe 2
1 mmol Fe 55.847 g Fe 
 0.8179 mmoL KMnO 4 



mmol KMnO 4 mmol Fe 2 1000 mmol 

 100%
0.7120 g sample
 32.08% Fe
(b)

5 mmol Fe 2 O 3 159.692 g Fe 2 O 3 
 0.8179 mmoL KMnO 4 


2 mmol KMnO 4
1000 mmol 

 100%  45.86% Fe 2 O 3
0.7120 g sample
20-36 3Sn 3  Cr2 O 7
2
 14 H 


3Sn 4  2Cr 3  7 H 2 O
1 mmol Cr2O72-  3 mmol Sn3+  3 mmol SnO2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
(a)
mmol Sn 
0.01735 mmol K 2 Cr2 O 7
3 mmol Sn 3
1 mmol Sn
 29.77 mL K 2 Cr2 O 7 

 1.54953 mmol Sn
mL
mmol K 2 Cr2 O 7 mmol Sn 3

118.71 g Sn 
1.54953 mmol Sn 

1000 mmol 

 100%  42.27% Sn
0.4352 g sample
(b)
mmol SnO 2 
0.01735 mmol K 2 Cr2 O 7
3 mmol SnO 2
 29.77 mL K 2 Cr2 O 7 
 1.54953 mmol SnO 2
mL
mmol K 2 Cr2 O 7

150.71 g SnO 2 
1.54953 mmol SnO 2 

1000 mmol 

 100%  53.66% SnO 2
0.4352 g sample
20-37
2H 2 NOH  4Fe3
Cr2 O 7
2


N 2 O(g)  4Fe 2  4H   H 2 O
 6Fe 2  14H 


2Cr 3  6Fe3  7H 2 O
1 mmol Cr2O72- 6 mmol Fe3+  3 mmol H2NOH
 0.0325 mmol K 2 Cr2 O 7
3 mmol H 2 NOH 


 19.83 mL K 2 Cr2 O 7 
mL
mmol K 2 Cr2 O 7 

 0.0387 M H 2 NOH
50.00 mL sample
20-38
Zn 2  H 2 C 2 O 4  ZnC 2 O 4 ( s )  2 H 

2 MnO 4  5H 2 C 2 O 4  6H   2 Mn 2  10CO2 ( g )  8H 2 O
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
2 mmol MnO4-  5 mmol Na2C2O4  5 mmol ZnO
 0.01508 mmol KMnO 4
5 mmol ZnO
81.139 g ZnO 


 37.81 mL KMnO 4 

mL
2 mmol KMnO 4
1000 mmol 

 100%
0.9280 g sample
 12.46% ZnO
20-39 (Note: In the first printing of the text, the answer in the back of the book was in error.)

ClO 3  6Fe2  6H   Cl  3H 2O  6Fe3
0.08930 mmol Fe 2
 50.00 mL Fe 2  4.4650 mmol Fe 2
mL
0.083610 mmol Ce4
2
4
mmol Fe titrated by Ce 
 14.93 mL Ce4 
mL
2
1 mmol Fe
 1.2483 mmol Fe 2
mmol Ce4
mmol Fe 2 
mmol Fe 2 reacted with KClO 3  4.4650  1.2483  3.2167 mmol Fe 2

1 mmol KClO 3 122.549 g KClO 3 
 3.2167 mmoL Fe 2 


6 mmol Fe 2
1000 mmol

  100%  51.37% KClO
3
0.1279 g sample
20-40
Pb(C2 H 5 ) 4  I 2  Pb(C2 H 5 ) 3 I  C2 H 5 I
I 2  2S2 O3
2
 2I   S 4 O 6
2
1 mmol I2  2 mmol S2O32-  1 mmol Pb(C2H5)4
0.02095 mmol I 2
 15.00 mL I 2  0.314250 mmol I 2
mL
0.03465 mmol Na 2S 2 O 3
2
mmol I 2 titrated by S2 O 3 
 6.09 mL Na 2S 2 O 3 
mL
1 mmol I 2
 0.10551 mmol I 2
2 mmol Na 2S 2 O 3
mmol I 2 
mmol I 2 reacted with Pb(C 2 H 5 ) 4  0.314250  0.10551  0.20874 mmol I 2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
1 mmol Pb(C 2 H 5 ) 4 323.4 mg Pb(C 2 H 5 ) 4

2.70  103 mg Pb(C 2 H 5 ) 4
mmol I 2
mmol

L
L sample
25.00 mL sample 
1000 mL
0.20874 mmol I 2 
20-41 H 3 AsO 3  I 2  H 2 O  H 3 AsO 4  2I   2 H 
1 mmol I2  1 mmol H3AsO3  ½ mmol As2O3
 0.01985 mmol I 2
1 mmol As 2 O 3 197.841 g As 2 O 3 


 24.56 mL I 2 

mL
2 mmol I 2
1000 mmol 

 100%
7.41 g sample
 0.651% As 2 O 3
20-42 Cr2O7
2
 3U 4  2H   3UO 2
2
 2Cr 3  H 2O
1 mmol Cr2O72-  3 mmol U4+  1 mmol NaCl
 0.100 mmol K 2 Cr2 O 7
1 mmol NaCl 58.442 g NaCl 


 19.9 mL K 2 Cr2 O 7 

mL
mmol K 2 Cr2 O 7
1000 mmol 

 100%
25.0 mL
0.800 g sample 
50.0 mL
 29.1% NaCl
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-43 2C 2 H 5SH  I 2  C 2 H 5SSC2 H 5  2I   2 H 
0.01293 mmol I 2
 50.0 mL I 2  0.6465 mmol I 2
mL
2
mmol I 2 titrated by S2 O 3 
mmol I 2 
0.01425 mmol Na 2S2 O 3
1 mmol I 2
 15.72 mL Na 2S2 O 3 
 0.1120 mmol I 2
mL
2 mmol Na 2S2 O 3
mmol I 2 reacted with C 2 H 5SH  0.6465  0.1120  0.5345 mmol I 2

2 mmol C 2 H 5SH 62.14 g C 2 H 5SH 
 0.5345 mmoL I 2 


mmol I 2
1000 mmol 

 100%  4.33% C 2 H 5SH
1.534 g sample
20-44
3TeO 2  Cr2 O 7
Cr2 O 7
2
2
 8H   3H 2 TeO 4  2Cr 3  H 2 O
 6Fe 2  14 H   2Cr 3  6Fe 3  7 H 2 O
0.03114 mmol K 2 Cr2 O 7
 50.00 mL K 2 Cr2 O 7  1.5570 mmol K 2 Cr2 O 7
mL
mmol K 2 Cr2 O 7 titrated by Fe 2 
mmol K 2 Cr2 O 7 
1 mmol K 2 Cr2 O 7
0.1135 mmol Fe 2
 10.05 mL Fe 2 
 0.1901 mmol K 2 Cr2 O 7
mL
6 mmol Fe 2
mmol K 2 Cr2 O 7 reacted with TeO 2  1.5570  0.1901  1.3668 mmol K 2 Cr2 O 7

3 mmol TeO 2 159.6 g TeO 2 
1.3668 mmoL K 2 Cr2 O 7 


mmol K 2 Cr2 O 7
1000 mmol 

 100%  13.16% TeO 2
4.971 g sample
20-45
2I   Br2  I 2  2 Br 

IO 3  5I   6H   3I 2  3H 2 O
I 2  2S 2 O 3
2
 2I   S 4 O 6
2
1 mmol KI  1 mmol IO3-  3 mmol I2  6 mmol S2O32-
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
 0.05982 mmol Na 2S 2 O 3
1 mmol KI
166.00 g KI 


 19.96 mL Na 2S 2 O 3 

mL
6 mmol Na 2S 2 O 3 1000 mmol 

 100%
1.309 g sample
 2.524% KI

20-46 MnO 4  5Fe 2  8H 
mmol Fe 


Mn 2  5Fe 3  4 H 2 O
0.01920 mmol KMnO 4
5 mmol Fe
500.0 mL
 13.72 mL KMnO 4 

mL
mmol KMnO 4 50.00 mL
 13.1712 mmol Fe
0.01920 mmol KMnO 4
5 mmol Fe
500.0 mL
mmol Fe and Cr 
 36.43 mL KMnO 4 

mL
mmol KMnO 4 100.0 mL
 17.4864 mmol Fe and Cr
mmol Cr  17.4864  13.1712   4.3152 mmol Cr
13.1712 mmol
Fe 
55.847 g Fe
1000 mmol
 100%  69.07% Fe
1.065 g sample
4.3152 mmol Cr   51.996 g Cr
1000 mmol
 100%  21.07% Cr
1.065 g sample
20-47 In the Walden reductor,

V(OH) 4  2 H   e 


VO 2  3H 2 O
In the Jones reductor,

V(OH ) 4  4 H   3e 


V 2  4 H 2 O
In the first titration,
Ce 4  Fe 2  Fe 3  Ce3

Ce 4  VO 2  3H 2 O  V(OH) 4  Ce3  2 H 
Fundamentals of Analytical Chemistry: 8th ed.
mmol Fe and V  mmol Ce4 
Chapter 20
0.1000 mmol Ce4
 17.74 mL Ce4  1.7740 mmol Fe and V
mL
In the second titration,
Ce 4  Fe 2  Fe 3  Ce3

3Ce 4  V 2  4 H 2 O  V(OH ) 4  3Ce3  4 H 
mmol Fe and 3  V  mmol Ce 4 
0.1000 mmol Ce 4
 44.67 mL Ce 4  4.4670 mmol Fe and 3  V
mL
Subtracting the first equation from the second equation gives
4.4670  1.7740  2.6930  2  mmol V
2.6930
mmol V 
 1.3465 mmol V
2
1.3465 mmol V
mmol V2 O 5 
 0.67325 mmol V2 O 5
2
mmol Fe  1.7740  1.3465  0.4275 mmol Fe
0.4275 mmol Fe
mmol Fe 2 O 3 
 0.21375 mmol Fe 2 O 3
2
0.67325 mmol
V2 O 5  
181.88 g V2 O 5
1000 mmol
 100%  47.85% V2 O 5
50.00 mL
2.559 g sample 
500.0 mL
0.21375 mmol Fe 2 O 3   159.69 g Fe 2 O 3
1000 mmol
 100%  13.34% Fe 2 O 3
50.00 mL
2.559 g sample 
500.0 mL
20-48
2Tl   CrO 4
2
 Tl 2 CrO 4 (s)
2Tl 2 CrO 4 (s)  2 H   4Tl   Cr2 O 7
Cr2 O 7
2
2
 H 2O
 6Fe 2  14 H   2Cr 3  6Fe 3  7 H 2 O
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
6 mmol Fe2+  1 mmol Cr2O72-  4 mmol Tl+
0.1044 mmol Fe 2
4 mmol Tl  204.38 g Tl 
 39.52 mL Fe 2 

 0.5622 g Tl 
mL
6 mmol Fe 2 1000 mmol
20-49
SO 2 ( g )  2OH   SO 3

2
 H 2O


IO 3  2 H 2SO 3  2Cl  ICl 2  SO 4
2
 2H 
1 mmol IO3-  2 mmol H2SO3  2 mmol SO2
2.50 L
 64.00 min  160 L of sample , there are
min
0.003125 mmol KIO3
2 mmol SO 2 64.065 g SO 2
 4.98 mL KIO3 

 1.99402  10 3 g SO 2
mL
mmol KIO3
1000 mmol
In


 1.99402  10 3 g SO 
2

  10 6 ppm  10.4 ppm SO 2
1
.
20
g
 160 L 



L


20-50
I 2 O5 ( s)  5CO( g )  5CO2 ( g )  I 2 ( g )
I 2  2S2 O3
2
 2I   S 4 O 6
2
1 mmol I2  5 mmol CO  2 mmol S2O320.00221 mmol S 2 O 3
mL
2
2
 7.76 mL S 2 O 3 
5 mmol CO
28.01 g CO

 1.2009  10 3 g CO
2
1000 mmol
2 mmol S 2 O 3


 1.2009  10 3 g CO 

  10 6 ppm  40.5 ppm CO
1
.
2
g
 24.7 L 



L


Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-51
S 2  I 2  S( s )  2I 
I 2  2S 2 O 3
2
 2I   S 4 O 6
2
1 mmol I2  1 mmol H2S  2 mmol S2O320.01070 mmol I 2
 10.00 mL I 2  0.1070 mmol I 2
mL
2
0.01344 mmol S 2 O 3
1 mmol I 2
2
mmol I 2 in excess 
 12.85 mL S 2 O 3 
2
mL
2 mmol S 2 O 3
mmol I 2 added 
 0.08635 mmol I 2
mmol I 2 reacted  mmol H 2S  0.1070  0.08635  0.02065 mmol H 2S
34.082 g H 2S 

 0.02065 mmol H 2S 

1000 mmol 

 106 ppm  19.5 ppm H 2S
1.2 g


30.00 L 


L


20-52 (a)
AgI  2S 2 O 3
2
 Ag (S 2 O 3 ) 2
3
 I

3Br2  I   3H 2 O  IO 3  6Br   6 H 

5I   IO 3  6H   3I 2  3H 2 O
I 2  2S 2 O 3
2
 2I   S4 O 6
2
(b) 1 mmol IO3-  3 mmol I2  1 mmol AgI  6 mmol S2O32-
0.0352 mmol S 2 O 3
mL
2
2
 13.7 mL S 2 O 3 
1 mmol AgI
234.77 mg AgI

2
mmol
6 mmol S 2 O 3
4.00 cm 2

4.72 mg AgI
cm 2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 20
20-53
O 2  4Mn (OH) 2 ( s )  2 H 2 O 
4 Mn (OH) 3 ( s )

4 Mn (OH) 3 ( s )  12 H   4I 
0.00942 mmol S 2 O 3
mL
2


4 Mn 2  2I 2  6H 2 O
2
 13.67 mL S 2 O 3 
1.03 mg O 2
0.0423 mg O 2

mL sample

150 mL 
 25 mL sample 

154 mL 

1 mmol O 2
32.0 mg O 2

 1.03 mg O 2
2
mmol
4 mmol S 2 O 3
Fundamentals of Analytical Chemistry: 8th ed.
A
B
Chapter 20
C
D
E
F
G
H
1
20-54 (a) Titration of 25.00 mL of 0.025 M SnCl2 with 0.050 M FeCl3
2
Reaction: Sn2+ + 2Fe3+  Sn4+ + 2Fe2+
3
For Fe3+/Fe2+, Eo
0.771
4
For Sn4+/Sn2+, Eo
0.154 Equivalence point will be at 25.00 x 0.025 x 2/0.050=25.00 mL
5
Initial concentration Sn2+
0.025
6
Concentration Fe3+
7
Volume SnCl2 solution
8
Equivalence point volume
9
0.05
25.00
25.00
Vol. Fe3+, mL
Percentages
[Sn4+]
[Sn2+]
[Fe3+]
[Fe2+]
E, V
10
10
2.50
0.0023
0.0205
0.126
11
20
5.00
0.0042
0.0167
0.136
12
30
7.50
0.0058
0.0135
0.143
13
40
10.00
0.0071
0.0107
0.149
14
50
12.50
0.0083
0.0083
0.154
15
60
15.00
0.0094
0.0063
0.159
16
70
17.50
0.0103
0.0044
0.165
17
80
20.00
0.0111
0.0028
0.172
18
90
22.50
0.0118
0.0013
0.182
19
95
23.75
0.0122
0.000641
0.192
20
99
24.75
0.0124
0.000126
0.213
21
99.9
24.98
0.0125
0.000013
0.243
22
100
25.00
23
101
25.25
0.0002
0.0249
0.653
24
105
26.25
0.0012
0.0244
0.694
25
110
27.50
0.0024
0.0238
0.712
26
120
30.00
0.0045
0.0227
0.730
27
Spreadsheet Documentation
28
B10=(A10/11)*$B$8
29
30
C10=($B$6*B10/2)/($B$7+B10)
D10=($B$5*$B$7-$B$6*B10.2)/($B$7+B10)
31
G10=$B$4-(0.0592/2)*LOG10(D10/C10)
32
33
34
35
36
37
38
39
40
41
42
43
44
45
0.360
G22=($B$3+2*$B$4)/3
E23=($B$6*B23-$B$5*$B$7*2)/($B$7+B23)
F23=($B$5*$B$7*2)/($B$7+B23)
G23=$B$3-0.0592*LOG10(F23/E23)
Fundamentals of Analytical Chemistry: 8th ed.
A
B
C
Chapter 20
D
E
1
20-54 (b) Titration of 25.00 mL of 0.08467 M Na2S2O3 with 0.10235 M I2 (I3-)
2
Reaction: 2S2O32- + I3-  S4O62- + 3I-
3
For S2O32-/S4O62-, Eo
4
For I3-/I-, Eo
5
Initial concentration S2O32-
F
G
H
0.08
Equivalence point will be at 25.00 x 0.08467 / 2 / 0.10235=10.34
0.536 mL
6
Concentration
7
Volume Na2S2O3 solution
8
Equivalence point volume
9
0.08467
I3-
0.10235
25.00
10.34
Vol. I3-, mL
Percentages
[S4O62-]
[S2O32-]
[I3-]
[I-]
E, V
10
10
1.03
0.0041
0.0732
0.076
11
20
2.07
0.0078
0.0626
0.089
12
30
3.10
0.0113
0.0527
0.098
13
40
4.14
0.0145
0.0436
0.106
14
50
5.17
0.0175
0.0351
0.114
15
60
6.20
0.0203
0.0271
0.123
16
70
7.24
0.0230
0.0197
0.132
17
80
8.27
0.0254
0.0127
0.145
18
90
9.31
0.0278
0.0062
0.165
19
95
9.82
0.0289
0.0030
0.183
20
99
10.24
0.0297
0.000605
0.225
21
99.9
10.33
0.0299
0.000064
0.283
22
100
10.34
23
101
10.44
0.000296
0.0896
0.525
24
105
10.86
0.0014736
0.0885
0.546
25
110
11.37
0.0029074
0.0873
0.555
26
120
12.41
0.00565611
0.0849
0.565
30
Spreadsheet Documentation
B10=(A10/11)*$B$8
C10=($B$6*B10/2)/($B$7+B10)
D10=($B$5*$B$7-$B$6*B10*2)/($B$7+B10)
31
G10=$B$3-(0.0592/2)*LOG10(D10^2/C10)
27
28
29
32
33
34
35
36
37
38
39
40
41
42
43
44
45
0.384
G22=($B$3+2*$B$4)/3
E23=($B$6*B23-$B$5*$B$7*2)/($B$7+B23)
F23=($B$5*$B$7*2)/($B$7+B23)
G23=$B$3-0.0592*LOG10(F23/E23)
Fundamentals of Analytical Chemistry: 8th ed.
A
B
Chapter 20
C
D
E
F
G
H
2
20-54 (c) Titration of 0.1250 g Na2S2O3 with 0.01035 M KMnO4
Reaction: 2MnO4- + 5H2C2O4 +6H+  2Mn2+ + 2CO2 +8H2O
3
For oxalate acid, Eo
4
For MnO4-, Eo
5
Initial mmol oxalate
0.93284 acid initially present. Every 5 mmol of oxalate requires 2 mmol
6
Concentration MnO4-
0.01035 MnO4-. Equivalence point will be at (0.93284 mmol X 2 / 5) / 0.01035
7
Initial volume solution
8
Equivalence point volume
1
9
-0.49
1.51 There are 0.1250 g X 1000 mg/g/133.999 = 0.93284 mmol oxalate
25.00 mmol/mL=36.05 mL KMnO4
36.05
Vol. MnO4-, mL
Percentages
pCO2
[H2C2O4]
[MnO4-]
[Mn2+]
[H+]
E, V
10
10
3.61
1.00
0.0294
1.00
-0.44
11
20
7.21
1.00
0.0232
1.00
-0.44
12
30
10.82
1.00
0.0182
1.00
-0.44
13
40
14.42
1.00
0.0142
1.00
-0.44
14
50
18.03
1.00
0.0108
1.00
-0.43
15
60
21.63
1.00
0.0080
1.00
-0.43
16
70
25.24
1.00
0.0056
1.00
-0.42
17
80
28.84
1.00
0.0035
1.00
-0.42
18
90
32.45
1.00
0.0016
1.00
-0.41
19
95
34.25
1.00 0.000788
1.00
-0.40
20
99
35.69
1.00 0.000154
1.00
-0.38
21
99.9
36.01
1.00 0.000016
1.00
-0.35
22
100
36.05
1.00
0.0061
1.00
0.94
23
101
36.41
6.05E-05
0.0061
1.00
1.49
24
105
37.85
2.97E-04
0.0059
1.00
1.49
25
110
39.66
5.77E-04
0.0058
1.00
1.50
26
120
43.26
0.0011
0.0055
1.00
1.50
27
28
Spreadsheet Documentation
B10=(A10/11)*$B$8
F22=$B$5*(2/5)/($B$8+B22)
E23=($B$6*B23-$B$5*2/5)/($B$7+B23)
H23=$B$4-(0.0592/5)*LOG10(F23/(E23*G23^8))
30
D10=($B$5-$B$6*B10*5/2)/($B$7+B10)
H10=$B$3-(0.0592/2)*LOG10(D10/(C10^2*G10^2))
31
H22=((2*$B$3+5*$B$4)/7)-(0.0592/7)*LOG10(1/(C22*2*G22^10))
29
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Fundamentals of Analytical Chemistry: 8th ed.
A
B
C
Chapter 20
D
E
2
20-54 (d) Titration of 20.00 mL 0.1034 M Fe2+ with 0.01500 M K2Cr2O7
Reaction: 6Fe2+ + Cr2O72- +14H+  6Fe3+ + 2Cr3+ +7H2O
3
For dichromate, Eo
4
For Fe3+/Fe2+, Eo
5
Initial Fe2+ concentration
1
G
H
1.33
0.771 There are 20.00 mL X 0.1034mmol/mL = 2.068 mmol Fe2+
0.1034 initial present. Every mmol of dichromate requires 6 mmol Fe2+.
Cr2O72-
6
Concentration
7
Initial volume solution
8
Equivalence point volume
9
F
0.015 Equivalence point will be at (20.00 mL / 0.93284 mmol / 6) / 0.01500
20.00 mmol/mL=22.98 mL
22.98
Vol. Cr2O72-, mL
Percentages
[Fe3+]
[Fe2+]
[Cr2O72-]
[Cr3+]
[H+]
E, V
10
10
2.30
0.0093
0.0835
1.00
0.715
11
20
4.60
0.0168
0.0673
1.00
0.735
12
30
6.89
0.0231
0.0538
1.00
0.749
13
40
9.19
0.0283
0.0425
1.00
0.761
14
50
11.49
0.0328
0.0328
1.00
0.771
15
60
13.79
0.0367
0.0245
1.00
0.781
16
70
16.09
0.0401
0.0172
1.00
0.793
17
80
18.38
0.0431
0.0108
1.00
0.807
18
90
20.68
0.0458
0.0051
1.00
0.828
19
95
21.83
0.0470
0.0025
1.00
0.847
20
99
22.75
0.0479
0.00047911
1.00
0.889
21
99.9
22.96
0.0481
4.349E-05
1.00
0.951
22
100
22.98
0.0160
1.00
1.26
23
101
23.21
8.05E-05
0.0160
1.00
1.33
24
105
24.13
3.91E-04
0.0156
1.00
1.33
25
110
25.28
7.62E-04
0.0152
1.00
1.34
26
120
27.58
0.0014
0.0145
1.00
1.34
27
Spreadsheet Documentation
28
B10=(A10/11)*$B$8
F22=($B$5*$B$7/3)/($B$7+B22)
29
C10=($B$6*B10*6)/($B$7+B10)
H22=(($B$4+6*$B$3)/7)-(0.0592/7)*LOG10(2*F22/G22^14)
30
D10=($B$5*$B$7-$B$6*B10*6)/($B$7+B10)
E23=($B$6*B23-($B$5*$B$7/6)/($B$7+B23)
31
H10=$B$4-0.0592*LOG10(D10/C10)
H23=$B$3-(0.0592/6)*LOG10(F23^2/(E23*G23^14))
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Fundamentals of Analytical Chemistry: 8th ed.
A
B
Chapter 20
C
D
2
20-54 (e) Titration of 35.00 mL 0.0578 M IO3- with 0.05362 M Na2S2O3
Reactions: IO3- + 5I- + 6H+  3I2 + 3H2O; I2 + 2S2O32-  2I- + S4O62-
3
For thiosulfate, Eo
1
4
For I2/I-, E
5
IO3-
Initial
F
G
0.08
0.615 There are 35.00 mL X 0.0578 mmol/mL = 2.023 mmol IO3-
o
0.0578 initially present. Every mmol of IO3- requires 6 mmol S2O32-.
concentration
S2O32-
6
Concentration
7
Initial volume solution
8
Equivalence point volume
9
E
0.05362 Equivalence point will be at ( 2.023 mmol * 6) /
35.00 0.05362 mmol/mL=226.37 mL
226.37
Vol. S2O32-, mL
Percentages
[I-]
[I2]
[S2O32-]
[S4O62-]
E, V
10
10
22.64
0.0211
0.0948
0.684
11
20
45.27
0.0302
0.0605
0.669
12
30
67.91
0.0354
0.0413
0.660
13
40
90.55
0.0387
0.0290
0.653
14
50
113.19
0.0410
0.0205
0.647
15
60
135.82
0.0426
0.0142
0.641
16
70
158.46
0.0439
0.0094
0.635
17
80
181.10
0.0449
0.0056
0.628
18
90
203.73
0.0458
0.0025
0.617
19
95
215.05
0.0461
0.0012
0.608
20
99
224.11
0.0464
2.34E-04
0.587
21
99.9
226.14
0.0464
2.33E-05
0.557
22
100
226.37
23
101
228.63
4.60E-04
0.0230
0.23
24
105
237.69
2.23E-03
0.0223
0.19
25
110
249.01
4.27E-03
0.0214
0.17
26
120
271.64
7.92E-03
0.0198
0.15
30
Spreadsheet Documentation
B10=(A10/100)*$B$8
C10=($B$6*B10)/($B$7+B10)
D10=(($B$5*$B$7*3)-($B$6*B10/2))/($B$7+B10)
31
H10=$B$4-(0.0592/2)*LOG10(D10^2/C10)
27
28
29
32
33
34
35
36
37
38
39
40
41
42
43
44
45
0.35
E23=(($B$6*B23)-($B$5*$B$7*6)/($B$7+B23)
F23=(($B$5*$B$7*3)/($B$7+B23)
H22=($B$3+$B$4)/2
H23=$B$3-(0.0592/2)*LOG10(E23^2/F23)