Skoog - Solucionário Capítulo 20

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 20

Chapter 20

20-1

(a) 2Mn2  5S2 O8

2

 8H 2 O  10SO 4

2



 2MnO4  16H 

(b) NaBiO 3 ( s)  2Ce3  4H   BiO   2Ce 4  2H 2O  Na  (c) H 2 O2  U 4  UO2

2

 2H 



(d) V(OH) 4  Ag( s)  Cl   2H   VO 2  AgCl ( s)  3H 2 O 

(e) 2MnO4  5H 2 O2  6H   5O2  2Mn2  8H 2 O 

(f) ClO 3  6I   6H   3I 2  Cl   3H 2 O

20-2

(a) 2Fe3  SO 2 ( g )  2H 2 O  2Fe2  SO 4

2

 4H 

(b) 2H 2 MoO4  3Zn( s)  12H   3Zn2  2MoO3  8H 2O 



(c) 2MnO4  5HNO2  H   2Mn2  5NO3  3H 2 O 

(d) BrO3  5Br   C6 H 5 NH 2  3H   Br3C6 H 5 NH 2  3Br   3H 2 O 2

(e) 2HAsO3  O2 ( g )  2HAsO4

2

(f) 2I   2HNO2  2H   I 2  2NO  2H 2O 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)  4H   2Cl 

 

U 4  2AgCl ( s)  H 2 O

20-5

UO2

20-6

2TiO2  Zn( s)  4H 

20-7

Standard solutions of reductants find somewhat limited use because of their susceptibility

 

2Ti3  Zn2  2H 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 2MnO4  3Mn2  2H 2 O  5MnO2 ( s)  4H  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 4MnO4  2H 2 O  4MnO2 ( 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 2 O 2



20-17 S2 O3  H   HSO 3  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 2 O 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 O 7

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 Fe2  4.03961 mmol Fe2 55.847 g

4.03961 mmol Fe2 1 mmol Ce 4   0.1142 M Ce 4 2 35.37 mL mmol Fe 2

4.03961 mmol Fe2 1 mmol Cr2 O 7 (b)  35.37 mL 6 mmol Fe2

 0.01904 M Cr2 O7

2



(c)

4.03961 mmol Fe2 1 mmol MnO4    0.02284 M MnO4 2 35.37 mL 5 mmol Fe 

4.03961 mmol Fe2 1 mmol V(OH) 4  (d)   0.1142 M V(OH) 4 2 35.37 mL mmol Fe 

4.03961 mmol Fe2 1 mmol IO3  (e)   0.02855 M IO3 2 35.37 mL 4 mmol Fe

20-24 0.02500 mmol K 2 Cr2 O7 294.185 g K 2 Cr2 O 7  500.0 mL   3.677 g K 2 Cr2 O7 mL 1000 mmol

Dissolve 3.677 g K2Cr2O7 in water and dilute to 500.0 mL. 20-25

0.02500 mmol KBrO3 1000 mL 167.001 g KBrO3  2.000 L    8.350 g KBrO3 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 KMnO4 1000 mL 158.034 g KMnO4  2.0 L    15.80 g KMnO4 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 2MnO4  5H 2 C2 O4  6H   2Mn2  10CO2 ( g )  8H 2 O 0.1756 g Na 2 C 2 O 4 1000 mmol Na 2 C2 O 4 2 mmol KMnO4    0.01636 M KMnO4 32.04 mL KMnO4 133.999 g 5 mmol Na 2 C 2 O 4

20-29 Ce 4  Fe2  Ce3  Fe3

0.1809 g Fe 1000 mmol Fe 1 mmol Fe2 1 mmol Ce 4     0.1034 M Ce 4 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  7H 2 O

I 2  2S2 O 3

2

 2I   S 4 O 6

2

1 mmol Cr2O72-  3 mmol I2  6 mmol S2O322

0.1259 g K 2 Cr2 O7 1000 mmol K 2 Cr2 O7 6 mmol S2 O3    0.06223 M Na 2S2 O3 41.26 mL Na 2S2 O3 294.185 g mmol K 2 Cr2 O7

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 20

20-31 

BrO 3  6I   6H   Br   3I 2  3H 2 O I 2  2S2 O 3

2

 2I   S 4 O 6

2

1 mmol BrO3-  3 mmol I2  6 mmol S2O322

0.1017 g KBrO3 1000 mmol KBrO3 6 mmol S2 O3    0.09192 M Na 2S2 O 3 39.75 mL Na 2S2 O3 167.001 g mmol KBrO3 20-32 1 mmol I2  1 mmol Sb3+  2 mmol Sb2S3 (a)

0.02870 mmol I 2 1 mmol Sb 3 121.760 g Sb 3  44.87 mL I 2   mL mmol I 2 1000 mmol  100%  16.03% Sb 3 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 MnO2  2I   4H   Mn2  I 2  2 H 2 O I 2  2S2 O 3

2

 2I   S 4 O 6

2

1 mmol MnO2  1 mmol I2  2 mmol S2O32-

 0.07220 mmol Na 2S2 O 3 1 mmol MnO2 86.937 g MnO2     32.30 mL Na 2S 2 O 3   mL 2 mmol Na 2S2 O 3 1000 mmol    100% 0.1344 g sample  75.42% MnO2

Fundamentals of Analytical Chemistry: 8th ed. 

20-34 3CS( NH 2 ) 2  4BrO3  3H 2 O  3CO( NH 2 ) 2  3SO 4

Chapter 20 2

 4Br   6H 

4 mmol BrO3-  3 mmol CS(NH2)2

 0.00833 mmol KBrO3 3 mmol CS( NH 2 ) 2 76.122 g CS( NH 2 ) 2     14.1 mL KBrO3   mL 4 mmol KBrO3 1000 mmol    100% 0.0715 g sample  9.38% CS( NH 2 ) 2



20-35 MnO4  5Fe2  8H   Mn2  5Fe3  4H 2 O 1 mmol MnO4-  5 mmol Fe2+  5/2 mmol Fe2O3 mmol KMnO4 

0.02086 KMnO4  39.21 mL KMnO4  0.8179 mmol KMnO4 mL

(a)

 5 mmol Fe2 1 mmol Fe 55.847 g Fe   0.8179 mmoL KMnO4     mmol KMnO4 mmol Fe2 1000 mmol    100% 0.7120 g sample  32.08% Fe (b)  5 mmol Fe2 O 3 159.692 g Fe2 O 3   0.8179 mmoL KMnO4    2 mmol KMnO4 1000 mmol    100%  45.86% Fe2 O 3 0.7120 g sample

20-36 3Sn 3  Cr2 O7

2

 14H 

 

3Sn 4  2Cr 3  7H 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

2 H 2 NOH  4 Fe3 Cr2 O 7

2

 

N 2 O( g)  4 Fe2  4 H   H 2 O

 6Fe2  14 H 

 

2Cr 3  6Fe3  7 H 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 Zn2  H 2 C 2 O 4  ZnC2 O 4 ( s )  2 H  

2MnO4  5H 2 C 2 O 4  6H   2Mn2  10CO 2 ( g )  8H 2 O

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 20

2 mmol MnO4-  5 mmol Na2C2O4  5 mmol ZnO

 0.01508 mmol KMnO4 5 mmol ZnO 81.139 g ZnO     37.81 mL KMnO4   mL 2 mmol KMnO4 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 2 O  6Fe3 0.08930 mmol Fe2  50.00 mL Fe2  4.4650 mmol Fe2 mL 0.083610 mmol Ce 4 2 4 mmol Fe titrated by Ce   14.93 mL Ce 4  mL 2 1 mmol Fe  1.2483 mmol Fe2 mmol Ce 4 mmol Fe2 

mmol Fe2 reacted with KClO3  4.4650  1.2483  3.2167 mmol Fe2

 1 mmol KClO3 122.549 g KClO3   3.2167 mmoL Fe2    6 mmol Fe2 1000 mmol    100%  51.37% KClO3 0.1279 g sample

20-40

Pb(C 2 H 5 ) 4  I 2  Pb(C 2 H 5 ) 3 I  C 2 H 5 I I 2  2S2 O 3

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 S 2 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 3AsO3  I 2  H 2 O  H 3AsO4  2I   2H  1 mmol I2  1 mmol H3AsO3  ½ mmol As2O3

 0.01985 mmol I 2 1 mmol As2 O 3 197.841 g As2 O 3     24.56 mL I 2   mL 2 mmol I 2 1000 mmol    100% 7.41 g sample  0.651% As2 O 3

20-42 Cr2 O7

2

 3U 4  2H   3UO2

2

 2Cr 3  H 2 O

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 2C2 H 5SH  I 2  C2 H 5SSC 2 H 5  2I   2H 

0.01293 mmol I 2  50.0 mL I 2  0.6465 mmol I 2 mL 2 mmol I 2 titrated by S 2 O 3  mmol I 2 

0.01425 mmol Na 2S 2 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

 6Fe2  14 H   2Cr 3  6Fe3  7H 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 Fe2  mmol K 2 Cr2 O 7 

1 mmol K 2 Cr2 O 7 0.1135 mmol Fe2  10.05 mL Fe2   0.1901 mmol K 2 Cr2 O 7 mL 6 mmol Fe2 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 2S2 O 3 1 mmol KI 166.00 g KI     19.96 mL Na 2S 2 O 3   mL 6 mmol Na 2S2 O 3 1000 mmol    100% 1.309 g sample  2.524% KI



20-46 MnO4  5Fe2  8H  mmol Fe 

 

Mn2  5Fe3  4H 2 O

0.01920 mmol KM nO4 5 mmol Fe 500.0 mL  13.72 mL KM nO4   mL mmol KM nO4 50.00 mL

 13.1712 mmol Fe mmol Fe and Cr 

0.01920 mmol KM nO4 5 mmol Fe 500.0 mL  36.43 mL KM nO4   mL mmol KM nO4 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  2H   e 

 

VO 2  3H 2 O

In the Jones reductor, 

V(OH) 4  4H   3e 

 

V 2  4 H 2 O

In the first titration, Ce 4  Fe2  Fe3  Ce 3 

Ce 4  VO 2  3H 2 O  V(OH) 4  Ce 3  2H 

Fundamentals of Analytical Chemistry: 8th ed.

mmol Fe and V  mmol Ce 4 

Chapter 20

0.1000 mmol Ce 4 17.74 mL Ce 4  1.7740 mmol Fe and V mL

In the second titration, Ce 4  Fe2  Fe3  Ce 3 

3Ce 4  V 2  4 H 2 O  V(OH) 4  3Ce 3  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 mmol Fe2 O 3 

0.4275 mmol Fe  0.21375 mmol Fe2 O 3 2

0.67325 mmol V2 O5   181.88 g V2 O5 1000 mmol 50.00 mL 2.559 g sample  500.0 mL

 100%  47.85% V2 O 5

0.21375 mmol Fe2 O 3   159.69 g Fe2 O 3 1000 mmol 50.00 mL 2.559 g sample  500.0 mL

 100%  13.34% Fe2 O 3

20-48 2Tl  CrO 4

2

 Tl2 CrO 4 (s)

2Tl2 CrO 4 (s)  2 H   4Tl  Cr2 O 7 Cr2 O 7

2

2

 H 2O

 6Fe2  14 H   2Cr 3  6Fe3  7 H 2 O

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 20

6 mmol Fe2+  1 mmol Cr2O72-  4 mmol Tl+

0.1044 mmol Fe2 4 mmol Tl 204.38 g Tl  39.52 mL Fe2    0.5622 g Tl mL 6 mmol Fe2 1000 mmol

20-49 SO 2 ( g )  2OH   SO 3 

2

 H 2O 



IO3  2H 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 )  5CO 2 ( g )  I 2 ( g ) I 2  2S2 O 3

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 S2 O 3 

5 mmol CO 28.01 g CO   1.2009  10 3 g CO 2 1000 mmol 2 mmol S2 O 3

   1.2009  10 3 g CO     106 ppm  40.5 ppm CO 1 . 2 g  24.7 L     L  

Fundamentals of Analytical Chemistry: 8th ed.

Chapter 20

20-51 S2  I 2  S( s )  2I  I 2  2S2 O 3

2

 2I   S 4 O 6

2

1 mmol I2  1 mmol H2S  2 mmol S2O32-

0.01070 mmol I 2  10.00 mL I 2  0.1070 mmol I 2 mL 2 0.01344 mmol S2 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   6H  

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  4 Mn(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 Mn2  2I 2  6H 2 O 2

 13.67 mL S2 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: Sn

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

2+

+ 2Fe

3+

 Sn

4+

+ 2Fe

2+

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

F

G

H

-

1

20-54 (b) Titration of 25.00 mL of 0.08467 M Na2S2O3 with 0.10235 M I2 (I3 )

2

Reaction: 2S2O3 + I3  S4O6

3

For S2O32-/S4O62-, Eo

4

For I3-/I-, Eo

5

Initial concentration S2O32-

2-

-

2-

+ 3I

-

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 + 2+ Reaction: 2MnO4 + 5H2C2O4 +6H  2Mn + 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

F

G

H

2+

2

20-54 (d) Titration of 20.00 mL 0.1034 M Fe with 0.01500 M K2Cr2O7 2+ 2+ 3+ 3+ Reaction: 6Fe + Cr2O7 +14H  6Fe + 2Cr +7H2O

3

For dichromate, Eo

4

For Fe3+/Fe2+, Eo

5

Initial Fe2+ concentration

6

Concentration Cr2O72-

0.015 Equivalence point will be at (20.00 mL / 0.93284 mmol / 6) / 0.01500

7

Initial volume solution

20.00 mmol/mL=22.98 mL

8

Equivalence point volume

1

9

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+.

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 3

For thiosulfate, Eo

4

For I2/I-, E

5

IO3-

F

G

Initial

0.08

o

0.615 There are 35.00 mL X 0.0578 mmol/mL = 2.023 mmol IO3 -

S2O32-

Concentration

7

Initial volume solution

8

Equivalence point volume

2-

0.0578 initially present. Every mmol of IO3 requires 6 mmol S2O3 .

concentration

6

9

E

-

20-54 (e) Titration of 35.00 mL 0.0578 M IO3 with 0.05362 M Na2S2O3 + 22Reactions: IO3 + 5I + 6H  3I2 + 3H2O; I2 + 2S2O3  2I + S4O6

1

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)