Solutions to Chapter 19

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


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 


[ 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.


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


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).


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).


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


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


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


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+]


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


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+]


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


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


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)


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


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 


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


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.


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


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


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


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.


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

Solutions to Chapter 19

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