Download - Solutions to Chapter 19

7/25/2019 Solutions to Chapter 19

1/24

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)Equilibriumis the state that a system assumes after each addition of reagent.

Equivalencerefers to a particular equilibrium state when a stoichiometric amount of

titrant has been added.

(b)Atrue oxidation/reduction indicatorowes its color change to changes in the electrode

potential of the system. Aspecific indicatorexhibits 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. Post-

equivalence 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.


2/24


ed. Chapter 19

19-7 (a)

V441.00511.0

1log

2

0592.0403.0right

E

V290.0)151.0(441.0

V151.01393.0

1log

2

0592.0126.0

leftrightcell

left

EEE

E

BecauseEcellis 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)

V04.2)806.0(23.1

V806.00364.0

1log

2

0592.0763.0

V23.11006.9

0620.0log

2

0592.025.1

leftrightcell

left

3right

EEE

E

E

BecauseEcellis positive, the reaction proceeds spontaneously in the direction considered.

(c)

V0620.0)237.0(299.0

V237.01000.1

760/765log

2

0592.0000.0

V299.00214.0

1log

2

0592.0250.0

leftrightcell

24left

right

EEE

E

E

BecauseEcellis 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)

V785.01059.4

1log

2

0592.0854.0

3right

E


3/24


ed. Chapter 19

2

92922

]I[

109.7]Pb[and109.7]I][Pb[

V04.1)252.0(785.0

V252.0

109.7

0120.0log

2

0592.0126.0

leftrightcell

9

2

left

EEE

E


(e)

10

10

3

310

4

33

1093.4438.0

379.01070.5]OH[

379.0

438.0]OH[1070.5

]NH[

]NH][OH[

V551.0)551.0(000.0

V551.01093.4

00.1log

2

0592.0000.0

V000.0

leftrightcell

210left

right

EEE

E

E


(f)

V286.0)063.0(223.0

V063.01047.10790.0

00918.0log0592.0099.0

V223.00538.01340.0

0784.0log0592.0359.0

leftrightcell

22left

2right

EEE

E

E



4/24


ed. Chapter 19

19-8 (a)

V452.0)793.0(341.0

V793.00955.0

1

log2

0592.0

763.0

V341.01078.6

1log

2

0592.0277.0

leftrightcell

left

3right

EEE

E

E

BecauseEcellis positive, the reaction proceeds spontaneously in the direction considered

(oxidation on the left, reduction on the right).

(b)

V819.00671.0

1

log2

0592.0

854.0right

E

V788.01310.0

0681.0log

2

0592.0771.0left

E

V031.0788.0819.0leftrightcell EEE



(c)

V414.0751.0165.1

V751.01544.0

1log0592.0799.0

V165.10794.012.1

1log

4

0592.0229.1

leftrightcell

left

4right

EEE

E

E




5/24


ed. Chapter 19

(d)

V401.0301.0100.0

V301.00601.0

1log

2

0592.0337.0

V100.01350.0log0592.0151.0

leftrightcell

left

right

EEE

E

E

BecauseEcellis negative, the reaction does not proceed spontaneously in the direction

considered (reduction on the left, oxidation on the right).

(e)

443

343

1007.30764.0

1302.01080.1]OH[

1302.0

0764.0]OH[1080.1

]HCOOH[

]HCOO][OH[

V000.0

V208.01007.3

00.1log

2

0592.0000.0

left

24right

E

E

V208.0000.0208.0leftrightcell EEE

BecauseEcellis negative, the reaction does not proceed spontaneously in the direction

considered (reduction on the left, oxidation on the right).

(f)

V724.0)040.0(684.0

V040.0

1016.11093.7

1037.6log

2

0592.0334.0

V684.0003876.0

1134.0log0592.0771.0

leftrightcell

4

33

2

left

right

EEE

E

E




6/24


ed. Chapter 19

19-9 (a)

V631.0)789.0(158.0

V789.01364.0

1

log2

0592.0

763.0

V158.00848.0

1log

2

0592.0126.0

leftrightcell

Zn

Pb

2

2

EEE

E

E

(b)

V286.0461.0747.0

V461.01564.0

00309.0log0592.036.0

V747.00301.0

0760.0log0592.0771.0

leftrightcell

)CN(Fe

Fe

36

3

EEE

E

E

(c)

V331.0)331.0(000.0

V331.0101046.1

02723.0log0592.0099.0

V000.0

leftrightcell

233T iO

SHE

2

EEE

E

E

19-10 (a)ZnZn2+(0.1364 M)Pb2+(0.0848 M)Pb

(b)PtFe(CN)64-

(0.00309 M), Fe(CN)63-

(0.1564 M)Fe3+(0.0301 M), Fe2+(0.0760 M)Pt

(c)PtTiO+(1.4610-3M), Ti3+(0.02723 M), H+(1.0010-3M)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)

256.0771.0VFeVFe oV

o

Fe

322333

EE


7/24


ed. Chapter 19

1717

eq23

32

eq23

32

3

2

3

2

102.21023.2]V][Fe[

]V][Fe[

348.17log]V][Fe[

]V][Fe[log

0592.0

256.0771.0

]V[

]V[log0592.0256.0

]Fe[

]Fe[log0592.0771.0

K

K

(b)

408.036.0Cr)CN(FeCr)CN(Fe oCr

o

)CN(Fe

34

6

23

6 336

EE

1212

eq23

6

34

6

eq23

6

34

6

3

2

3

6

4

6

109104.9]Cr][)CN(Fe[

]Cr][)CN(Fe[

973.12log]Cr][)CN(Fe[

]Cr][)CN(Fe[log

0592.0

408.036.0

]Cr[

]Cr[log0592.0408.0

])CN(Fe[

])CN(Fe[log0592.036.0

K

K

(c)

334.000.1OH4UOVO2U)OH(V2 oUO

o

)OH(V2

2

2

24

4 224

EE

2222

eq42

4

2

2

22

eq42

4

2

2

22

42

2

4

42

4

22

103102.3]U[])OH(V[

]UO[]VO[

50.22log]U[])OH(V[

]UO[]VO[log

0592.0

2334.000.1

]H][UO[

]U[log

2

0592.0334.0

]H[])OH(V[

]VO[log

2

0592.000.1

K

K

(d)

25.1771.0Fe2TlFe2Tl oT l

o

Fe

3233

EE


8/24


ed. Chapter 19

1616

eq223

23

eq223

23

23

22

3

102105.1]Fe][Tl[

]Fe][Tl[

18.16log]Fe][Tl[

]Fe][Tl[log

0592.0

2771.025.1

]Fe[

]Fe[log

2

0592.0771.0

]Tl[

]Tl[log

2

0592.025.1

K

K

(e)

577.070.1)HClOM1in(

H2AsOHCe2OHAsOHCe2

o

AsOH4

o

Ce

43

3

233

4

434

EE

3737

eq

43

24

2

33

23

eq

43

24

2

33

23

2

33

43

24

23

109109.8]AsOH[]Ce[

]H][AsOH[]Ce[

94.37log]AsOH[]Ce[

]H][AsOH[]Ce[log

0592.0

2577.070.1]H][AsOH[

]AsOH[log

2

0592.0577.0

]Ce[

]Ce[log

2

0592.070.1

K

K

(f)

172.000.1OH5SOVO2SOH)OH(V2 oSO

o

)OH(V2

2

4

2

324 244

EE

2727

eq

32

2

4

2

4

22

eq

32

2

4

2

4

22

42

4

32

42

4

22

109104.9]SOH[])OH(V[

]SO[]VO[

97.27log]SOH[])OH(V[

]SO[]VO[log

0592.0

2172.000.1

]H][SO[

]SOH[log

2

0592.0172.0

]H[])OH(V[

]VO[log

2

0592.000.1

K

K

(g)

256.0359.0OHV2H2VVO oV

o

VO2

3223@

EE


9/24


ed. Chapter 19

10

eq222

23

eq222

23

3

2

22

3

104.2]V[]H][VO[

]V[

389.10log]V[]H][VO[

]V[log

0592.0

256.0359.0

]V[

]V[log0592.0256.0

]H][VO[

]V[log0592.0359.0

K

K

(h)

369.0099.0OHTi2H2TiTiO oT

o

T iO2

32232

EE

7

eq222

23

eq222

23

3

2

22

3

100.8]Ti[]H][TiO[

]Ti[

9054.7log]Ti[]H][TiO[

]Ti[log

0592.0

369.0099.0

]Ti[

]Ti[log0592.0369.0

]H][TiO[

]Ti[log0592.0099.0

K

K

19-12 (a)

At equivalence, [Fe2+

] = [V3+

] and [Fe3+

] = [V2+

]

V258.02

256.0771.0

]V][Fe[

]V][Fe[log0592.0256.0771.02

]V[

]V[log0592.0256.0

]Fe[

]Fe[log0592.0771.0

eq

33

22

eq

3

2

eq

3

2

eq

E

E

E

E

(b)

At equivalence, [Fe(CN)63-

] = [Cr2+

] and [Fe(CN)64-

] = [Cr3+

]


10/24


ed. Chapter 19

V024.02

408.036.0

]Cr][)CN(Fe[

]Cr][)CN(Fe[log0592.0408.036.02

]Cr[

]Cr[log0592.0408.0

])CN(Fe[

])CN(Fe[log0592.036.0

eq

33

6

24

6eq

3

2

eq

3

6

4

6eq

E

E

E

E

(c)

At equivalence, [VO2+

] = 2[UO22+

] and [V(OH)4+] = 2[U

4+]

V444.03333.1

V333.1355.0688.1100.0

1log0592.0344.0200.13

]H][UO][)OH(V[

]U][VO[log0592.0344.0200.13

]H][UO[

]U[log0592.0344.022

]H][)OH(V[]VO[log0592.000.1

eq

6eq

62

24

42

eq

42

2

4

eq

2

4

2

eq

E

E

E

E

E

(d)

At equivalence, [Fe2+

] = 2[Tl3+

] and [Fe3+

] = 2[Tl+]

V09.13

27.3

]Tl][Tl[2

]Tl][Tl[2log0592.027.3

]Tl][Fe[

]Tl][Fe[log0592.025.12771.03

]Tl[

]Tl[

log0592.025.122

]Fe[

]Fe[log0592.0771.0

eq

3

3

33

2

eq

3eq

3

2

eq

E

E

E

E


11/24


ed. Chapter 19

(e)

At equivalence, [Ce3+

] = 2[H3AsO4], [Ce4+

] = 2[H3AsO3] and [H+] = 1.00

V951.03

854.2

00.1

1log0592.0854.2

]H][AsOH][AsOH[2

]AsOH][AsOH[2log0592.0854.2

]H][AsOH][Ce[

]AsOH][Ce[log0592.0577.0270.13

]H][AsOH[

]AsOH[log0592.0577.022

]Ce[

]Ce[log0592.070.1

eq

22

4333

3343

2

43

4

33

3

eq

2

43

33eq

4

3

eq

E

E

E

E

(f)

At equivalence, [V(OH)4+] = 2[H2SO3] and [VO

2+] = 2[SO4

2-]

V330.03

1089.9

V1089.9355.0344.1100.0

1log0592.0172.0200.13

]H][SO][)OH(V[

]SOH][VO[log0592.0172.0200.13

]H][SO[

]SOH[log0592.0172.022

]H][)OH(V[

]VO[log0592.000.1

1

eq

1

6eq

62

44

32

2

eq

42

4

32eq

2

4

2

eq

E

E

E

E

E

(g)

At equivalence, [VO+] = [V

2+]


12/24


ed. Chapter 19

V008.02

154.0

V154.0118.0103.0100.0

1log0592.0103.02

]H][V[

]V[log0592.0103.0

]H][VO[

]V[log0592.0256.0359.02

]V[

]V[log0592.0256.0

]H][VO[

]V[log0592.0359.0

eq

2eq

22

2

2

2

eq

3

2

eq

22

3

eq

E

E

E

E

E

(h)

At equivalence, [Ti2+

] = [TiO2+

]

V194.02

388.0

V388.0118.0270.0100.0

1log0592.0270.02

]H][Ti[

]Ti[log0592.0103.0

]H][TiO[

]Ti[log0592.0369.0099.02

]Ti[

]Ti[log0592.0369.0

]H][TiO[

]Ti[log0592.0099.0

eq

2eq

22

2

2

2

eq

3

2

eq

22

3

eq

E

E

E

E

E

19-13 (a)In the solution to Problem 19-11(a) we find

17

32

23

eq 1023.2]Fe][V[

]Fe][V[

K

At the equivalence point,

0500.02

1000.0]Fe[]V[

]Fe[]V[

23

32

x

Substituting into the first equation we find


13/24


ed. Chapter 19

M0500.0]Fe[]V[

and

M1006.1]Fe[]V[,Thus

1006.11023.2

00250.0

0500.01023.2

23

1032

10

17

2

2

17

x

x

(b)Proceeding in the same way, we find

M0500.0]Cr[])CN(Fe[ 34

6

M107.1][Cr][Fe(CN) 8236

(c)At equivalence

M0333.03

1000.0]UO[

M0667.03

2000.02

3

1000.02]UO[2]VO[

]U[2])OH(V[

2

2

2

2

2

2

4

4

x

x

From the solution for Problem 19-11(c)

M100.12

101.2]U[andM101.2])OH(V[

M1010.21062.42

102.3

0333.00667.0

2

2

0333.00667.0102.3

]U[])OH(V[

]UO[]VO[

99

49

4

93 27

22

23

2

2

22

42

4

2

2

22

2

x

x

xx

(d)

M0333.0]Tl[andM0667.0]Fe[

]Tl[2]Fe[

3

32

x


14/24


ed. Chapter 19

From the solution for Problem 19-11(d)

]Tl[M104.12

1070.2

2

]Fe[M1070.21088.92

105.10333.00667.0

2

2

0667.00333.0105.1

]Fe][Tl[

]Fe][Tl[

377

273 21

16

23

2

2

16

223

23

x

x

x

xx

(e)Proceeding as in part (c), we find

2

0333.00667.0

00.1]AsOH[]Ce[

]AsOH[]Ce[109.8

]H][AsOH[]Ce[

]AsOH[]Ce[3

2

2

33

24

43

2337

2

33

24

43

23

x

When ]AsOH[2]Ce[ 334 x

M033.0]AsOH[

M067.0]Ce[

M105.3]AsOH[

M109.6]Ce[

43

3

14

33

144

(f)Proceeding as in part (c)

M106.1]SOH[

M102.3])OH(V[

M033.0]SO[

M067.0]VO[

11

32

11

4

2

4

(g)

100.02

200.0]V[

]VO[]V[

3

22

x

Assume [H+] = 0.1000. From the solution to Problem 19-11(g)


15/24


ed. Chapter 19

M100.0]V[

M105.6]V[]VO[

00.1104.2

100.0]V][VO[

100.0104.2

]H][V][VO[

]V[

3

622

2

10

222

2

10

222

23

x

x

(h)Proceeding as in part (g), we find

M101.1]Ti[]TiO[

M100.0]H[

M100.0]Ti[

52

2

3

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)

2342 SnV2SnV2

At 10.00 mL,


16/24


ed. Chapter 19

V292.00167.0

0667.0log0592.0256.0

]V[

]V[log0592.0256.0

M0667.0M1067.1mL00.60

mL00.50mL

Vmmol1000.0

]V[

M0167.0mL00.60

Snmmol

Vmmol2mL00.10

mL

Snmmol0500.0

]V[

3

2

2

2

2

4

34

3

E

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

]Sn][V[

]Sn][V[log0592.0154.02256.03

]Sn[

]Sn[log0592.0154.022

]V[

]V[log0592.0256.0

43

22

4

2

3

2

E

E

E

At equivalence, [V2+

] = 2[Sn4+

] and [V3+

] = 2[Sn2+

]. Thus,

V017.0

3

00.1log0592.0

3

154.02256.0eq

E

At 50.10 mL,

V074.0100.5

105.2log

2

0592.0154.0

]Sn[

]Sn[log

2

0592.0154.0

]Sn[M100.5M025.0mL10.100

mL10.50mL

Snmmol0500.0

]Sn[M025.0mL10.100

Vmmol2

Snmmol1mL00.50

mL

Vmmol1000.0

5

2

4

2

45

4

Sn

2

2

22

Sn

4

2

E

c

c


17/24


ed. Chapter 19

The remaining post-equivalence points are obtained in the same way and are found in the

table that follows.

A B C D E F G

1 19-15 (a) Titration of 50.00 mL of 0.1000 M V2+

with 0.0500 M Sn4+

2 Reaction: 2V2+

+ Sn4+

2V3+

+ Sn2+

3 For V3+

/V2+

, E0 -0.256

4 For Sn4+

/Sn2+

, E0 0.154

5 Initial conc. V2+

0.1000

6 Conc. Sn4+

0.0500

7 Volume solution, mL 50.00

8 Vol. Sn4+

, mL [V3+

] [V2+

] [Sn4+

] [Sn2+

] E, V

9 10.00 0.0167 0.0667 -0.292

10 25.00 0.0333 0.0333 -0.256

11 49.00 0.0495 0.0010 -0.15612 49.90 0.0499 0.0001 -0.096

13 50.00 0.017

14 50.10 5.00E-05 0.0250 0.074

15 51.00 4.95E-04 0.0248 0.104

16 60.00 4.55E-03 0.0227 0.133

17 Spreadsheet Documentation

18 B9=$B$6*A9*2/($B$7+A9)

19 C9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)

20 F9=$B$3-0.0592*LOG10(C9/B9)

21 F13=($B$3+2*$B$4)/3

22 D14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)

23 E14=$B$7*$B$5/(2*($B$7+A14))

24 F14=$B$4-(0.0592/2)*LOG10(E14/D14)

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41


18/24


ed. Chapter 19

(b)

34

6

23

6 Cr)CN(FeCr)CN(Fe

The pre-equivalence point data are obtained by substituting concentrations into the

equation

])CN(Fe[

])CN(Fe[log0592.036.0

3

6

4

6E

The post-equivalence point data are obtained with the Nernst expression for the Cr2+

/ Cr3+

system. That is,

]Cr[

]Cr[log0592.00408.0

3

2

E

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.


19/24


ed. Chapter 19

A B C D E F

1 19-15 (b) Titration of 50.00 mL of 0.1000 M Fe(CN)63-

with 0.1000 M Cr2+

2 Reaction: Fe(CN)63-

+ Cr2+

Fe(CN)64-

+ Cr3+

3 For Fe(CN)6

4-, E

0 0.36

4 For Cr3+

/Cr2+

, E0 -0.408

5 Initial conc. Fe(CN)63-

0.1000

6 Conc. Cr2+

0.1000


8 Vol. Cr2+

, mL [Fe(CN)64-

] [Fe(CN)63-

] [Cr3+

] [Cr2+

] E, V

9 10.00 0.0167 0.0667 0.40

10 25.00 0.0333 0.0333 0.36

11 49.00 0.0495 0.0010 0.26

12 49.90 0.0499 0.0001 0.20

13 50.00 -0.0214 50.10 0.0500 9.99E-05 -0.248

15 51.00 0.0495 0.0010 -0.307

16 60.00 0.0455 0.0091 -0.367


18 B9=$B$6*A9/($B$7+A9)

19 C9=($B$5*$B$7-$B$6*A9)/($B$7+A9)

20 F9=$B$3-0.0592*LOG10(B9/C9)

21 F13=($B$3+$B$4)/2

22 D14=($B$5*$B$7/($B$7+A14)

23 E14=($B$6*A14-$B$5*$B$7)/($B$7+A14)

24 F14=$B$4-0.0592*LOG10(E14/D14)25

26

27

28

29

30

31

32

33

34

3536

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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).

A B C D E F

1 19-15 (c) Titration of 50.00 mL of 0.1000 M Fe(CN)64-

with 0.0500 M Tl3+

2 Reaction: 2Fe(CN)64-

+ Tl3+

2Fe(CN)63-

+ Tl+

3 For Fe(CN)64-

, E0 0.36

4 For Tl3+

/Tl+, E

0 1.25

5 Initial conc. Fe(CN)64-

0.1000

6 Conc. Tl3+

0.0500


8 Vol. Tl3+

, mL [Fe(CN)64-

] [Fe(CN)63-

] [Tl3+

] [Tl+] E, V

9 10.00 0.0667 0.0167 0.32

10 25.00 0.0333 0.0333 0.36

11 49.00 0.0010 0.0495 0.4612 49.90 0.0001 0.0499 0.52

13 50.00 0.95

14 50.10 5.00E-05 0.0250 1.17

15 51.00 4.95E-04 0.0248 1.20

16 60.00 4.55E-03 0.0227 1.23


18 B9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)

19 C9=$B$6*A9*2/($B$7+A9)

20 F9=$B$3-0.0592*LOG10(B9/C9)

21 F13=($B$3+2*$B$4)/3

22 D14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)23 E14=($B$5*$B$7/2)/($B$7+A14)

24 F14=$B$4-(0.0592/2)*LOG10(E14/D14)

25

26

27

28

29

30

31

32

3334

35

36

37

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21/24


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).

A B C D E F

1 19-15 (d) Titration of 0.1000 M Fe3+

with Sn2+

2 Reaction: 2Fe3+

+ Sn2+

2Fe2+

+ Sn4+

3 For Fe3+

/Fe2+

E0 0.771

4 For Sn4+

/Sn2+

, E0 0.154

5 Initial conc. Fe3+

0.1000

6 Conc. Sn2+

0.0500


8 Vol. Sn2+

, mL [Fe3+

] [Fe2+

] [Sn4+

] [Sn2+

] E, V

9 10.00 0.0667 0.0167 0.807

10 25.00 0.0333 0.0333 0.771

11 49.00 0.0010 0.0495 0.67112 49.90 0.0001 0.0499 0.611

13 50.00 0.360

14 50.10 0.0250 5.00E-05 0.234

15 51.00 0.0248 4.95E-04 0.204

16 60.00 0.0227 4.55E-03 0.175


18 B9=($B$5*$B$7-$B$6*A9*2)/($B$7+A9)

19 C9=$B$6*A9*2/($B$7+A9)

20 F9=$B$3-0.0592*LOG10(B9/C9)

21 F13=($B$3+2*$B$4)/3

22 D14=(($B$5*$B$7/2)/($B$7+A14)

23 E14=($B$6*A14-$B$5*$B$7/2)/($B$7+A14)

24 F14=$B$4-(0.0592/2)*LOG10(E14/D14)

25

26

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28

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22/24


ed. Chapter 19

(e)

H4UO5Mn2OH2U5MnO2 2

2

2

2

4

4

At 10.00 mL,

V316.0

00.11033.8[

1033.3log

2

0592.0334.0

]H][UO[

]U[log

2

0592.0334.0

UM0333.01033.8mL00.60

Ummol5000.2]U[

UOM1033.8solutionmL00.60

MnOmmol2

UOmmol5MnOmmol2000.0

]UO[

Ummol500.2UmL00.50mL

Ummol05000.0

MnOmmol2000.0MnOmL00.10mL

MnOmmol02000.0

43

2

42

2

4

434

4

U

2

2

34

2

24

2

2UO

444

444

4

22

E

c

c

Additional pre-equivalence point data, obtained in the same way, are given in the

spreadsheet that follows.

At 50.00 mL,

42

2

4

eq

8

4

2

eq

]H][UO[

]U[log0592.0334.022

]H][MnO[

]Mn[log0592.051.155

E

E

Adding the two equations gives

122

24

42

eq]H][UO][MnO[

]U][Mn[log0592.0334.0251.157E

At equivalence, [MnO4-] = 2/5[U

4+] and [Mn

2+] = 2/5[UO2

2+]

Substituting these equalities and [H+] = 1.00 into the equation above gives

V17.1

7

218.8

7

00.1log0592.0

7

218.8eq E


23/24


ed. Chapter 19

At 50.10 mL,

V48.100.110.100/0020.1

10.100/000.1log

5

0592.051.1

]H][MnO[

]Mn[log

5

0592.051.1

MnOmmol100.2000.10020.1remainingMnOmmol

Mnmmol000.1Ummol5

Mnmmol2

UmL00.50mL

Ummol05000.0

formedMnmmol

MnO0020.1MnOmL10.50mL

MnO02000.0addedMnOmmol

88

4

2

4

3

4

2

4

24

42

444

4

E

The remaining post equivalence point data are derived in the same way and are given in

the spreadsheet that follows.


24/24


ed. Chapter 19

A B C D E F G

1 19-15 (e) Titration of 50.00 mL 0.05000 M U4+

with 0.02000 M MnO4-

2 Reaction: 2MnO4-+ 5U

4++2H2O 2Mn

2++ 5UO2

2++ 4H

+

3 For U4+

/UO22+

, E0 0.334

4 For MnO4-, E0 1.51

5 Initial conc. U4+

0.0500

6 Conc. MnO4- 0.0200


8 Vol. MnO4-, mL [U

4+] [UO2

2+] [MnO4

-] [Mn

2+] [H

+] E, V

9 10.00 0.0333 0.0083 1.00 0.316

10 25.00 0.0167 0.0167 1.00 0.334

11 49.00 0.0005 0.0247 1.00 0.384

12 49.90 0.0001 0.0250 1.00 0.414

13 50.00 1.00 1.17

14 50.10 2.00E-05 0.0100 1.00 1.48

15 51.00 0.0002 0.0099 1.00 1.49

16 60.00 0.0018 0.0091 1.00 1.50


18 B9=$B$5*$B$7-$B$6*A9*5/2)/($B$7+A9)

19 C9=($B$6*A9*5/2)/($B$7+A9)

20 F9=1.00 (entry)

21 G9=$B$3-(0.0592/2)*LOG10(B9/(C9*F9^4))

22 G13=((5*$B$4+2*$B$3)/7)-(0.0592/7)*LOG10(F13)

23 D14=($B$6*A14-$B$5*$B$7*2/5)/($B$7+A14)

24 E14=($B$5*$B$7*2/5)/($B$7+A14)

25 G14=$B$4-(0.0592/5)*LOG10(E14/(D14*F14^8))

2627

28

29

30

31

32

33

34

35

36

3738

39

40

41

42

Download - Solutions to Chapter 19

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