### m 10 33.7 m 78.54 997.0 0.0500 10 9.211 ε kg mol I

```Chem340
Physical Chemistry for Biochemists
Dr. Yoshitaka Ishii
Homework 10
Please solve the following questions. This is a good part of the final Exam.
P9.13-14, P9.18, P9.21, P9.24, P9.28-9, P9.32, P9.40, P9.44 P9.48, P9.50
P10.1-10.2, P10.4-10.7, P10.9
P9.13) In the Debye-Hückel theory, the counter charge in a spherical shell of radius r
and thickness dr around the central ion of charge +q is given by –q2re–r dr. Calculate
the most probable value of r , rmp, from this expression. Evaluate rmp for a 0.050 m
solution of NaCl at 298 K.
d   q 2 re  r 
dr
  q 2 e
1   rmp  0; rmp 

  rmp
 q 3 rmp e
 rmp
0
1


I/I mol kg-11  solvent
8 0.05000.997
1
8
1
0.050
mol kg

9.211

m
1
1
1 7.33  10 m
8 10
7
m
=
9.211
10
m
2.32
10
m



78.54 78.54
 εr
10
κ1 29.211
.9110
 108
rmp
r
r1mp 14.30
10 m
. 36 10
m =43.0
1 . 36nm
nm
8
7
κ
P9.14) Calculate the Debye–Hückel screening length 1/ at 298 K in a 0.00100 m
solution of NaCl.
For water, the screening length at 298 K in m-1 can be calculated as:
κ  2.91  1010


I/ mol kg -1  solvent
 9.211  10 8
εr
0.00100.997
78.54
m 1  1.04  10 8 m 1
1
 9.65  10 9 m  9.65 nm
κ
P9.18) Calculate I, ±, and a± for a 0.0250 m solution of K2SO4 at 298 K. Assume
complete dissociation. How confident are you that your calculated results will agree with
experimental results?
K 2SO 4  v  2, z  1, v  1, z  2
0.0250
 2  4  0.0750 mol kg 1
2
ln    1.173  2  0.0750  0.6425
I
   0.523
m 
a      
m 
m3   0.050  0.0250  6.25  105
2
m  0.03969 mol kg 1
a  0.03969  0.5260  0.0209
Not very confident. Figures 10.5 and 10.6 show significant deviations from predicted
behavior for I > 0.01 mol kg–1.
P9.21) Dichloroacetic acid has a dissociation constant of Ka = 3.32  10–2. Calculate the
degree of dissociation for a 0.125 m solution of this acid (a) using the Debye–Hückel
limiting law and (b) assuming that the mean ionic activity coefficient is one.
a) We first consider the case when  is given by the Debye–Hückel limiting law. The
ionic strength is given by
m
 2  m  0.04992 mol kg 1
2
ln    1.173  1  0.04992  0.2621
I
   0.7694
We recalculate the ionic strength and iterate several times.
m 2 2
when  = 0.7694,
 3.32  102
0.125  m
m 2  7.0097  103  0.05608 m
m 2  0.05608 m  7.0097  103  0
m
0.05608 
 0.05608 2  4 1  7.0097  103 
2
 0.06025 mol kg 1
I  0.06025 mol kg 1
ln    1.173  1  0.06025  0.2879
   0.7498
when  = 0.7498
m2
 0.05905
0.125  m
m 2  0.05905 m  7.3815  103  0
m
0.05905 
 0.05905 2  4 1  7.3815  103 
2
 0.06132 mol kg 1
I  0.06132 mol kg 1
ln    0.7479
   0.7479
when  = 0.7479
m2
 0.05935
0.125  m
m 2  0.05935 m  7.4191  103  0
m
0.05935 
 0.05935 2  4 1  7.4191  103 
I  0.06143 mol kg 1
ln    0.2907
   0.7477
2
 0.06143 mol kg 1
when  = 0.7477
m2
 0.05938
0.125  m
m 2  0.05938 m  7.4228  103  0
m
0.05938 
 0.05938 2  4 1  7.4228  103 
2
 0.06144 mol kg 1
This is a sufficiently good result, and we calculate the degree of dissociation to be
0.06144
 100%  49%.
0.125
b)  = 1
m 2  4.15  103  3.32  102 m
m 2  3.32  102 m  4.15  103  0
m
3.32  102 
3.32  10 
2 2

 4 1 4.15  103
2
  0.04992 mol kg
1
0.04992
 100%  40%
0.125
This result is 20% smaller than that calculated using the Debye–Hückel limiting law.
P9.24) From the data in Table 9.3 (see Appendix B, Data Tables), calculate the activity
of the electrolyte in 0.100 m solutions of
a. KCl
b. H2SO4
c. MgCl2
a) KCl
   0.770
0.100 m
m 
a      
m 
m  m  m 
m2   0.100 0.100
m  0.100 mol kg 1
a   0.100 0.770  0.0770
b) H 2SO 4
0.100 m
   0.2655
m3   2  0.100  0.100  4.0  103
2
1
m  0.15874 mol kg 1
a   0.15874   0.2655  0.0422
c) MgCl2
0.100 m
   0.529
m3   0.100 2  0.100  4.0  103
2
m  0.15874 mol kg 1
a   0.15874   0.529  0.0840
P9.28) The principal ions of human blood plasma and their molal concentrations are
mNa   0.14 m, mCl   0.10 m, mHCO   0.025 m.
3
Calculate the ionic strength of blood plasma.
I

 
 

1
1
2
mi z i  0.14 m  12  0.10 m  12  0.025 m  12  0.1325 m

2 i
2
P9.29) Estimate the degree of dissociation of a 0.100 m solution of acetic acid (Ka = 1.75
 10–5) that is also 0.500 m in the strong electrolyte given in parts (a)–(c). Use the data
tables in Appendix B to obtain ±, because the electrolyte concentration is too high to use
the Debye–Hückel limiting law.
a. Ca(Cl)2
b. KCl
c. MgSO4
In each case, the activity of the ions produced in the dissociation of acetic acid is
determined by the strong electrolyte.
a) Ca  Cl 2
 +  1,
Cl
Ca 2+
0.500 m 1.00 m
  2
z  2,
z  1
From Table 10.3,   = 0.448
 2 m 2
5
1.75  10 
0.100  m
m  0.002910 mol kg 1
2
0.448 m 2


0.100  m
0.002910
 100%  2.91%
0.100
 +  1,    1, z  1, z  1
From Table 10.3,   = 0.649
b) KCl
1.75  105 
 0.649 2 m2
0.100  m
m  2.018  103 mol kg 1
2.018  103
 100%  2.02%
0.100
 +  1,
c) MgSO 4
z  2,
  1
z  2
From Table 10.3,   = 0.0675
5
1.75  10 
 2 m 2
2
0.0675 m 2


0.100  m
0.100  m
1
m  0.01777 mol kg
0.01777
 100%  17.7%
0.100
P9.32) Calculate Gr and the equilibrium constant at 298.15K for the reaction
 

 
 

aq  3H 2 g  8H  aq  2Cr 3 aq  7H2 O l .
Cr2 O2
7
Using the half cell reactions:
Oxidation:
3 H2 
 6 H   6 e 
Reduction:
Cr2 O 7
2
E0  0 V
 6 e 
 2 Cr2
3
E 0  1.232 V
E cell  E red  E ox  1.232V


ΔG reaction   n F Δφ  6   96485 C mol -1  1.232 V   713.2 kJ mol -1




6  96485 C mol -1
nF

K eq  Exp 
E cell   Exp 
 1.232 V   Exp287.85
-1
-1
R T

 8.314472 J mol K  298.15 K 



P9.40) By finding appropriate half-cell reactions, calculate the equilibrium constant at
298.15 K for the following reactions:
a. 2Cd(OH)2  2Cd + O2 + 2H2O

b. 2MnO2
4  2H 2 O  2MnO 2  4OH  O 2
a) The half cell reactions are
Cd(OH)2 + 2e– → Cd + 2 OH–
Eº = –0.809V
4OH– → O2 + 2H2O + 4e–
Eº = –0.401V
The overall reaction is
2 Cd(OH)2 → 2 Cd + O2 + 2H2O
Eº = –1.21V
nF 
4  96485 C mol1 1.21V
ln K 
E 
8.314 J K 1mol1  298.15 K
RT
 188.391
K  1.52 1082
b) The half cell reactions are
MnO42– + 2H2O + 2e– → MnO2 + 4 OH–
Eº = + 0.60V
4OH– → O2 + 2H2O + 4e–
Eº = –0.401V
The overall reaction is
2MnO42– + 2H2O → 2MnO2 +4 OH– + O2
Eº = + 0.20V
nF 
4  96485 C mol1  0.20 V
ln K 
E 
8.314 J K 1mol1  298.15 K
RT
 31.139
K  3.34 1013
P9.44) Determine Ksp for AgBr at 298.15 K using the electrochemical cell described by







Ag s AgBr s Br  aq, a Br  Ag aq, a Ag  Ag s .
The half cell and overall reactions are
AgBr + e– → Ag + Br–
Eº = +0.07133 V
Ag → Ag+ + e–
Eº = –0.7996 V
AgBr → Ag+ + Br–
Eº = –0.72827 V
nE 
0.729 V

 12.310
0.05916
0.05916 V
K sp  4.90  1013
log10 K sp  
P9.48) Consider the half-cell reaction O2(g) + 4H+(aq) + 4e–  2H2O. By what factor
are n, Q, E, and E° changed if all of the stoichiometric coefficients are multiplied by a
Eo = 1.03 V
aO2  1
E  E 
RT
1
ln
nF  a  4
H
a) aH+  0.5
8.3145 J mol K    298.15 K  ln  1
E  1.03 V 

 2    96, 485 C mol 
 0.5
1
1
4
1
4

  1.0122 V

b) aH+   0.757    0.5 
8.3145 J mol K    298.15 K  ln  0.757  0.5 
  

 2    96, 485 C mol 
1
E  1.03 
1
1
4
 1.0050 V
The relative error is given by
1.0122 V  1.0050 V
100%  0.72%
1.0050 V
If rounding off to the correct number of significant figures were done at intermediate
stages of the calculation, the relative error would be zero.
P9.50) Consider the couple Ox  e  
 Red with the oxidized and reduced species at
unit activity. What must be the value of E° for this half-cell if the reductant R is to
liberate hydrogen at 1 atm from
a. an acid solution with aH   1
b. water at pH = 7?
c. Is hydrogen a better reducing agent in acid or basic solution?
a) H+ + e–  ½ H2
Red  Ox + e–
E = 0
E  =?
Overall: H+ + Red  ½ H2 + Ox
Because the activities of H+ and H2 are one, E = E  = 0 V. The same is true of the OxRed couple.
To liberate H2, E  for the reaction must be > 0.
Therefore, E  for Ox + e–  Red must be < 0.
b) For a pH of 7, aH   107
E  E 
aH2
RT
ln
 0.05916 log10 107   0.414 V
nF
aH 
For the overall reaction to be spontaneous, E  for Red  Ox + e– must be > 0.414 V.
c) It is a better reducing agent in basic solution, because the oxidation potential for the
reaction ½ H2  H+ + e– becomes more positive as the pH decreases
P10.1) In anaerobic cells glucose C6H12O6 is converted to lactic acid:
C6H12O6  2CH3CH(OH)COOH. The standard enthalpies of formation of glucose and
lactic acid are –1274.45 and –694.04 kJ/mol, respectively. The molar heat capacities (at
constant pressure) for glucose and lactic acid are 218.86 and 127.6 J/mol K, respectively.
a. Calculate the molar enthalpy associated with forming lactic acid from glucose at 298
K.
ΔH reaction,298   i ΔH i,reaction,298  products    i ΔH i,reaction,298  reactants 
i
i
  2    694.04 kJ    1   1274.45 kJ   113.64 kJ
For one mol of lactic acid:
ΔH reaction,298 
 113.64 kJ mol   56.82 kJ mol
1
2mol 
1
b. What would the quantity in part (a) be if the reaction proceeded at a physiological
temperature of 310 K?
At 310 K the enthalpy is given by:
ΔH reaction,310  ΔH i, reaction,298  ΔC p ΔT

 




  56.82 kJ mol -1  2   127.6 J mol -1 K -1   1  218.86 J mol -1 K -1  12 K   56.38 kJ mol -1
c. Based on your answers in parts (a) and (b) how sensitive is the enthalpy of this
reaction to moderate temperature changes?
Based on the results of a) and b), the enthalpy of reaction is not very sensitive to
moderate temperature changes.
P10.2) Using the data in Table 10.1, calculate the equilibrium constant for the
phosphorylation of fructose by ATP to form fructose-6-phosphate.
To use the data in Table 10.1 we have to consider the following pair of reactions:
ATP  H 2 O 
G 1  -31.5 kJ mol -1
fructose  Pi 
 fructose - 6 - phosphate   H 2 O
G 2  13.8 mol -1
That means that the standard Gibbs energy for the phosphorylation of fructose is:
G phosphorylation  -31.5 kJ mol -1  13.8 mol -1  -16.7 kJ mol -1
The equilibrium constant is then:
K phosphorylation
   ΔG phosphorylation  


  16700 J mol -1 
 Exp 
 845.6
  Exp 

-1
-1
R T


 8.314472 J mol K    298 K  
P10.4) Glutamine (G-ine) is an amino acid formed from the reaction of ammonium ion
with glutamate (G-ate–):
a. At pH 7 and T = 298 K, the equilibrium constant is 0.003. Calculate the standard
Gibbs energy change for the formation of glutamine.
b. Glutamine synthesis can be coupled to ATP hydrolysis as follows:
Assuming the equilibrium constant for this reaction is 9600, calculate the standard
Gibbs energy change.
c. From the data in parts (a) and (b), calculate the equilibrium constant for ATP
hydrolysis and the Gibbs energy of hydrolysis of ATP.
a) The standard Gibbs energy change for the formation of glutamine is given by:
ΔG reaction   R T ln  K reaction   8.314472 J mol-1 K -1    298 K   Exp  0.003  14.39 kJ mol-1
b) The standard Gibbs energy change for Glutamine synthesis is given by:
ΔG reaction   R T ln  K reaction    8.314472 J mol-1 K -1    298 K   Exp  9600  22.72 kJ mol -1
c) To calculate the equilibrium constant for ATP hydrolysis and the Gibbs energy of
hydrolysis of ATP we have to consider the chemical equations from parts a) and b):

G - ine  H 2 O 
 NH 4  G - ate 
NH 4  G - ate -  ATP 
 G - ine  H 2 O  ADP  Pi
G reaction  14.39 kJ mol -1
G reaction  22.72 mol -1
 ATP
G reaction  37.11 kJ mol -1
K hydrolysis
  ΔG hydrolysis 


  37.11 kJ mol -1 
6
 Exp 
  Exp 
  3.20  10
-1
-1


R
T






 8,314472 J mol K  298 K 
P10.5) Calculate the reversible work required to transport 1 mol of sucrose from a region
where Csucrose = 0.140 M to a region where Csucrose = 0.340 M. Assume T = 298 K.
The reversible work is given by:
~  R T ln  c out   8.314472 J mol -1 K -1   298 K   ln  0.340    2.20 kJ mol -1
Δμ


 0.140  


 c in 


P10.6) Calculate the reversible work required to transport 1mol of K+ from a region
where to a region where if the potential change accompanying this movement is  =
0.055 V. Assume T = 298 K.
The reversible work is given by:
c 
Δ  R T ln  out   n F 
 cin 
  35.5 mM  
-1
 8.314472 J mol-1 K -1    298K   ln 
  1   96485 C mol    0.055 V 
  5.25 mM  
 10.0 kJ mol -1
P10.7) Suppose the concentration of protons inside a cell membrane is and similarly the
concentration outside is . Assume the electrostatic potential difference is  = out – in.
a. Obtain an expression for the electrochemical potential difference in terms of the pH
difference and the electrostatic potential difference .
b. Suppose the pH inside the inner mitochondrial membrane is 0.75 units higher than
outside the membrane and  = out – in = –0.168V. Calculate the electrochemical
potential difference between protons inside versus outside the membrane.
a) We can derive the following expression:
      n F Δφ

c 
H
Δμ  R T ln  out   n F Δφ  R T ln  cout   ln  cin   n F Δφ  R T ln c out
 ln cinH
 cin 
 R T ln 10  xΔpH  n F Δφ
b) Using the result from part a), the electrochemical potential difference between protons
inside versus outside the membrane is calculated as:
Δμ  R T ln 10  ΔpH  n F Δφ
 8.314472 J mol-1 K -1    298 K   ln 10   0.75  1   96485.3 C mol-1    0.168 V 
 20.49 kJ mol -1
P10.9) Suppose a membrane is passively permeable to water and to Cl– ion, but not to
H+. The electrostatic potential  and the equilibrium concentrations of Cl– and H+ inside
and outside the membrane are given here. Assume that the ionic strength inside the
membrane is sufficiently small so that the activity coefficients of univalent ions may be
taken to be 1 inside the membrane. Assume T = 298 K.
Outside Membrane
Inside Membrane
a. What is the equilibrium concentration of Cl– inside the membrane ?
b. Calculate the difference between the chemical potentials of [H+] inside and outside
the membrane.
c. The reaction is coupled to the transport of H+ with an efficiency of 50% (i.e., half
the free energy from ATP hydrolysis is recovered and available to drive the active
transport of H+). Assuming the concentration of inorganic phosphate is 0.01 M, what
ratio of ATP to ADP is required to drive the transport of H+?
a) We are using the equilibrium condition and solve for cin:
Δφ 
ln
R T  c out 
ln 

zF
 cin 
c out  zFΔφ 

cin  R T 
c out
 zFΔφ 
 Exp 
cin
 R T 
cin 
c out
 zFΔφ 
Exp 
 R T 

 0.05 M 
  0.150 V    96485.3 C mol    1 
Exp 

-1
-1
 8.314472 J mol K    298 K  
-1
 1.45  104 M
b) The difference between the chemical potentials of [H+] inside and outside the
membrane is given by:
c 
Δ  R T ln  out   z F Δφ
 cin 
  5  106 M  
 8.314472 J mol K    298 K   ln 
   96485.3 C mol -1   1   0.150 V 
7
 1  10 M  
 24.16 kJ mol-1
-1
-1
c) To obtain the ratio of ATP to ADP we use the fact that 50% of the energy produced by
the hydrolysis of ATP is available:
ΔG transport  0.5 ΔG ATPADP  ΔG ATPADP  R T ln  Q
  0.01 M   c ADP 
2   24.16 kJ mol-1    31 kJ mol -1   8.314472 J mol-1 K -1    298 K   ln 

c ATP


 0.01 M  c 
 17.32 kJ mol-1    2477.71 J mol-1   ln   c  ADP 
ATP


Exp  7 
 0.01 M   c ADP
c ATP