University of Washington Department of Chemistry

Chemistry 452/456 Summer Quarter 2010

Homework Assignment #7: Due at 500 pm Wednesday 18 August

Text Problems: 9.58, 9.60, 10.6, 10.8, 11.11, 11.19

9.58) Using the results for the Donnan effect with a = 3.00 x 10-1 M and b = 1.00 x 10-2 M

cL= x =

a2

b + 2a= 0.15M

c+

L= b + x = 0.01M + 0.15M = 0.16M

c+

R= c

R= a ! x = 0.30M ! 0.15M = 0.15M

" =2b

2+ 2ab

b + 2a

#

$%

&

'(RT =

0.0062M2

0.61M

#

$%

&

'(1000L

m3

)*+

,-.8.314Jmol

!1K

!1( ) 298K( ) = 25kPa

NOTE: need value of T to calculate osmotic pressure 9.60) Using results for the Donnan effect with z = 2, a = 1.00 x 10-1 M and b = 1.00 x 10-2 M

cL= x =

a2

zb + 2a= 0.045M

c+

L= zb + x = 0.02M + 0.045M = 0.065M

c+

R= c

R= a ! x = 0.1M ! 0.045M = 0.055M

rD=c+

L

c+

R=0.065M

0.055M= 1.2

or alternatively,

rD=a + zb

a=0.12

0.1= 1.2

!D= "

RT

F

#$%

&'(ln r

D= "

8.314Jmol"1K

"1( ) 298K( )

96485Cmol"1

#

$%

&

'( ln1.2 = "4.7mV

10.6)

( )

( )( ) ( )( )( )1 1 1

1

ln

35.58.31 298 ln 1 96485 0.055

5.25

10.0

final

final initial

initial

CRT z

C

JK mol K Cmol V

kJmol

µ ! !

" " "

"

# = + $ "

% &= +' (

) *=

10.8) At equilibrium:

( )

( )

( )( )

( )( )

1 1

1

ln 0

ln

8.31 298 0.120ln ln 0.082

0.0051 96485

out

out in

in

out

out in

in

out

in

CRT z

C

CRT z z

C

JK mol KCRTV

z C Cmol

µ ! !

! ! !

!" "

"

# = + $ " =

= " $ " = " $#

% &'# = " = " =( )$ " * +

11.11) a) Generate Scatchard plot of ν/c vs. ν Fitting to straight line gives slope = -1.0x104 and y intercept = 4.0x104 Thus equilibrium constant K293 = -slope = 1.0x104 and KN = y intercept = 4.0x104 so N = KN/K = 4 b)

ln2 = lnK293

K310

!

"#$

%&= '

(H !

R

!"#

$%&

1

293K'

1

310K

!"#

$%&

(H ! = '8.314Jmol

'1K

'1( ) ln2( )

1

293K'

1

310K

!"#

$%&

= 31kJmol'1

c)

!G

293

!

= "RT lnK293

= 8.314Jmol"1K

"1( ) 293K( ) ln 1.0 #104( ) = "22kJmol"1

d)

!S293

!

=!H

!

" !G!

T=

"31kJmol"1( ) " "22kJmol

"1( )293K

= "31Jmol"1

NOTE: assuming ΔHo does not vary between 293 and 310 K.

11.19) Create a Langmuir plot of V/P vs. V. The data roughly follow a straight line, especially for the lower P values.

0

0.2

0.4

0.6

0.8

1

1.2

0 0.05 0.1 0.15 0.2

V

V/P

Fitting all data to a line gives slope = -15.2 atm-1 and y intercept of 2.91mL/atm

Thus P0=

1

15.2atm!1= 6.6 "10

!2atm , and V

m=2.91mLatm

!1

15.2atm!1

= 0.19mL

Fitting four lowest P data to a line gives slope = -18.2 and y intercept of 3.34

Thus P0=

1

18.2atm!1= 5.5 "10

!2atm , and V

m=3.34mLatm

!1

18.2atm!1

= 0.18mLVm = 3.34/18.2

= 0.18 mL

1) The standard free energy of reaction is related to the standard cell potential by the equation 0

G n! = " #Є0, where ! is Faraday’s constant and Є0 is the standard cell potential.

a) Prove that the standard entropy change is given by 0S n

T

!" = #

! Є0, at

constant pressure. Solution: G n E H T S

G E Hn S S

T T T

ES n

T

! = " # = ! " !

$! $! $!% = " # = "! & "!

$ $ $

$!%! = #

$

! ! ! !

! ! !

! !

!

!

b) Prove also that 0H n! = " # Є0

n TT

!+ "

! Є0

Solution: G n E H T S

EH G T S n E n T

T

! = " # = ! " !

$!%! = ! + ! = " # + #

$

! ! ! !

!

! ! ! !

c) For the redox reaction ( ) ( ) ( ) ( )2 22 2 2Ag s Hg Cl s AgCl s Hg s+ ! + , the

standard cell potential as a function of temperature is T(K) 291 298 303 311 Є0(mV) 43.0 45.4 47.1 50.1

• Write out the oxidation and reduction half reactions Solution:

( ) ( )2

2 2 2 2

2 2 2

Ag s Cl AgCl s e

Hg e Hg

! !

+ !

+ " +

+ "

• Using the data in the table calculate ΔG0, ΔH0, and ΔS0. Make a plot of E0 versus T.

Solution:

0.042

0.043

0.044

0.045

0.046

0.047

0.048

0.049

0.05

0.051

290 295 300 305 310 315

T(K)

Cell P

ote

nti

al (V

)

The plot is virtually a straight line with slope 4 13.55 10

EVK

T

! !"#= $

"

!

Assume T=298K. Then ( )( )( )

( )( )( )

( )( )

1

0

1 4 1 1

1 1

2 96485 0.0454 8761

2 96485 3.55 10 68.5

8761 298 68.5 8761 20413 29174

G n E Cmol V J

ES n Cmol VK JK

T

H G T S J K JK mol J J J

!

! ! ! !

! !

" = ! #" = ! =

$"" = # = % =

$

" = " + " = + = + =

!

!

!

! ! !

2) Nicotine adenine dinucleotide (NAD) is cellular redox reagent. The reduced form of NAD is abbreviated NADH and the oxidized for is NAD+. In the cell, oxygen O2 is reduced by NADH according to : 1

2 22NADH H O NAD H O

+ ++ + ! + .

For this reaction at T=298K ΔG0=-259.83kJ/mole. Assume [NADH]=0.035M, [NAD+]=0.004M, pH=4.8, and PO2=0.05 bars. Assume the standard oxygen pressure is 1 bar. Note the standard Gibbs energy change assumes a standard H+ concentration of 1M.

a) Calculate the standard cell potential Є0 assuming the standard state for H+ is [H+]0=1M. Repeat the calculation assuming [H+]0=10-7 M.

Solution:

!G!

= "n#!E!

$ !E!

= "!G

!

n#= "

"259.83kJmol-1( )

1( ) 96485Cmol"1( )

= 2.69V

To use the alternative standard state notation to get the cell voltage we need to know the definition of the standard free energy in the new convention. Recall:

K =

CNAD

+

C0

NAD+

!"#

$%&

CNADH

C0

NADH

( )C

H+

CH+

0

!"#

$%&

PO

2

PO

2

0

!"#

$%&

1/ 2=

CNAD

+

1M( )C

NADH

1M( )C

H+

CH+

0

!"#

$%&

PO

2

1bar( )1/ 2

=C

NAD+

CNADH

CH+

CH+

0

!"#

$%&

PO

2

1/ 2

=C

NAD+

CNADH

CH+

CH+

0

!"#

$%&

PO

2

1/ 2

=C

NAD+

CNADH

CH

+P

O2

1/ 2C

H+

0

Note if C

H+

0= 1M then

K =

CNAD

+

CNADH

CH

+P

O2

1/ 2. We assume this definition of K in what

follows…

!G!" = #RT ln "K = #RT ln

CNAD

+

CNADH

CH

+P

O2

1/ 2C

H+

0= #RT ln

CNAD

+

CNADH

CH

+P

O2

1/ 2

$

%&&

'

())# RT lnC

H+

0

= #RT lnC

NAD+

CNADH

CH

+P

O2

1/ 2

$

%&&

'

())# RT lnC

H+

0= #RT ln K # RT ln10

#7= !G

! # RT ln10#7

= !G!

+ 7RT ln10 = !G!

+ 7 2.3( )RT = #259.83kJmol#1+ 16.1( ) 8.31JK

#1mol

#1( ) 298K( )= #219.96kJmol

#1

!"E!# = $

"G!#

n%= $

$219.96kJmol-1( )

1( ) 96485Cmol$1( )

= 2.28V

b) Calculate the cell voltage Є assuming the concentrations given above.

Will the cell voltage depend upon the standard state definition for H+? Explain.

Solution:

The cell voltages in the two conventions are obtained from lnRT

E E Qn

! = ! +"

! and

lnRT

E E Qn

!! !" = " #$

!

So we have to calculate the two reaction quotients:

Q =

CNAD

+

C0

NAD+

!"#

$%&

CNADH

C0

NADH( )

CH+

CH+

0

!"#

$%&

PO

2

PO

2

0

!"#

$%&

1/ 2=

0.004

1( )

0.035

1( ) 10

'4.8

1( ) 0.05

1( )

1/ 2=

0.004

0.035( ) 1.58 (10'5( ) 0.224( )

= 3.23(104

)Q =

CNAD

+

C0

NAD+

!"#

$%&

CNADH

C0

NADH( )

CH+

CH+

0

!"#

$%&

PO

2

PO

2

0

!"#

$%&

1/ 2=

0.004

1( )

0.035

1( ) 10

'4.8

10'7( ) 0.05

1( )

1/ 2=

0.004

0.035( ) 1.58 (102( ) 0.224( )

= 3.23(10'3= Q (10

'7

In the ‘1M” convention:

( )( )

( )( )( )

1 1

4

1

8.31 298ln 2.69 ln 3.23 10

1 96485

2.69 0.27 2.42

JK mol KRTE E Q V

n Cmol

V V V

! !

!" = " ! = ! #

$

= ! =

!

In the “10-7M” convention: ( )( )

( )( )( )

1 1

3

1

8.31 298ln 2.28 ln 3.23 10

1 96485

2.28 0.16 2.42

JK mol KRTE E Q V

n Cmol

V V V

! !

!

!

"" "# = # + = ! $%

= + =

!

Therefore E E!" = " . The cell potential is independent of the standard state definitions.

3) A protein complex called the sodium/potassium pump uses the free energy of hydrolysis of ATP to pump sodium ions Na+ out of the cell and potassium ions K+ into the cell. The net reaction for active transport of sodium and potassium ions is thought to be:

3Na+inside( )

+

2K+outside( )

!

"##

$##

+ ATP% ADP + phosphate +

3Na+outside( )

+

2K+inside( )

&

'##

(##

The diagram below shows the concentrations of sodium and potassium ions inside and outside a cell. The electrical potential E inside and outside the cell is also given.

a) Calculate the change in the electrochemical potential involved in transporting 1 mole of sodium ion out of the cell. Assume the activity coefficients of sodium ion inside and outside the cell are unity. Assume the temperature is 310K.

Solution:

( ) ( )

( )( ) ( )( )

( )( ) ( )( )

ln

1408.31 / 310 ln 96485 / 0 ( 0.07 )

10

1408.31 / 310 ln 96485 / 0.07

10

6798 6754 13552

outNa in out in out out in

in

Na

G G G n RT zNa

J mol K K C mole V

J mol K K C mole V

J J J

µ ! !+

+

"+

# $% &' () *+ = , = + = + - ,) *% &' (. /

# $= 0 + , ,) *

. /# $

= 0 +) *. /

= + =

!

b) Calculate the free energy change involved in transporting 1 mole of potassium ion into the cell. Assume the activity coefficients of potassium ion inside and outside the cell are unity. Assume the temperature is 310K.

Solution:

!GK+out"in( ) = Gin #Gout = RT ln

K+$% &'in

K+$% &'out

(

)*

+

,- + z. /

in#/

out( )

= 8.31J / mol 0K( ) 310K( ) ln100

5

()*

+,-+ 96485C / mole( ) #0.70V( ) = 1000J / mole

c) Calculate the total free energy change involved in transporting 3 moles of

sodium ion out of the cell and two moles of potassium into the cell at T=310K. Assume, as in parts a and b, that all activity coefficients are unity.

Solution: !G

total= 3!G Na

+out( ) + 2!G K

+in( ) = 3( ) 13,600J / mole( ) + 2 1000J / mole( ) = 42,800J

d) The standard free energy change for the hydrolysis of ATP i.e.

ATP!"

ADP + phosphate#

$%&

'( at 310K is DG kJ mole

0313= - . / . If the

total concentration of inorganic phosphate is 0.01M, calculate the ratio of ATP to ADP (i.e. in the reaction quotient Q) which will furnish the work required to accomplish the transport described in part c.

Solution: !Gtotal = !Gtotal

transport+ !GATP"ADP # 0

!Gtotal

transport+ !GATP"ADP = 42,600J + !GATP"ADP

0+ RT lnQ # 0

!Gtotal

transport+ !GATP"ADP = 42,600J # 31,300J + 8.31J / K( ) 310K( ) ln

Pi[ ] ADP[ ]ATP[ ]

$

%&'

()* 0

!Gtotal

transport+ !GATP"ADP = 11,500J + 2576J( ) ln

Pi[ ] ADP[ ]ATP[ ]

#

$%&

'() 0

lnPi[ ] ADP[ ]

ATP[ ]

!

"#$

%&' (4.464)

Pi[ ] ADP[ ]

ATP[ ]' e(4.464 = 0.01152

Pi[ ] ADP[ ]ATP[ ]

! 0.01152"0.01( ) ADP[ ]

ATP[ ]! 0.01152"

ADP[ ]ATP[ ]

! 1.152

Therefore, ATP[ ]ADP[ ]

! 0.868 .

4) The data below are for the binding of oxygen to squid hemacyanin. The percent

saturation is the parameter 100% 100%fN

!" = " .

PO2 (mmHg) Percent saturation 1.13 0.30 7.72 1.92 31.71 8.37 100.5 32.9 136.7 55.7 203.2 73.4 327.0 83.4 566.9 89.4 736.7 91.3

a) Construct a Hill plot of the data shown

Solution:

logv

1! v versus logP

O2

log v/n-v

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.5 1 1.5 2 2.5 3

log PO2

b) From the plot, determine whether the binding is cooperative or independent. Explain.

Solution: The slope =1.3 so the binding is not independent.

c) Estimate the number of oxygen molecules that can be attached to a single hemacyanin molecule.

Solution: The slope of this ‘line’ is 1.3, therefore N ≥ 2.

5) Carbon disulfide adsorbs onto a finely powdered dust. The pressure of the gas in equilibrium with its adsorbed form is listed in the table below as a function of temperature. From a graph of these data, calculate the isosteric heat of adsorption.

Pressure (mmHg)

0.41 0.80 1.50 2.65 4.52

Temperature (K)

263 273 283 293 303

Solution:

Integrating : ln stq

P cRT

= ! +

where c is a constant. This means that a plot of lnP versus 1/T will be linear with a slope of –qst/R. Convert the data in the table to lnP and 1000/T: lnP -0.89 -0.22 0.41 0.98 1.51 1000/T 3.80 3.66 3.53 3.41 3.30 A plot of lnP versus 1000/T appears below

-1.5

-1

-0.5

0

0.5

1

1.5

2

3.2 3.4 3.6 3.8 4

1000/T

lnP

Correcting for the factor of 1000 the slope is about -5000K. This mean the isosteric heat is ( )( )1 1 1

5000 8.31 41550stq K Jmol K Jmol! ! != ! ! =

6) From an isothermal titration calorimetry study, a protein P was found to have four binding sites for ligand L (i.e. N=4). It was also found that the equilibrium constant for ligand binding was Kb=5.00x108 and the binding enthalpy was

175

bH kJmol

!" = !

! .

a) Determine the fraction of ligand bound and determine the heat of binding qb in a 100mL solution where the protein concentration is cP=0.0001M, and the total ligand concentration is cL(total)=0.0005M.

Solution: ( )b b Lq V H c bound= !

! . So we have to determine ( ) ( ) ( )L L Lc bound c total c free= ! where

cL

free( ) =! NK

bc

P! K

bc

Ltotal( ) +1( ) ± NK

bc

P! K

bc

Ltotal( ) +1( )

2

+ 4Kbc

Ltotal( )"

#$%

1/ 2

2Kb

NKbc

P! K

bc

Ltotal( ) +1= 4( ) 5&10

8( ) 10!4 M( ) ! 5&10

8( ) 5&10!4 M( ) +1' !5&10

4 M

(cL

free( ) =5&10

4 M ± 25&108 M 2

+ 4 5&108( ) 5&10

!4 M( )( )1/ 2

2 5&108( )

=5&10

4 M ± 25&108 M 2

+100 &104( )

1/ 2

109

'5&10

4 M ± 5&104 M

109

= 10!4 M

Therefore

cL

bound( ) = cL

total( ) ! cL

free( ) = 0.0005M ! 0.0001M = 0.0004

"qb=V#H

b

!cL

bound( ) = 0.1L( ) !75kJmol!1( ) 0.0004M( ) = !0.003kJ

80% of the ligand is bound.

b) What is the heat of binding when the protein sites are fully saturated with ligand? Based on this number calculate the percent saturation.

Solution:

q

b=V!H

b

!NcP= 0.1L( ) "75kJmol"1( ) 4( ) 0.0001M( ) = 0.003kJ . The protein is fully

saturated (100%).