CHM310S Spring 2010 Homework#2 Answer Key
1. A
O
O
NH
Properties
MP (oC)
MW (g/mol)
142
201.22
Literature Values
2.36
logKOW
sat
2.53
-logC w (mol/L)
n/a
logKH (Latm/mol)
n/a
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
fC aromatic
fH
fC aromatic between rings
fC
fCOO aromatic
f-NHFCH
#
8
10
2
1
1
1
1
logKOW
Constant
0.13
0.23
0.23
0.20
-0.56
-2.15
(3-1)*(-0.12)
Csatw
a = 0.87; b = 0.68
Polyaromatic hydrocarbons
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
sat
C w (mol/L) = 0.376 mol/L
= (0.376 mol/L)(201.22 g/mol)(1000mg/1g)
Csatw (ppm)
= 7.56x104 mg/L
= 7.56x104 ppm
sat
= (0.376 mol/L)(1 L/10-3 m3)
C w
= 376 mol/m3
(mol/m3)
Contribution
1.04
2.30
0.46
0.20
-0.56
-2.15
-0.24
1.05
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH7
-0.46
-3.22
=CX3
-0.63
-1.89
X-OR
1
-0.70
-0.70
X-C(=O)R
1
-1.63
-1.63
X-NHR
1
-0.86
-0.86
X-CH3
1
-0.28
-0.28
Σaifi
-8.58
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -8.58 – [1.56(142–25)]/100 + 4.42
= -4.16
-4.16
= 10
VP
= (6.92x10-5 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 9.22x10-3 Pa
= 9.22x10-3 Pa/[(376 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
sat
= 9.91x10-9
KAW = VP/(C wRT)
1. B
Cl
Cl
Properties
o
Cl
Liquid
262.78
MP ( C)
MW (g/mol)
Literature Values
Cl
Cl
Cl
4.90
logKOW
4.90
-logCsatw (mol/L)
1.44
logKH (Latm/mol)
3.46
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
fC
fH
fCl
FII
FCH
Fpolyhalo, 3 on same C
#
4
2
6
1
1
2
logKOW
Constant
0.20
0.23
0.06
-0.09
(8-1)*(-0.12)
1.59
Csatw
a = 0.81; b = -0.20
Contribution
0.80
0.46
0.36
-0.09
-0.84
3.18
3.87
Alkanes
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 1.33x10-5 mol/L
= (1.33x10-5 mol/L)(262.78 g/mol)(1000mg/1g)
Csatw (ppm)
= 3.49 mg/L
= 3.49 ppm
= (1.33x10-5 mol/L)(1 L/10-3 m3)
Csatw
3
= 1.33x10-2 mol/m3
(mol/m )
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH2
-0.46
-0.92
X-CR3
2
-0.38
-0.76
X-Cl
6
-0.79
-4.74
Σaifi
-6.42
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -6.42 + 4.42 (liquid)
= -2.00
-2.00
= 10
VP
= (1.00x10-2 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 1.33 Pa
= 1.33 Pa/[(1.33x10-2 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
sat
= 4.05x10-2
KAW = VP/(C wRT)
1. C
o
Properties
218
178.23
MP ( C)
MW (g/mol)
Literature Values
4.57
logKOW
4.46
-logCsatw (mol/L)
-1.45
logKH (Latm/mol)
6.05
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
#
Constant
Contribution
fC aromatic
10
0.13
1.30
fH
10
0.23
2.30
fC aromatic between rings
4
0.23
0.92
logKOW
4.52
sat
C w
Polyaromatic hydrocarbons a = 0.87; b = 0.68
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 3.86x10-5 mol/L
= (3.86x10-5 mol/L)(178.23 g/mol)(1000mg/1g)
Csatw (ppm)
= 6.87 mg/L
= 6.87 ppm
= (3.86x10-5 mol/L)(1 L/10-3 m3)
Csatw
3
= 3.86x10-2 mol/m3
(mol/m )
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH10
-0.46
-4.60
=CX4
-0.63
-2.52
Σaifi
-7.12
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -7.12 - [1.56(218-25)]/100 + 4.42
= -5.71
= 10-5.71
VP
= (1.95x10-6 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 2.59x10-4 Pa
=2.59x10-4Pa/[(3.86x10-2mol/m3)(8.314m3Pa/Kmol)(298 K)]
KAW = H/RT
KAW = VP/(CsatwRT) = 2.72x10-6
1. D
O
Properties
o
O
O
O
Liquid
278.34
MP ( C)
MW (g/mol)
Literature Values
4.57
logKOW
4.47
-logCsatw (mol/L)
-2.89
logKH (Latm/mol)
7.02
-logP (atm)
Ref: Schwarzenbach et al. Environmental
Organic Chemistry
logKOW
Fragment Type
fC aromatic
fH
fC
fCOO aromatic
FCH
Fnearby polar on adj C
#
6
22
8
2
2
1
logKOW
Constant
0.13
0.23
0.20
-0.56
(5-1)*(-0.12)
-0.16*(-0.56-0.56)
Csatw
a = 1.06; b = -0.22
Phthalates
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 3.69x10-6 mol/L
= (3.69x10-6 mol/L)(278.34 g/mol)(1000mg/1g)
Csatw (ppm)
= 1.03 mg/L
= 1.03 ppm
= (3.69x10-6 mol/L)(1 L/10-3 m3)
Csatw
3
= 3.69x10-3 mol/m3
(mol/m )
OR
Subst. bz with polar subst.
a = 0.72; b = 1.18
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 8.82x10-7 mol/L
= (8.82x10-7 mol/L)(278.34 g/mol)(1000mg/1g)
Csatw (ppm)
= 0.246 mg/L
= 0.246 ppm
sat
= (8.82x10-7 mol/L)(1 L/10-3 m3)
C w
= 8.82x10-4 mol/m3
(mol/m3)
Contribution
0.78
5.06
1.60
-1.12
-0.96
0.1792
5.54
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH4
-0.46
-1.84
=CX2
-0.63
-1.26
X-COOR ar
2
-1.19
-2.38
X-CH2-R
6
-0.45
-2.70
X-CH3
2
-0.28
-0.56
Σaifi
-8.74
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -8.74 + 4.42
= -4.32
-4.32
= 10
VP
= (4.79x10-5 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 6.38x10-3 Pa
If you used the Csatw for phthalates:
= 6.38x10-3 Pa/[(3.69x10-3 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
sat
= 6.98x10-4
KAW = VP/(C wRT)
If you used the Csatw for subst. bz with polar subst.:
= 6.38x10-3 Pa/[(8.82x10-4 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
= 2.92x10-3
KAW = VP/(CsatwRT)
1. E
O
Properties
o
N
NH
164
232.20
MP ( C)
MW (g/mol)
Literature Values
F
F
F
1.34
logKOW
1.39
-logCsatw (mol/L)
n/a
logKH (Latm/mol)
n/a
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
fC aromatic
fH
fC
fCΦ
fN(H)C(O)N-Φ
fF
Fpolyhalo, 3 on same C
#
6
10
2
1
1
3
1
logKOW
Constant
0.13
0.23
0.20
0.20
-2.29
-0.38
1.59
Csatw
a = 0.84; b = 0.12
Misc. pesticides
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 0.700 mol/L
= (0.700 mol/L)(232.20 g/mol)(1000mg/1g)
Csatw (ppm)
= 1.63x105 mg/L
= 1.63x105 ppm
= (0.700 mol/L)(1 L/10-3 m3)
Csatw
3
= 700 mol/m3
(mol/m )
OR
Subst. bz with polar subst.
a = 0.72; b = 1.18
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 19.6 mol/L
= (19.6 mol/L)(232.20 g/mol)(1000mg/1g)
Csatw (ppm)
= 4.54x106 mg/L
= 4.54x106 ppm
= (19.6 mol/L)(1 L/10-3 m3)
Csatw
3
= 1.96x104 mol/m3
(mol/m )
Contribution
0.78
2.30
0.40
0.20
-2.29
-1.14
1.59
1.84
Vapour Pressure and KAW
#
Constant
Contribution
4
-0.46
-1.84
2
-0.63
-1.26
1
0.32
0.32
1
-6.47
-6.47
2
-0.28
-0.56
Σaifi
-9.81
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -9.81 – [1.56(164-25)]/100 + 4.42
= -7.56
= 10-7.56
VP
= (2.76x10-8 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 3.69x10-6 Pa
If you used the Csatw for misc. pesticides:
= 3.69x10-6 Pa/[(700 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
= 2.12x10-12
KAW = VP/(CsatwRT)
sat
If you used the C w for subst. bz with polar subst.:
= 3.69x10-6 Pa/[(1.96x104 mol/m3)(8.314 m3Pa/Kmol)(298 K)]
KAW = H/RT
sat
= 7.60x10-14
KAW = VP/(C wRT)
Fragment Type
=CH=CXX-CF3 ar
X-NHCONR2 ar
X-CH3
1. F
Cl
MP (oC)
MW (g/mol)
Properties
305
321.97
Literature Values
6.64
logKOW
sat
7.50
-logC w (mol/L)
O
Cl
-1.30
logKH (Latm/mol)
8.80
-logP (atm)
Cl
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
#
Constant
Contribution
fC aromatic
12
0.13
1.56
fH
4
0.23
0.92
Φ
fCl
4
0.94
3.76
fOΦ
2
0.53
1.06
Fnearby polar on adj C (Cl-Cl)
2
-0.16*(0.94+0.94)
-0.6016
Fnearby polar on adj C (O-O)
2
-0.16*(0.53+0.53)
-0.3392
Fnearby polar on C’s sep by 1
4
-0.08*(0.94+0.53)
-0.4704
logKOW
5.89
Note: If anyone lost 0.5 marks for not accounting for Fnearby polar effects of the
aromatic Cl’s, you may come by LM318 and look for Holly Lee to get that
mark back.
Csatw
Subst. bz with polar subst.
a = 0.72; b = 1.18
sat
logKOW = alog(1/C w) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 8.71x10-9 mol/L
= (8.71x10-9 mol/L)(178.23 g/mol)(1000mg/1g)
Csatw (ppm)
= 2.80x10-3 mg/L
= 2.80x10-3 ppm
= (8.71x10-9 mol/L)(1 L/10-3 m3)
Csatw
= 8.71x10-6 mol/m3
(mol/m3)
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH4
-0.46
-1.84
=CX8
-0.63
-5.04
X-Cl ar
4
-0.53
-2.12
X-OAr
2
-0.33
-0.66
Σaifi
-9.66
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -9.66 - [1.56(305-25)]/100 + 4.42
= -9.61
Cl
O
= 10-9.61
= (2.47x10-10 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 3.29x10-8 Pa
=3.29x10-8Pa/[(8.71x10-6mol/m3)(8.314m3Pa/Kmol)(298 K)]
KAW = H/RT
KAW = VP/(CsatwRT) = 1.52x10-6
VP
1. G
Cl
o
OH
Properties
62
197.45
MP ( C)
MW (g/mol)
Literature Values
2.67
logKOW
n/a
-logCsatw (mol/L)
n/a
logKH (Latm/mol)
Cl
Cl
n/a
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
#
Constant
Contribution
fC aromatic
6
0.13
0.78
fH
2
0.23
0.46
fClΦ
3
0.94
2.82
Φ
fOH
1
-0.44
-0.44
Fnearby polar on adj C (Cl-OH)
2
-0.16*(0.94-0.44)
-0.16
Fnearby polar on C’s sep by 1
3
-0.08*(0.94+0.94)
-0.4512
logKOW
3.01
Note: If anyone lost 0.5 marks for not accounting for Fnearby polar effects of the
aromatic Cl’s, you may come by LM318 and look for Holly Lee to get that
mark back.
Csatw
Subst. bz with polar subst.
a = 0.72; b = 1.18
sat
logKOW = alog(1/C w) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 2.88x10-3 mol/L
= (2.88x10-3 mol/L)(197.45 g/mol)(1000mg/1g)
Csatw (ppm)
= 569 mg/L
= 569 ppm
= (2.88x10-3 mol/L)(1 L/10-3 m3)
Csatw
3
= 2.88 mol/m3
(mol/m )
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH2
-0.46
-0.92
=CX4
-0.63
-2.52
X-Cl ar
3
-0.53
-1.59
X-OH ar
1
-1.79
-1.79
Σaifi
-6.82
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -6.82 - [1.56(62-25)]/100 + 4.42
= -2.98
= 10-2.98
VP
= (1.05x10-3 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 0.141 Pa
KAW = H/RT
KAW = VP/(CsatwRT)
=0.141 Pa/[(2.88 mol/m3)(8.314m3Pa/Kmol)(298 K)]
= 1.97x10-5
1. H
OH
o
Properties
31
108.14
MP ( C)
MW (g/mol)
Literature Values
1.96
logKOW
1.59
-logCsatw (mol/L)
n/a
logKH (Latm/mol)
n/a
-logP (atm)
Ref: Schwarzenbach et al. Environmental Organic
Chemistry
logKOW
Fragment Type
#
Constant
Contribution
fC aromatic
6
0.13
0.78
fH
7
0.23
1.61
fC
1
0.20
0.20
Φ
fOH
1
-0.44
-0.44
logKOW
2.15
Csatw
Subst. bz with polar subst.
a = 0.72; b = 1.18
logKOW = alog(1/Csatw) + b
Csatw =
1
.
10^[(logKOW – b)/a]
Csatw (mol/L) = 0.0450 mol/L
= (0.0450 mol/L)(108.14 g/mol)(1000mg/1g)
Csatw (ppm)
= 4.86x103 mg/L
= 4.86x103 ppm
sat
= (0.0450 mol/L)(1 L/10-3 m3)
C w
= 45.0 mol/m3
(mol/m3)
Vapour Pressure and KAW
Fragment Type
#
Constant
Contribution
=CH4
-0.46
-1.84
=CX2
-0.63
-1.26
X-CH3 ar
1
-0.34
-0.34
X-OH ar
1
-1.79
-1.79
Σaifi
-5.23
logVP = Σaifi – [1.56(MP – 25)]/100 + 4.42 = -5.23 - [1.56(31-25)]/100 + 4.42
= -0.904
-0.904
= 10
VP
= (0.125 mm Hg)(1 torr/1 mm Hg)(1 atm/760 torr)(101325 Pa/1 atm)
= 16.6 Pa
= 16.6 Pa/[(45.0 mol/m3)(8.314m3Pa/Kmol)(298 K)]
KAW = H/RT
sat
KAW = VP/(C wRT) = 1.49x10-4
Summary of physical properties
logKOW
VP (Pa)
Csatw (ppm)
-3
1.05
9.22x10
7.56x104
3.75
1.33x100
3.49
-4
4.52
2.59x10
6.87
5.54
6.38x10-3
1.03
or 0.246
0.25
3.69x10-6
4.54x106
or 1.63x105
6.98
3.29x10-8
2.80x10-3
3.01
1.41x10-1
569
2.15
1.66x101
4.86x103
Compound
A
B
C
D
E
F
G
H
Metabolism of compounds A, C, and G:
KAW
9.91x10-9
4.05x10-2
2.72x10-6
6.98x10-4
or 2.92x10-3
7.60x10-14
or 2.12x10-12
1.52x10-6
1.97x10-5
1.49x10-4
Multiple pathways possible.
O
A
OH
O
OGlu
NH
C
OH
O
OH
H2O
MFO
OGlu
OGlu
G
Cl
Cl
OGlu
OH
Cl
Cl
Cl
Cl
2.
O
o
N
O
O
-
P
O
n/a
291.26
MP ( C)
MW (g/mol)
+
S
Properties
O
Given Values
logKOW
Mass of lake trout (g)
[Parathion]lake (ng/L)
Assume density of fish (g/mL)
3.81
700
15
1
Calculations:
logBCF = 0.76logKOW – 0.23
logBCF = 0.76(3.81) – 0.23
logBCF = 2.67
BCF = 102.67
BCF = 463
BCF = [Parathion]fish/[Parathion]lake
[Parathion]fish = (463)(15 ng/L)(1 g/109 ng)(1 L/1 kg)
= 6.95x10-6 g/kg
Amount in fish = (6.95x10-6 g/kg)(0.700 kg)(1000 mg/1 g)(1000 ug/1 mg)
= 4.86 ug
Therefore, there is 4.86 ug of parathion in the lake trout.
Multimedia Partitioning Model Exercise
Use: VA = 105 m3 VW = 104 m3
VO = 1 m3
Sample Calculation for Toluene
Properties
Calculations
92.13
MW (g/mol)
3800
VP (Pa)
sat
3
515
C w (g/m )
sat
3
= (515 g/m3)(1 mol/92.13 g)
C w (mol/m )
= 5.59 mol/m3
2.69
logKOW
= 102.69
KOW
= 490
= VP/CsatW
H (Pa·m3/mol)
= 3800 Pa/5.59 mol/m3
= 680 Pa·m3/mol
= H/RT
KAW
= (680 Pa·m3/mol)/[(8.314 Pa·m3/molK)(298 K)]
= 0.274
Mtotal = MA + MW + MO
Mtotal = CAVA + CWVW + COVO
Mtotal = CWKAWVA + CWVW + CWKOWVO
Mtotal = CW(KAWVA + VW + KOWVO)
Part I:
Assign arbitrary value to Mtotal, eg. Mtotal = 100 g
=
100g
.
CW (g/m3)
(0.274)(105 m3) + 104 m3 + (490)(1 m3)
= 2.64x10-3 g/m3
3
= KAWCW
CA (g/m )
= (0.274)(2.64x10-3 g/m3)
= 7.23x10-4 g/m3
= KOWCW
CO (g/m3)
= (490)(2.64x10-3 g/m3)
= 1.29 g/m3
= VWCW
MWater (g)
= (104 m3)(2.64x10-3 g/m3)
= 26.4 g
= VACA
MAir (g)
= (105 m3)(7.23x10-4 g/m3)
= 72.3 g
= VOCO
MOctanol (g)
= (1 m3)(1.29 g/m3)
= 1.29 g
Therefore, 26.4% of toluene is partitioning into water, 72.3% into air, and
1.3% into octanol.
Use: VA = 107 m3 VW = 105 m3
VO = 1 m3
Sample Calculation for Lindane at 15oC or 288K
Properties
Calculations
290.9
MW (g/mol)
0.00472 (see reference for VP’s at different temperatures)
VP (Pa)
7.31 (see reference for Csatw’s at different temperatures)
Csatw (g/m3)
sat
3
= (7.31 g/m3)(1 mol/290.9 g)
C w (mol/m )
= 0.0251 mol/m3
3.80
logKOW
= 103.80
KOW
= 6.31x103
= VP/CsatW
H (Pa·m3/mol)
= 0.00472 Pa/0.0251 mol/m3
= 0.188 Pa·m3/mol
= H/RT
KAW
= (0.188 Pa·m3/mol)/[(8.314 Pa·m3/molK)(288 K)]
= 7.84x10-5
Mtotal = MA + MW + MO
Mtotal = CAVA + CWVW + COVO
Mtotal = CWKAWVA + CWVW + CWKOWVO
Mtotal = CW(KAWVA + VW + KOWVO)
Part II:
Assign arbitrary value to Mtotal, eg. Mtotal = 100 g
=
100g
.
CW (g/m3)
-5
7
3
5
3
3
3
(7.84x10 )(10 m ) + 10 m + (6.31x10 )(1 m )
= 9.34x10-4 g/m3
= KAWCW
CA (g/m3)
= (7.84x10-5)(9.34x10-4 g/m3)
= 7.32x10-8 g/m3
= KOWCW
CO (g/m3)
= (6.31x103)(9.34x10-4 g/m3)
= 5.89 g/m3
= VWCW
MWater (g)
= (105 m3)(9.34x10-4 g/m3)
= 93.4 g
= VACA
MAir (g)
= (107 m3)(7.32x10-8 g/m3)
= 0.7 g
= VOCO
MOctanol (g)
= (1 m3)(5.89 g/m3)
= 5.9 g
Therefore, 93.4% of lindane is partitioning into water, 0.7% into air, and
5.9% into octanol at 15oC.