Chapter 7 organic solvent-water
partitioning the solvent - water
partitioning constant
Near the end of the 19thcentury people
interested in pharmaceuticals discovered
that organic drugs accumulated in
organisms in way that was proportional to
their partitioning in octanol.
More recently environmental chemist have
found similar correlations with partitioning to
solid humus and other naturally occurring
organic phases.
by analogy to the dimentionless Henry’s law
constant, Kiaw = Cia/Ciw
we could define partitioning between and
organic phase (solvent) and the water
phase as:
Kisw = Cis / Ciw
1
Water is somewhat soluble in the octanol
phase, such that 1 out of every 4
molecules in the organic phase or octanol
phase will be water.
In the aqueous phase the octanol activity
coef. is
iw = 3.7x103
1/iw = xiw = 2.7x10-4
only 27 molecules out of 100,000 will be
soluble in the water phase.
first, we must be concerned about the
effects of intermolecular interactions of
solute molecules with solvent molecules
second, we must be aware that the high
amount of water in the octanol phase can
influence the molar volume of the octanol
phase
_
pure n-octanol V o = 0.16L/mol (at 25oC)
2
Water saturated octanol
Vs =(0.79)(0.16) + (0.21)(0.018) = 0.12 L mol-1
What about the molar vol. of water
saturated with octanol?
What if we used hexane?
What happens when we add a high water.
organic to an octanol-water system?
3
From the thermodynamics of Chapter 3,
at equilibrium
isxis = iw xiw
going to molar concentrations Ci = Xi / Vmix
is Cis Vs = iw Ciw Vw
solving for Cis/Ciw and defining a
partitioning coefficient Kisw
Kisw = Cis/Ciw gives
Kisw
Cis
 iw Vw


Ciw
 is Vs
ln Kisw = ln iw - ln is + ln
Vw
Vs
If our solvent is octanol
An octanol-water partitioning coef. is
defined; when is it hexane, a hexane-water
Ksw is defined
Kisw = Cis/Ciw
4
Comparing partitioning in Octanol-water and
hexane water; octanol is an amphiphilic
solvent, e.g., it has both a non polar and
polar group.
Kiow
Kihw
N hexane
13000
52,000
Benzene
130
170
Toluene
490
569
Chlorobenzene
830
810
Naphthalene
2300
2400
Benzaldehyde
30
13
Nitrobenzene
68
29
1-hexanol
34
2.8
aniline
7.9
0.8
phenol
28
0.1
water
0.04
5e-5
5
For weakly apolar or polar compounds
Kiow and Kios are related
Log Kihw = 1.21 log Kiow -0.43
Figure 7.1 new book
6
Basic assumptions of partitioning into an
organic and an aqueous phase
1. the activity coefficient of compound i in
water is independent of its amount in the
aqueous phase; ie. even at saturation
the probability of two solute molecules
“seeing” one another is small…This is
especially true for compounds with low
solubilities and large wsat.
2. The organic phase molecules in the
water phase (ie. octanol in water) do not
affect the wsat of compound i.
This may not actually be the case for
some for some very hydrophobic
molecules.
Ignoring this possibility, for a dilute
compound i, we can say
iw Vw = wsat Vw
and
iwsat Vw = 1/Ciwsat
7
Thus in Kisw  Cis   iw Vw
Ciw
 is Vs
Kisw= 1/( Ciwsat is Vs)
log Kiow= -log Ciwsat -log io - log VO
8
new book Fig 7.2 , plots Kiow vs. iw
before we said ln Kisw = ln iw - ln is + ln Vw /Vs
9
Going back to
ln Kiow= ln iw - ln iO + ln Vw /VO
Kiow is related to the activity coef. of a
compound in both the water and octanol
phases
We might expect that the activity coef. iO
of organics in octanol is reasonably constant
for a given compound class such as PAHs or
alkanes.
Vw /Vo is also constant and
iw is proportional to 1/ Ciw (sat)
so
ln Kiow= -a ln Ciw(sat) + b
If we have a number of compounds from
given classes of compounds with known
Kiow and Csatiw values we can generate
regression coefficients for a and b
10

ln Kow = -a ln Cw (sat) + b
known Kiow
known Ciw (sat)
PAH1
PAH2
PAH3
PAH4
=============================================
log Kiow= -a ln Csatiw + b’
a
Alkanes
0.85
PAHs
0.75
 alkylbenzenes 0.94
chlorobenzens 0.90
PCBs
0.85
phthalates
1.09
Alcohols
0.94

b’
0.62
1.17
0.60
0.62
0.78
-0.26
0.88
r2
0.98
0.99
0.99
0.99
0.92
1.00
0.98
If you have the Csatiw for anthracene, could you
calculate its Kiow?
11
Remember in Chapter 5 we
estimated Csatiw from molar volumes
ln Csatiw = -a (size) +b
12
From combinatorial methods you can
also estimate a Kiow for many compounds
log Kiow =  nk fk + nj cj+0.23
where n is the frequency of a fragment type (fk)
and specific interaction (cj) for adjacent
functional groups.
13
14
15
If we have a Kiow we can add or subtract
functional units to get another Kiow for a
similar compound. Say we have the Kiow
for DDT and want to estimate the Kiow for
methoxychlor.
16
17
To Review
We have estimation techniques for relationships
for Csatiw
Csatiw = -a (size) +b
18
from the boiling point you can
estimate its vapor pressure
from Henry’s law structural units
you can estimate a Kiaw
 Kiaw x RT = P*i(L) / Csatiw
Csatiw can then be used in
 ln Kiow = -a ln Csatiw + b’
-----------------------------------------------Kiow from Chromatographic data
Total
HPLC
uv or
fluorescence
detector
Total
PAH 2 PAHs> Total 19
&3 rings 3 rings -PAHs
The column in many HPLC systems is
reverse phase, (column is non-polar and
mobile phase is polar, often ACN-water or
MeOH-H2O
MeOH-H2O
Partitioning of a hydrophobic organic may be
view as sorbing into the non-polar C18
alkanes on the LC packing and then back
into the mobile polar phase. Generally nonpolar Kiow and Ki hex-w correlate well
20
Since the time that a partitioning
compound spends in the mobile
phase will depend on the partitioning
coefficient (Kiow or Kism)
log Kiow = a log t +b;
where t is the retention time of the non-polar compound
log Kiow = a log (t-to )/to +b
where to is the r.t. of some non-retained species
21
Bioaccumulation and octanol water, Kiow
It has long been recognized that nonpolar organic compounds accumulate in
organisms in a way that is directly
related to insolubility in water (iw). Kiow is
related to related to Ciwsat, and
bioaccumulation in lipids is related to
Kiow. Hence, Veinth et al (Water Res, 13,
43-47, 1979) have related bio
accumulation factors (BCF) to Kiow.
log BCF {
moli / g wet fish
}  0.85log K iow  0.70
moli / mlwater
(a)
n = 59, r2 = 0.90
-------------------Chou et al (ES&T , 19, 57-62, 1985) does it
for fish lipids
log BCFL {
moli / g fishlipids
}  0.89 log Kiow  0.61
moli / mlwater
comparing eq (a) to (b) lipids are about 5%
of the fish weight.
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(b)
Going back to
log BCF {
moli / g wet fish
}  0.85log K iow  0.70
moli / mlwater
(a)
and assuming that a trout (fish) was
exposed to 10 nmolar trichlorobenzene,
and assuming a Kiow = 104 for
trichlorobenzene,
log BCF = 0.85 log 104 – 0.7 = 2.7
BCF = 500 ml water/g wet fish
BCF= [trout]/ Ciw
[trout] = BCF x Ciw
Ciw = 10x10-9 moles/liter = 10x10-12 moles/ml
[trout] =10x10-12 x 500 = 5x10-9 moles /g
wet fish
In lipid tissue of the fish
23
log BCF {
moli / g fishlipids
}  0.89 log K iow  0.61
moli / mlwater
(b )
[ trout lipid] = 1.5x10-7 moles /g fish lipid
What does this say about mother’s
milk??
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An Atmospheric gas particle example
using Kiow (Finizio, MacKay, Bidleman and Harner,
Atmos. Environ, 31,2289-2296,1997)
Is there a way to get Kiow into gasparticle partitioning and avoid using
or estimating vapor pressure??
What happens if you divide Kiow by
Kiaw or (KH’old book)???
Kiow = Cio/Ciw and Kiaw = Cia/Ciw (conc. In
mol/m3)
Kiow /Kiaw = {Cio/Ciw }/ {Cia /Ciw} =
Cio/ Cia = Kioa
(an octanol-air partitioning coef)
pi
octanol
water
Cio
Ciw
25
In the gas phase,
pi V= nRT and pi =Cia RT;
(R=8.314 Pa m3 mol-1 K-1)
pi is really the fugacity, fia ,in the gas phase
as well, under ideal conditions
The fugacity in the octanol phase is
fio = pI = xi i p* iL
The mole fraction, xi = Cio Vmix,O
And Vmix,O = Mwo / (o x(1000))
Where Mwo (g/mol) and o are the molecular
weight and density (kg m-3) of octanol, so
fio = pi = Cio Mwo p*iL / ((1000)o)
at equilibrium the fugacities in the air and
octanol phases are equal, so
26
i
fa =iCa RT
Cia RT = Cio Mwo  p*iL / (1000o )
and solving for p*iL
p*iL = Cia RT x o { Cio Mwo }
Let’s define Cio /Cia as Kioa or an octanolair partitioning coefficient- equilibrium
constant.
If we now substitute for p*iL into the Pankow
equation
Kip = RT fomx10-6/ {p*iLom Mwom}
Kip = fomx10-9 Cio Mwom i/{ Cia x om iom
Mwom}
Mackay, and students make the assumption
that Mwo, fom ,Mwom, o iom tend to be
constant for a class of compounds
27
And ultimately conclude
Kip = Kioa x const.,
So a plot of log Kip vs log Kiao should give a
straight line with a slope of 1
Since Kioa = Kiow/Kiaw’ and there these can
either be calculated or are in the literature it
is possible for get a comparison for Kip and
Kiao
28
To evaluate the theory Finizio, MacKay et al
calculated Kip from p*iL for a number of PAHs that
had logKp vs p*iL regression plots from previous
studies; they then computed Kioa from Kiow/Kiaw , and
plotted Kip oa
29
H = KiH, R = 8.31 Pa m3/(mol oK)
KiH needs to be changed into Kiaw
30
ARE we surprised
Recall chapter 3 where we related partitioning
to refractive indices
31
Figure 3.6 page 71 air-hexane, top, air-water,
bottom
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