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

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 = 0.16L/mol (at 25oC)

Water saturated octanol
Vmix=(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?

From the thermodynamics discussion,

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

ln Kisw = ln iw - ln is +ln

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

Comparing partitioning in Octanol-water and hexane water; octanol is an amphiphilic solvent, e.g., it has both a non polar and polar group.

KiowKihw

N hexane13000 52,000

Benzene 130 170

Toluene 490 569

Chlorobenzene 830810

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


For weakly apolar or polar compounds Kiowand Kiosare related

Log Kihw= 1.21 log Kiow -0.43


for a dilute compound i, we can say

iw Vw = wsat Vw

and

iwsat Vw = 1/Ciwsat

Thus in

Kisw= 1/( Ciwsatis Vs)

log Kiow= -log Ciwsat -log io -log VO

Another form of this is:

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

 ln Kow= -a ln Cw(sat) + b
known Kiow known Ciw(sat)

PAH1
PAH2
PAH3
PAH4
======

log Kiow= -a ln Csatiw + b’
ab’r2

Alkanes0.850.620.98

PAHs0.751.170.99

 alkylbenzenes0.940.600.99

chlorobenzens0.900.620.99

PCBs0.850.780.92

phthalates1.09 -0.26 1.00

Alcohols0.940.880.98

If you have the Csatiw for anthracene, could you calculate its Kiow?

It can also be show that Csatiwvalues can be related to molar volumes

ln Csatiw = -a (size) +b


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.

If we have a Kiow we can add or subtract functional units to get another Kiowfor a similar compound. Say we have the Kiow for DDT and want to estimate the Kiow for methoxychlor.

To Review

We have estimation techniques for relationships for Csatiw

Csatiw = -a (size) +b

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

Csatiwcan then be used in

 ln Kiow = -a ln Csatiw+ b’
------

Kiow from Chromatographic data

The column in many HPLC systems is reverse phase, (column is non-polar and mobile phase is polar, often ACN-water or 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 non-polar Kiow and Ki hex-w correlate well


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


Bioaccumulation and octanol water, Kiow

It has long been recognized that non-polar 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.

n = 59, r2 = 0.90

------

Chou et al (ES&T , 19, 57-62, 1985) does it for fish lipids

comparing eq (a) to (b) lipids are about 5% of the fish weight.

Going back to

and assuming that a trout (fish) was exposed to 10 nmolar trichlorobenzene, and assuming a Kiow = 104for 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

[trout lipid] = 1.5x10-7moles /g fish lipid

What does this say about mother’s milk??

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 gas-particle partitioning and avoid using or estimating vapor pressure??

What happens if you divide Kiowby 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 Cio

water Ciw

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ip* 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

ifa =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 octanol-air partitioning coefficient- equilibrium constant.

If we now substitute for p*iL into the Pankow equation

Kip = RT fomx10-6/ {p*iLom Mwom}

Kip = fomx10-9Cio 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

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

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 Kipoa

H = KiH, R = 8.31 Pa m3/(mol oK)

KiH needs to bechanged intoKiaw

ARE we surprised

Recall chapter 3 where we related partitioning to refractive indices

Figure 3.6 page 71air-hexane, top, air-water, bottom

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