Chapter 5: Solubility and Activity
Coefficients in Water
First let’s get a feeling for the “driving
force of mixing” in terms of the partial
molar free energy G or chemical potential,
i of a compound in a phase, which is:

i =  i* +RT ln f i/f i *
if we choose f i * as the fugacity
of the pure liquid and
f i =
 i X if i *pure liquid
 i =  i* +RT ln  i X i
 i w =  i* +RT ln  i w X i w
the organic phase
the water phase
The difference in chemical potentials for our
compound in interest, i, in the two phases is
1
 i w -  i= RT ln  i w X i w - RT ln  i X i
where  i w -
 i= solG i the molar free energy of
solution or “the driving force” for phase
transfer
Initially if we start with pure organic and water and just
consider the X i w (organic in water),
X i w is -->
zero
solG i = RT ln  i w X i w - RT ln  i X i

at some small time, dt, after the phase transfer
starts
let say  i.999 and X i= 0.999;
lni Xi= -0.002
At this same time, dt, let’s say two organic
molecules have gone into the water phase; X iw
2
will be very small, eg., 2 divided by 6.02 x 10
23
molecules for one mole of water, or X iw= ~10-24
and say  iw >> 1 for a very dilute solution of
the organic in water, remember toluene
 iw in water = 1x10+4
the product of
Xiw and iw is still << than 1
so ln iw Xiw = ln [1x10-24 x 10+4] = - 46
multiplying by RT gives, RT ln
-114 kJ mole-1 for solG
 iw X iw =
This makes G negative (it’s chemical potential
difference between the two phases); the sign tells
the direction of the desired transfer.
This process continues until
 iw X iw = i X i
or
f iw = f i
3

where
and
f iw =  iw X iw f io
f i = i X i f io
going back to

 solGi = RT ln  iw X iw - RT ln i Xi
 For the majority of compounds Xi, the mole
fraction of the organic in the organic layer is
essentially one i.e. there is almost no water in
the organic phase;
 we will also assume that the activity coef. in the

organic phase is essentially ideal and is close
to one

From Chapter 3
RT lni hx Xi hx = RT lniH2OXi H2O

xiH2O
RT ln
 (RT ln iH2O  RT ln ihx ) 
xihx
4
xiH2O
'
 KiH
2O / hx = Ki12;
xihx

1,2G = - RT ln KH2O/hx; (Free energy of transfer)
12Gi = RT ln iw + RT ln
where
Xiw
 GE = RT ln iw ;
 RT ln Xiw is called the entropy of ideal
mixing.
the RT ln iw term is the molar excess free
energy, GiE, of the liquid compound in water
due to the non-ideally of solution of the organic
in water
5
Chapter 5 compares
saturated and infinitely dilute activity coefficients
6
Table 5.2 page 80 old book (see page 141 new book)
7
Solubility of solids and gases in liquids
for liquid -liquid interactions (organic-water)
12Gi = RT ln iw + RT ln
Xiw
and
we have shown over and over again
that
1/Xiw = iw
In dissolving a solid into a liquid we need to also
account for melting
From the Gibbs Duhem equation
for gases in equilibrium with a liquid
vapGi = vapi = RT ln p*iL / po
if po is one bar (or atm)
vapGi = vapi = RT ln p*iL
for gas-solids by analogy
subGi = RT ln p*iS
and fusGi = subGi - vapGi= RTln {p*iL/ p*iS}
to account for melting
8
12Gi = RT lnXiw + RT
ideal
mixing
*
piL
lniw - RT ln *
pis
nonideal
effects
melting
so
Xiwsat =
1/iwsat
(liquids)
*
sat
sat pis
Xiw = 1/iw
*
piL
(solids)
Note that
Xsatiw/
Vmix =
Xsatiw(L) /
Csatiw = Csatiw(L)
Vmix
*
pis
*
piL
*
pis
*
piL
*
p
Xiwsat = 1/iwsat ig*
piL
(gases)
As far as computing  from Ciwsat values,
9
the new book gives log Cwsat corrected for
solid -- liquid interactions; the old book gives
both
new book example page 140
Estimate Csatiw (L), satiwand GEiw for di-n-butyl
phthalate
1st
Csatiw (L),= Csatiw
on page 1206, -log Csatiw= 4.36
Csatiw = 4.37x10-5
Csatiw = Xi / Vmix= 1 /i Vmix
So i = 1/( 4.37x10-5 x 0.018) = 1.27x106
GEiw= RTlnI = 3483 J/ (molK)
What about solid hexa-chlorocyclohexane???
10
If we can estimate p*is/ p*iw

sat
sat
iw = 1/X
iw
*
pis
*
piL
(solids)
and plug into
iwsat =
1/(
Csatiw
*
p
Vmix) is
*
piL
(solids)
*
pis
a ”poor man’s” estimate of ln * is
piL
ln
*
pis
0
=
-56.5/R
(T
m/Tamb –1); for Tm= 113 C
*
piL
*
pis
*
piL
 0.076 ;
iwsat
Csatiw= 2.5x10-5 moles/L
= 1.69 x105
and
GEiw= RTlnI
11
Heats of Solution relationships
Figure 5.3 page 83 (old book)
Hcav= H1+H2 + H3
 H1used to break
orgainc-organic bonds
 H2 used to break H2O H2O bonds and forming
a cavity
 H3 heat released from
organic-H2O bonds
 H1&H2>1; H3 <1
 Hice = water molecules
around organic are attracted
to outside water molecules
and “solidified” in place
 HsE = Hcav+Hice  molecular size
12
Enthalpies of solution appear to be related to surface
area of the molecule
13
14
Is there a relationship between
solubility and molar volumes with in
a compound class?
15
16
This suggests a generalized relationship
ln iw = a (size) +b
or
ln Csatiw = -a (size) +b
17
Entropy of Dissolution
Gs = RT ln w + RT ln
Xw (entropy term)
It is difficult to derive an exact analog between
excess enthalpies of solution, Hes and excess
entropies of solution Ses
Since entropy is an indicator of randomness, for an
ideal solution,
Sidealmix= -R (nsoluteln Xsolute + nsolvent ln Xsolvent)
here it is assumed that each molecule has
approximately the same size and shape
The non-ideal mixing, of large organic molecules
results in the displacement of many water
molecules. It is suggested (old book) that a better
description of the displacement of water molecules
is the volume fraction
Srealmix= -R (norgln X ’org + nH2O ln X ’H2O)
where X’’ is the volume fraction
18
Srealmix= -R (norgln X ’org + nH2O ln X’’H2O)
the volume fraction of X ’H2O is almost 1
so Srealmix= -R norgln X ’org
if we represent X’org as vol. In the organic phase/
totalvol
S
real
mix
  Rnorgln
norg Vorg
norg Vorg  nH2O VH2O
because nH2O >> norg
S
real
mix
  Rnorgln
norg Vorg
nH2O VH2O
separating the ln term and, norg/nH2O = Xorg
S
real
mix
 Rnorglnx org  Rnorgln
Sidealmix
per mole
T S
e
mix
Vorg
VH2O
Semix
 RT ln
Vorg
VH2O
volume is important
19
RT ln wGes = Hes + TSes
=
Hcav+Hice –T(Scav+Sice+Semix)
Contribution of molecular size to entropy of dissolution
of an organic compound in water (Figure 5.4 p 87 old
book)
20
Table 5.3 page 87 (old book)
21
GEi = HEi + TSEi= RT ln w
Enthalpy and Entropy contributions to Excess Free
Energies of solution (Table 5.4 p 88, old book)
22
Effect of Temperature and Solution Composition
on Aqueous Solubility and Activity Coefficients
*
ln p i pure liq .
sat
ln iw
 vapHi 1

 constp
R
T
HE
1
 i
 const
R T
assuming a const. HEi ,and Csatiw = Xsatiw / Viw
where Viw does not change with temperature
sat
ln Ciw
E
since H
i
HE
  i  const
RT
is small and negative for most organics in water is
reasonable that ln X does not change with temperature
23
Solubility vs. temp. (Figure 5.6, p 91, see page 155
new book)
24
Effects of temperature on activity coef.
25
Effects of Salts; Figure 5.7 page 94
26
Csat
log[ satw ]  K s [ salt ]t
Cw,salt
in sea water where salt = 30,000 ppm or
= 30,000x10-6g/ml = 30,000x10-3g/L
if we take NaCl at a Mw of 58.5g/mol
30,000x10  3 g/L
#moles /L of NaCl 
 0.5 M
58.5g/mol
and If Ks is ~0.3, sea water will have the effect of
C sat
K
[ satw ] 10 s
C w,salt
[ salt ]t
 10.3 x0.5  14
.
salting out phenomenon may be viewed as
polar ions Na+ and Cl- being hydrated and reducing the
availability water to dissolve into or less and less water
to form cavities
27
If the effects of individual salts are additive
Ks 
 K is xi
i
Effects of different Salts; Table 5.7 page 97(old book)
28
Dissolved Organics Solutes and Solvents
 effects on a large amt.
dissolved organic in an
organic/water solution;
say MeOH and H2O
 other dissolve organics
but a lower conc.
 low levels of other
dissolved organics
29
Effects on a large amt. dissolved organic in an organic/water
solution; say MeOH and H2O
Yalkowsky and co-workers reasoned that the excess free energy should
be the sum of the solution free energies in each solvent
Ges:mix (1 fc )Ges:w  fc Ges:c
fc= vol faction of co-solvent
Ges: w  excess free energy in H2O
Ges:c  excess free energy in co  solvent organic.
recalling that Gef = + RT ln

 


since =1/X,
lnmix= (1-fc) lnw + fcln(c)
lnXmix= (1-fc) lnXw + fcln(Xc)
30
Yalkowsky then reasons that the excess free energy of the dissolved
organic in water and a co-organic solvent is the sum of the free energies
in water and the co-organic. This free energy is a function of the
interfacial energy in J/cm-2 where the organic of interest contacts the
water and similarly the organic co-solvent
Everything we have seen ---> the importance of molecule surface area
in water

Ges:w = (h:w) ( HSA)(N)+ (p:w) (PSA)(N)
h:w= interfacial energy where the hydrophobic part of the
solute molecule contacts water
p:w= interfacial energy where the polar part of the solute molecule
contacts water
HSA and PSA = solute molecule hydrophobic and polar surf.area for
For the organic

Ges:c = (h:c) ( HSA)(N)+ (p:c) (PSA)(N)
31
since G = + RT ln and
substituting

Geh:w = (h:w) ( HSA)(N)+ (p:w) (PSA)(N) and

Ges:c = (h:c) ( HSA)(N)+ (p:c) (PSA)(N)
into
-RT lnXmix= -(1-fc) RT lnXw -fc RT ln(Xc)
sat
 log Xmix
N( h: w )HSA N( p: w )PSA 
 

2.303RT 
 2.303RT
 f N(
fcN( p: w   p:c )PSA 
c
h: w   s: c )HSA



2.303RT
2.303RT


if the polar surface area of the solute molecule is very small
32
N( h: w )HSA  fcN( h: w   s:c )HSA 
sat
 log Xmix
  



2
.
303
RT
2
.
303
RT

 

rearranging
sat
log Xmix
N( h: w )HSA 
N( s:c )HSA 
 

(
1

f
)

f

c
c

 2.303RT 
 2.303RT

recalling
lnXmix= (1-fc) lnXw + fcln(Xc) and going back to our equation above
sat
log X mix

log X sat
w
f c N( h:w   s:c )HSA 

 and
2.303RT


sat 
fcN(h: w   s:c )HSA 
 Xmix
log sat    

2
.
303
RT
 Xw 


33
Yalkowsky et al
solute in a water organic mixture
sat 
fcN( h: w   h:c )HSA 
 Xmix
log sat    

2
.
303
RT


 Xw 
HSA = Hydrophobic surface area
h:w=hydrophobic interfacial energy
where the solute contacts the water
h:w=hydrophobic interfacial energy
where the solute contacts the organic
fc = volume fract. of organic
for water sea water
 X sat
fcN( t: w   t:sw )HSA 
s, w 
log sat    

2
.
303
RT

 X w 

Looks like a Setschenow
34
c N( air: water   air:salt water )HSA 

K [salt ]   
2.303RT


s
air:water=surface tensions against air
where the solute contacts the air-water
surface
35
increased hydrophobic surface area
Increased fraction of co-solvent (propylene glycol)
36
37