Chapter 5:
Aqueous Solubility
equilibrium
partitioning of
a compound
between its
pure phase
and water
KH = PoL/Csatw
Kow = Csato/Csatw
Koa = Csato/PoL
Air
A gas is a gas is a gas
T, P
Koa
KH
Octanol
Po L
Water
Fresh, salt, ground, pore
T, salinity, cosolvents
Csat
w
Kow
Pure Phase
(l) or (s)
Ideal behavior
NOM, biological lipids,
other solvents
T, chemical composition
Csato
water
• covers 70% of the earth’s surface
• is in constant motion
• is an important vehicle for transporting
chemicals through the environment
Solubility
• is important in its own right
• will lead us to Kow and Kaw
Relationship between solubility and activity
coefficient
Consider an organic liquid dissolving in water:
 iL  
 iw  
*
iL
*
iL
for the organic
liquid phase
 RT ln  iL  x iL
 RT ln  iw  x iw
at equilibrium (maximum solubility):
for the organic
chemical in the
aqueous phase
 iw   i L  0  RT ln  iw  x iw  RT ln  iL  x iL
ln
x
sat
iw
x iL

RT ln  iL  RT ln 
RT
sat
iw
At saturation!
RT ln  iL  RT ln  iw
sat
ln
x iw
sat

x iL
RT
Assume: xiL = 1 and iL = 1
RT ln  iw
sat
ln x iw  
sat
E , sat

RT
G iw
RT
Solubility = excess free energy of solubilization (comprised
of enthalpy and entropy terms) over RT
The relationship between solubility and activity coefficient is:
x
sat
iw

1

sat
iw
or
C
sat
iw

1
V w
sat
iw
for liquids
The activity coefficient is the inverse of the mole fraction solubility
Solids
must account for the effect of “melting” of
solid
i.e. additional energy is needed to melt the
solid before it can be solubilized:
C
sat
iw
(L)  C
sat
iw
(s)  e

fu s G i
/ RT
At any given
temperature
Substitute activity
coefficient for liquid
solubility and rearrange:
*

fus
G i  RT ln
p iL

p is
sat
sat

1
V  C iw ( s )
sat
e

fu s G i
/ RT
Use this for HW 5.5
*
C iw ( L )  C iw ( s ) 
sat
iw
p iL
p is
Recall Prausnitz:
0
ln
ps
0
pL

 S fus (T m )  T m

 1

R
 T

Phase
change costs
or
Why bother
with the
hypothetical
liquid?
Melting point vs. boiling point
Naphthalene
Fluorene
Phenanthrene
Anthracene
Fluoranthene
Pyrene
Benz[a]anthracene
Benzo[a]pyrene
Tm and Tb vs. MW
700
y = 2.768x - 152.94
R2 = 0.9524
600
Tm or Tb (C)
500
400
300
200
100
0
100
y = 0.6573x + 13.002
R2 = 0.3016
120
140
160
180
MW (g/m ol)
200
220
240
260
MW
128.2
166.2
178.2
178.2
202.3
202.3
228.3
252.3
Tm
80.6
113
99.5
217.5
110.8
156
159.8
176.5
Tb
217.9
295
340.2
342
435
584
Gases
• solubility commonly reported at 1 bar or 1
atm (1 atm = 1.013 bar)
• O2 is an exception
• the phase change “advantage” of condensing the
gas to a liquid are already incorporated.
• the solubility of the hypothetical superheated
liquid (which you might get from an estimation
technique) may be calculated as:
C
sat
iw
(L)  C
pi
iw

p
*
iL
pi
theoretical “partial” pressure of
the gas at that T (i.e. > 1 atm)
Actual partial pressure of the
gas in your system
concentration
dependance of 
In reality,
 at saturation   at
infinite dilution
However, for
compounds with
 > 100 assume:
 at saturation =  at
infinite dilution
i.e. solute molecules
do not interact,
even at saturation
Molecular picture of the
dissolution process
The two most important driving forces in
determining the extent of dissolution of a
substance in any liquid solvent are
• an increase in disorder (entropy) of the
system
• compatability of intermolecular forces of
attraction.
Ideal liquids
The solubility of ideal liquids is determined by energy
lowering from mixing the two substances. For ideal
liquids in dilute solution in water, the intermolecular
attractive forces are identical, and Hmix = 0. The molar
free energy of solution is:
Gs = Gmix= -TSmix = RT ln (Xf/Xi)
Gs ,Gmix = Gibbs molar free energy of solution, mixing (kJ/mol)
-TSmix = Temperature  Entropy of mixing (kJ/mol)
R = gas law constant (8.414 J/mol-K)
T = temperature (K)
Xf, Xi = solute mole fraction concentration final, initial
Note: mole fraction of solvent  1 for dilute
solutions (dilute solution has solute conc <10-3 M)
dissolution
solute
solvent
two-phase form
- low disorder
solution form high disorder
The greater the dilution, the smaller (i.e., more
negative) the value of Gs and the more
spontaneous in the dissolution process
Nonideal liquids
The intermolecular attractive forces are not normally equal in
magnitude between organics and water.
Gs  Gmix (no longer equal)
Instead:
Gs = Gmix + Ge
Ge = Excess Gibbs free energy (kJ/mol)
Gs = Hs - TSs = He - T(Smix + Se)
He, Se = Excess enthalpy and excess entropy (kJ/mol)
He = intermolecular attractive forces; cavity formation
(solvation)
Se = cavity formation (size); solvent restructuring; mixing
Entropy:
Enthalpy:
For small
molecules,
enthalpy term
is small (± 10
kJ/mol)
Only for
large
molecules is
enthalpy
significant
(positive)
Entropy term
is generally
favorable
Except for
large
compounds,
for which
water forms a
“flickering
crystal”,
which fixes
both the
orientation of
the water and
of the
organic
molecule
Solubility Process
A mechanistic perspective of
solubilization process for organic
solute in water involves the following
steps:
a. break up of solute-solute
intermolecular bonds
b. break up of solvent-solvent
intermolecular bonds
c. formation of cavity in solvent phase
large enough to accommodate solute
molecule
d. vaporization of solute into cavity of
solvent phase
e. formation of solute-solvent
intermolecular bonds
f. reformation of solvent-solvent bonds
with solvent restructuring
Estimation technique
Activity coefficients and water solubilities can be estimated a
priori using molecular size, through molar volume (V,
cm3/mol).
Molar volumes in cm3/mol can be approximated:
Vi 
 (N
ij
)( a ij ) 
 (n
ij
)( 6 . 56 )
Ni = number of atoms of type i in jth molecule
ai = atomic volume of ith atom in jth molecule (cm3/mol)
nj = number of bonds in jth molecule (all types)
a values: see p. 149
Solubility can approximated using a LFER of the type:
ln  iw ( L )  a  ( size )  b
ln C iw ( L )   c  ( size )  d
sat
Molar volume here must be estimated by the atom
fragment technique (see p. 149)
This type of LFER is only applicable within a group of
similar compounds:
Another estimation technique
molar volume
describes vdW
forces
VP describes
self:self
interactions
ln  iw   ln p iL
*
refractive index
describes polarity
2


n
 1 
2/3
Di
   5 . 78 ( i )
 0 . 572  V ix   2

n

2
 Di


 8 . 77 ( i )  11 . 1(  i )  0 . 0472 V ix  9 . 49
H-bonding
cavity term
additional
polarizability
term
Note that this is similar to the equation we used to estimate vapor pressure,
but is much more complicated! Also, introduced , the polarizability term.
This approach is universal – valid for all compounds/classes/types
This approach can also be used (with different coefficients) to predict other
physical properties (for example, solubility in solvents other than water).
Factors Influencing Solubility in Water
•
•
•
•
•
Temperature
Salinity
pH
Dissolved organic matter (DOM)
Co-solvents
Temperature effects on solubility
Generally:
• as T , solubility  for solids.
• as T , solubility can  or  for liquids and gases.
• BUT For some organic compounds, the sign of Hs changes;
therefore, opposite temperature effects exist for the same compound!
The influence of temperature on water solubility can
be quantitatively described by the van't Hoff
equation as:
ln Csat = -H/(RT) + Const.
ln
C sat T 2
C sat T 1

H
R
 1
1 


T  T 
2 
 1
recall from
thermodynamic
lecture
What H is this?
H
E
iw
the energy (enthalpy) needed to get the liquid (real or
hypothetical) compound into aqueous solution
Liquids:
 wL H  H iw
Solids:
 ws H  H iw  
OR
 wa H  H iw   vap H i
E
E
fus
Hi
E
gas
gas
water
 wa H i
Pure liquid
 wL H i
Pure solid

fus
Hi
aqueous
 vap H i
 ws H i
 sub H i
liquid
solid
Note: sometimes energy states are higher/lower, so some of these enthalpy terms could be negative!
Solids, liquids, gases…
Solids
ln C
(s)  
sat
iw

H i  H iw
E
fus
 cst
RT
E
Liquids
ln C
sat
iw
(L)  
H iw
 cst
RT
  vap H i  H iw
E
ln C iw ( g )  
sat
Gases
RT
liquid
gas
 cst
Parameters for this plot:
lnCsat

fus
H i  20 kJ/mol
H iw  10 kJ/mol
E
solid
Tb
1/T
Tm
 vap H i  20 kJ/mol
Salinity effects on solubility
As salinity increases, the solubility of neutral organic
compounds decreases (activity coefficient increases)
 iw , salt   iw  10
s
 K i [ salt ] to t
typical seawater
[salt] = 0.5M
Ks = Setschenow salt constant (depends on the compound and
the salt)
[salt] = molar concentration of total salt.
K i , seawater 
s

K i , salt k  x k
s
k
The addition of salt makes it more difficult for the organic
compound to find a cavity to fit into, because water molecules
are busy solvating the ions.
pH can increase apparent solubity
pH effect depends on the structure of the solute.
If the solute is subject to acid/base reactions then pH
is vital in determining water solubility.
The ionized form has much higher solubility than the
neutral form.
The apparent solubility is higher because it
comprises both the ionized and neutral forms.
The intrinsic solubility of the neutral form is not
affected.
We will talk about this more when we look at
acid/base reactions
Dissolved organic matter (DOM) can
increase apparent solubility
DOM increases the apparent water solubility for sparingly
soluble (hydrophobic) compounds. DOM serves as a site
where organic compounds can partition, thereby
enhancing water solubility. Solubility in water in the
presence of DOM is given by the relation:
Csat,DOM = Csat (1 + [DOM]KDOM)
[DOM] = concentration of DOM in water, kg/L
KDOM = DOM/water partition coefficient
Again, the intrinsic solubility of the compound is not
affected.
Co-solvent effect on solubility
• the presence of a co-solvent can increase the
solubility of hydrophobic organic chemicals
• co-solvents can completely change the solvation
properties of “water”
• examples:
–
–
–
–
industrial wastewaters
“gasohol”
engineered systems for soil or groundwater remediation
HPLC
focus on
• sparingly soluble solutes
• completely water-miscible organic solvents
– methanol, ethanol, propanol, acetone, dioxane,
acetonitrile, dimethylsulfoxide,
dimethylformamide, glycerol, and more
What do these solvents have in common?
In general
• solubility increases exponentially as
cosolvent fraction increases.
• need 5-10 volume % of cosolvent to see an
effect.
• extent of solubility enhancement depends
on type of cosolvent and solute
– effect is greatest for large, nonpolar solutes
– more “organic” cosolvents have greater effect
propanol>ethanol>methanol
Bigger, more non-polar
compounds are more
affected by co-solvents
Different co-solvents behave
differently, behavior is not
always linear
We can develop linear
relationships to describe the
affect of co-solvents on
solubility. These relationships
depend on the type and size of
the solute
Quantifying cosolvent effect can be complex, so
assume log-linear relationship between solubility and
volume fraction of cosolvent (fv)
log x il ( f v )  log x il ( f v )   i  ( f v  f v )
sat
sat
1
c
1
if fv1 = 0, then we are describing the solubility
enhancement relative to the standard aqueous solubility:
log x il ( f v )  log x il   i  f v
sat
sat
c
ic is the slope term, which depends on the both the
cosolvent and solute
Problem 5.4
• estimate the solubilities of 1-heptene and
isooctane (2,2,4 trimethylpentane)
compound
isoctane:
r = 0.692 g/mL
1-heptene
r = 0.697 g/mL
Characteristic
volumes:
H = 8.71
C = 16.35
-per bond = 6.56
1-pentene
2-me-1-pentene
1-hexene
4-me-1-pentene
2,2-dimethylbutane
2,2-dimepentane
2,2,3-trime butane
3-me hexane
1-octene
2-me heptane
1-nonene
3-me octane
2,2,5-trimethylhexane
MW
g/mol
70.1
84.2
84.2
84.2
86.2
100.2
100.2
100.2
112.2
114.2
126.3
128.3
128.3
Tb
C
30
60.7
63.4
53.9
49.7
79.2
80.9
92
121.3
117.6
146.9
143
124
Csat @ 25C
mg/L
148
78
50
48
12.8
4.4
4.4
3.3
2.7
0.85
1.12
1.42
1.15