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 DG or chemical potential, mi of a compound in a phase,f, which is:

mif = m i* +RT ln f i f /f i *

if we choose f i * as the fugacity
of the pure liquid and

f i f = g i f X i f f i *pure liquid

m i f = m i* +RT ln g i f X i f the organic phase

m i w = m i* +RT ln g i w X i w the water phase

The difference in chemical potentials for our compound in interest, i, in the two phases is

m i w - m i f= RT ln g i w X i w - RT ln g i f X i f

where m i w - m i f= DsolG 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

DsolG i = RT ln g i w X i w - RT ln g i f X i f

at some small time, dt, after the phase transfer starts

let say g i f = .999 and X i f= 0.999;

lngif Xi f = -0.002

At this same time, dt, let’s say two organic molecules have gone into the water phase; X iw will be very small, eg., 2 divided by 6.02 x 1023 molecules for one mole of water, or X iw= ~10-24

and say g iw > 1 for a very dilute solution of
the organic in water, remember toluene

g iw in water = 1x10+4

the product of
Xiw and giw is still < than 1

so ln giw Xiw = ln [1x10-24 x 10+4] = - 46

multiplying by RT gives, RT ln giw Xiw =

-114 kJ mole-1 for DsolG

This makes DG negative (it’s chemical potential difference between the two phases); the sign tells the direction of the desired transfer.

This process continues until

g iw X iw =g if X if

or f iw = f if

where f iw = g iw X iw f io

and f if =g if X if f io

going back to

DsolGi = RT ln g iw X iw - RT ln gif Xif

· For the majority of compounds Xif, 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 lngi hx Xi hx = RT lngiH2OXi H2O

= Ki12;

D1,2G = - RT ln KH2O/hx; (Free energy of transfer)

D12Gi = RT ln giw + RT ln Xiw

where D GE = RT ln giw ;

· RT ln Xiw is called the entropy of ideal mixing.
the RT ln giw term is the molar excess free energy, DGiE, of the liquid compound in water due to the non-ideally of solution of the organic in water

Chapter 5 compares

saturated and infinitely dilute activity coefficients

Table 5.2 page 80 old book (see page 141 new book)

Solubility of solids and gases in liquids

for liquid -liquid interactions (organic-water)

D12Gi = RT ln giw + RT ln Xiw and we have shown over and over again that

1/Xiw = giw

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

DvapGi = Dvapmi = RT ln p*iL / po

if po is one bar (or atm)

DvapGi = Dvapmi = RT ln p*iL

for gas-solids by analogy

DsubGi = RT ln p*iS

and DfusGi = DsubGi - DvapGi= RTln {p*iL/ p*iS}

to account for melting

D12Gi = RT lnXiw + RT lngiw -

ideal nonideal melting
mixing effects

Xiwsat = 1/giwsat (liquids)

Xiwsat = 1/giwsat (solids)

Note that

Xsatiw/ Vmix = Xsatiw(L) / Vmix

Csatiw = Csatiw(L)

Xiwsat = 1/giwsat (gases)

As far as computing g from Ciwsat values,

the new book gives log Cwsat corrected for solid -- liquid interactions; the old book gives both

new book example page 140

Estimate Csatiw (L), gsatiwand 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 /gi Vmix

So gi = 1/( 4.37x10-5 x 0.018) = 1.27x106

GEiw= RTln gI = 3483 J/ (molK)

What about solid hexa-chlorocyclohexane???

If we can estimate p*is/ p*iw

gsatiw = 1/Xsat iw (solids)

and plug into

giwsat = 1/( Csatiw Vmix) (solids)

a ”poor man’s” estimate of is

= -56.5/R (Tm/Tamb –1); for Tm= 1130C

; Csatiw= 2.5x10-5 moles/L

giwsat = 1.69 x105

and

GEiw= RTln gI

Heats of Solution relationships Figure 5.3 page 83 (old book)

DHcav= DH1+DH2 + DH3

· DH1used to break
orgainc-organic bonds

· DH2 used to break H2O -
H2O bonds and forming
a cavity

· DH3 heat released from
organic-H2O bonds

· DH1 DH2>1; DH3 <1

· DHice = water molecules
around organic are attracted
to outside water molecules
and “solidified” in place

· DHsE = DHcav+ DHice µ molecular size

Enthalpies of solution appear to be related to surface area of the molecule

Is there a relationship between solubility and molar volumes with in a compound class?

This suggests a generalized relationship

ln giw = a (size) +b

ln Csatiw = -a (size) +b

Entropy of Dissolution

DGs = RT ln gw + RT ln Xw (entropy term)

It is difficult to derive an exact analog between excess enthalpies of solution, DHes and excess entropies of solution DSes

Since entropy is an indicator of randomness, for an ideal solution,

DSidealmix= -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

DSrealmix= -R (norgln X ’org + nH2O ln X ’H2O)

where X’’ is the volume fraction

DSrealmix= -R (norgln X ’org + nH2O ln X’’H2O)

the volume fraction of X ’H2O is almost 1

so DSrealmix= -R norgln X ’org

if we represent X’org as vol. In the organic phase/ totalvol

because nH2O > norg

separating the ln term and, norg/nH2O = Xorg

DSidealmix DSemix

per mole volume is important

RT ln gw = DGes = DHes + TDSes

= Hcav+DHice –T(DScav+DSice+DSemix)

Contribution of molecular size to entropy of dissolution of an organic compound in water (Figure 5.4 p 87 old book)

Table 5.3 page 87 (old book)

DGEi = DHEi + TDSEi= RT ln gw

Enthalpy and Entropy contributions to Excess Free Energies of solution (Table 5.4 p 88, old book)

Effect of Temperature and Solution Composition on Aqueous Solubility and Activity Coefficients

assuming a const. DHEi ,and Csatiw = Xsatiw / Viw

where Viw does not change with temperature

since DHEi is small and negative for most organics in water is reasonable that ln X does not change with temperature
Solubility vs. temp. (Figure 5.6, p 91, see page 155 new book)

Effects of temperature on activity coef.

Effects of Salts; Figure 5.7 page 94

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

and If Ks is ~0.3, sea water will have the effect of

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

If the effects of individual salts are additive

Effects of different Salts; Table 5.7 page 97(old book)

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

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

fc= vol faction of co-solvent

recalling that DGef = + RT lng

` lngmix= (1-fc) lngw + fcln(gc)

since g=1/X, lnXmix= (1-fc) lnXw + fcln(Xc)

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

DGes:w = (sh:w) ( HSA)(N)+ (sp:w) (PSA)(N)

sh:w= interfacial energy where the hydrophobic part of the solute molecule contacts water
sp: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

DGes:c = (sh:c) ( HSA)(N)+ (sp:c) (PSA)(N)

since DG = + RT lng, and

substituting

DGeh:w = (sh:w) ( HSA)(N)+ (sp:w) (PSA)(N) and

DGes:c = (sh:c) ( HSA)(N)+ (sp:c) (PSA)(N)

into

-RT lnXmix= -(1-fc) RT lnXw -fc RT ln(Xc)

if the polar surface area of the solute molecule is very small

rearranging

recalling

lnXmix= (1-fc) lnXw + fcln(Xc) and going back to our equation above

and

Yalkowsky et al

solute in a water organic mixture

HSA = Hydrophobic surface area

sh:w=hydrophobic interfacial energy
where the solute contacts the water

sh:w=hydrophobic interfacial energy
where the solute contacts the organic

fc = volume fract. of organic

for water sea water

Looks like a Setschenow

sair:water=surface tensions against air
where the solute contacts the air-water
surface

increased hydrophobic surface area

Increased fraction of co-solvent (propylene glycol)