Chapter 5: Solubility and Activity Coefficients in Water

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the organic phase

 i w =  i* +RT ln  i w X i wthe water phase

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

 i w -  i= RT ln  i w X i w - RT ln  iX i

where  i w -  i= solG ithemolar 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 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  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 iw Xiw =

-114 kJ mole-1 for solG

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

orf iw = f i

where f iw =  iw X iw f io

and 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 hxXi hx= RT lniH2OXi H2O



= Ki12;

1,2G = - RT ln KH2O/hx; (Free energy of transfer)

12Gi = RT ln iw + RT ln Xiw

where 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

Chapter 5 compares

saturatedand 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)

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

12Gi = RT lnXiw + RT lniw -

ideal nonideal melting
mixing effects

so

Xiwsat = 1/iwsat (liquids)

Xiwsat = 1/iwsat (solids)

Note that

Xsatiw/ Vmix = Xsatiw(L)/ Vmix

Csatiw = Csatiw(L)

Xiwsat = 1/iwsat (gases)

As far as computing  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), satiwand GEiwfor 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???

If we can estimate p*is/ p*iw

satiw= 1/Xsat iw (solids)

and plug into

iwsat =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

iwsat = 1.69 x105

and

GEiw= RTlnI

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Heats of Solution relationshipsFigure 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

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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 iw = a (size) +b

or

ln Csatiw = -a (size) +b

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)

whereX’’ is the volume fraction

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

because nH2O > norg

separating the ln term and, norg/nH2O = Xorg

SidealmixSemix

per mole volume is important

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)

Table 5.3 page 87 (old book)

GEi = HEi + TSEi= RT ln w

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. HEi ,andCsatiw = Xsatiw / Viw

where Viw does not change with temperature

since HEi 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)

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Effects of temperature on activity coef.

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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)

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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

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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 Gef = + RT ln

lnmix= (1-fc) lnw + fcln(c)

since =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

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)

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)

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

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Yalkowsky et al

solute in a water organic mixture

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

Looks like a Setschenow

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

increased hydrophobic surface area

Increased fraction of co-solvent (propylene glycol)

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