Physical Chemistry , 1st ed.

solubility is (grams of solute)/(100 mL of solvent).] Most of the solutions we

work with are unsaturated, having less than the maximum amount of solute that

can dissolve. Occasionally, it is possible to dissolve more than the maximum. This

is typically done by heating the solvent, dissolving more solute, then cooling the

solution carefully so that the excess solute does not precipitate. These are super-

saturatedsolutions. However, they are not thermodynamically stable.

For an ideal liquid/solid solution, it is possible to calculate the solubility of

the solid solute. If a saturated solution exists, then saturated solution is in equi-

librium with excess, undissolved solute:

solute (s) solvent () solute (solv) (7.32)

where the solute (solv) refers to the solvated solute, that is, the dissolved

solid.

If this equilibrium does exist, then the chemical potential of the undissolved

solid equals the chemical potential of the dissolved solute:

°pure solute (s)dissolved solute (7.33)

The undissolved solute’s chemical potential has a ° superscript because it is a

pure material, whereas the chemical potential of the dissolved solute is part of

a solution. However, if the dissolved solute can be considered as one compo-

nent of a liquid/liquid solution (with the other liquid being the solvent itself ),

then the chemical potential of the dissolved solute is

dissolved solute°dissolved solute ()RTln xdissolved solute (7.34)

Substituting for dissolved solutein equation 7.33:

°pure solute (s)°dissolved solute ()RTln xdissolved solute (7.35)

This can be rearranged to find an expression for the mole fraction of the dis-

solved solute in solution:

ln xdissolved solute (7.36)

The expression in the numerator of equation 7.36 is the chemical potential

of the solid minus the chemical potential of the liquid for a pure solute,

which equals the change in the molar Gibbs free energy for the following

process:

solute () →solute (s) (7.37)

That is, the numerator refers to the change in free energy for a solidification

process. The Gibbs free energy for this process would equal zero if it occurred

at the melting point. IfTis not the melting-point temperature, then (^) fusGis
not zero. Equation 7.37 is the reverse of the melting process, so the change in
Gcan be represented as  fusG. Therefore, equation 7.36 becomes
ln xdissolved solute



R

f

T

usG (7.38)

Here, we are substituting for (^) fusG, again noting that (^) fusHand (^) fusSrepre-
sent the changes in enthalpy and entropy at some temperature T, which is not
the melting point.
Now we will add zero to the last expression in equation 7.38, but in an un-
usual way: by adding ( (^) fusGMP)/RTMP,where (^) fusGMPis the Gibbs free en-
ergy of fusion and TMPis the melting point of the solute. At the melting
( (^) fusHT (^) fusS)

RT
°pure solute (s)°pure solute ()

RT

JQPJ

186 CHAPTER 7 Equilibria in Multiple-Component Systems