Solubility of Organic Molecules in Water - Surface and Colloid Chemistry: Principles and Applications - page 236

Free energy of solubility = Δ G o sol = R T log (solubility) (B.2)

= (γ cavity ) (S area alkane) (B.3)

= 25.5 (S area alkane) (B.4)

For solubility of alkanes in water, the total surface area, TSA, gives the solubility:

ln (sol) = −0.043 (TSA) + 11.78 (B.5)

where solubility (sol) is in molar units and TSA in Å 2 . For example,

alkane (sol) tSa Predicted (sol)

N -Butane 0.00234 255 0.00143

n -Pentane 0.00054 287 0.0004

n -Hexane 0.0001 310 0.0001

n -Butanol 1.0 272 0.82

N -Pentanol 0.26 304 0.21

N -Hexanol 0.06 336 0.05

Note: The constant 0.043 is equal to γ cavity/RT =
25.5/600.

The surface areas of each group in n -nonanol (C 9 H 19 OH) were estimated by different
methods. These data are as follows:

ch 3 ch 2 ch 2 ch 2 ch 2 ch 2 ch 2 ch 2 ch 2 oh

85 43 32 32 32 32 32 40 45 59

It is seen that, if one needs the TSA of n -decanol, then it will be [TSA of nonanol +
TSA of CH 2 ] 431 + 32 = 463 Å 2 .
The data for solubility of homologous series of n -alcohols is of interest, as shown
in the following table.

alcohol Solubility (mol/l) log (S)

C 4 OH 0.97 0.013

C 5 OH 0.25 −0.60

C 6 OH 0.06 −1.22

C 7 OH 0.015 −1.83

C 8 OH 0.004 −2.42

C 9 OH 0.001 −3.01

C 10 OH 0.00023 −3.63

Accordingly, this algorithm allows one to estimate the solubility of water of an
organic substance. The estimated solubility of cholesterol was almost in accord with
the experimental data (Birdi, 2008).
It is seen that log (S) is a linear function of number of the carbon atoms in the
alcohol. Each -CH 2 - group reduces log (S) with a 0.06 unit.