Computational Chemistry

was"7.72 kcal mol"^1 or"32.3 kJ mol"^1. Note that the calculated solvation free
energies, terms (1) and (3), are free energydifferences, gas to solvent, estimated by
the program, and do not require time-consuming frequency calculations as is the
case for statistical mechanical calculation of free energies.
Term (2) calculated by the high-accuracy multistep CBS-APNO method
(Section 5.5.2.2b) was 341.2 kcal mol"^1 or 1426 kJ mol"^1. The Sackur-Tetrode
equation for the gas-phase entropy of the proton was mentioned in this regard, but
in fact the algorithm automatically handles this.
Term (3), calculated as for term (1) was"77.58 kcal mol"^1 or"324.6 kJ mol"^1.
Term (4), the solvation free energy of the proton was taken from experiment
as"264.61 kcal mol"^1 or"1,107 kJ mol"^1.
The free energy of dissociation in water follows (Fig.8.4):

DGdiss;s¼"DGsðÞþRCOOH DGhighþDGsðÞþRCOO" DGsðÞHþ

¼"ð" 32 : 3 Þþ 1426 " 324 : 6 "1107 kJ mol"^1 ¼ 26 :70 kJ mol"^1 :

From the usual relation of free energy to pKa(e.g. [ 46 ]), withRTat 298 K¼
2.478 kJ mol"^1 we get pKa¼26.70/2.303RT¼4.68. The experimental value for
acetic acid was reported [ 46 ] to be 4.75, for an error of only"0.07 pKaunit.
As Liptak and Shields point out, accurate values of gas phase deprotonation and
solvation energies are needed for reasonably accurate pKavalues. An error of 1 pKa
unit results from an error inDGof 1.36 kcal mol"^1 or 5.7 kJ mol"^1 , and an error of
0.5 pKaunit corresponds to an error inDGof only 2.9 kJ mol"^1. For some purposes
such an energy-difference error would be considered small, 1 kcal mol"^1 or 4 kJ

∆Gs(RCOO–) + ∆Gs(H+)

RCOOHg

from gas phase

calcs

RCOOHs

wanted

RCOO– g + H+^ g RCOO–s + H+s

∆Gdiss, s

−∆Gs(RCOOH)

free E of solvation of RFCOOH

∆Ghigh

Fig. 8.4 The principle behind the absolute method of calculating pKa. In this thermodynamic cycle
we wantDGfor dissociation of RCOOH in water (gdenotes gas phase andssolvent phase, water;
we refer to standard temperature and pressure free energy differences). The other terms are: (1)
"DGs(RCOOH), the negative of the solvation free energy of RCOOH (the solvation free energy
itself is negative); we takeDGof solvation as the free energy that one mustput in(a negative
quantity) to solvate a species, so going from solution to gas requires input of"DG(a positive
quantity). This quantity is calculated by a continuum method. (2)DGhigh, the gas-phase ionization
free energy of RCOOH, calculated by a high-level multistep method. (3þ4)DGs(RCOO")+
DGs(H+), the free energy of solvation of the anion plus the free energy of solvation of the proton.
The first term is calculated by a continuum method method and the second is an experimental value.
For conservation of energy:DGdiss,s¼"DGs(RCOOH) +DGhigh+DGs(RCOO")+DGs(H+)

532 8 Some “Special” Topics