Essentials of Computational Chemistry

412 11 IMPLICIT MODELS FOR CONDENSED PHASES
AH n (gas)
AH n (sol)
∆G o (gas)
∆G o (sol)
A n –1 (gas) + H + (gas)
∆G o S(AH n ) ∆G o S(A n–1 ) ∆G o S(H + )
A n –1 (sol) + H + (sol)
Figure 11.9 Free energy cycle for computation of pKa values (where n is an integer). This cycle is
sometimes referred to as a Born–Haber cycle
potentials. Computation of the former is accomplished by employing the free-energy cycle
depicted in Figure 11.9. Thus, gas-phase free energies of AHn and An−1 may be computed
at arbitrarily high levels of theory to establish as accurately as possible the deprotonation
free energy of AHn . Note that if n and/or n − 1 are negative numbers then the basis set will
need to include diffuse functions in order to obtain even modest quantitative accuracy. As
for the proton, its electronic energy is obviously zero, and its gas-phase free energy derives
entirely from a PV enthalpy term and a translational free energy that may be computed from
Eqs. (10.16) and (10.17). At 298 K in the usual 1 atm standard state the free energy of the
proton is −0.00999 a.u.
To compute the deprotonation free energy in solution, we take the gas-phase free energy
change, add the free energies of solvation of An−1 and H + (see above for the latter), and
subtract the free energy of solvation of AHn . However, note that most continuum solvation
models compute the free energy of solvation assuming the same standard-state concentration
in the gas phase as in solution. As most pKa values adopt a standard-state concentration
of 1 M, we need then to compute the free energy change associated with adjusting the
concentrations of all of the gas-phase species from 1 mol per 24.5 L (the concentration of
an ideal gas at 1 atm pressure and 298 K) to 1 mol per 1 L. As described in Section 10.5.4,
this change is RT ln(24.5) for every species. As there are two products and only one reactant
in the deprotonation reaction, the net effect is to make deprotonation less favorable by
1.9 kcal mol−1 in the 1 M standard state compared to the 1 atm standard state at 298 K.
Having computed the free-energy change in solution by this protocol, we may then
compute Ka as
and pKa as
Ka = e −Go
(sol) /RT
pKa =−log e −Go
/RT (sol)
(11.24)
(11.25)
As errors in ionic solvation free energies are often on the order of 5 kcal mol −1 ,andas
errors in the gas-phase deprotonation free energies may be of similar magnitude even with
reasonably good levels of theory, errors in predicted absolute pKa values of 5 or more pK
units are not terribly unusual, which is not particularly satisfying insofar as experimental
measurements can be accurate to 0.01 pK units.