For gas phase of acid and acetate anion I used opt+freq in CBS-QB3 and found the Gibbs free energy(sum of electronic and thermal free energies)
GgasCH3COOH= -229.122052 Hartree/particle
GgasCH3COO- = -228.555849 Hartree/particle
for calculating solvation energies I used opt+freq with HF/6-31G(d) for gas phase and CPCM HF/6-31G(d) for the solvated phase in water and took the difference of respective values for solvation energies.The values are:
GgasCH3COOH= -227.770611 Hartree/particle
GgasCH3COO-= -227.200572 Hartree/particle
GsCH3COOH= -227.779303 Hartree/particle
GsCH3COO-= -227.305984 Hartree/particle
So that the solvation energies becomes
GsolvationCH3COOH= -8.692*10^-3 Hartree or -5.4543 Kcal/mol
GsolvationCH3COO- = -0.105412 Hartree or -66.1470 Kcal/mol
I took values for Ggas(H+)= -6.28 Kcal/mol and Gsolvation(H+)= -264.61 Kcal/mol
Using
ΔG°solv = ΔG°gas + ΔG°s(CH3COO-) + ΔG°s(H+) − ΔG°s(CH3COOH)
and Pka = ΔG°solv / ln (10) RT Here T=298K
I am getting Pka of acetic acid to be 16.0 which is too much away from experimental value of 4.75.
Any help would be appreciated.
Mr. Parashar,
Have you involved a state-correction, using
pKa = DGo/2.3003RT,
and a GP reference state via following equation
DGg = DGg + RT.ln(24.46)?
In this context, you can pay attention to:
Phys. Chem. Chem. Phys., 2015,17, 6174-6191 (http://pubs.rsc.org/en/Content/ArticleLanding/2015/CP/C5CP00288E)
The secon aspect is associated with DGg(H+) = -6.28 kcal.mol-1:
1. Try to compute pKa value using from Sackur-Tetrode entropy with/without rotational constituent along with SoTtrans.
In particular, for isolated H+ I have received:
Utran = 3.718 kJ.mol-1 0.889 kcal.mol-1
Htran = 6.197 kJ.mol-1 1.481 kcal.mol-1
Urot = 2.479 kJ.mol-1 0.592 kcal.mol-1
Hrot = 2.479 kJ.mol-1 0.592 kcal.mol-1
Or
Htot = 0.00330473 Hartree (tran + rot)
Htot = 0.00236012 Hartree (tran)
DHtot = 0.00094461 Hartree or ca. 0.001 Hartree
Because of as lastly shown depending on the theoretical methods the error contributions computing the energy have comparable magnitude orders.
2. Compare the data with a theoretical DGg(H+) obtained via computing of a system H+/H2O. The value corresponding to isolated H+ is obtained by subtraction from this one of H2O (in GP). You should perform two subsequent computations, where in the second case the coordinates of H2O in H+/H2O-file should only remain.
The values, which you should receive are:
HF=-76.0654396 (H2O)
HF=-76.0892829 (H+/H2O)
Or HF (H+) = -0.0238433 (MP2/cc-pvdz) (NB! There are not any intermolecular/ionic interactions between H2O and H+ in H+/H2O ensemble).
Other data about H+: S (Cal/Mol-Kelvin): 26.014
Simply speaking, you do not need to derive the value -6.28, according to GDo(GP) = 2.5RT-TDS = 1.48 - 7.76 [in kcal.mol-1] from the experiment.
Mr. Parashar,
You should perform set of computations with various perturbations of the conformation of the CH3COOH. then you should take an average value for DG. Your results corresponds to microscopic state/s.
What you try to do is rather hopeless. You want to calculate the Gibbs energies of water, hydronium ions (whatever their structure may be), acetic acid, and acetate ions. These energies are statistical averages of configurations dominated by hydrogen bonding. You would have to run Gaussian on sets of molecule clusters to obtain them. Furthermore, there are also dispersion forces present, which can be computed with post-Hartree–Fock methods only.
P.S.: NEVER use kcal! It is an obsolete unit, and there exist at least three different definitions in the literature.
I tried with the other conformer of acetic acid and the Pka for average of the two came out to be 14.71 which is still way too off 4.75.
You need to run Gaussian not on single conformers only, but on clusters of water molecules surrounding an acid molecule. It is not sufficient to study one cluster in a configutation of minimal energy; instead, one has to study a large set of clusters having a thermal energy distribution.
What you propose is a good topic for a doctoral thesis—provided that you have a generous allowance of CPU time on a supercomputer. Also remember that the HF/6-31G(d) method is insufficient for calculation of intermolecular interactions.
What you might attempt instead is a relative calculation: You can try to compute the difference of the dissociation energies of acetic acid and, e.g., propanoic acid. Then you can check whether this matches the difference of the observed pKa values. But do expect miracles: Quantum-mechanical programs tend to have a rather limited grasp of concepts of statistical thermodynamics.
Actually I am trying to reproduce the results as given in (http://scitation.aip.org/content/aip/journal/jcp/114/10/10.1063/1.1337862). They have used Absolute Pka calculation method with implicit solvent and got Pka around 4 using the same methods that I am using.The calculations were done in Gaussian 98(instead of Gaussian 09).Can results really vary that much with just the change in the version of the program.Or I am doing some mistake but where?
Thank you for the literature reference. Please note that the authors state “Relative values were calculated for each acid using one of the acids as a reference point.” in their abstract: You can perform a quantum-mechanical calculation of the dissociation energy (not the Gibbs energy!) in the vapour phase. You can perhaps correct that for the liquid phase by some semiempirical tricks. But at present you cannot get the absolute value of the Gibbs energy of dissociation from a quantum mechanics program alone.
Mr. Parashar,
Have you involved a state-correction, using
pKa = DGo/2.3003RT,
and a GP reference state via following equation
DGg = DGg + RT.ln(24.46)?
In this context, you can pay attention to:
Phys. Chem. Chem. Phys., 2015,17, 6174-6191 (http://pubs.rsc.org/en/Content/ArticleLanding/2015/CP/C5CP00288E)
The secon aspect is associated with DGg(H+) = -6.28 kcal.mol-1:
1. Try to compute pKa value using from Sackur-Tetrode entropy with/without rotational constituent along with SoTtrans.
In particular, for isolated H+ I have received:
Utran = 3.718 kJ.mol-1 0.889 kcal.mol-1
Htran = 6.197 kJ.mol-1 1.481 kcal.mol-1
Urot = 2.479 kJ.mol-1 0.592 kcal.mol-1
Hrot = 2.479 kJ.mol-1 0.592 kcal.mol-1
Or
Htot = 0.00330473 Hartree (tran + rot)
Htot = 0.00236012 Hartree (tran)
DHtot = 0.00094461 Hartree or ca. 0.001 Hartree
Because of as lastly shown depending on the theoretical methods the error contributions computing the energy have comparable magnitude orders.
2. Compare the data with a theoretical DGg(H+) obtained via computing of a system H+/H2O. The value corresponding to isolated H+ is obtained by subtraction from this one of H2O (in GP). You should perform two subsequent computations, where in the second case the coordinates of H2O in H+/H2O-file should only remain.
The values, which you should receive are:
HF=-76.0654396 (H2O)
HF=-76.0892829 (H+/H2O)
Or HF (H+) = -0.0238433 (MP2/cc-pvdz) (NB! There are not any intermolecular/ionic interactions between H2O and H+ in H+/H2O ensemble).
Other data about H+: S (Cal/Mol-Kelvin): 26.014
Simply speaking, you do not need to derive the value -6.28, according to GDo(GP) = 2.5RT-TDS = 1.48 - 7.76 [in kcal.mol-1] from the experiment.
Calculating the solvation energy of H+ as a single Hydronium ion (or any other molecule ion based on the main solvent) using an implicit solvent is a "very dangerous" thing to do, unless it is for relative comparison. This can cause a huge error in your calculations. You
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