# Hybrid Models¶

Abstract

## Equilibria¶

### $$pK_a$$ (Born-Haber cycle)¶

$$pK_a$$ is typically calculated using the Born-Haber cycle with the equation:

Where $$\Delta G^{\circ\prime}=\Delta G^{\circ\prime}_{aq}$$

\begin{align} pK_a&=-\log\bigg[\exp{\bigg(\frac{-\Delta G^{\circ\prime}}{RT}\bigg)}\bigg]\\ &=\frac{\Delta G^{\circ\prime}}{2.303RT} \end{align}

Using the free energy cycle:

### Process:¶

1. For the anion, I need diffuse functions, a big basis set and a good level of theory
2. QM packages won’t let you calculate the electronic energy of $$\ce{H+}$$, since there’s no electron to calculate
3. $$\Delta G^\circ_{(s)}=-264.0\:kcal\cdot mol^{-1}$$ (experimentally derived)
4. Standard-state concentration-change free energy must be included
5. each non-cancelled error of $$1.4\:kcal\cdot mol^{-1}$$ in any step will lead to an error of 1 $$pK_a$$ unit
• Errors in ionic solvation free energies can be much larger than that
6. Function-group systematic errors can be corrected for

## The overall equation¶

We can condense this all into the one line equation:

$2.303RT\:pK_a=\Delta G_g^\circ(AH)-\Delta G^*_{aq}(AH)+\Delta G^*_{aq}(A^-)+\Delta G_{aq}^*(H^+)$

Using this free energy cycle:

## Free energy cycles and ion structures¶

Here we’re treating the ion as a cluster, in an attempt to try and reduce the amount error on the cluster:

$2.303RT \: pK_a = \Delta G_g^\circ (AH) - \Delta G^*_{aq}(AH) - \Delta G^*_{aq}(\ce{H2O})+\Delta G^*_{aq}(\ce{H2O}\cdot A^-)+\Delta G_{aq}^*(H^+)$

Utilising this free energy cycle:

## Comparison (Experimental Data $$pK_a=15.5$$)¶

### Experimental data¶

Method 1 $$\ce{MeOH/MeO-}$$ Method 2 $$\ce{MeOH/H2O.MeO-}$$
$$\Delta G^\circ_g=375.0\:kcal\cdot mol^{-1}$$ $$\Delta G^\circ_g=358.0\:kcal\cdot mol^{-1}$$
$$\Delta G_{aq}^*(\ce{H+})=-265.9\:kcal\cdot mol^{-1}$$ $$\Delta G_{aq}^*(\ce{H+})=-265.9\:kcal\cdot mol^{-1}$$
$$\Delta G_{aq}^*(\ce{MeOH})=-5.11\:kcal\cdot mol^{-1}$$ $$\Delta G_{aq}^*(\ce{MeOH})=-5.11\:kcal\cdot mol^{-1}$$
$$\Delta G_{aq}^*(\ce{H2O})=-6.32\:kcal\cdot mol^{-1}$$

### Computed data (SM6)¶

Method 1 $$\ce{MeOH/MeO-}$$ Method 2 $$\ce{MeOH/H2O.MeO-}$$
$$\Delta G^*_{aq}(\ce{MeO-})=-88.3\:kcal\cdot mol^{-1}$$ $$\Delta G^*_{aq}(\ce{H2O.MeO-})=-81.8\:kcal\cdot mol^{-1}$$
$$pK_a=20.4$$ $$pK_a=16.0$$

This is the experimental $$pK_a$$ data
$\ce{H2CO3 <=>[pK_{a_1}=6.4] HCO3- <=>[pK_{a_2}=10.3] CO3^{2-}}$
Without adding explicit waters of solvation, the results are insanely inaccurate, however when we add water molecules in, the structure becomes much more stabilised and the $$pK_a$$ is much more in line with experiment.
No. $$\ce{H2O}$$ $$pK_{a_1}$$ $$pK_{a_2}$$