We're working on pKa prediction using semiempirical methods and need to compute pKa values for molecules with several ionizable groups. Here are my current thoughts so far.
Background: one ionizable group
If there is only one tritrateable site
\mathrm{BH \rightleftharpoons B + H^+} \ \ \ K=\mathrm{\frac{[B][H^+]}{[BH]}}
then the fraction of \mathrm{BH} molecules f_{\mathrm{BH}} is \begin{split} f_{\mathrm{BH}} & =\mathrm{\frac{[BH]}{[B]+[BH]} } \\ & = \mathrm{\frac{[B]}{[B]}\frac{[BH]/[B]}{1+[BH]/[B]} } \\ & = \mathrm{\frac{[H^+]/K}{1+[H^+]/K} } \\ & = \mathrm{\frac{10^{p\textit{K}-pH}}{1+10^{p\textit{K}-pH}} } \end{split}
and similarly for K.
From this we can see that the pK value is the pH value for which f_{\mathrm{BH}} = 1/2. So you compute the pK value from the standard free energy difference
\text{p}K =\left( \Delta G^\circ(\mathrm{B})+\Delta G^\circ(\mathrm{H^+})- \Delta G^\circ(\mathrm{BH})\right)/RT\ln(10)
and you're done.
Two ionizable groups
For a molecule with two titrateable groups ( \mathrm{HB_\alpha B_\beta H}) and the following equilibrium constants \mathrm{HBBH \rightleftharpoons BBH + H^+} \ \ K_{\alpha1}
If one of the groups (say \alpha) titrates at a significantly lower pH than the other (\mathrm{p}K_{\alpha1} << \mathrm{p}K_{\beta1}) then \mathrm{p}K_{\alpha}=\mathrm{p}K_{\alpha1} and \mathrm{p}K_{\beta}=\mathrm{p}K_{\beta0} and it is not necessary to compute the free energy of \mathrm{HBB}, but it can be hard to determine this in advance. Similarly, if there is no significant interaction between the sites then \mathrm{p}K_{\alpha}=\mathrm{p}K_{\alpha1}=\mathrm{p}K_{\alpha0} and \mathrm{p}K_{\beta}=\mathrm{p}K_{\beta1}=\mathrm{p}K_{\beta0} and one can skip one of the protonation states.
For N ionizable groups one has to determine 2^N microscopic pKa values, which quickly gets out of hand if one has to do a conformational search for each protonation state and the molecule is large.
Related post
Generating protonation states and conformations
This work is licensed under a Creative Commons Attribution 4.0
Background: one ionizable group
If there is only one tritrateable site
\mathrm{BH \rightleftharpoons B + H^+} \ \ \ K=\mathrm{\frac{[B][H^+]}{[BH]}}
then the fraction of \mathrm{BH} molecules f_{\mathrm{BH}} is \begin{split} f_{\mathrm{BH}} & =\mathrm{\frac{[BH]}{[B]+[BH]} } \\ & = \mathrm{\frac{[B]}{[B]}\frac{[BH]/[B]}{1+[BH]/[B]} } \\ & = \mathrm{\frac{[H^+]/K}{1+[H^+]/K} } \\ & = \mathrm{\frac{10^{p\textit{K}-pH}}{1+10^{p\textit{K}-pH}} } \end{split}
where \mathrm{pH = -log[H^+] \implies [H^+] = 10^{-pH}}
and similarly for K.
From this we can see that the pK value is the pH value for which f_{\mathrm{BH}} = 1/2. So you compute the pK value from the standard free energy difference
\text{p}K =\left( \Delta G^\circ(\mathrm{B})+\Delta G^\circ(\mathrm{H^+})- \Delta G^\circ(\mathrm{BH})\right)/RT\ln(10)
and you're done.
Two ionizable groups
For a molecule with two titrateable groups ( \mathrm{HB_\alpha B_\beta H}) and the following equilibrium constants \mathrm{HBBH \rightleftharpoons BBH + H^+} \ \ K_{\alpha1}
\mathrm{HBBH \rightleftharpoons HBB + H^+} \ \ K_{\beta1}
\mathrm{HBB \rightleftharpoons BB + H^+} \ \ K_{\alpha0}
\mathrm{BBH \rightleftharpoons BB + H^+} \ \ K_{\beta0}
The probability of, for example, \mathrm{BBH} is f_{\mathrm{BBH}} =\mathrm{\frac{[BBH]}{[BB]+[BBH]+[HBB]+[HBBH]}= \frac{[BBH]}{\textit{P}}}
f_{\mathrm{BBH}} can be rewritten in terms of pK values f_{\mathrm{BBH}} = \mathrm{\frac{[BBH]/[BB]}{\textit{P}/[BB]} = \frac{10^{p\textit{K}_{\beta0}-pH}}{\textit{P}/[BB]}}
where \mathrm{ \textit{P}/[BB] = 1+10^{p\textit{K}_{\alpha0}-pH}+10^{p\textit{K}_{\beta0}-pH}+ 10^{p\textit{K}_{\alpha0}+p\textit{K}_{\beta1}-2pH}}
Similarly, f_{\mathrm{HBB}} = \mathrm{\frac{10^{p\textit{K}_{\alpha0}-pH}}{\textit{P}/[BB]}}
and f_{\mathrm{HBBH}} = \mathrm{\frac{10^{p\textit{K}_{\alpha0}+p\textit{K}_{\beta1}-2pH}}{\textit{P}/[BB]}}
The apparent pK value of the \alpha group (\mathrm{p}K_{\alpha}) is the pH at which its protonation probability f_\alpha =f_{\mathrm{HBB}} + f_{\mathrm{HBBH}}
is 1/2 and similarly for the \beta group. So compute the microscopic pK values (Eq 4-7), then f_\alpha and f_\beta as a function of pH, and then \mathrm{p}K_{\alpha} and \mathrm{p}K_{\beta}
If one of the groups (say \alpha) titrates at a significantly lower pH than the other (\mathrm{p}K_{\alpha1} << \mathrm{p}K_{\beta1}) then \mathrm{p}K_{\alpha}=\mathrm{p}K_{\alpha1} and \mathrm{p}K_{\beta}=\mathrm{p}K_{\beta0} and it is not necessary to compute the free energy of \mathrm{HBB}, but it can be hard to determine this in advance. Similarly, if there is no significant interaction between the sites then \mathrm{p}K_{\alpha}=\mathrm{p}K_{\alpha1}=\mathrm{p}K_{\alpha0} and \mathrm{p}K_{\beta}=\mathrm{p}K_{\beta1}=\mathrm{p}K_{\beta0} and one can skip one of the protonation states.
For N ionizable groups one has to determine 2^N microscopic pKa values, which quickly gets out of hand if one has to do a conformational search for each protonation state and the molecule is large.
Related post
Generating protonation states and conformations
This work is licensed under a Creative Commons Attribution 4.0
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