How to determine and derive the protonation state of a functional group based on its pKa?

Question

In our biochemistry class we were introduced to a rule of thumb, which goes something like this:

"If the pH of a solution is one or more units below the $\mathrm{p}K_\mathrm{a}$ of a group, then said group is protonated and if it is one or more unit above, then the group is de-protonated"

However, we weren't provided with a mathematical derivation of this fact. I attempted one by myself using:

$$\ce{HA + H2O <=> A- + H3O+}$$ thus, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{A-}}{\ce{HA}}$$ or, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{1-\alpha}{\alpha}$$

and then define $\mathrm{p}K_\mathrm{a}$ as the point where ${\alpha=0.5}$ i.e proton occupancy is 50%

$$\alpha(\ce{pH}) = \frac{1}{1+\exp\left\{-\ln10(\mathrm{p}K_\mathrm{a}-\ce{pH})\right\}}$$

where $\alpha$:= degree of proton occupancy.

By looking at this function, I can sort of get that if pH goes up, $\alpha$ goes down i.e case of protonation, and if pH goes down $\alpha$ goes up i.e case of deprotonation.

This seems a little bit "back of the envelope" to me. Is this sufficient, or am I missing something.

Answer 1

You're on the right track. The Henderson-Hasselbalch equation is where this rule of thumb comes from; the key thing to remember is that, in biochemistry, your $\mathrm{pH}$ is fixed by the environment (there's always something acting as a buffer). This is in contrast with typical H.H. problems in general chemistry where the amount of molecule and its $K_\mathrm{a}$ is changing the $\mathrm{pH}$ of the solution.

Anyhow, given that $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ are fixed values, we're free to simply compare the ratio of $[\ce{A-}]$ and $[\ce{HA}]$:

\begin{align} \mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\\ \log\frac{[\ce{A-}]}{[\ce{HA}]} &= \mathrm{pH} - \mathrm{p}K_\mathrm{a}\\ \frac{[\ce{A-}]}{[\ce{HA}]} &= 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}\\ \end{align}

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} + 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{-1}$. In other words, we have 10 times as much $\ce{HA}$ as we have $\ce{A-}$ (i.e., we have $\approx91\%\ \ce{HA}$) as we should expect for something that is mildly basic.

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{1}$. In other words, we have 10 times as much $\ce{A-}$ as we have $\ce{HA}$ (i.e., we have $\approx91\%\ \ce{A-}$) as we should expect for something that is mildly acidic.

Clearly, if the $\mathrm{pH}$ differs from the $\mathrm{p}K_\mathrm{a}$ by only 1 $\mathrm{pH}$ unit, we don't really have a completely protonated or deprotonated molecules, but it's certainly mostly protonated or deprotonated. As the $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ get further apart, then we continue to change this ratio by orders of magnitude.

Let's look at a concrete example. Physiological $\mathrm{pH}$ is 7.4 and the side chain of aspartic acid has a carboxylic acid group with a $\mathrm{p}K_\mathrm{a} = 3.9$, so we predict

$$\frac{[\ce{RCOO^-}]}{[\ce{RCOOH}]} = 10^{7.4 - 3.9} = 10^{3.5} = 3200$$

In other words, we still have some protonated sidechain, but we have >3000 times more deprotonated sidechain than protonated sidechain (did you notice that? the difference was >3 so we have >1000 difference in concentration ratio). For all practical purposes, the aspartic acid side chain is "completely deprotonated" at physiological $\mathrm{pH}$.