# Effect of pH on charge of jasmonic acid

It has been such a long time, that I totally forgot how it works.

I am interested in the charge of jasmonic acid at different concentrations. More specifically, I want to know at what pH my molecule has no charge (or I should rather say jasmonic acid would occur as a zwitter ion).

Only the decaboxylic group can be deprotonated at higher pH.

1. Is it correct to assume that most of the time, when you have an acidic compound it would predominantly occur in a nonionic form?

2. Does the $$\log{D}$$ say anything about this? does the maximum $$\log{D}$$ value corresponds to the pH at the iso electric point?

If we would take aspirin for example

https://disco.chemaxon.com/apps/demos/logd/

at what pH should I expect the uncharged version?

Jasmonic acid has one group that is significantly ionizable, the carboxylic group:

Most molecules in the sample will be negatively charged when the $$\mathrm{pH}$$ is higher than the $$\mathrm{p}K_\mathrm{a}$$ of the molecule and an equal number will be neutral or charged when $$\mathrm{pH} = \mathrm{p}K_\mathrm{a}$$. The same will hold for aspirin, which also has a carboxylic acid group. Below a $$\mathrm{pH}$$ equal to the $$\mathrm{p}K_\mathrm{a}$$ it will be mostly uncharged, and above this $$\mathrm{pH}$$ it will mostly be charged.

If you look up $$\mathrm{p}D$$, you might encounter the following equation:

$$-\mathrm{p}D = \log D_\mathrm{oct/wat} = \log\left( \frac{[C]^\mathrm{ionized}_\mathrm{octanol} + [C]^\mathrm{unionized}_\mathrm{octanol}}{[C]^\mathrm{ionized}_\mathrm{water} + [C]^\mathrm{unionized}_\mathrm{water}}\right)$$

Since the concentration $$[C]$$ of the ion is typically higher in water than in octanol, increased ionization at high $$\mathrm{pH}$$ will increase the solubility in the water phase, and $$\log D_\mathrm{oct/wat}$$ will decrease (become more negative) at high $$\mathrm{pH}$$, that is, $$\mathrm{p}D$$ will increase (become less negative). You can expect the solubility in octanol to continue to decrease as you near and pass the isoelectric point.

For more complex molecules with more ionizable groups you will need to consider individual contributions from different groups. Basic groups will respond in the opposite manner to acidic ones.

• Thanks for clarifying this. So basically, the ionisation graph would be a more or less a mirror of the Log D graph – Annelies Roussou Feb 28 at 17:45
• Hmm, that's an interesting relation. I'll have to think about that a little more. – Buck Thorn Feb 28 at 18:08
• @AnneliesRoussou I agree with your analysis. logP should be measured when the molecule is not ionized, that is low pH. So the result makes sense for aspirin, not so for jasmonic acid. It may have to do with how the programs make predictions. They are probably smart enough to predict logP at a point where the molecule is not ionized (if this is possible). – Buck Thorn Mar 1 at 10:22
• Hi Night writer, i deleted my comment because i realized when i uploaden the structure of JA in chemaxon logD predictor I made a minor error in its structure. Actually the Log D (at pH far below the pka), when the compound is unionized, does correspond to the LogP. My mistake. The chemistry part has just been such a long time ago since i switched to a different field – Annelies Roussou Mar 1 at 10:25
• Yes, its completely clear now. Thanks a lot, you've been really helpfull! – Annelies Roussou Mar 1 at 10:28