Effect of pH on charge of jasmonic acid
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.