Carboxylic acids, esters and soaps

The $\ce{-COOH}$ group combines a $\ce{C=O}$ and an $\ce{-OH}$ on the same carbon. In water it gives up its proton only partially: $$\ce{R-COOH + H2O <=> R-COO^- + H3O+}$$ For most of these acids, the $\text{p}K_a$ of the $\ce{R-COOH}/\ce{R-COO^-}$ couple lies between $4$ and $5$: the equilibrium lies far to the left.