Question

How can phthalic acid not be soluble in water in both hot or cold temperatures?

Answer 1

There are a few features to the pthalic acid of importance here, namely that it is a weak acid, a benzene ring, and a carboxyl group. What do these indicate?

• Weak Acid - means benzoic acid will not fully dissociate in water, unlike HCl (H3O+/H+ and Cl- in water) or H2SO4 (H3O+/H+ and HSO4-) do. Another way of putting it would be strong acids balance heavily towards the right side of that equation (they dissociate, or are more solvent in water) while weak acids stay further to the left than their strong counterparts (they do not dissociate as easily, and are less solvent in water). Thus, without adjusting energy (adding heat, introducing catalysts), phthalic acid is not fully soluble in neutral solutions.
• Benzene Ring - is hydrophilic in nature, but is unable to form strong hydrogen bonds with water, relying instead on Van Der Waals interactions, which are the weakest of the various intermolecular interactions we study. The benzene ring structurally is the largest portion of this molecule, so it’s ability to form good bonds within the solution plays a huge role in determining solubility. This reinforces the above - in a neutral solution, benzene (and in turn phthalic acid) will not fully dissolve. This also factors into behaviour in an acidic solution. A solution with more readily available H30+/H+ ions is adding more product to the right side of the equation for benzoic acid. This will, in turn, push the equilibrium to the left, making phthalic acid less soluble in an acid than a neutral solution.
• Carboxyl Group - will readily react with an easily accessible OH- molecule (benzoic acid dissociates to C6H5COO- and H30+/H+). Thus, putting benzoic acid in an alkaline (basic) solution will cause a normal acid/base reaction, consuming H3O+/H+ produced by dissociation, and drawing equilibrium further in that direction by creating an H3O+/H+ deficit that is balanced by more acid-dissociating as the equilibrium shifts further to the right side of the equation.