Nicotinic acid, HC6H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4 × 10-5. Calculate [H+] and the pH of a 0.041 M solution of HC6H4NO2.

[H+] = 7.576x10⁻⁴M

pH = 3.12

Explanation:

Based on the equilibrium of the nicotinic acid in water:

HC6H4NO2(aq) + H2O(l) ⇄ C6H4NO2-(aq) + H3O+(aq)

Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]

As both C6H4NO2-(aq) and H3O+(aq) comes from the same equilibrium, we can approximate their concentration as X and replace:

Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]

1.4x10⁻⁵ = [X] [X] / [0.041M]

5.47x10⁻⁷ = X²

And as pH is defined as -log [H⁺]

the answer is c. nacl

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.