Question

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.

Answer 1

[H+] = 7.576x10⁻⁴M

pH = 3.12

Step-by-step 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

7.576x10⁻⁴M = X = [H+]

And as pH is defined as -log [H⁺]

pH = 3.12