Animated Solution for Chemistry - Ionic Equilibrium: Three reactions involving ${\text{H}}_2{\text{PO}}_4^{-}$ are given below I. ${\text{H}}_3{\text{PO}}_4+{\text{H}}_2\text{O}⟶{\text{H}}_3{\text{O}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}$ II. ${\text{H}}_2{\text{PO}}_4^{-}+{\text{H}}_2\text{O}⟶{\text{HPO}}_4^{2-}+{\text{H}}_3{\text{O}}^{+}$ III. ${\text{H}}_2{\text{PO}}_4^{-}+{\text{OH}}^{-}⟶{\text{H}}_3{\text{PO}}_4+{\text{O}}^{2-}$ In which of the above does ${\text{H}}_2{\text{PO}}_4^{-}$ act as an acid?

Bronsted-Lowry Acids and Bases

To solve this problem, we must first anchor ourselves in the Bronsted-Lowry theory of acids and bases. According to this fundamental concept, an acid is defined as any chemical species that can donate a proton (an ${\text{H}}^{+}$ ion) to another species. Conversely, a base is a species that accepts a proton.

Our objective is to examine three distinct chemical reactions and determine in which of them the dihydrogen phosphate ion, ${\text{H}}_2{\text{PO}}_4^{-}$, is acting as a proton donor.

Let's break down the reactions one by one to see exactly where the protons are moving.

Reaction I: In this first reaction, phosphoric acid (${\text{H}}_3{\text{PO}}_4$) is the reactant. It donates one of its protons to the water molecule, forming the hydronium ion (${\text{H}}_3{\text{O}}^{+}$) and leaving behind the ${\text{H}}_2{\text{PO}}_4^{-}$ ion. Here, ${\text{H}}_2{\text{PO}}_4^{-}$ is merely a product (specifically, the conjugate base of ${\text{H}}_3{\text{PO}}_4$). It is not acting as an acid in the forward reaction.

Reaction II: Now look closely at the second reaction. The ${\text{H}}_2{\text{PO}}_4^{-}$ ion is on the reactant side. As the reaction proceeds, it loses a hydrogen atom and an associated positive charge, transforming into the hydrogen phosphate ion, ${\text{HPO}}_4^{2-}$. Because it has donated a proton to water, it is unequivocally acting as a Bronsted-Lowry acid in this scenario.

Reaction III: There is a catch here. In this unusual reaction, the ${\text{H}}_2{\text{PO}}_4^{-}$ ion starts with two hydrogen atoms and ends up with three, becoming ${\text{H}}_3{\text{PO}}_4$. This means it has accepted a proton from the hydroxide ion (${\text{OH}}^{-}$). Since it is accepting a proton, it is acting as a Bronsted-Lowry base, not an acid.

This problem beautifully illustrates a favorite concept in chemistry: Amphoterism.

Polyprotic acids like phosphoric acid (${\text{H}}_3{\text{PO}}_4$) lose their protons in stages. The intermediate ions formed during these stages, such as ${\text{H}}_2{\text{PO}}_4^{-}$ and ${\text{HPO}}_4^{2-}$, have the unique ability to either donate another proton (acting as an acid) or accept a proton back (acting as a base), depending on the chemical environment they are placed in.

Because ${\text{H}}_2{\text{PO}}_4^{-}$ only donates a proton in Reaction II, that is our final answer.