Three reactions involving $$\text{H}_2\text{PO}_4^-$$ are given below: (i) $$\text{H}_3\text{PO}_4 + \text{H}_2\text{O} \to \text{H}_3\text{O}^+ + \text{H}_2\text{PO}_4^-$$ (ii) $$\text{H}_2\text{PO}_4^- + \text{H}_2\text{O} \to \text{HPO}_4^{2-} + \text{H}_3\text{O}^+$$ (iii) $$\text{H}_2\text{PO}_4^- + \text{OH}^- \to \text{H}_3\text{PO}_4 + \text{O}^{2-}$$. In which of the above does $$\text{H}_2\text{PO}_4^-$$ act as an acid?

According to the Brønsted-Lowry concept,

  • an acid is a species that donates a proton ( $$\text{H}^+$$ ) to some other species.
  • a base is a species that accepts a proton from some other species.

We examine each reaction to see whether $$\text{H}_2\text{PO}_4^-$$ donates or accepts a proton.

Case (i):

$$\text{H}_3\text{PO}_4 + \text{H}_2\text{O} \;\rightarrow\; \text{H}_3\text{O}^+ + \text{H}_2\text{PO}_4^-$$

Here $$\text{H}_3\text{PO}_4$$ loses a proton to water, producing $$\text{H}_3\text{O}^+$$ and $$\text{H}_2\text{PO}_4^-$$. The species $$\text{H}_2\text{PO}_4^-$$ is formed after proton loss, so in the forward direction it has accepted that proton in the reverse reaction. Thus, in this forward reaction $$\text{H}_2\text{PO}_4^-$$ behaves as the conjugate base of $$\text{H}_3\text{PO}_4$$, i.e. it acts as a base, not as an acid.

Case (ii):

$$\text{H}_2\text{PO}_4^- + \text{H}_2\text{O} \;\rightarrow\; \text{HPO}_4^{2-} + \text{H}_3\text{O}^+$$

Compare the left- and right-hand sides:

  • $$\text{H}_2\text{PO}_4^-$$ loses one proton and becomes $$\text{HPO}_4^{2-}$$.
  • Water gains that proton and becomes $$\text{H}_3\text{O}^+$$.

Since $$\text{H}_2\text{PO}_4^-$$ donates a proton to water, it acts as a Brønsted-Lowry acid in this reaction.

Case (iii):

$$\text{H}_2\text{PO}_4^- + \text{OH}^- \;\rightarrow\; \text{H}_3\text{PO}_4 + \text{O}^{2-}$$

Here $$\text{H}_2\text{PO}_4^-$$ accepts a proton from $$\text{OH}^-$$ to form $$\text{H}_3\text{PO}_4$$. Accepting a proton classifies $$\text{H}_2\text{PO}_4^-$$ as a Brønsted-Lowry base in this reaction, not as an acid.

Therefore, $$\text{H}_2\text{PO}_4^-$$ behaves as an acid only in reaction (ii).

Option A which is: (ii) only