JEE Main 2010 — Ionic Equilibrium Question with Solution

To determine in which reactions $H_{2}P{O}_4^{-}$ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases. According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.

Let's analyze each reaction :

1. 1. $H_{3}P{O}_4+H_{2}O\to H_3{O}^{+}+H_{2}P{O}_4^{-}$

   
   
   • In this reaction, $H_{3}P{O}_4$ is donating a proton to $H_{2}O$, forming $H_3{O}^{+}$ and $H_{2}P{O}_4^{-}$.
   
   
   • $H_{2}P{O}_4^{-}$ is the product of this reaction and is not acting as an acid here.



2. 2. $H_{2}P{O}_4^{-}+H_{2}O\to HP{O}_4^{2-}+H_3{O}^{+}$

   
   
   • In this reaction, $H_{2}P{O}_4^{-}$ donates a proton to $H_{2}O$, resulting in $HP{O}_4^{2-}$ and $H_3{O}^{+}$.
   • 
     $H_{2}P{O}_4^{-}$ is acting as an acid in this reaction.
3. 
   

   3. $H_{2}P{O}_4^{-}+O{H}^{-}\to H_{3}P{O}_4+O^{2-}$

   • 
     This reaction is not correctly balanced and does not follow standard chemical reaction rules. The product $O^{2-}$ is highly unlikely in aqueous solutions due to its reactivity.
   • 
     A more correct reaction would be $H_{2}P{O}_4^{-}+O{H}^{-}\to HP{O}_4^{2-}+H_{2}O$. In this corrected reaction, $H_{2}P{O}_4^{-}$ is donating a proton to $O{H}^{-}$, which makes it an acid.

Given the information, the correct option is :

Option A : (ii) only

This is because in reaction (ii), $H_{2}P{O}_4^{-}$ is clearly acting as an acid by donating a proton to water. In reaction (i), $H_{2}P{O}_4^{-}$ is not acting as an acid but rather is formed as a product. Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $H_{2}P{O}_4^{-}$ acting as an acid.