QJEEM10Ch987c: JEE Main 2010 Chemistry - Class 11 JEEM10Ch987c Solution: Three reactions involving are given below : (i) H 3 PO 4 + H 2…

To determine in which reactions $H_2PO_4^-$ 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. $H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^-$

   
   
   • In this reaction, $H_3PO_4$ is donating a proton to $H_2O$ , forming $H_3O^+$ and $H_2PO_4^-$ .
   
   
   • $H_2PO_4^-$ is the product of this reaction and is not acting as an acid here.



2. 2. $H_2PO_4^- + H_2O \to HPO_4^{2−} + H_3O^+$

   
   
   • In this reaction, $H_2PO_4^-$ donates a proton to $H_2O$ , resulting in $HPO_4^{2−}$ and $H_3O^+$ .
   • 
     $H_2PO_4^-$ is acting as an acid in this reaction.
3. 
   

   3. $H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-}$

   • 
     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 $H_2PO_4^- + OH^- \to HPO_4^{2−} + H_2O$ . In this corrected reaction, $H_2PO_4^-$ is donating a proton to $OH^-$ , which makes it an acid.

Given the information, the correct option is :

Option A : (ii) only

This is because in reaction (ii), $H_2PO_4^-$ is clearly acting as an acid by donating a proton to water. In reaction (i), $H_2PO_4^-$ 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 $H_2PO_4^-$ acting as an acid.