Problem 214 - A Complete Resource Book in Chemistry for JEE Main

Question

Three reactions involving \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)are given below: (i) \(\mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HPO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\) (iii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{OH}^{2} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{O}^{3}\) In which of the above does \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)act as an acid? (a) (ii) only (b) (i) and (ii) (c) (iii) only (d) (i) only

Step-by-Step Solution

Answer

(a) (ii) only

Step 1: Understanding the role of an acid

An acid is a substance that donates a proton (\(\text{H}^+\)) to another species in a chemical reaction. The aim is to identify where \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) donates a proton.

Step 2: Evaluating Reaction (i)

In reaction (i), \(\mathrm{H}_{3} \mathrm{PO}_{4}\) reacts with water to form \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\). Here, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is donating a proton to water to form \(\mathrm{H}_{3} \mathrm{O}^{+}\). \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is not donating a proton here.

Step 3: Evaluating Reaction (ii)

In reaction (ii), \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) reacts with water to form \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\). Here, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) donates a proton to water, forming \(\mathrm{H}_{3} \mathrm{O}^{+}\). This is where \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) acts as an acid.

Step 4: Evaluating Reaction (iii)

In reaction (iii), \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) reacts with \(\mathrm{OH}^{2}\), however, the reaction equation provided is not chemically accurate as \(\mathrm{OH}^{2}\) does not exist as an ion in this form. Therefore, it is impossible for \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to act as an acid in this reaction.

Step 5: Final conclusion

Based on the analysis, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) acts as an acid only in reaction (ii).

Key Concepts

Proton Donor

A proton donor is any substance that can give up a proton (\( ext{H}^+\)) to another substance. In the realm of acid-base chemistry, acids are typically known to be proton donors. This is a simple yet crucial concept because it helps to understand why certain substances behave as acids.

• Characteristics of Proton Donors: They donate protons in a chemical reaction, increasing the concentration of \( ext{H}_3 ext{O}^+\) (hydronium ions) in water.
• Examples: For instance, in the reaction between hydrochloric acid (HCl) and water, HCl donates a proton to water, forming hydronium ion and chloride ion.

The ability to donate protons is what categorizes a substance as an acid in various chemical reactions. In reaction (ii) mentioned in the exercise, \( ext{H}_2 ext{PO}_4^-\) is a proton donor, making it act as an acid.

Chemical Reactions

Chemical reactions are processes in which one set of chemicals is transformed into another. To understand how \( ext{H}_2 ext{PO}_4^-\) can act as an acid, it is important to understand the nature of chemical reactions.

• Reactants and Products: Every chemical reaction involves reactants (starting materials) and products (the substances formed).
• Types of Reactions: These include acid-base reactions, oxidation-reduction reactions, and more. In acid-base reactions, one species donates a proton and another one accepts it.

In the given exercises, reaction (ii) is an example of such a process where \( ext{H}_2 ext{PO}_4^-\) donates a proton, transforming to \( ext{HPO}_4^{2-}\). This demonstrates the role of chemical reactions in changing the identity and properties of the substances involved.

Dihydrogen Phosphate Ion

The dihydrogen phosphate ion, \( ext{H}_2 ext{PO}_4^-\), is an important player in various chemical reactions, especially in acid-base chemistry. It is a versatile ion, capable of either accepting or donating a proton, depending on the circumstances.

• Role in Buffer Solutions: \( ext{H}_2 ext{PO}_4^-\) plays a crucial role in buffer solutions, which are used to maintain a stable pH in a solution. It can react to neutralize added acids or bases.
• Participates in Equilibrium: In reaction (ii) given above, \( ext{H}_2 ext{PO}_4^-\) donates a proton and forms \( ext{H}_3 ext{O}^+\), thus lowering the pH. The reverse reaction would involve \( ext{HPO}_4^{2-}\) accepting a proton, emphasizing its dual role.

Understanding the behavior of \( ext{H}_2 ext{PO}_4^-\) in various reactions helps in comprehending its significance in both biological processes and industrial applications.