Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following is true for any diprotic acid, \(\mathrm{H}_{2} \mathrm{X} ?\) (a) \(\mathrm{Ka}_{2}=\mathrm{Ka}_{1}\) (b) \(\mathrm{Ka}_{2}>\mathrm{Ka}_{1}\) (c) \(\mathrm{Ka}_{1}>\mathrm{Ka}_{2}\) (d) \(\mathrm{Ka}_{2}=1 / \mathrm{Ka}_{1}\)

Step-by-Step Solution

Answer

Option (c) \(\mathrm{Ka}_1 > \mathrm{Ka}_2\) is true for any diprotic acid.

Step 1: Understanding Diprotic Acids

A diprotic acid, such as \( \mathrm{H}_2 \mathrm{X} \), can donate two protons (hydrogen ions) in a stepwise manner. It dissociates in two stages: the first dissociation \( \mathrm{H}_2 \mathrm{X} \rightarrow \mathrm{H}^+ + \mathrm{HX}^- \) and the second \( \mathrm{HX}^- \rightarrow \mathrm{H}^+ + \mathrm{X}^{2-} \). Each stage has its own ionization constant: \( \mathrm{Ka}_1 \) for the first dissociation and \( \mathrm{Ka}_2 \) for the second.

Step 2: Comparing Ionization Constants

In general, \( \mathrm{Ka}_1 \) is greater than \( \mathrm{Ka}_2 \) for diprotic acids. The first proton is more easily ionized compared to the second proton, as the negative charge on \( \mathrm{HX}^- \) makes it harder to remove the second proton, thus reducing \( \mathrm{Ka}_2 \).

Step 3: Concluding the Correct Option

The relationship \( \mathrm{Ka}_1 > \mathrm{Ka}_2 \) is characteristic of any diprotic acid due to the increasing difficulty of removing the second proton. Therefore, option (c), \( \mathrm{Ka}_1 > \mathrm{Ka}_2 \), is true for any diprotic acid.

Key Concepts

Ionization Constant

The ionization constant is a crucial concept in understanding diprotic acids. It represents the equilibrium constant for the ionization of an acid. For a diprotic acid, such as \( \text{H}_2\text{X} \), there are two ionization constants: \( \text{K}_{a1} \) and \( \text{K}_{a2} \). Each constant corresponds to one of the acid's dissociation steps. The first ionization constant, \( \text{K}_{a1} \), is related to the ionization of the first hydrogen ion.

• The equation for this dissociation is: \( \text{H}_2\text{X} \rightarrow \text{H}^+ + \text{HX}^- \)
• Likewise, the second ionization constant, \( \text{K}_{a2} \), pertains to the second ionization: \( \text{HX}^- \rightarrow \text{H}^+ + \text{X}^{2-} \)

Typically, \( \text{K}_{a1} \) is greater than \( \text{K}_{a2} \), reflecting that the first proton is more readily lost than the second.

Acid Dissociation

Acid dissociation is the process through which an acid breaks down into its ions in a solution. For diprotic acids, this happens in two separate steps. Each step involves the donation of a proton (H+) and has its own unique dissociation equation.

• In the first dissociation, the acid releases one proton: \( \text{H}_2\text{X} \rightarrow \text{H}^+ + \text{HX}^- \).
• The second dissociation step involves the remaining \( \text{HX}^- \) ion losing its proton: \( \text{HX}^- \rightarrow \text{H}^+ + \text{X}^{2-} \).

Understanding these dissociations helps in studying the strength and behavior of diprotic acids. The lower the value of the ionization constant, the weaker the acid in that particular dissociation step.

Proton Donation

Proton donation is a key feature of the behavior of acids in solution. Diprotic acids can donate two protons, and this occurs step-by-step. Initially, a diprotic acid like \( \text{H}_2\text{X} \) will release the first proton into the solution when in contact with water or another solvent.

• The release of this first proton is relatively easy and is characterized by the first ionization constant, \( \text{K}_{a1} \).
• Once the first proton is released, the acid becomes \( \text{HX}^- \) and holds onto the second proton more tightly.

The second proton is harder to remove due to the negative charge of \( \text{HX}^- \), which makes \( \text{K}_{a2} \) smaller compared to \( \text{K}_{a1} \). This emphasizes the sequential nature of proton donation and impacts the strength and pH of the solution.