Correct relation between dissociation constants of a di-basic acid

Options

A. `K_(a_1)=K_(a_2)`
B. `K_(a_1) gt K_(a_2)`
C. `K_(a_1) lt K_(a_2)`
D. `K_(a_1) =1/(K_(a_2))`

Answer

To find the correct relation between the dissociation constants \( K_a1 \) and \( K_a2 \) of a dibasic acid \( H_2A \), we can follow these steps:

### Step 1: Understand the Dissociation of a Dibasic Acid
A dibasic acid, such as \( H_2A \), can donate two protons (hydrogen ions). The dissociation can be represented in two steps:
1. The first dissociation:
\[ H_2A \rightleftharpoons HA^- + H^+ \]
The dissociation constant for this reaction is \( K_{a1} \).

2. The second dissociation:
\[ HA^- \rightleftharpoons A^{2-} + H^+ \]
The dissociation constant for this reaction is \( K_{a2} \).

### Step 2: Analyze the Speed of Proton Donation
- The first proton is donated quickly because the acid \( H_2A \) is neutral and readily loses the first proton.
- The second proton is donated more slowly because the conjugate base \( HA^- \) has a negative charge, which makes it less favorable to lose another proton.

### Step 3: Compare the Values of \( K_{a1} \) and \( K_{a2} \)
Since the first proton is lost more readily than the second, we can conclude that:
\[ K_{a1} > K_{a2} \]
This indicates that the dissociation constant for the first dissociation is greater than that for the second dissociation.

### Conclusion
The correct relation between the dissociation constants of a dibasic acid is:
\[ K_{a1} > K_{a2} \]

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