For a diprotic acid, which of the following is true for `1^(st)` and `2^(nd)` ionization constants (`K_(a_(1))` and `K_(a_(2))`)

Options

A. `K_(a_(1)) = K_(a_(2))`
B. `K_(a_(1)) gt K_(a_(2))`
C. `K_(a_(2)) gt K_(a_(1))`
D. `K_(a_(2)) ge K_(a_(1))`

Explanation

To determine the relationship between the first and second ionization constants (\(K_{a1}\) and \(K_{a2}\)) for a diprotic acid, we can analyze the ionization process step by step.

### Step 1: Understanding Diprotic Acids
A diprotic acid is an acid that can donate two protons (H⁺ ions). For example, carbonic acid (\(H_2CO_3\)) can dissociate in two steps:
1. \(H_2CO_3 \rightleftharpoons H^+ + HCO_3^-\) (First ionization)
2. \(HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}\) (Second ionization)

### Step 2: Defining Ionization Constants
The ionization constants for these steps are defined as:
- \(K_{a1}\) for the first ionization:
\[ K_{a1} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} \]
- \(K_{a2}\) for the second ionization:
\[ K_{a2} = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]} \]

### Step 3: Comparing the Ionization Steps
In general, the first ionization of a diprotic acid is more favorable than the second ionization. This is because:
- The first ionization involves the release of the first proton, which is typically easier due to the stronger bond of the remaining protons to the molecule.
- The second ionization involves the removal of a proton from a negatively charged ion (bicarbonate in this case), which is less favorable.

### Step 4: Conclusion on Ionization Constants
Since the first ionization is more favorable, it will have a larger equilibrium constant compared to the second ionization. Therefore, we can conclude:
\[ K_{a1} > K_{a2} \]

### Final Answer
For a diprotic acid, the relationship between the first and second ionization constants is:
\[ K_{a1} > K_{a2} \]