`H_(2) SO_(4)` in aqueous medium ionises in two steps
`H_(2) SO_(4) (aq) + H_(2) O (i) to H_(3) O^(+) (aq) + HSO_(4)^(-) (aq) , K_(a_(1)) = x`
`HSO_(4)^(-) (aq) + H_(2) O(l) to H_(3) O^(+) (aq) + SO_(4)^(2-) , K_(a_(2)) = y`
What is relation between x and y ?

Options

A. ` x lt y`
B. ` x = y`
C. `x gt y`
D. `x gt gt y`

Answer

To find the relationship between the dissociation constants \( K_{a1} \) (denoted as \( x \)) and \( K_{a2} \) (denoted as \( y \)) for the ionization of sulfuric acid \( H_2SO_4 \) in aqueous medium, we can analyze the two steps of the ionization process.

### Step-by-Step Solution:

1. **Identify the Ionization Steps**:
- The first ionization step is:
\[ H_2SO_4 (aq) + H_2O (l) \rightarrow H_3O^+ (aq) + HSO_4^- (aq) \]
This step has a dissociation constant \( K_{a1} = x \).

- The second ionization step is:
\[ HSO_4^- (aq) + H_2O (l) \rightarrow H_3O^+ (aq) + SO_4^{2-} (aq) \]
This step has a dissociation constant \( K_{a2} = y \).

2. **Understand Acid Strength**:
- \( H_2SO_4 \) is a strong acid, meaning it completely dissociates in the first step. Therefore, the value of \( K_{a1} \) (or \( x \)) is very large.
- \( HSO_4^- \) is a weaker acid compared to \( H_2SO_4 \). It does not dissociate as completely as the first step, leading to a smaller value for \( K_{a2} \) (or \( y \)).

3. **Compare the Values of \( K_{a1} \) and \( K_{a2} \)**:
- Since \( H_2SO_4 \) is a strong acid, it dissociates significantly in the first step, resulting in a larger \( K_{a1} \).
- In contrast, \( HSO_4^- \) being a weaker acid means it dissociates less completely, resulting in a smaller \( K_{a2} \).

4. **Establish the Relationship**:
- From the above analysis, we conclude that:
\[ K_{a1} > K_{a2} \quad \text{or} \quad x > y \]

5. **Final Conclusion**:
- Therefore, the relationship between \( x \) and \( y \) is:
\[ x > y \]