Which of the following is right for diprotic acid ?

Q: Which of the following is right for diprotic acid ?

Which of the following is right for diprotic acid ?

Question

Accepted Answer

Diprotic acids are acids that can donate two protons (H⁺ ions) in two separate steps. The dissociation of a diprotic acid can be represented by the following two equilibrium reactions:

1. First dissociation: $ HA \rightleftharpoons H^+ + A^- $ with equilibrium constant $ Ka_1 $.
2. Second dissociation: $ A^- \rightleftharpoons H^+ + A^{2-} $ with equilibrium constant $ Ka_2 $.

**Explanation for Option 2: $ Ka_1 > Ka_2 $**

1. **Justification for Option 2:**
   - In diprotic acids, the first dissociation involves the removal of a proton from a neutral molecule, which is generally easier than removing a proton from a negatively charged ion. Therefore, the first dissociation constant $ Ka_1 $ is typically larger than the second dissociation constant $ Ka_2 $.
   - The first proton is more readily donated because it is being removed from a neutral molecule, whereas the second proton is removed from an already negatively charged ion, which requires more energy and is less favorable.
   - This is a common characteristic of diprotic acids, where the first dissociation is stronger (higher $ Ka_1 $) than the second dissociation (lower $ Ka_2 $).

2. **Explanation for why other options are incorrect:**

- **Option 1: $ Ka_2 > Ka_1 $**
     - This option suggests that the second dissociation is stronger than the first, which contradicts the typical behavior of diprotic acids. As explained, the first dissociation is generally more favorable, making $ Ka_1 $ greater than $ Ka_2 $.

- **Option 3: $ Ka_2 > \frac{1}{Ka_1} $**
     - This option implies a relationship between $ Ka_2 $ and the reciprocal of $ Ka_1 $ that is not typically relevant or meaningful in the context of acid dissociation constants. The comparison between $ Ka_2 $ and $ \frac{1}{Ka_1} $ does not reflect the inherent properties of diprotic acids.

- **Option 4: $ Ka_2 = Ka_1 $**
     - This option suggests that both dissociation steps have the same equilibrium constant, which is not the case for diprotic acids. The first dissociation is always more favorable than the second, leading to $ Ka_1 $ being greater than $ Ka_2 $.

In conclusion, Option 2 ($ Ka_1 > Ka_2 $) correctly describes the typical behavior of diprotic acids, where the first dissociation constant is greater than the second.