STATEMENT-1: When 0.1 M weak diprotic acid `H_(2)A` dissociated with its dissociation constants `K_(a_1)=10^(-3)` and `K_(a_2)=10^(-8)` , then `[A^(2-)]` is almost equal to `10^(-3)` M
STATEMENT-2: Since `K_(a_2)ltltK_(a_1)` for 0.1 M `H_(2)A,so[A^(2-)]` is negligible w.r.t. `[HA^(-)]`

Options

A. If both the statements are TRUE and STATEMENT-2 is the correct explation of STATEMENT-1
B. If both the statements are TRUE AND STATEMENT-2 is NOT the correct explanation of STATEMENT-1
C. If STATEMENT-1 is TRUE and STATEMETN-2 is FLASE
D. If STATEMENT-1 is FLASE and STATEMENT-2 is TRUE

Answer

To solve the problem, we need to analyze the dissociation of the weak diprotic acid \( H_2A \) and its dissociation constants \( K_{a1} \) and \( K_{a2} \).

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Acid**:
The weak diprotic acid \( H_2A \) dissociates in two steps:
\[ H_2A \rightleftharpoons HA^- + H^+ \quad (K_{a1} = 10^{-3}) \]
\[ HA^- \rightleftharpoons A^{2-} + H^+ \quad (K_{a2} = 10^{-8}) \]

2. **Setting Up the Initial Concentration**:
The initial concentration of \( H_2A \) is given as 0.1 M. Let \( x \) be the concentration of \( H^+ \) ions produced from the first dissociation. Therefore, at equilibrium:
- Concentration of \( HA^- \) will be \( x \)
- Concentration of \( H_2A \) will be \( 0.1 - x \)

3. **Applying the First Dissociation Constant**:
For the first dissociation, we can write the expression for \( K_{a1} \):
\[ K_{a1} = \frac{[HA^-][H^+]}{[H_2A]} = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]
Setting this equal to \( 10^{-3} \):
\[ \frac{x^2}{0.1 - x} = 10^{-3} \]

4. **Assuming \( x \) is Small**:
Since \( K_{a1} \) is relatively large, we can assume \( x \) is small compared to 0.1 M. Thus, \( 0.1 - x \approx 0.1 \):
\[ \frac{x^2}{0.1} = 10^{-3} \implies x^2 = 10^{-4} \implies x = 10^{-2} \text{ M} \]

5. **Calculating the Concentration of \( HA^- \)**:
From the first dissociation, we find that:
\[ [HA^-] = x \approx 10^{-2} \text{ M} \]

6. **Applying the Second Dissociation Constant**:
For the second dissociation, let \( y \) be the concentration of \( A^{2-} \) produced. The equilibrium expression for the second dissociation is:
\[ K_{a2} = \frac{[A^{2-}][H^+]}{[HA^-]} = \frac{y \cdot x}{x} = y \]
Setting this equal to \( 10^{-8} \):
\[ y = 10^{-8} \text{ M} \]

7. **Conclusion**:
- From the calculations, we find that \( [A^{2-}] = y = 10^{-8} \text{ M} \).
- Since \( K_{a2} \ll K_{a1} \), the concentration of \( A^{2-} \) is negligible compared to \( [HA^-] \).

### Final Evaluation of Statements:
- **Statement 1**: Incorrect, as \( [A^{2-}] \) is not \( 10^{-3} \) M but \( 10^{-8} \) M.
- **Statement 2**: Correct, since \( [A^{2-}] \) is negligible compared to \( [HA^-] \).