In atmosphere, `SO_(2)` and `NO` are oxidised to `SO_(3)` and `NO_(2)`, respectively,w hcih react with water to given `H_(2)SO_(4)` and `HNO_(3)`. The resultant solution is called acid rain. `SO_(2)`dissolves in water to form diprotic acid.
`SO_(2)(g) +H_(2)O(l) hArr HSO_(3)^(Theta) + H^(o+), K_(a_(1)) = 10^(-2)`.
`HSO_(3)^(Theta) hArr SO_(3)^(2-) + H^(o+), K_(a_(2)) = 10^(-7)`
and for equilibrium,
`SO_(2)(aq) + H_(2)O (l) hArr SO_(3)^(2-)(aq) +2H^(o+)(aq)`
`K_(a) = K_(a_(1)) xx K_(a_(2)) = 10^(-9) at 300K`.
The dominant equilibrium in an aqueous solution of sodium hydrogen sulphite `(NaHSO_(3))` is
`2HSO_(3)^(Theta) (aq) hArr SO_(2) (aq) +SO_(3)^(2-) (aq) + H_(2)O(l)`
The equilibrium constant for the above reaction is

Options

A. `10^(-3)`
B. `10^(-5)`
C. `10^(-6)`
D. `10^(-9)`

Explanation

To find the equilibrium constant for the reaction \(2 \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_2 (aq) + \text{SO}_3^{2-} (aq) + \text{H}_2O (l)\), we can utilize the provided dissociation constants for the diprotic acid \(H_2SO_3\).

### Step-by-Step Solution:

1. **Identify the Reactions and Their Constants:**
- The first dissociation of \(H_2SO_3\):
\[ \text{SO}_2 (g) + \text{H}_2O (l) \rightleftharpoons \text{HSO}_3^- (aq) + \text{H}^+ (aq) \quad K_{a_1} = 10^{-2} \]
- The second dissociation:
\[ \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_3^{2-} (aq) + \text{H}^+ (aq) \quad K_{a_2} = 10^{-7} \]

2. **Write the Overall Reaction:**
- The overall reaction we are interested in is:
\[ 2 \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_2 (aq) + \text{SO}_3^{2-} (aq) + \text{H}_2O (l) \]

3. **Relate the Equilibrium Constants:**
- The equilibrium constant for the overall reaction can be derived from the two dissociation reactions:
- For the first reaction, we can express it as:
\[ \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_2 (aq) + \text{H}^+ (aq) \quad K = \frac{1}{K_{a_1}} = 10^2 \]
- The second reaction gives:
\[ \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_3^{2-} (aq) + \text{H}^+ (aq) \quad K = K_{a_2} = 10^{-7} \]

4. **Combine the Reactions:**
- When we combine these two reactions, we get:
\[ 2 \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_2 (aq) + \text{SO}_3^{2-} (aq) + \text{H}_2O (l) \]
- The equilibrium constant for this combined reaction is:
\[ K = K_{a_1} \times K_{a_2} = (10^{-2}) \times (10^{-7}) = 10^{-9} \]

5. **Final Calculation:**
- The equilibrium constant for the reaction \(2 \text{HSO}_3^- (aq) \rightleftharpoons \text{SO}_2 (aq) + \text{SO}_3^{2-} (aq) + \text{H}_2O (l)\) is:
\[ K = K_{a_1} \times K_{a_2} = 10^{-2} \times 10^{-7} = 10^{-9} \]

### Final Answer:
The equilibrium constant for the reaction is \(K = 10^{-9}\).