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 `pH` of `0.01M` aqueous solutioon of sodium sulphite `(Na_(2)SO_(3))`

Options

A. `4.5`
B. `8.5`
C. `9.0`
D. `9.5`

Explanation

To calculate the pH of a 0.01 M aqueous solution of sodium sulfite (Na₂SO₃), we can follow these steps:

### Step 1: Identify the relevant equilibria
Sodium sulfite (Na₂SO₃) dissociates in water to form sodium ions (Na⁺) and sulfite ions (SO₃²⁻):
\[ \text{Na}_2\text{SO}_3 \rightarrow 2 \text{Na}^+ + \text{SO}_3^{2-} \]

The sulfite ion (SO₃²⁻) can act as a weak base and can react with water:
\[ \text{SO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HSO}_3^{-} + \text{OH}^- \]

### Step 2: Calculate the pKa values
We are given the dissociation constants (Ka) for the sulfurous acid (H₂SO₃):
- \( K_{a1} = 10^{-2} \)
- \( K_{a2} = 10^{-7} \)

To find the pKa values, we use the formula:
\[ pK_a = -\log(K_a) \]

Calculating \( pK_{a1} \):
\[ pK_{a1} = -\log(10^{-2}) = 2 \]

Calculating \( pK_{a2} \):
\[ pK_{a2} = -\log(10^{-7}) = 7 \]

### Step 3: Calculate the average pKa
The average pKa for the diprotic acid can be calculated as:
\[ pH = \frac{pK_{a1} + pK_{a2}}{2} \]

Substituting the values:
\[ pH = \frac{2 + 7}{2} = \frac{9}{2} = 4.5 \]

### Step 4: Conclusion
The pH of the 0.01 M aqueous solution of sodium sulfite (Na₂SO₃) is 4.5.

### Final Answer
The pH of the 0.01 M aqueous solution of sodium sulfite (Na₂SO₃) is **4.5**.

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