Selenious acid `(H_(2)SeO_(3))`,a diprotoc acid has `Ka_(1)=10^(-6)` and `Ka_(2)=10^(-8)` respectively. Appoximate pH of 0.01M `NaHSeO_(3)` is given by :-

Options

A. `pH=7+(pKa_(1))/(2)+(logC)/(2)`
B. `pH=7-(pKa_(1))/(2)-(logC)/(2)`
C. `pH = (pKa_(1)+pKa_(2))/(2)`
D. `pH=7+(pKa_(1))/(2)-(pKa_(2))/(2)`

Explanation

To approximate the pH of a 0.01 M solution of `NaHSeO3`, we can follow these steps:

### Step 1: Understand the nature of `NaHSeO3`
`NaHSeO3` is the sodium salt of selenious acid (`H2SeO3`). In solution, it dissociates to give `Na+` and `HSeO3-`. The `HSeO3-` ion can act as both an acid and a base (amphiprotic), which means it can donate a proton to form `SeO3^2-` or accept a proton to form `H2SeO3`.

### Step 2: Write the dissociation reactions
1. The first dissociation of selenious acid:
\[ H2SeO3 \rightleftharpoons H^+ + HSeO3^- \]
with dissociation constant \( K_{a1} = 10^{-6} \).

2. The second dissociation of selenious acid:
\[ HSeO3^- \rightleftharpoons H^+ + SeO3^{2-} \]
with dissociation constant \( K_{a2} = 10^{-8} \).

### Step 3: Set up the equilibrium expressions
For the first dissociation:
\[ K_{a1} = \frac{[H^+][HSeO3^-]}{[H2SeO3]} \]

For the second dissociation:
\[ K_{a2} = \frac{[H^+][SeO3^{2-}]}{[HSeO3^-]} \]

### Step 4: Relate the concentrations
Assuming that the concentration of `HSeO3^-` is approximately equal to the initial concentration of `NaHSeO3` (0.01 M), we can denote:
- \([HSeO3^-] \approx 0.01 \, M\)
- Let \([H^+] = x\) and \([SeO3^{2-}] = y\).

From the first dissociation:
\[ K_{a1} = \frac{x \cdot 0.01}{0.01 - x} \approx \frac{x \cdot 0.01}{0.01} = x \quad \text{(since } x \text{ is small)} \]
Thus, \( x \approx K_{a1} = 10^{-6} \).

### Step 5: Calculate the pH
Now, we can find the pH:
\[ pH = -\log[H^+] = -\log(10^{-6}) = 6 \]

### Step 6: Consider the second dissociation
Now we consider the second dissociation:
\[ K_{a2} = \frac{x \cdot y}{0.01 - x} \approx \frac{x \cdot y}{0.01} \]
Substituting \( x \approx 10^{-6} \):
\[ 10^{-8} \approx \frac{10^{-6} \cdot y}{0.01} \]
Solving for \( y \):
\[ y \approx \frac{10^{-8} \cdot 0.01}{10^{-6}} = 10^{-4} \]

### Step 7: Update the concentration of H+
The concentration of \( H^+ \) will now be:
\[ [H^+] = x + y \approx 10^{-6} + 10^{-4} \approx 10^{-4} \]
Thus, the new pH will be:
\[ pH = -\log(10^{-4}) = 4 \]

### Final Answer
The approximate pH of a 0.01 M solution of `NaHSeO3` is **4**.

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