`H_(2)S` behaves as a weak diprotic acid in aqueous solution. Which of the following is the correct explanantion for `pH` of a solution of `H_(2)S` in terms of its `pK_(1), pK_(2), [H_(2)S]` and `[S^(2-)]`

Options

A. `pH = (1)/(2) (pK_(1) + pK_(2))`
B. `pH = (1)/(2) (pK_(1) + pK_(2) - "log"([S^(2-)])/([H_(2)S]))`
C. `pH = (1)/(2)(pK_(1) + pK_(2) + "log"([S^(2-)])/([H_(2)S]))`
D. `pH = (1)/(2) (pK_(1) - pK_(2) + "log"([H_(2-)S])/([S^(2-)]))`

Explanation

To solve the question regarding the behavior of \( H_2S \) as a weak diprotic acid in aqueous solution and its relationship with \( pK_1 \), \( pK_2 \), \( [H_2S] \), and \( [S^{2-}] \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of \( H_2S \)**:
\( H_2S \) can dissociate in two steps:
- First dissociation:
\[ H_2S \rightleftharpoons H^+ + HS^- \]
This is characterized by the equilibrium constant \( K_1 \).
- Second dissociation:
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
This is characterized by the equilibrium constant \( K_2 \).

2. **Write the Equilibrium Constants**:
The equilibrium constants for the above reactions can be expressed as:
- For the first dissociation:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]
- For the second dissociation:
\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]

3. **Relate \( K_1 \) and \( K_2 \)**:
If we multiply \( K_1 \) and \( K_2 \):
\[ K_1 \cdot K_2 = \frac{[H^+]^2 [S^{2-}]}{[H_2S]} \]

4. **Express \( [H^+] \) in terms of \( K_1 \), \( K_2 \), \( [H_2S] \), and \( [S^{2-}] \)**:
Rearranging the equation gives:
\[ [H^+]^2 = K_1 \cdot K_2 \cdot \frac{[H_2S]}{[S^{2-}]} \]

5. **Take the Negative Logarithm**:
To find the pH, we take the negative logarithm of both sides:
\[ -\log[H^+] = \frac{1}{2}(-\log(K_1 \cdot K_2)) + \frac{1}{2} \log\left(\frac{[H_2S]}{[S^{2-}]}\right) \]
This simplifies to:
\[ pH = \frac{1}{2}(pK_1 + pK_2) + \frac{1}{2} \log\left(\frac{[H_2S]}{[S^{2-}]}\right) \]

6. **Final Expression for pH**:
Thus, the final expression for the pH of the solution of \( H_2S \) is:
\[ pH = \frac{1}{2}(pK_1 + pK_2) + \log[S^{2-}] - \log[H_2S] \]

### Conclusion:
The correct explanation for the pH of a solution of \( H_2S \) in terms of \( pK_1 \), \( pK_2 \), \( [H_2S] \), and \( [S^{2-}] \) is given by the derived formula.