The pH of a buffer solution containing equal moles of acetic acid and sodium acetate is 4.8. `(pK_(a)` of acetic acid is 4.8)
Statement 2: The ionic product of water at `25^@C` is `10^(-14)mol^(2)L^(-2)`.

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Explanation

To solve the problem, we need to analyze the given statements regarding the buffer solution and the ionic product of water.

### Step-by-Step Solution:

1. **Understanding the Buffer Solution**:
- We have a buffer solution containing equal moles of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa).
- The pH of this buffer solution is given as 4.8.
- The pKa of acetic acid is also given as 4.8.

2. **Using the Henderson-Hasselbalch Equation**:
- The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
- Here, [A⁻] is the concentration of the base (sodium acetate) and [HA] is the concentration of the acid (acetic acid).

3. **Substituting Values**:
- Since the moles of acetic acid and sodium acetate are equal, their concentrations will also be equal:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 1 \]
- Therefore, the logarithm term becomes:
\[ \log(1) = 0 \]

4. **Calculating pH**:
- Substituting into the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + 0 \]
- Thus,
\[ \text{pH} = \text{pKa} = 4.8 \]

5. **Conclusion for Statement 1**:
- Statement 1 is true because the pH of the buffer solution is indeed equal to the pKa when the concentrations of the acid and its conjugate base are equal.

6. **Analyzing Statement 2**:
- The ionic product of water (Kw) at 25°C is given as \(10^{-14} \, \text{mol}^2 \, \text{L}^{-2}\).
- This is a known constant and can be derived from the dissociation of water:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
- At equilibrium, the concentrations of H⁺ and OH⁻ ions are both \(10^{-7} \, \text{mol/L}\), leading to:
\[ K_w = [\text{H}^+][\text{OH}^-] = (10^{-7})(10^{-7}) = 10^{-14} \, \text{mol}^2 \, \text{L}^{-2} \]

7. **Conclusion for Statement 2**:
- Statement 2 is also true as it correctly states the ionic product of water at 25°C.

### Final Conclusion:
- Both statements are true, but they are not directly related to each other.