The values of Kw in 0.1 M NaOH and 0.1 M NaCl are
(a) same (b) same (c) only at `25^@C`

Answer

To solve the question regarding the values of \( K_w \) in 0.1 M NaOH and 0.1 M NaCl, we need to analyze the properties of these solutions and how they relate to the ion product of water (\( K_w \)).

### Step-by-Step Solution:

1. **Understanding \( K_w \)**:
- \( K_w \) is the ion product of water, which is defined as:
\[ K_w = [H^+][OH^-] \]
- The value of \( K_w \) is temperature-dependent and is typically \( 1.0 \times 10^{-14} \) at \( 25^\circ C \).

2. **Effect of NaOH**:
- Sodium hydroxide (NaOH) is a strong base that completely dissociates in water:
\[ NaOH \rightarrow Na^+ + OH^- \]
- In a 0.1 M NaOH solution, the concentration of \( OH^- \) ions will be 0.1 M. The concentration of \( H^+ \) ions can be calculated using the relationship:
\[ [H^+] = \frac{K_w}{[OH^-]} \]
- Substituting the values:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{0.1} = 1.0 \times 10^{-13} \, M \]

3. **Effect of NaCl**:
- Sodium chloride (NaCl) is a neutral salt that does not affect the \( H^+ \) or \( OH^- \) concentrations significantly. It dissociates into:
\[ NaCl \rightarrow Na^+ + Cl^- \]
- In a 0.1 M NaCl solution, the concentrations of \( H^+ \) and \( OH^- \) remain at their equilibrium values at \( 25^\circ C \), which means:
\[ [H^+] = 1.0 \times 10^{-7} \, M \quad \text{and} \quad [OH^-] = 1.0 \times 10^{-7} \, M \]

4. **Comparing \( K_w \) Values**:
- In both solutions, at \( 25^\circ C \):
- For NaOH: \( K_w = [H^+][OH^-] = (1.0 \times 10^{-13})(0.1) = 1.0 \times 10^{-14} \)
- For NaCl: \( K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \)
- Therefore, the value of \( K_w \) remains the same in both solutions at \( 25^\circ C \).

5. **Conclusion**:
- The values of \( K_w \) in 0.1 M NaOH and 0.1 M NaCl are the same at \( 25^\circ C \). Thus, the correct answer is:
\[ \text{(c) only at } 25^\circ C \]