At a certain temperature the value of `pK_(w)` is `13.4` and the measured `pH` of soln is `7`. The solution is

Options

A. Acidic
B. Basic
C. Neutral
D. Unpredictable

Explanation

To determine the nature of the solution based on the given pK_w and pH, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Values:**
- pK_w = 13.4
- pH of the solution = 7

2. **Calculate the pOH:**
- We know that pK_w is related to pH and pOH by the formula:
\[ pK_w = pH + pOH \]
- Rearranging this gives:
\[ pOH = pK_w - pH \]
- Substituting the values:
\[ pOH = 13.4 - 7 = 6.4 \]

3. **Calculate the Concentration of Hydroxide Ions (OH⁻):**
- We can find the concentration of OH⁻ ions using the formula:
\[ pOH = -\log[OH^-] \]
- Rearranging this gives:
\[ [OH^-] = 10^{-pOH} \]
- Substituting the value of pOH:
\[ [OH^-] = 10^{-6.4} \approx 3.98 \times 10^{-7} \, \text{M} \]

4. **Calculate the Concentration of Hydrogen Ions (H⁺):**
- We can find the concentration of H⁺ ions using the relationship:
\[ K_w = [H^+][OH^-] \]
- At 25°C, \( K_w \) is \( 1.0 \times 10^{-14} \). However, we will use the given pK_w to calculate:
\[ K_w = 10^{-pK_w} = 10^{-13.4} \approx 3.98 \times 10^{-14} \]
- Now we can find [H⁺]:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{3.98 \times 10^{-14}}{3.98 \times 10^{-7}} \approx 1.0 \times 10^{-7} \, \text{M} \]

5. **Determine the pH of Neutral Water at the Given Temperature:**
- In a neutral solution, [H⁺] = [OH⁻]. Therefore, we can calculate the pH of neutral water at this temperature:
\[ pH_{neutral} = -\log[H^+] = -\log(1.0 \times 10^{-7}) = 7 \]

6. **Compare the Given pH with the Neutral pH:**
- Since the pH of the solution is 7, which is equal to the pH of neutral water at this temperature, we conclude that the solution is neutral.

### Conclusion:
The solution is neutral.