The value of `K_w `in neutral solution is ` 5.474 xx 10 ^(-14) " at " 50 ^(@) C ` . Calculate the pH of the solution at this temperature.

Answer

To calculate the pH of a neutral solution at 50°C where the value of \( K_w \) is given as \( 5.474 \times 10^{-14} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of \( K_w \)**:
\[ K_w = [H^+][OH^-] \]
In a neutral solution, the concentration of hydrogen ions \([H^+]\) is equal to the concentration of hydroxide ions \([OH^-]\). Therefore, we can express this as:
\[ K_w = [H^+]^2 \]

2. **Substitute the given value of \( K_w \)**:
\[ [H^+]^2 = 5.474 \times 10^{-14} \]

3. **Solve for \([H^+]\)**:
To find the concentration of hydrogen ions, take the square root of both sides:
\[ [H^+] = \sqrt{5.474 \times 10^{-14}} \]

4. **Calculate \([H^+]\)**:
Performing the calculation:
\[ [H^+] = \sqrt{5.474 \times 10^{-14}} \approx 2.34 \times 10^{-7} \, \text{M} \]

5. **Calculate the pH**:
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substitute the value of \([H^+]\):
\[ pH = -\log(2.34 \times 10^{-7}) \]

6. **Break down the logarithm**:
Using the properties of logarithms:
\[ pH = -\log(2.34) - \log(10^{-7}) = -\log(2.34) + 7 \]
Since \(\log(10^{-7}) = -7\).

7. **Calculate \(-\log(2.34)\)**:
Using a calculator or logarithm table:
\[ -\log(2.34) \approx -0.370 \]

8. **Combine the results**:
\[ pH \approx -0.370 + 7 \approx 6.63 \]

### Final Answer:
The pH of the solution at 50°C is approximately **6.63**.