At `100^@C` the `K_w` of water is 55 times its value at `25^@C` . What will be the pH of neutral solution ?
(log 55=1.74 )

Options

A. 6.13
B. `7.00`
C. `7.87`
D. `5.13`

Answer

To find the pH of a neutral solution at 100°C where the ion product of water (K_w) is 55 times its value at 25°C, we can follow these steps:

### Step 1: Determine K_w at 100°C
The value of K_w at 25°C is \( 1.0 \times 10^{-14} \). At 100°C, K_w is given as 55 times this value:
\[ K_w = 55 \times 10^{-14} \]

### Step 2: Calculate K_w
Calculating K_w:
\[ K_w = 55 \times 10^{-14} = 5.5 \times 10^{-13} \]

### Step 3: Use the relationship for neutral solutions
In a neutral solution, the concentration of hydrogen ions \([H^+]\) is equal to the concentration of hydroxide ions \([OH^-]\). Therefore, we can express K_w as:
\[ K_w = [H^+] \times [OH^-] = [H^+]^2 \]

### Step 4: Solve for [H^+]
To find the concentration of hydrogen ions, we take the square root of K_w:
\[ [H^+] = \sqrt{K_w} = \sqrt{5.5 \times 10^{-13}} \]

### Step 5: Calculate [H^+]
Calculating the square root:
\[ [H^+] = \sqrt{5.5} \times 10^{-6.5} \]
Using the approximation \(\sqrt{5.5} \approx 2.35\):
\[ [H^+] \approx 2.35 \times 10^{-6.5} \approx 7.41 \times 10^{-7} \text{ M} \]

### Step 6: Calculate pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ pH = -\log(7.41 \times 10^{-7}) \]

### Step 7: Simplify the logarithm
Using properties of logarithms:
\[ pH = -\log(7.41) - \log(10^{-7}) = -\log(7.41) + 7 \]
Given that \(\log(7.41) \approx 0.87\):
\[ pH \approx -0.87 + 7 = 6.13 \]

### Final Answer
Thus, the pH of the neutral solution at 100°C is:
\[ \text{pH} \approx 6.13 \]