At `100^@C`, `K_w =10^(-12)` . PH of pure water at `100^@C` will be

Options

A. `7.0`
B. `6.0`
C. `8.0`
D. `12.0`

Answer

To find the pH of pure water at 100°C where \( K_w = 10^{-12} \), we can follow these steps:

### Step 1: Understand the relationship between \( K_w \), \( [H^+] \), and \( [OH^-] \)
At any temperature, the ion product of water (\( K_w \)) is defined as:
\[ K_w = [H^+][OH^-] \]
For pure water, the concentration of hydrogen ions \([H^+]\) is equal to the concentration of hydroxide ions \([OH^-]\):
\[ [H^+] = [OH^-] \]

### Step 2: Set up the equation using \( K_w \)
Since \([H^+] = [OH^-]\), we can denote the concentration of hydrogen ions as \( x \):
\[ K_w = x^2 \]
Given that \( K_w = 10^{-12} \), we can write:
\[ x^2 = 10^{-12} \]

### Step 3: Solve for \( x \)
To find \( x \), take the square root of both sides:
\[ x = \sqrt{10^{-12}} = 10^{-6} \]
Thus, the concentration of hydrogen ions \([H^+]\) in pure water at 100°C is \( 10^{-6} \, \text{mol/L} \).

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(10^{-6}) \]
Using the properties of logarithms:
\[ \text{pH} = -(-6) \cdot \log(10) = 6 \cdot 1 = 6 \]

### Conclusion
The pH of pure water at 100°C is \( 6 \).