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 when \( K_w = 10^{-12} \), we can follow these steps:

### Step 1: Understand the relationship between \( K_w \), \( [H^+] \), and \( [OH^-] \)
The ion product of water (\( K_w \)) is given by the equation:
\[ K_w = [H^+][OH^-] \]
For pure water, the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) are equal. Therefore, we can express this as:
\[ K_w = [H^+]^2 \]

### Step 2: Substitute the value of \( K_w \)
Given that \( K_w = 10^{-12} \), we can set up the equation:
\[ [H^+]^2 = 10^{-12} \]

### Step 3: Solve for \([H^+]\)
To find \([H^+]\), take the square root of both sides:
\[ [H^+] = \sqrt{10^{-12}} = 10^{-6} \]

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

### Conclusion
Thus, the pH of pure water at 100°C is 6.

### Final Answer
The pH of pure water at 100°C is **6**.

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