`pH` value of pure water at `0^(@)C` will be?

Options

A. Greater than `7`
B. Less than `7`
C. `7`
D. All of these

Explanation

To find the pH value of pure water at \(0^\circ C\), we need to follow these steps:

### Step 1: Understand the Ionization of Water
Pure water undergoes self-ionization, which can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
In pure water, the concentration of hydrogen ions \([H^+]\) is equal to the concentration of hydroxide ions \([OH^-]\).

### Step 2: Define the Ion Product of Water (\(K_w\))
The ion product of water at a certain temperature is given by:
\[ K_w = [H^+][OH^-] \]
At \(25^\circ C\), \(K_w\) is \(1.0 \times 10^{-14}\). However, we need the value of \(K_w\) at \(0^\circ C\).

### Step 3: Determine \(K_w\) at \(0^\circ C\)
At temperatures below \(25^\circ C\), the value of \(K_w\) decreases. Specifically, at \(0^\circ C\), \(K_w\) is approximately \(0.1 \times 10^{-14}\) (or less than \(1.0 \times 10^{-14}\)).

### Step 4: Calculate the Concentration of \(H^+\)
Since \([H^+] = [OH^-]\) in pure water, we can express \(K_w\) as:
\[ K_w = [H^+]^2 \]
Thus, we can find \([H^+]\) as:
\[ [H^+] = \sqrt{K_w} \]
Substituting the value of \(K_w\) at \(0^\circ C\):
\[ [H^+] = \sqrt{K_w} < \sqrt{1.0 \times 10^{-14}} \]
This implies:
\[ [H^+] < 1.0 \times 10^{-7} \text{ mol/L} \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Since \([H^+] < 1.0 \times 10^{-7}\), it follows that:
\[ pH > 7 \]

### Conclusion
Thus, the pH value of pure water at \(0^\circ C\) is greater than 7.