What is pH of pure water whose is degree of dissociation `18xx10^(-9)`?

Answer

To find the pH of pure water with a degree of dissociation of \( 18 \times 10^{-9} \), we can follow these steps:

### Step 1: Understand the dissociation of water
Water dissociates into hydrogen ions (\( H^+ \)) and hydroxide ions (\( OH^- \)). The degree of dissociation (\( \alpha \)) indicates the fraction of water molecules that dissociate.

### Step 2: Set up the equilibrium expression
Let the initial concentration of water be \( C \). At equilibrium, the concentrations will be:
- \( [H^+] = C \alpha \)
- \( [OH^-] = C \alpha \)
- \( [H_2O] = C(1 - \alpha) \)

### Step 3: Determine the concentration of water
For pure water, we consider 1 kg of water. The number of moles of water in 1 kg is:
\[ \text{Moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ mol} \]
Since the density of water is approximately 1 g/mL, this corresponds to a volume of about 1 liter. Therefore, the concentration \( C \) of water is approximately:
\[ C \approx 55.56 \text{ M} \]

### Step 4: Calculate the concentration of \( H^+ \)
Using the degree of dissociation \( \alpha = 18 \times 10^{-9} \):
\[ [H^+] = C \alpha = 55.56 \times (18 \times 10^{-9}) \approx 1 \times 10^{-6} \text{ M} \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ \text{pH} = -\log(1 \times 10^{-6}) = 6 \]

### Final Answer
The pH of pure water with a degree of dissociation of \( 18 \times 10^{-9} \) is **6**.

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