The degree of dissociation of 100 mL of pure water at `25^(@)C` is

Options

A. `1.8xx10^(-16)`
B. `1xx10^(-14)`
C. `1.8xx10^(-9)`
D. `1.0`

Answer

To find the degree of dissociation of 100 mL of pure water at 25°C, we can follow these steps:

### Step 1: Calculate the number of moles of water
The density of pure water is approximately 1 g/mL. Therefore, 100 mL of water has a mass of 100 g.

To calculate the number of moles of water, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Number of moles of water} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.56 \text{ moles} \]

### Step 2: Calculate the concentration of water
Now, we need to convert the volume of water from mL to L:
\[ 100 \text{ mL} = 0.1 \text{ L} \]

The concentration (C) of water can be calculated as:
\[ C = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
\[ C = \frac{5.56 \text{ moles}}{0.1 \text{ L}} = 55.56 \text{ M} \]

### Step 3: Write the dissociation equation
The dissociation of water can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
Let the degree of dissociation be represented by \( \alpha \). If \( \alpha \) is the fraction of water that dissociates, then at equilibrium:
- The concentration of \( H^+ \) ions = \( \alpha \times C \)
- The concentration of \( OH^- \) ions = \( \alpha \times C \)
- The concentration of undissociated water = \( C(1 - \alpha) \)

### Step 4: Apply the ion product of water (K_w)
At 25°C, the ion product of water \( K_w \) is given by:
\[ K_w = [H^+][OH^-] = (C \alpha)(C \alpha) = C^2 \alpha^2 \]
Substituting the value of \( C \):
\[ K_w = (55.56)^2 \alpha^2 \]
Given that \( K_w = 1 \times 10^{-14} \):
\[ 1 \times 10^{-14} = (55.56)^2 \alpha^2 \]

### Step 5: Solve for \( \alpha \)
First, calculate \( (55.56)^2 \):
\[ (55.56)^2 \approx 3086.5 \]
Now substitute back into the equation:
\[ 1 \times 10^{-14} = 3086.5 \alpha^2 \]
\[ \alpha^2 = \frac{1 \times 10^{-14}}{3086.5} \approx 3.24 \times 10^{-18} \]
Taking the square root:
\[ \alpha \approx \sqrt{3.24 \times 10^{-18}} \approx 1.8 \times 10^{-9} \]

### Final Answer
The degree of dissociation of 100 mL of pure water at 25°C is approximately:
\[ \alpha \approx 1.8 \times 10^{-9} \]