At 277 K, degree of dissociation water is`1xx10^(-7)%`. The value of ionic product of water is

Options

A. `3.0xx10^(-14)`
B. `3.085xx10^(-15)`
C. `1xx10^(-16)`
D. `1xx10^(-14)`

Answer

To find the ionic product of water (Kw) at 277 K given the degree of dissociation of water, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Ionization Equation for Water:**
The ionization of water can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]

2. **Determine the Concentration of Water:**
We know that the molar mass of water (H2O) is approximately 18 g/mol. For 1000 g of water:
\[ \text{Number of moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.5 \text{ moles} \]
Since we are considering 1 liter of solution, the concentration (C) of water is:
\[ C = 55.5 \text{ moles/L} \]

3. **Convert the Degree of Dissociation to a Decimal:**
The degree of dissociation (α) is given as \(1 \times 10^{-7}\%\). To convert this percentage to a decimal:
\[ \alpha = \frac{1 \times 10^{-7}}{100} = 1 \times 10^{-9} \]

4. **Calculate the Concentration of Ions:**
For every mole of water that dissociates, it produces one mole of \(H^+\) and one mole of \(OH^-\). Therefore, the concentrations of \(H^+\) and \(OH^-\) ions at equilibrium can be expressed as:
\[ [H^+] = [OH^-] = C \cdot \alpha = 55.5 \cdot 1 \times 10^{-9} \]

5. **Calculate the Ionic Product of Water (Kw):**
The ionic product of water is given by:
\[ K_w = [H^+][OH^-] = (C \cdot \alpha)(C \cdot \alpha) = C^2 \cdot \alpha^2 \]
Substituting the values:
\[ K_w = (55.5 \cdot 1 \times 10^{-9})^2 \]
\[ K_w = 55.5^2 \cdot (1 \times 10^{-9})^2 \]
\[ K_w = 3080.25 \cdot 1 \times 10^{-18} \approx 3.085 \times 10^{-15} \]

### Final Answer:
The value of the ionic product of water (Kw) at 277 K is approximately:
\[ K_w \approx 3.085 \times 10^{-15} \]