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, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Degree of Dissociation**:
The degree of dissociation (α) of water is given as \(1 \times 10^{-7}\%\). To convert this percentage into a fraction, we divide by 100:
\[ \alpha = \frac{1 \times 10^{-7}}{100} = 1 \times 10^{-9} \]

2. **Calculate the Concentration of Water**:
The concentration of water (C) in moles per liter is approximately 55.56 mol/L (since the density of water is about 1 g/mL and the molar mass of water is 18 g/mol):
\[ C = 55.56 \, \text{mol/L} \]

3. **Calculate the Concentration of H\(^+\) and OH\(^-\) Ions**:
The concentration of H\(^+\) ions and OH\(^-\) ions produced from the dissociation of water can be calculated using the formula:
\[ [H^+] = C \cdot \alpha \]
Substituting the values:
\[ [H^+] = 55.56 \times 1 \times 10^{-9} = 5.556 \times 10^{-8} \, \text{mol/L} \]

4. **Calculate the Ionic Product of Water (Kw)**:
The ionic product of water is given by the product of the concentrations of the H\(^+\) and OH\(^-\) ions:
\[ K_w = [H^+] \times [OH^-] \]
Since the concentrations of H\(^+\) and OH\(^-\) are equal in pure water:
\[ K_w = [H^+]^2 = (5.556 \times 10^{-8})^2 \]
Calculating this gives:
\[ K_w = 3.086 \times 10^{-15} \, \text{mol}^2/\text{L}^2 \]

5. **Final Answer**:
Therefore, the value of the ionic product of water (Kw) at 277 K is approximately:
\[ K_w \approx 3.085 \times 10^{-15} \, \text{mol}^2/\text{L}^2 \]