The dissoctiation of water of `25^(@)C` is `1.9 xx 10^(-7)%` and the density of water is `1g//cm^(3)` the ionisation constant of water is

Options

A. `1.0 xx 10^(-14)`
B. `1.8 xx 10^(-6)`
C. `1.0 xx 10^(-6)`
D. `1.0 xx 10^(-8)`

Answer

To find the ionization constant of water (K_w) at 25°C given the dissociation percentage and density of water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of Water:**
Water dissociates into hydrogen ions (H⁺) and hydroxide ions (OH⁻):
\[ H_2O \rightleftharpoons H^+ + OH^- \]

2. **Given Data:**
- Dissociation of water: \(1.9 \times 10^{-7}\%\)
- Density of water: \(1 \, \text{g/cm}^3\)

3. **Convert Dissociation Percentage to Fraction:**
The dissociation percentage can be converted to a fraction (α):
\[ \alpha = \frac{1.9 \times 10^{-7}}{100} = 1.9 \times 10^{-9} \]

4. **Calculate the Concentration of Water:**
Since the density of water is \(1 \, \text{g/cm}^3\), we know that:
- 1 liter of water weighs 1000 grams.
- Molar mass of water (H₂O) = 18 g/mol.
\[ \text{Number of moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]
Therefore, the concentration of water (C) is:
\[ C = 55.56 \, \text{mol/L} \]

5. **Calculate the Ionization Constant (K_w):**
The ionization constant of water can be expressed as:
\[ K_w = [H^+][OH^-] = C \cdot \alpha^2 \]
Since at equilibrium, \([H^+] = [OH^-] = C \cdot \alpha\):
\[ K_w = (C \cdot \alpha)(C \cdot \alpha) = C \cdot \alpha^2 \]
Substituting the values:
\[ K_w = 55.56 \cdot (1.9 \times 10^{-9})^2 \]

6. **Calculate K_w:**
First, calculate \((1.9 \times 10^{-9})^2\):
\[ (1.9 \times 10^{-9})^2 = 3.61 \times 10^{-18} \]
Now calculate \(K_w\):
\[ K_w = 55.56 \cdot 3.61 \times 10^{-18} \approx 2.0 \times 10^{-16} \]

### Final Answer:
The ionization constant of water (K_w) at 25°C is approximately:
\[ K_w \approx 2.0 \times 10^{-16} \]