If the degree of ionization of water is `1.8xx10^(-9)` at `298K`. Its ionization constant will be

Options

A. `1.8xx10^(-16)`
B. `1xx10^(-14)`
C. `1xx10^(-16)`
D. `1.67xx10^(-14)`

Explanation

To find the ionization constant of water (Kw) given the degree of ionization, we can follow these steps:

### Step 1: Understand the Degree of Ionization
The degree of ionization (α) is given as \(1.8 \times 10^{-9}\). This represents the fraction of water molecules that ionize into hydrogen ions (H⁺) and hydroxide ions (OH⁻).

### Step 2: Use the Ionization Constant Formula
The ionization constant for water (Kw) can be expressed as:
\[ K_w = [H^+][OH^-] \]
Since water ionizes in the following manner:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
At equilibrium, the concentration of \(H^+\) and \(OH^-\) will both be equal to the degree of ionization (α) multiplied by the initial concentration of water.

### Step 3: Calculate the Concentration of Water
The concentration of pure water in molarity is approximately \(55.5 \, \text{mol/L}\).

### Step 4: Substitute Values into the Ionization Constant Equation
Since the concentration of \(H^+\) and \(OH^-\) at equilibrium is equal to α times the concentration of water:
\[ [H^+] = [OH^-] = \alpha \times 55.5 \]
Thus, we can express Kw as:
\[ K_w = (α \times 55.5)^2 \]

### Step 5: Substitute the Value of α
Substituting the value of α:
\[ K_w = (1.8 \times 10^{-9} \times 55.5)^2 \]

### Step 6: Calculate the Value
Calculating the inside of the parentheses first:
\[ 1.8 \times 10^{-9} \times 55.5 \approx 1.0 \times 10^{-7} \]
Now squaring this value:
\[ K_w \approx (1.0 \times 10^{-7})^2 = 1.0 \times 10^{-14} \]

### Step 7: Final Result
Thus, the ionization constant \(K_w\) at 298K is approximately:
\[ K_w = 1.0 \times 10^{-14} \]

### Conclusion
The ionization constant of water at 298K is \(1.0 \times 10^{-14}\).