`2H_(2)O hArr H_(3)O^(o+) + overset(Theta)OH,K_(w) = 10^(-14)` at `25^(@)C`, hence `K_(a)` is

Options

A. `10^(-7)`
B. `5.55 xx 10^(-13)`
C. `10^(-14)`
D. `18 xx 10^(-17)`

Explanation

To find the acid dissociation constant \( K_a \) of water, we can follow these steps:

### Step 1: Understand the relationship between \( K_w \) and \( K_a \)
The ion product of water \( K_w \) is defined as:
\[ K_w = [H^+][OH^-] \]
For pure water at 25°C, \( K_w = 10^{-14} \).

### Step 2: Define \( K_a \)
The acid dissociation constant \( K_a \) for water can be expressed as:
\[ K_a = \frac{[H^+][OH^-]}{[H_2O]} \]
Since water is in a large excess compared to the ions produced, we can consider the concentration of water to be constant.

### Step 3: Calculate the concentration of water
To find the concentration of water, we can use the density of water. The density of water is approximately \( 1 \, \text{g/mL} \), which means:
- 1 liter of water weighs 1000 grams.
- The molar mass of water \( H_2O \) is approximately \( 18 \, \text{g/mol} \).

Thus, the number of moles of water in 1 liter is:
\[ \text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 4: Substitute values into the \( K_a \) equation
Now, substituting the values into the \( K_a \) expression:
\[ K_a = \frac{K_w}{[H_2O]} = \frac{10^{-14}}{55.56} \]

### Step 5: Calculate \( K_a \)
Now, we perform the calculation:
\[ K_a = \frac{10^{-14}}{55.56} \approx 1.8 \times 10^{-16} \]

### Conclusion
Thus, the value of \( K_a \) for water at 25°C is approximately:
\[ K_a \approx 1.8 \times 10^{-16} \]