Dissociation constant of water at `25^(@)` C is

Options

A. `1.0xx10^(-14)`
B. `1xx10^(-9)`
C. 14
D. `1.8xx10^(-16)`

Answer

To find the dissociation constant of water at 25°C, we can follow these steps:

### Step 1: Understand the dissociation of water
Water (H₂O) dissociates into hydrogen ions (H⁺) and hydroxide ions (OH⁻). The dissociation can be represented by the following equilibrium reaction:
\[ H_2O \rightleftharpoons H^+ + OH^- \]

### Step 2: Write the expression for the dissociation constant
The dissociation constant (also known as the ionic product of water, \( K_w \)) is defined as:
\[ K_w = \frac{[H^+][OH^-]}{[H_2O]} \]
Where:
- \([H^+]\) is the concentration of hydrogen ions.
- \([OH^-]\) is the concentration of hydroxide ions.
- \([H_2O]\) is the concentration of water.

### Step 3: Consider the concentration of pure water
Since water is a pure liquid, its activity is considered to be 1. Therefore, we can simplify the expression for \( K_w \):
\[ K_w = [H^+][OH^-] \]

### Step 4: Use known values at 25°C
At 25°C, it has been experimentally determined that the product of the concentrations of hydrogen ions and hydroxide ions is:
\[ K_w = 1.0 \times 10^{-14} \]

### Step 5: Conclusion
Thus, the dissociation constant of water at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

### Final Answer
The dissociation constant of water at 25°C is \( 1.0 \times 10^{-14} \).

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