Dissociation constant of `H_2O" at "25^@C` is

Options

A. `1 xx 10^(-14)`
B. `1 xx 10^(14)`
C. 14
D. `1.8 xx 10^(-16)`

Answer

To find the dissociation constant of water (H₂O) at 25°C, we need to understand the dissociation reaction of water and how to express its dissociation constant.

### Step-by-Step Solution:

1. **Write the Dissociation Reaction of Water:**
The dissociation of water can be represented as:
\[ H_2O \rightleftharpoons H^+ + OH^- \]

2. **Define the Equilibrium Constant (Kw):**
The dissociation constant for water, known as the ion product of water (Kw), is defined as:
\[ K_w = [H^+][OH^-] \]
where \([H^+]\) and \([OH^-]\) are the molar concentrations of hydrogen ions and hydroxide ions at equilibrium.

3. **Determine the Values at 25°C:**
At 25°C, the concentrations of \([H^+]\) and \([OH^-]\) in pure water are equal due to the self-ionization of water. Each concentration is \(1.0 \times 10^{-7} \, M\).

4. **Calculate Kw:**
Substituting the values into the equation for Kw:
\[ K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \]

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

### Final Answer:
The dissociation constant of \(H_2O\) at \(25°C\) is \(1.0 \times 10^{-14}\).