The ionic product of water at `25^@C` is `10^(-14)` its ionic product at `90^@C` will be

Options

A. `1 xx 10^(-14)`
B. `1 xx 10^(-16)`
C. `1 xx 10^(-20)`
D. `1 xx 10^(-12)`

Answer

To solve the problem of finding the ionic product of water (K_w) at 90°C, given that K_w at 25°C is 10^(-14), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Ionic Product of Water (K_w)**:
The ionic product of water (K_w) is defined as the product of the concentrations of hydrogen ions (H⁺) and hydroxide ions (OH⁻) in water:
\[ K_w = [H^+][OH^-] \]

2. **Given Information**:
We know that at 25°C, K_w = 10^(-14).

3. **Effect of Temperature on K_w**:
The dissociation of water into H⁺ and OH⁻ ions is an endothermic reaction. This means that as the temperature increases, the equilibrium shifts to favor the formation of more ions (H⁺ and OH⁻).

4. **Le Chatelier's Principle**:
According to Le Chatelier's Principle, if we increase the temperature of an endothermic reaction, the system will shift to absorb the added heat. This results in an increase in the concentrations of H⁺ and OH⁻ ions.

5. **Conclusion about K_w at Higher Temperature**:
Since the concentrations of H⁺ and OH⁻ increase with temperature, K_w at 90°C will be greater than K_w at 25°C. Therefore, K_w at 90°C will be greater than 10^(-14).

6. **Options Analysis**:
We need to find the correct value of K_w at 90°C from the given options. The options are:
- Option 1: 10^(-14) (same as at 25°C)
- Option 2: 10^(-16) (less than at 25°C)
- Option 3: 10^(-14) (same as at 25°C)
- Option 4: 10^(-12) (greater than at 25°C)

Since we established that K_w increases with temperature, the only option that reflects an increase is:
- **Option 4: 10^(-12)**

7. **Final Answer**:
Therefore, the ionic product of water at 90°C is **10^(-12)**.