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 find the ionic product of water at 90°C, we will follow these steps:

### Step 1: Understand the Ionic Product of Water (Kw)
The ionic product of water (Kw) is defined as the product of the concentrations of hydrogen ions (H⁺) and hydroxide ions (OH⁻) in water at a given temperature. At 25°C, Kw is known to be \(1 \times 10^{-14}\).

### Step 2: Recognize the Effect of Temperature on Kw
It is important to note that the ionic product of water increases with an increase in temperature. This is because the dissociation of water into H⁺ and OH⁻ ions is an endothermic process. Therefore, as the temperature rises, the degree of dissociation increases, leading to a higher concentration of H⁺ and OH⁻ ions.

### Step 3: Analyze the Given Information
We know that at 25°C, Kw = \(1 \times 10^{-14}\). We need to find Kw at 90°C. Since we established that Kw increases with temperature, we can conclude that:
\[ Kw \text{ at } 90°C > 1 \times 10^{-14} \]

### Step 4: Evaluate the Options
The given options for Kw at 90°C are:
1. \(1 \times 10^{-14}\)
2. \(1 \times 10^{-16}\)
3. \(1 \times 10^{-20}\)
4. \(1 \times 10^{-12}\)

Since we established that Kw at 90°C is greater than \(1 \times 10^{-14}\), we can eliminate options 1, 2, and 3, as they are all less than or equal to \(1 \times 10^{-14}\).

### Step 5: Conclusion
The only option that is greater than \(1 \times 10^{-14}\) is:
\[ \text{Option 4: } 1 \times 10^{-12} \]

Thus, the ionic product of water at 90°C will be \(1 \times 10^{-12}\).

### Final Answer:
The ionic product of water at 90°C is \(1 \times 10^{-12}\).

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