At `90^(@)C`, pure water has `[H_(3)O^(o+)] = 10^(-6)M`. What is the value of `K_(w)` at `90^(@)C`

Options

A. `10^(-6)`
B. `10^(-8)`
C. `10^(-12)`
D. `10^(-14)`

Explanation

To find the value of \( K_w \) at \( 90^\circ C \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ion Product of Water**:
The ion product of water (\( K_w \)) is defined as:
\[ K_w = [H_3O^+] \times [OH^-] \]
where \([H_3O^+]\) is the concentration of hydronium ions and \([OH^-]\) is the concentration of hydroxide ions.

2. **Given Information**:
From the problem, we know that at \( 90^\circ C \):
\[ [H_3O^+] = 10^{-6} \, M \]

3. **Assuming Water is Neutral**:
In pure water, the concentration of hydronium ions is equal to the concentration of hydroxide ions:
\[ [OH^-] = [H_3O^+] = 10^{-6} \, M \]

4. **Calculating \( K_w \)**:
Now, we can substitute the values into the equation for \( K_w \):
\[ K_w = [H_3O^+] \times [OH^-] = (10^{-6}) \times (10^{-6}) \]
\[ K_w = 10^{-12} \]

5. **Final Answer**:
Therefore, the value of \( K_w \) at \( 90^\circ C \) is:
\[ K_w = 10^{-12} \]