At `90^(@)C` , pure water has `[H_(3)O^(+)]` as `10^(-6)` mol `L^(-1)`. What is the value of `K_(w)` at `90^(@)C` ?

Options

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

Explanation

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

### Step 1: Understand the relationship in pure water
In pure water, the concentration of hydronium ions \([H_3O^+]\) is equal to the concentration of hydroxide ions \([OH^-]\). This is because water dissociates into hydronium and hydroxide ions in equal amounts.

### Step 2: Given data
We are given that the concentration of hydronium ions \([H_3O^+]\) at \( 90^\circ C \) is \( 10^{-6} \, \text{mol L}^{-1} \).

### Step 3: Determine hydroxide ion concentration
Since the solution is neutral, the concentration of hydroxide ions \([OH^-]\) will also be equal to \( 10^{-6} \, \text{mol L}^{-1} \).

### Step 4: Calculate the ionic product of water \( K_w \)
The ionic product of water \( K_w \) is calculated using the formula:
\[ K_w = [H_3O^+][OH^-] \]
Substituting the values we have:
\[ K_w = (10^{-6})(10^{-6}) = 10^{-12} \, \text{mol}^2 \text{L}^{-2} \]

### Conclusion
Thus, the value of \( K_w \) at \( 90^\circ C \) is \( 10^{-12} \, \text{mol}^2 \text{L}^{-2} \).

---