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

Explanation

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

### Step 1: Understand the dissociation of water
Water dissociates into hydronium ions (\( H_3O^+ \)) and hydroxide ions (\( OH^- \)):
\[ H_2O \rightleftharpoons H_3O^+ + OH^- \]

### Step 2: Write the expression for \( K_w \)
The equilibrium constant (\( K_w \)) for the dissociation of water is given by:
\[ K_w = [H_3O^+][OH^-] \]

### Step 3: Identify the concentration of \( H_3O^+ \)
According to the problem, the concentration of hydronium ions (\( [H_3O^+] \)) at \( 90^\circ C \) is:
\[ [H_3O^+] = 10^{-6} \, \text{mol L}^{-1} \]

### Step 4: Determine the concentration of \( OH^- \)
In pure water, the concentration of \( H_3O^+ \) is equal to the concentration of \( OH^- \):
\[ [OH^-] = [H_3O^+] = 10^{-6} \, \text{mol L}^{-1} \]

### Step 5: Substitute the values into the \( K_w \) expression
Now we can substitute the values of \( [H_3O^+] \) and \( [OH^-] \) into the equation for \( K_w \):
\[ K_w = (10^{-6})(10^{-6}) = 10^{-12} \]

### Step 6: Conclusion
Thus, the value of \( K_w \) at \( 90^\circ C \) is:
\[ K_w = 10^{-12} \]

---