At `90^(@)C`, pure water has `[H_(3)O^(o+)] = 10^(-7)mol L^(-1)`. What is the value of `K_(w)`at `90^(@)C`?

Options

A. `10^(-6)`
B. `10^(-12)`
C. `10^(-13)`
D. `10^(-14)`

Explanation

To find the value of \( K_w \) at \( 90^\circ C \) given that the concentration of hydronium ions \([H_3O^+]\) is \( 10^{-7} \, \text{mol L}^{-1} \), 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 \([OH^-]\). Therefore, we can write:
\[ [H_3O^+] = [OH^-] \]

### Step 2: Substitute the given concentration
Since it is given that \([H_3O^+] = 10^{-7} \, \text{mol L}^{-1}\), we can also say:
\[ [OH^-] = 10^{-7} \, \text{mol L}^{-1} \]

### Step 3: Write the expression for \( K_w \)
The ionic product of water, \( K_w \), is defined as:
\[ K_w = [H_3O^+][OH^-] \]

### Step 4: Substitute the concentrations into the \( K_w \) expression
Now, substituting the values we have:
\[ K_w = (10^{-7})(10^{-7}) \]

### Step 5: Calculate \( K_w \)
Now we can perform the multiplication:
\[ K_w = 10^{-7} \times 10^{-7} = 10^{-14} \]

### Conclusion
Thus, the value of \( K_w \) at \( 90^\circ C \) is:
\[ K_w = 10^{-14} \]