At ` 80 ^(@ ) C ,` pure distilled water has `[H_3O^(+) ]= 1 xx 10 ^(-6) " mol " L^(-1) ` The value of `K_w` at this temperature will be

Options

A. `1 xx 10 ^(-6)`
B. ` 1 xx 10 ^(-12)`
C. ` 1 xx 10 ^(-14)`
D. ` 1 xx 10 ^(-5)`

Answer

To find the value of \( K_w \) (the ion product of water) at \( 80^\circ C \), we can follow these steps:

### Step 1: Understand the dissociation of water
Water dissociates into hydronium ions (\( H_3O^+ \)) and hydroxide ions (\( OH^- \)):
\[ H_2O \rightleftharpoons H_3O^+ + OH^- \]
The equilibrium constant for this reaction is given by:
\[ K_w = [H_3O^+][OH^-] \]

### Step 2: Identify the concentration of \( H_3O^+ \)
From the problem, we are given that the concentration of \( H_3O^+ \) at \( 80^\circ C \) is:
\[ [H_3O^+] = 1 \times 10^{-6} \, \text{mol L}^{-1} \]

### Step 3: Determine the concentration of \( OH^- \)
In pure water, the concentration of \( H_3O^+ \) is equal to the concentration of \( OH^- \) because water is neutral. Therefore:
\[ [OH^-] = [H_3O^+] = 1 \times 10^{-6} \, \text{mol L}^{-1} \]

### Step 4: Calculate \( K_w \)
Now we can substitute the concentrations into the expression for \( K_w \):
\[ K_w = [H_3O^+][OH^-] = (1 \times 10^{-6})(1 \times 10^{-6}) \]
\[ K_w = 1 \times 10^{-12} \, \text{mol}^2 \, \text{L}^{-2} \]

### Conclusion
Thus, the value of \( K_w \) at \( 80^\circ C \) is:
\[ K_w = 1 \times 10^{-12} \]

### Final Answer
The correct option is:
**Option 2: \( 1 \times 10^{-12} \)**

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