The pH of pure water at `50^@C` is 6.63. What is the value of `K_w` at `50^@C`?

Options

A. `1.8xx10^(-15)`
B. `1.0xx10^(-14)`
C. `5.5xx10^(-14)`
D. `2.2xx10^(-13)`

Answer

To find the value of \( K_w \) at \( 50^\circ C \) given that the pH of pure water at that temperature is 6.63, we can follow these steps:

### Step 1: Calculate the concentration of \( H^+ \) ions
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 6.63, we can find the concentration of \( H^+ \) ions using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-6.63} \]

### Step 2: Calculate \( [H^+] \)
Now, we calculate \( [H^+] \):
\[ [H^+] = 10^{-6.63} \approx 2.34 \times 10^{-7} \, \text{mol/L} \]

### Step 3: Use the relationship for \( K_w \)
In pure water, the concentration of \( OH^- \) ions is equal to the concentration of \( H^+ \) ions:
\[ [OH^-] = [H^+] = 2.34 \times 10^{-7} \, \text{mol/L} \]

The ionic product of water \( K_w \) is given by:
\[ K_w = [H^+][OH^-] \]

### Step 4: Calculate \( K_w \)
Substituting the values we found:
\[ K_w = (2.34 \times 10^{-7})(2.34 \times 10^{-7}) = (2.34 \times 10^{-7})^2 \]
\[ K_w = 5.47 \times 10^{-14} \, \text{mol}^2/\text{L}^2 \]

### Final Answer
Thus, the value of \( K_w \) at \( 50^\circ C \) is approximately:
\[ K_w \approx 5.47 \times 10^{-14} \]