The pH of pure water or neutral solution at `50^(@)C` is …. `(pK_(w) = 13.26 = 13.26 " at " 50^(@)C)`

Options

A. `7.0`
B. `7.13`
C. `6.0`
D. `6.63`

Explanation

To find the pH of pure water at 50°C, we can follow these steps:

### Step 1: Understand the relationship between pK_w and pH
At any temperature, the relationship between the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) in water can be expressed as:
\[ K_w = [H^+][OH^-] \]
where \(K_w\) is the ion product of water. The pK_w is defined as:
\[ pK_w = -\log K_w \]
At 50°C, we are given that \(pK_w = 13.26\).

### Step 2: Calculate the ion product of water (K_w)
Using the given pK_w, we can find \(K_w\):
\[ K_w = 10^{-pK_w} = 10^{-13.26} \]

### Step 3: Determine the concentrations of hydrogen and hydroxide ions
For pure water, the concentrations of hydrogen ions and hydroxide ions are equal:
\[ [H^+] = [OH^-] \]
Thus, we can express the ion product as:
\[ K_w = [H^+]^2 \]
This gives us:
\[ [H^+]^2 = 10^{-13.26} \]

### Step 4: Solve for [H^+]
Taking the square root of both sides, we find:
\[ [H^+] = \sqrt{10^{-13.26}} = 10^{-13.26/2} = 10^{-6.63} \]

### Step 5: Calculate the pH
The pH is defined as:
\[ pH = -\log[H^+] \]
Substituting the value we found:
\[ pH = -\log(10^{-6.63}) = 6.63 \]

### Conclusion
Therefore, the pH of pure water at 50°C is **6.63**.

---