The value of `K_(w)` at a certain temperature is `6.25xx10^(-14)`. Calculate the pH of water.

Answer

To calculate the pH of water given the value of \( K_w \) at a certain temperature, follow these steps:

### Step 1: Understand the relationship of \( K_w \)
The ion product of water, \( K_w \), is defined as:
\[ K_w = [H^+][OH^-] \]
At neutral pH, the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) are equal. Therefore, we can express this as:
\[ K_w = [H^+]^2 \]

### Step 2: Substitute the given value of \( K_w \)
Given that:
\[ K_w = 6.25 \times 10^{-14} \]
We can substitute this into the equation:
\[ [H^+]^2 = 6.25 \times 10^{-14} \]

### Step 3: Solve for \([H^+]\)
To find \([H^+]\), take the square root of both sides:
\[ [H^+] = \sqrt{6.25 \times 10^{-14}} \]
Calculating the square root:
\[ [H^+] = \sqrt{6.25} \times \sqrt{10^{-14}} = 2.5 \times 10^{-7} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ pH = -\log(2.5 \times 10^{-7}) \]

### Step 5: Use logarithmic properties
Using properties of logarithms:
\[ pH = -(\log(2.5) + \log(10^{-7})) \]
\[ pH = -\log(2.5) + 7 \]
Calculating \(-\log(2.5)\):
\[ -\log(2.5) \approx -0.3979 \]
Thus:
\[ pH \approx 7 - 0.3979 \approx 6.6021 \]

### Final Answer
The pH of water at the given temperature is approximately:
\[ \text{pH} \approx 6.60 \]

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