Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature?

Explanation

To find the pH of neutral water at 310 K, given that the ionic product of water (Kw) is \(2.7 \times 10^{-14}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ionic Product of Water (Kw)**:
The ionic product of water at a certain temperature is defined as:
\[ K_w = [H^+][OH^-] \]
where \([H^+]\) is the concentration of hydrogen ions and \([OH^-]\) is the concentration of hydroxide ions.

2. **Set Up the Equation for Neutral Water**:
In neutral water, the concentration of hydrogen ions is equal to the concentration of hydroxide ions:
\[ [H^+] = [OH^-] \]
Therefore, we can express \(K_w\) as:
\[ K_w = [H^+]^2 \]

3. **Substitute the Given Value of Kw**:
We know that:
\[ K_w = 2.7 \times 10^{-14} \]
So we can set up the equation:
\[ [H^+]^2 = 2.7 \times 10^{-14} \]

4. **Calculate the Concentration of Hydrogen Ions**:
To find \([H^+]\), take the square root of both sides:
\[ [H^+] = \sqrt{2.7 \times 10^{-14}} \]
Performing the calculation:
\[ [H^+] = 1.64 \times 10^{-7} \, \text{M} \]

5. **Calculate the pH**:
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ pH = -\log(1.64 \times 10^{-7}) \]
Performing the calculation gives:
\[ pH \approx 6.78 \]

### Final Answer:
The pH of neutral water at 310 K is approximately **6.78**.