The ionic product of water at 310 K is `2.7xx10^(-14)`. What is the pH of neutral 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**:
The ionic product of water (Kw) is defined as:
\[
K_w = [H^+][OH^-]
...