Question

The ionic product of water at 310 K is 2.7 × $10^{-14}$. What is the pH of neutral water at this temperature?

Answer 1

Ionic product, Kw = [H+][OH-]

Let's assume that [H+] = x

Now, as we know that [H+] = [OH-]

We get, Kw = x2 = 2.7 × 10–14

Therefore, we get, x = 1.64 × 10–7

Also, pH = -log[H+]

=-log(1.64 × 10–7)

= 6.78