Given:

Ionic product of water(K_w) = 2.7 × 10^–14

As we know that K_w = [H⁺][OH^-]

For neutral water [H⁺]=[OH^-]

∴ K_w = [H⁺][H⁺]

⇒[H⁺]² = 2.7 × 10^–14

⇒[H⁺] =

⇒[H⁺] = 1.643 × 10^–7

Since, pH= -log[H⁺]

⇒pH = -log (1.643 × 10^–7)

⇒pH = -log 1.643 + 7log 10

⇒pH = 7- 0.1256

⇒pH = 6.78

Thus, the pH of the neutral water is 6.78