Like all equilibrium constants, the value of Kw depends on temperature. At body temperature (37°C), Kw = 2.4 * 10-14. What are the [H3O+] and pH of pure water at body temperature?

pH = 6.8124

Explanation:

We know pH decreases with increase in temperature.

At room temperature i.e. 25⁰c pH of pure water is equal to 7

We know

Kw = [H⁺][OH⁻](1)

where Kw = water dissociation constant

At equilibrium [H⁺] = [OH⁻]

So at 37⁰c i.e body temperature Kw = 2.4 × 10⁻¹⁴

From equation (1)

[H⁺]² = 2.4 × 10⁻¹⁴

[H⁺] = √2.4 × 10⁻¹⁴

[H⁺] = 1.54 × 10⁻⁷

pH  = - log[H⁺]

      = - log{1.54 × 10⁻⁷}

      = 6.812