Question

at normal body temperature, 37 degree C, Kw = 2.4x10^-14. Calculate [H+] and [OH-] for a neutral solution at this temperature.

Answer 1

Answer:

1.55x10⁻⁷ = [H⁺] = [OH⁻]

Explanation:

The water equilibrium at a certain temperature is represented as follows:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

Where Kw is represented as:

Kw = [H⁺] [OH⁻]

In a neutral solution, [H⁺] = [OH⁻], thus, at 37°C:

2.4x10⁻¹⁴ = [H⁺]²

1.55x10⁻⁷ = [H⁺]

As The water equilibrium at a certain temperature is represented as follows:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

Where Kw is represented as:

Kw = [H⁺] [OH⁻]

In a neutral solution, [H⁺] = [OH⁻], thus, at 37°C:

2.4x10⁻¹⁴ = [H⁺]²

1.55x10⁻⁷M = [H⁺]

As [H⁺] = [OH⁻], [OH⁻] = 1.55x10⁻⁷M