Ask Question
28 February, 13:53

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

+3
Answers (1)
  1. 28 February, 15:52
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “at normal body temperature, 37 degree C, Kw = 2.4x10^-14. Calculate [H+] and [OH-] for a neutral solution at this temperature. ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers