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14 November, 03:34

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

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  1. 14 November, 04:49
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    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
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