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10 December, 01:26

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?

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  1. 10 December, 01:39
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    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
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