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11 April, 11:36

Find the h3o + and the oh - from the following values

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  1. 11 April, 11:43
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    Missing question on internet:

    a) pH = 5.

    pH = - log[H₃O⁺ ].

    [H₃O⁺] = 10∧ (-pH).

    [H₃O⁺] = 10⁻⁵ M.

    The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 10⁻¹⁴ M².

    Kw = [H ₃O⁺] · [OH⁻].

    [OH⁻] = Kw : [H₃O⁺].

    [OH⁻] = 10⁻¹⁴ M² : 10⁻⁵ M = 10⁻⁹ M.

    b) pH = 1.82.

    [H₃O⁺] = 10∧ (-1.82) = 1.5·10⁻² M.

    [OH⁻] = 10⁻¹⁴ M² : 1.5·10⁻² M = 6.6·10⁻¹³ M.

    c) pH = 10.65.

    [H₃O⁺] = 10∧ (-10.65) = 2.23·10⁻¹¹ M.

    [OH⁻] = 10⁻¹⁴ M² : 2.23·10⁻¹¹ M = 4.46·10⁻⁴ M.
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