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24 March, 21:20

The ionization constant for water (kw) is 9.311 * 10-14 at 60 °c. calculate [h3o+], [oh-], ph, and poh for pure water at 60 °c.

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  1. 24 March, 21:32
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    As,

    Kw = [H+] [OH-]

    For water, [H+] = [OH-]

    Therefore we can write

    Kw = [H+]²

    9.311 * 10-14 = [H+]²

    [H+] = 3.04 * 10-7 = [OH-]

    Ph = - log [H+]

    = - log (3.04 * 10-7)

    = 6.52

    Thus, Ph = PoH = 6.52
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