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22 March, 00:19

The ionization constant for water (kw) is 9.614*10-14 at 60°c. calculate [h3o+], [oh-], ph, and poh for pure water at 60°c.

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  1. 22 March, 03:29
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    The Kw (the ionic product for water) at 60°C is 9.614·10⁻¹⁴ mol²/dm⁶.

    Kw = [H ₃O⁺] · [OH⁻ ].

    [ H₃O⁺] = [OH⁻] = √9.614·10⁻¹⁴ mol²/dm⁶.

    [H ⁺] = [OH⁻] = 3.1·10⁻⁷ mol/L.

    pH = - log[H ₃O⁺].

    pH = - log (3.1·10⁻⁷ mol/L).

    pH = 6.5; potential of hydrogen of neutral solution at 60°C.

    pOH = - log [OH⁻].

    pOH = 6.5.
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