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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12-05-2018
Chemistry

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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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Dejanras Dejanras · Answer 1 · 2018-05-24T14:56:31+02:00

Dejanras Dejanras

24-05-2018

The Kw (the ionic product for water) at 60°C is 9.614·10⁻¹⁴ mol²/dm⁶.
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = [OH⁻] = √9.614·10⁻¹⁴ mol²/dm⁶.
[H⁺] = [OH⁻] = 3.1·10⁻⁷ mol/L.
pH = -log[H₃O⁺].
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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