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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Chemistry

Theresa Rice

18 February, 08:45

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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Cierra Figueroa · Answer 1

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.