Question

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.

Answer 1

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.