Question

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.

Answer 1

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