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.

Question

Chemistry

Keshawn Estrada

14 March, 00:16

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.

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Semaj · 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