Chemistry · Ch 3 — Ionic Equilibria

Relationship between pH and pOH

3.6.1

The ionic product of water is

K_w = [\mathrm{H_3O^{+}}][\mathrm{OH^{-}}]

Now, $K_w = 1 \times 10^{-14}$ at 298 K and thus

[\mathrm{H_3O^{+}}][\mathrm{OH^{-}}] = 1.0 \times 10^{-14}

Taking logarithm of both the sides, we write

\log_{10}[\mathrm{H_3O^{+}}] + \log_{10}[\mathrm{OH^{-}}] = -14

$-\log_{10}[\mathrm{H_3O^{+}}] + \{-\log_{10}[\mathrm{OH^{-}}]\} = 14$ …