Hitunglah pH larutan yang di buat dari 0,01 mol NaOH dalam 1 liter air

M=mol/volum=0,01/1
=0,01 M =10^-2M =[OH-]
pOH=-log[OH-]
=-log10^-2
=2
pH=14-pOH
=14-2
=12

[tex]\displaystyle PH \ LARUTAN \ [XI] \\ \\ M = \frac{mol}{v \ (L)} \\ M = \frac{0,01}{1} \\ M = 0,01 \ M \\ \\ OH^- = M \ \times \ b = 0,01 \ \times \ 1 = 0,01 \ M \\ pOH = - log \ OH^- \\ pOH = - log \ 0,01 \\ pOH = 2 \xrightarrow \ pH = 14 - pOH = 14 - 2 = 12 \xrightarrow \ Answer \\ [/tex]