Sebanyak 100 mL KOH 0,02 M memiliki harga pH larutan sebesar

ASAM BASA

KOH => K+ + OH-

[OH-] = b × Mb
[OH-] = 1 × 0,02
[OH-] = 2 × 10^(-2)

pOH = – log [OH-]
pOH = – log ( 2 × 10^-2)
pOH = 2 – log 2

pH = 14 – pOH
pH = 14 – (2 – log 2)
pH = 12 + log 2