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[OH⁻] = 10⁻⁵ . mol NH₄OH / mol NH₄⁺
[OH⁻] = 10⁻⁵ , 0,1/0,05
[OH⁻] = 2 x 10⁻⁵
pOH = -log [OH⁻]
pOH = -log [2 x 10⁻⁵]
pOH = 5 - log 2
pH = 14 - pOH
pH = 14 - (5 - log 2)
pH = 9 + log 2
Jadi, pH campuran adalah 9 + log 2