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POH = -log [OH⁻] = -log[1*10⁻¹] = - (log 1 + log 10⁻¹) = -log 1 - (-1) = 1 - log 1
PH = 14 - POH = 14 - 1 - log 1 = 13 log 1
maaf klo salah
karena NaOH merupakan basa kuat, maka rumusnya [OH-] = b. Mb
NaOH --> Na+ + OH-
0,1M 0,1M (karena koefisiennya sama)
maka: pOH = -log[OH-]
pOH = -log0,1
pOH = 1
pH = 14 - 1 =13