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pOH = 5 - log2
pH = 9 + log 2
POH = - log Kb . a/g
POH = - log 10⁻⁵ . 0,4/0,2
= - log10⁻⁵ 2
=-log 2.10⁻⁵
= 5 - log 2
PH = 14 - (5-log2)
= 9 + log 2
ingat meskipun ada volume dari larutan tp yg kita butuhkan cukupdg molaritas , dan Ka/kb jika tdk diket dlm soal kita gunakan 10-5