berapa gram natrium asetat ( mr = 82 ) harus ditambah kan ke dalam 200 ml larutan asam asetat 0,1 M ( ka = 10^-5 ) agar diperoleh ph = 4

CH3COOH + NaOH --> CH3COONa + H2O
mol CH3COOH = 0,1 x 200 = 20 mmol = 0,02 mol
mol CH3COONa = gram / 82
Larutan Buffer Asam :
H+ = ka x mol asam lemah / mol basa
konjugasi
10^-4 = 10^-5 x 0,02 / (gram/82)
10 x gram = 0,02 x 82
gram = 0,164 gram

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