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Mol KOH = 50 x 0,1 = 5 mmol
Mol CH3COOH = 10 X 0,1 = 1 mmol
Mol KOH sisa = 5 - 1 = 4 mmol
[OH-] = b x Mb = 1 x (4/60) = 1/15 = 0,06 = 6 x 10⁻²
pOH = 2 - log 6
pH = 14 - (2-log6)
pH = 12 + log 6
50 mL CH3COOH 0,1 M + 10 mL KOH 0,1 M
mol CH3COOH yang bereaksi = 10.01 = 1 mmol
mol CH3COOH yang tersisa = 50.0,1 - 1 = 4 mmol (a)
mol CH3COONa yang terbentuk = 1 mmol (g)
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