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= 10 mmol = 0,01 mol
pH ( OH-) = 0,2 x 50
= 10 mmol = 0,01 mol
CH3COOH + NaOH => CH3COONa + CH3COOH2
Mula 0,01 0,01 - -
reaksi 0,01 0,01 0,01 0,01
sisa 0 0 0,01 0,01
lw salah gx tanggung y ..
hehehe