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mol ch3coona = 1000 × 0,1 = 100mmol
mol hcl = 10 × 0,1 = 1 mmol
setelah penambahan hcl:
CH3COO- + H+ ⇒ CH3COOH
mula 100 1 100
reaksi 1 1 1
sisa 99 101
H+= 1,8 10^-5×101÷99=1,83 ×10^-5
pH = - log (1,83 ×10^-5) = 5 - log 1,83
Kelas: 2 SMA
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