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H+ = 10^-5
CH3COOH + NaOH ⇒ CH3COONa + H2O
m 4 0,2V - -
r 0,2 V 0,2V 0,2V 0,2V --
s 4-0,2V - 0,2V 0,2V
H+ = Ka (asam lemah/basa konjugasi)
10^-5 = 10^-5 (4 - 0,2V)/0,2V
0,2V = 4 - 0,2V
0,4V = 4
V = 4/0,4 = 10ml