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*n C2H5COOH = 100 mL x 0,04 M = 4 mmol
*n KOH = 150 mL x 0,02 M = 3 mmol
*Ka = 1,2 x 10^-5
*pH ?
C2H5COOH + KOH --> C2H5COOK + H2O
m 4 3
b -3 -3 +3
s 1 - 3
Bersisa asam dan garam, maka ada peristiwa buffer:
[H+] = Ka x C2H5COOH / C2H5COOK
= 1,2 x 10^-5 x 1 / 3
= 4 x 10^-6
pH = 6 - log 4