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n asam = 25.0,2 = 5mmol --> 0,5mol
n basa = 25.0,1 = 2,5mmol --> 0,25mol
lalu setelah di reaksikan muncul garam
ch3cooh + naoh ----> ch3coona + h2o
n garam yg terbentuk 0,25 mol
n asam yang sisa adalah 0,25 mol
karena asam yang bersisa maka ari [H+]
[H+] = n sisa asam/n garam dikali ka
= 0,25/0,25 dikali 1.10^-5
= 1.10^-5
pH = -log [H+]
= -log [1.10^-5]
= 5-log 1
= 5