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mol NaOH = V x M = 0 x 0.1 = 0
mol ch3cooh = 0,05x0,1 = 0,005
ch3cooh + naoh ch3coona + h2o
0,005 0 - -
-0,005 -0,005 +0,005 -
0 -0,005 0,005 -
[H+] = ka x a/g
= 1,8 x 10^-5 x 0/0,005
= 1,8 x 10^-5
[H+] = ka
pH = -log [H+]
= -log 1,8 x 10^-5
= 5 - log 1,8
= 4,74
kalo anda mengerti jawaban saya di atas, untuk nomer selanjutnya juga sama seperti cara itu.