Question

Jika diketahui 0,2 asam etonoat (ka. 2.10^-5. tentukan ph

Answer 1

Dik: Ma=0,2M , ka=2x10^-5
dit : pH?
jwb : [H+] = √ka.Ma
= √0,2.2x10^-5
=√4x10^-6
=2x10^-3
pH= -log [H+]
= - log 2x10^-3
= 3 - log 2

Answer 2

Dik : Molar = 0,2 M = 2 x 10^-1. Ka asam etanoat = 2 x 10^-5