larutan CH3COOH 0,1M dg Ka 10-5

[H+] = √Ka x Ma
[H+] = √10^-5 x 10^-1
[H+] = √10^-6
[H+] = 10^-3

pH = -log[H+]
pH = -log10^-3
pH = 3

Semoga membantu.

pH asam lemah

[H+]=√Ma.Ka
[H+]=√10^-1 x 10^-5
[H+]=√10^-6
[H+]=10^-3

pH=-loh[H+]
pH=-log10^-3
pH=3

Semoga membantu^^