Larutan ch2cooh 0,001 M dengan ka 4 x 10^-5. Tentukan pH Dan derajat ionisasi tersebut

[H^+] = √ Ka.M
= √ 4.10^-5 . 10^-3
= 2.10^-4 M
CH3COOH → CH3COO^- + H^+

pH = -log[H^+]
= -log[2.10^-4]
= 4-log2
α = terionisasi / awal
= [H^+] / [CH3COOH]
= 2.10^-4 / 10^-3
= 0,2