larutan HA 0,1 M mempunyai pH=3. hitunglah nilai Ka dan derajat ionisasi (x) asam tersebut

pH = -log[H^+]
3 = -log[H^+]
[ H^+ ] = 10^-3

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

α = [H^+]/Ma
= 10^-3/10^-1
= 10^-2