Larutan asam asetat (Ka = 2 x 10^5) mempunyai pH = 3-log 2

PH = 3 -log 2
3- log 2 = – log (H+)
log H+ = log 2- log 10^3
H+ = 2.10^3

H+ = √Ka.M
2.10^-3 = √2.10^-5.M
M = 4.10^-6/2.10^-5
M = 0,2 M

semoga membantu ^^

H+ = √2 x 10^-5.M
Maka M = 0,000006/0,00002
Jawaban nya 0,20 M

Maaf kalau salah ya