ada yang tau caranya? fast respond yaa thankss

[H+] = (mol asam / mol garam) X Ka = (0,2 mol/02 mol) . 6,8 x 10^-4 mol/L = 6,8 x 10^-4 M

pH = - log [H+] = - log 6,8 x 10^-4 = 4 - log 6,8 = 4 - 0,83 = 3,167 = 3,17