hitunglah pH larutan dari Ba(OH)2 0,0005 M

Ba(OH)2 -> Ba+ + 2OH-
= 2x0,0005 = 0,001
[OH-] = 1 x 10^-3
pOH = - log [OH-] 
= - log 1 x 10^-3
= 3
pH = pKw - pOH
= 14 - 3 = 11
semoga bermanfaat