sebanyak 3,42 gram Ba(OH)2 Mr=171 dilarutkan dalam 250ml larutan. pH larutan tersebut

Ba(OH)2 → Ba^2+ + 2OH^-
8.10^-2M ________ 16.10^-2M

M Ba(OH)2 = n / V(liter)
M = gr/Mr * 1000/ml
M = 3,42/171 * 1000/250
M = 1/500 * 4
M = 8.10^-2 M

pOH = -log[16.10^-2] = 2-log16
pH = 14 - ( 2-log16 ) = 12+ log 16