The pH of a 1.0 x 10^-3 M Ba(OH)2 solution at 25 degrees C is

Question

how can 11.3 be the right answer.

Answer 1

Ba(OH)2 ==> Ba^+2 + 2OH^-
[Ba(OH)2] = 1 x 10^-3 = x
(OH^-) = 2x = 2 x 10^-3
pOH = 2.7
pH = 14 - pOH = 11.3