Barium Hydroxide is a strong base, so it completely dissociates into its ions when hydrated. The reaction would be
Ba(OH)₂ ⇒ Ba⁺² + 2 OH⁻
However, it would also follow the Law of Definite Proportions. So, the concentration of OH⁻ would be twice that of Ba(OH)₂, Therefore, the concentration of OH⁻ is 2(0.1 M) = 0.2 M OH⁻.
We use the concentration of OH⁻ to determine pOH. Then, we use the relationship between pOH and pH to determine the pH of the strong base. The equations are:
pOH = -log[OH⁻]
pH + pOH = 14
Therefore,
pOH = -log[0.2]
pOH = 0.6989
0.6989 + pH = 14
pH = 13.3
