Ba(OH)₂ is a strong base and completely ionises into its corresponding ions Ba(OH)₂ ---> Ba²⁺ + 2OH⁻ 1 mol of Ba(OH)₂gives 2 mol of OH⁻ ions. molarity of Ba(OH)₂ is 4.82 x 10⁻³ g/L / 171.3 g/mol - 2.81 x 10⁻⁵ mol/L since 1 mol of Ba(OH)₂gives 2 mol of OH⁻ ions. therefore [OH⁻] = 2[Ba(OH)₂] [OH⁻] = 2 x 2.81 x 10⁻⁵ mol/L = 5.62 x 10⁻⁵ M pOH can be calculated using the OH⁻ concentration pOH = -log [OH⁻] pOH = - log (5.62 x 10⁻⁵ M) pOH = 4.25 pH can be calculated as follows pH + pOH = 14 pH = 14 - 4.25 pH = 9.75 pH of solution is 9.75
