We need to first find the molarity of Ba(OH₂) solution. A mass of 3.24 mg is dissolved in 1 L solution. Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol dissociaton of Ba(OH)₂ is as follows; Ba(OH)₂ --> Ba²⁺ + 2OH⁻ 1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions. Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M pOH = -log[OH⁻] pOH = -log (3.8 x 10⁻⁵) pOH = 4.42 pH + pOH = 14 therefore pH = 14 - 4.42 pH = 9.58
