This is an acid base reaction and the chemical equation for the above reaction is as follows; KOH + HClO₄ ---> KClO₄ + H₂O the stoichiometry of acid to base is 1:1 KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base. The excess base remaining is - 0.018 - 0.013 = 0.005 mol total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL [OH⁻] = 0.005 mol/0.075 L = 0.067 M pOH = -log[OH⁻] pOH = -log(0.067 M) pOH = 1.17 pOH + pH = 14 Therefore pH = 14 - 1.17 = 12.83 by knowing pH we can calculate the [H₃O⁺] pH = -log [H₃O⁺] [H₃O⁺] = antilog[-12.83] [H₃O⁺]= 1.47 x 10⁻¹³ M
