The reaction between NaOH and H₂SO₄ is as follows; 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O stoichiometry of base to acid is 2:1 NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place. number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol Number of NaOH moles in 25.0 mL - 0.0088 mol Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L Therefore molarity of NaOH - 0.352 M
