A 34.53 ml sample of a solution of sulfuric acid, h2s04, reacts with 27.86 ml of 0.08964 m naoh solution. calculate the molarity of the sulfuric acid solution
the balanced equation between NaOH and H₂SO₄ is as follows 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O stoichiometry of NaOH to H₂SO₄ is 2:1 number of moles of NaOH moles reacted = molarity of NaOH x volume number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol according to molar ratio of 2:1 2 mol of NaOH reacts with 1 mol of H₂SO₄ therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄ number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol molarity of H₂SO₄ is 0.03617 M
