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