Determine the volume (liters) of 0.750 m naoh solution required to neutralize 1.50 l of 0.500 m h2so4. the neutralization reaction is: h2so4 (aq) + 2naoh (aq) → na2so4 (aq) + 2 h2o (l)

The balanced equation for the neutralisation reaction is as follows

2NaOH + H₂SO₄ - - > Na₂SO₄ + 2H₂O

stoichiometry of NaOH to H₂SO₄ is 2:1

number of H₂SO₄ moles reacted - 0.500 mol/L x 1.50 L = 0.750 mol

1 mol of H₂SO₄ reacts with 2 mol of NaOH

therefore 0.750 mol of H₂SO₄ reacts with - 2 x 0.750 = 1.50 mol of NaOH

molarity of NaOH solution is 0.750 M

there are 0.750 mol in 1 L of solution

therefore 1.50 mol in - 1.50 mol / 0.750 mol/L = 2 L

2 L of NaOH is required for neutralisation