What is the molarity of a naoh solution if 39.1 ml of a 0.112 m h2so4 solution is required to neutralize a 25.0-ml sample of the naoh solution?

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