Question

In a reaction, aqueous NaOH is titrated against sulfuric acid solution, H2SO4, according to the equation:

2 NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O (l)

If 18.62 ml of sulfuric acid are neutralized with 25.00 mL of 0.100 M NaOH, what is its molar concentration (Molarity, M) of sulfuric acid solution?

Answer 1

Number of moles of NaOH , n = Molarity x volume in L

                                           = 0.100 M x 0.025 L

                                          = 0.0025 mol

2 NaOH (aq) + H₂SO₄ (aq) Na₂SO₄ (aq) + 2H₂O (l)

According to the balanced equation ,

2 moles of NaOH reacts with 1 mole of H₂SO₄

0.0025 moles of NaOH reacts with M mole of H₂SO₄

M = (0.0025 x 1) / 2

    = 0.00125 mol of H₂SO₄

So Molarity of H₂SO₄ is , M = number of moles / volume in L

                                         = 0.00125 mol / 0.01862 L

                                         = 0.067 M

Therefore the Molarity of H₂SO₄ is 0.067 M