Asample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. what volume in ml of 0.235 m h2so4 will neutralize this solution?

2.809 L of H₂SO₄

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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