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

Explanation:

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

First: We need to write the balanced equation for the reaction.The reaction between NaOH and H₂SO₄ is a neutralization reaction.The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Second: We calculate the umber of moles of NaOH used Number of moles = Mass ÷ Molar mass Molar mass of NaOH is 40.0 g/mol Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

Third: Determine the number of moles of the acid, H₂SO₄From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

Fourth: Determine the Volume of the acid, H₂SO₄ usedWhen given the molarity of an acid and the number of moles we can calculate the volume of the acid.That is; Volume = Number of moles ÷ Molarity

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.

Google it that might work

Because a model would make the investigation easier to understand