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?
i need lots of

702 mL

We are given;

The mass of sodium hydroxide  = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We are required to calculate the volume of the acid required to neutralize the solution of NaOH

The equation for the reaction between H₂SO₄ and NaOH is given by;

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

Number of moles is given by dividing mass by molar mass

Molar mass of NaOH = 40.0 g/mol

Therefore;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

                                             = 0.33 moles of NaOH

From the equation 2 moles of NaOH reacts with 1 mole H₂SO₄

Therefore, the mole ratio of NaOH : H₂SO₄ is 2 : 1

Thus, number of moles of H₂SO₄ = Moles of NaOH ÷ 2

                                                       = 0.33 moles ÷ 2

                                                       = 0.165 moles

Molarity refers to the concentration of a solution in moles per liter

Molarity = Number of moles ÷ Volume

Rearranging the formula;

Volume = Number of moles ÷ Molarity

            = 0.165 moles ÷ 0.235 M

            = 0.702 L

           = 702 mL

Therefore, the volume of 0.235 M acid solution required is 702 ml