<h3>
Answer:</h3>
702 mL
<h3>
Explanation:</h3>
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
<h3>Step 1: Balanced equation for the reaction</h3>
The equation for the reaction between H₂SO₄ and NaOH is given by;
2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
<h3>Step 2: Calculate the number of moles of NaOH</h3>
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
<h3>Step 3: Calculate the number of moles of H₂SO₄ used during the reaction </h3>
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
<h3>Step 4: Calculate the volume of the H₂SO₄</h3>
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
