A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Response:

702 mL

To elaborate:

Given the following:

Mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We're tasked with finding the volume of acid necessary to neutralize the sodium hydroxide solution

Step 1: Write the balanced reaction equation

The reaction between H₂SO₄ and NaOH can be summarized as follows:

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

Step 2: Calculate the moles of NaOH

Moles are calculated by dividing mass by molar mass

The molar mass of NaOH is 40.0 g/mol

Hence;

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

= 0.33 moles of NaOH

Step 3: Determine moles of H₂SO₄ that react

According to the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄

Therefore, the ratio giving moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

Step 4: Find the volume of H₂SO₄

Molarity indicates the concentration of the solution in moles per liter

Molarity = Moles ÷ Volume

By rearranging the formula, we find volume = Moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Thus, the volume of the 0.235 M H₂SO₄ acid solution required equals 702 mL