Question

What is the molarity of a NaOH solution, if 25.0 mL of this NaOH solution is required to neutralize 20.0 mL of 0.300 M H₂SO₄? (Write a balanced reaction to ascertain the stoichiometry of this reaction.)

a. 4.17 M
b. 0.480 M
c. None of these
d. 0.120 M
e. 0.240 M

Answer 1

The molarity of the NaOH solution is calculated using the balanced chemical equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O and the given volumes and concentration of H₂SO₄. The correct molarity of NaOH needed to neutralize the acid is 0.480 M.

Step-by-step explanation:

To determine the molarity of a NaOH solution that neutralizes a given volume of H₂SO₄, we must first write the balanced chemical equation and use stoichiometry.

The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) is:

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

This equation shows that it takes 2 moles of NaOH to neutralize 1 mole of H₂SO₄. With the information given, that 25.0 mL of NaOH is required to neutralize 20.0 mL of 0.300 M H₂SO₄, we can calculate the molarity of the NaOH solution

• Moles of H₂SO₄ = Volume (in L) x Molarity = 0.020 L x 0.300 M = 0.006 moles
• Moles of NaOH required = 2 x Moles of H₂SO₄ = 2 x 0.006 moles = 0.012 moles
• Molarity of NaOH = Moles of NaOH / Volume of NaOH (in L)
• Molarity of NaOH = 0.012 moles / 0.025 L = 0.480 M

Therefore, the molarity of the NaOH solution is 0.480 M, which corresponds to option (b).