Question

A 50.00 mL sample of a sodium hydroxide solution is titrated with a 1.605 M solution of sulfuric acid. The titration requires 24.09 mL of the acid solution to reach the equivalence point. What is the molarity of the base solution

Answer 1

The final result is a molarity of 1.54658 M for the NaOH solution.

Step-by-step explanation:

To find the molarity of the sodium hydroxide (NaOH) solution, we can use the stoichiometry of the titration reaction between NaOH and sulfuric acid (H₂SO₄). The balanced chemical equation for this reaction is:

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

Given that 24.09 mL (0.02409 L) of 1.605 M H₂SO₄ is needed to reach the equivalence point with 50.00 mL (0.05000 L) of NaOH, the molarity of NaOH can be calculated by using the molarity of H2SO4 and the volume of H₂SO₄ solution used, taking into account the stoichiometry of the reaction (1 mole of H₂SO₄ reacts with 2 moles of NaOH).

The moles of H₂SO₄ used are:

Moles of H₂SO₄ = Molarity of H₂SO₄ * Volume of H₂SO₄ in liters

= 1.605 M * 0.02409 L

= 0.0386645 moles

Since we have a 1:2 molar ratio of H₂SO₄ to NaOH:

Moles of NaOH = 2 * Moles of H₂SO₄

= 2 * 0.0386645

= 0.077329 moles

Now, we can calculate the molarity (concentration) of the NaOH solution:

Molarity of NaOH = Moles of NaOH / Volume of NaOH in liters

= 0.077329 moles / 0.05000 L

= 1.54658 M

The molarity of the NaOH solution is therefore 1.54658 M.