Question

The balanced equation for the neutralization reaction of aqueous H₂SO₄ with aqueous KOH is shown. H₂SO₄(aq) + 2KOH(aq) = 2H₂O(l) + K₂SO₄(aq). What is the volume of 0.360 M KOH needed to react completely with 12.9 mL of 0.160 M H₂SO₄?

Answer 1

Approximately 11.5 mL of 0.360 M KOH is required to neutralize 12.9 mL of 0.160 M H₂SO₄, based on stoichiometric calculations stemming from the balanced neutralization reaction.

Step-by-step explanation:

To calculate the volume of a 0.360 M KOH solution needed to neutralize 12.9 mL of a 0.160 M H₂SO₄ solution, we must first use the stoichiometry of the balanced chemical equation H₂SO₄(aq) + 2KOH(aq) → 2H₂O(l) + K₂SO₄(aq). From this, we know that 2 moles of KOH react with 1 mole of H₂SO₄. Next, we calculate the moles of H₂SO₄ present:

• 0.160 M H₂SO₄ x 0.0129 L = 0.00207 mol H₂SO₄

Then, using the mole ratio, we find the moles of KOH required:

• 0.00207 mol H₂SO₄ x (2 mol KOH/1 mol H₂SO₄) = 0.00414 mol KOH

To find the volume of the 0.360 M KOH solution needed:

• Volume = moles of KOH / concentration of KOH = 0.00414 mol / 0.360 M = 0.0115 L

This converts to 11.5 mL. Therefore, approximately 11.5 mL of the 0.360 M KOH solution is needed to neutralize 12.9 mL of 0.160 M H₂SO₄.