High School

What is the molarity of the [tex]H_2SO_4[/tex] solution, 18.6 mL of which neutralizes 30.5 mL of 1.55 M KOH solution?

Answer :

Final answer:

To determine the molarity of the H₂SO₄ solution that neutralizes a given volume of KOH solution, the balanced chemical equation is used to find the stoichiometry, from which the moles of H₂SO₄ are calculated, leading to the final molarity of 1.27 M.

Explanation:

To find the molarity of the H₂SO₄ solution, we first need to write the balanced chemical equation for the titration between H₂SO₄ and KOH: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. This shows that one mole of sulfuric acid reacts with two moles of potassium hydroxide. Given that 30.5 mL of 1.55 M KOH solution is used to neutralize 18.6 mL H₂SO₄, we calculate the moles of KOH used: moles of KOH = 1.55 mol/L * 0.0305 L = 0.047275 moles. Since the stoichiometry of the reaction is 1:2, the moles of H₂SO₄ = 0.047275 moles / 2 = 0.0236375 moles. Finally, to find the molarity of the H₂SO₄ solution, we divide the moles of H₂SO₄ by the volume of the H₂SO₄ solution in litres: molarity = 0.0236375 moles / 0.0186 L = 1.27 M.