Answer to Question #105823 in Chemistry for Alaysha

Question

If you used 61.5mL of 3.80 M KOH to titrate 15.0 ML of H2SO4, what was the concentration of the H2SO4?.

Answer 1

Solution:

Consider the option of complete neutralization.

The balanced chemical equation of the reaction of complete neutralization:

2KOH + H₂SO₄ = K₂SO₄ + 2H₂O.

At the equivalence point:

n(KOH) / 2 = n(H₂SO₄);

C(KOH) * V(KOH) = 2 * C(H₂SO₄) * V(H₂SO₄);

The molarity of the H₂SO₄ is calculated as follows:

C(H₂SO₄) = [C(KOH) * V(KOH)] / [2 * V(H₂SO₄)] = [3.80 M * 0.0615 L] / [2 * 0.0150 L] = 7.79 M

C(H₂SO₄) = 7.79 M

Answer: 7.79 M was the concentration of the H₂SO₄.