Titration & PH?

Question

Calculate the molarity of 35.0 ml of H2SO4 titrated to the equivalence point with 36.6 mL of 3.00 M NaOH

Answer 1

2NaOH + H₂SO₄ ==> Na₂SO₄ + 2H₂O ... balanced equation

moles NaOH used = 36.6 m x 1 L/1000 ml x 3.00 mol/L = 0.1098 moles NaOH

moles H₂SO₄ present = 0.1098 moles NaOH x 1 mole H₂SO₄/2 moles NaOH = 0.0549 moles H₂SO₄

molarity H₂SO₄ = 0.0549 moles/0.0350 L = 1.57 M (to 3 significant figures)

Answer 2

Balanced chemical reaction:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

moles of NaOH = Molarity x Volume = 3.00 mol/L x 36.6 mL x (1 L / 1000 mL) = 0.110 mol NaOH

From the balanced chemical reaction, 2 moles of NaOH are needed to neutralize 1 mole of H₂SO₄.

mol H₂SO₄ = 0.110 mol NaOH x (1 mol H₂SO₄ / 2 mol NaOH) = 0.0550 mol H₂SO₄

Molarity of H₂SO₄ = (0.0550 mol / 35.0 mL) x (1000 mL / 1 L) = 1.57 M H₂SO₄