The amount of sulfuric acid produced can be determined by titration with sodium hydroxide of known concentration. A 631.0 g sample of air that is known to contain sulfur dioxide was reacted with excess hydrogen peroxide. Given that 18.50 mL of 0.00250 M NaOH (aq) was required to neutralize the H 2 SO 4 (aq) produced, calculate the mass percentage of SO 2 (g) in the air sample.

0.000234 %

Explanation:

Equation of reaction of SO₂ and H₂O₂

SO₂ + H₂O₂ → H₂SO₄

Equation of neutralization of NaOH and H₂SO₄

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Molar mass of NaOH = 29.997 g/mol

molar mass of H₂SO₄ = 98.079 g/mol

molar mass of SO₂ = 64.066 g/mol

number of moles of NaOH = molarity * volume in L = 0.00250 M * (18.50 / 1000) = 0.00004625 mol

comparing the equation

2 moles of NaOH requires 1 mole of H₂SO₄

0.00004625 mol will require 0.000023125 mole of H₂SO₄

1 mole of SO₂ produced 1 mole of H₂SO₄

0.000023125 mole of SO₂ will produce 0.000023125 mole of H₂SO₄

Mass of SO₂ = 0.000023125 mole * 64.066 g/mol (mole x molar mass) = 0.00148 g

mass percentage of SO₂ = 0.00148 g / 631 g * 100 = 0.000234 %