An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500 mL sample of air at 700 torr and 38C, she adds 20.00 mL of 0.01017 M aqueous iodine, which reacts as follo

Question

An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500 mL sample of air at 700 torr and 38°C, she adds 20.00 mL of 0.01017 M aqueous iodine, which reacts as follows:

SO₂(g) + I₂(aq) + 2H₂O(l) --> HSO₄^1-(aq) + 2I^1-(aq) + 3H⁺(aq)

The excess I₂(aq) left over from the above reaction was titrated with 13.17 mL of 0.0105 M thiosulfate (S₂O₃^2-) as follows:

I₂(aq) + 2S₂O₃^2-(aq) --> 2I^1-(aq) + S₄O₆^2-(aq)

What is the volume % of SO₂ in the original air sample closest to?

a) 1.59%

b) 1.48%

c) 2.20%

d) 0.95%

Please provide an explanation and/or show the steps on how to get to the answer. Thanks in advance.

Answer 1

I think your reaction is incorrectly written as it isn't balanced. Note that you have 6 oxygens on the left side of the equation and 8 oxygens on the right side. I think the equation you are probably supposed to use might be

I₂(aq) + SO₂(g) + 2H₂O(l) ==> 2I^-(aq) + SO₄^2-(aq) + 4H⁺(aq)

Before doing a lengthy series of calculations, it would be beneficial if you check the equation and then submit the question again. This is a back titration procedure.

EDIT (after including a correctly balanced equation):

I₂ left over: 0.01317 L S₂O₃^2- x 0.0105 mol/L = 0.0001383 moles S₂O₃^2-

0.0001383 moles S₂O₃^2- x 1 mole I₂/2 mol S₂O₃^2- = 0.00006914 moles I₂ left over

I₂ initially added = 0.020 L x 0.01017 mol/L = 0.0002034 moles I₂ initial

moles I₂ used in rx with SO₂ = 0.0002034 - 0.00006914 = 0.0001343 moles I₂ used

moles SO₂ present = 0.0001343 moles I₂ x 2 mole SO₂/mole I₂ = 0.0002685 moles SO₂

Volume SO₂: PV = nRT; V = nRT/P

V = (0.0002685)(62.36)(311)/700 = 0.007439 liters = 7.439 mls

Volume percent SO₂ = 7.439 mls/500 mls (x100%) = 1.48% (answer b)

I hope this makes sense.