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    The concentration of iodine in solution can be measured by titration with
    sodium thiosulfate solution.
    I2(aq) + 2S2O3^2–(aq) → 2I–(aq) + S4O6^2–(aq)


    The amount of sulfur dioxide in the atmosphere can be measured by passing a known volume of air through iodine solution. Sulfur dioxide converts iodine to iodide ions.
    SO2(g) + I2(aq) + 2H2O(l) → SO4^2–(aq) + 4H+(aq) + 2I–(aq)
    In an experiment, 100 m3 of air were passed through 100 cm3 of iodine,
    concentration 0.0100 mol dm–3. The remaining iodine was titrated with sodium thiosulfate solution and reacted with 12.60 cm3 of sodium thiosulfate, concentration 0.100 mol dm–3.
    (i) How many moles of iodine were present in the solution of the iodine at the start of the experiment?
    (ii) How many moles of iodine remained in the solution at the end of the
    experiment?

    iii) Calculate the number of moles of iodine which reacted with the sulphur dioxide and hence calculte the number of moles of sulfur dioxide in 100cm^3 of air.
    ok so for i) i did 100/1000 * 0.01 = 1*10^-3
    can someone please explain 2 and 3 ........
    this a pp question from 2010 jan q 19
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    (Original post by NoorL)
    The concentration of iodine in solution can be measured by titration with
    sodium thiosulfate solution.
    I2(aq) + 2S2O3^2–(aq) → 2I–(aq) + S4O6^2–(aq)


    The amount of sulfur dioxide in the atmosphere can be measured by passing a known volume of air through iodine solution. Sulfur dioxide converts iodine to iodide ions.
    SO2(g) + I2(aq) + 2H2O(l) → SO4^2-(aq) + 4H+(aq) + 2I–(aq)
    In an experiment, 100 m3 of air were passed through 100 cm3 of iodine,
    concentration 0.0100 mol dm–3. The remaining iodine was titrated with sodium thiosulfate solution and reacted with 12.60 cm3 of sodium thiosulfate, concentration 0.100 mol dm–3.
    (i) How many moles of iodine were present in the solution of the iodine at the start of the experiment?
    (ii) How many moles of iodine remained in the solution at the end of the
    experiment?

    iii) Calculate the number of moles of iodine which reacted with the sulphur dioxide and hence calculte the number of moles of sulfur dioxide in 100cm^3 of air.
    ok so for i) i did 100/1000 * 0.01 = 1*10^-3
    can someone please explain 2 and 3 ........
    this a pp question from 2010 jan q 19
    After you reacted the iodine with the air, you're left with some remaining iodine. This remaining iodine reacts with the sodium thiosulfate.

    The question is asking how much of this remaining iodine you've got left. If the sodium thiosulfate completely reacted with this remaining iodine, then what you can do is calculate the moles of sodium thiosulfate and use the mole ratios to calculate the moles of iodine it reacted with.
    Spoiler:
    Show
     \mathrm {Moles \ of} \ S_2O_3^{2-} = \dfrac{12.6}{1000} \times 0.100 = 1.26 \times 10^{-3}\  \mathrm {mol}
     1:2 \ \mathrm {ratio\ therefore \Moles \ of } \ I_2 = \dfrac{1}{2} \times 1.26 \times 10^{-3} = 6.3 \times 10^{-4} \ \mathrm {mol}
    You know the moles of Iodine that reacted with the Thiosulfate, you also know the original moles. If you deduct one away from the other, that must be the moles that reacted with the Sulfur Dioxide.
    Spoiler:
    Show
     \mathrm{Moles \ of} \ SO_2 = (1 \times 10^{-3}) - (6.3 \times 10^{-4}) = 3.7 \times 10^{-4} \ \mathrm {mol}
 
 
 
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