A 1.857 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.540 grams of KI and 50.00 mL of a 0.00912 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?

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  1. The idea here is that I2 reacts with As and oxidizes it to As^5+, then you titrate the excess I2 wit S2O3^2- to see how much of the I2 initially there was used.
    As^3+ + I2 ==> As^5+ + 2I^-
    1 mol As = 1 mol I2.

    The I2 came from the reaction of
    5I^- + IO3^- + 6H^+ ==> 3I2 + 3H2O

    Then I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
    1 mol I2 = 1 mols S2O3^2-
    mmols IO3^- = 50 x 0.00912 = 0.456
    mmols I2 initially = 0.456 x 3 = about 1.37

    mmols S2O3^2- = 50 x 0.02 = 1 mmol or 1/2 mmols I2; therefore,
    1.368-0.500 = about 0.8 mmol I2 used in the As step or 8.68E-4 mols AsCl3
    Convert that to grams AsCl3 and
    %AsCl3 = (mass AsCl3/1.857g sample)*100 = ?

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