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A 1.147 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 1.690 grams of KI and 50.00 mL of a 0.00821 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?

  • chemistry -

    Note: I don't use I^3-; it's the I2 that does the job.
    As^3+ + I2 ==> 2I^- + As^5+
    Then mols I2 initially = 0.050L x 0.0821 = 0.004105
    mols I2 titrated with S2O3^2- = 0.050 x 0.02 = 0.001 which means 0.0005 mols I2 were used.
    I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
    mols I2 used for the As/I2 rxn = 0.004105-0.0005 = 0.003605

    Then g As = mols As x atomic mass As
    %As = (g As/g sample)*100 = ?

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