Question

A 2.358 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO_3 and HCI aqueous solution. To this solution was added 1830 grams of Kl and 50.00 mL of a 0.00889 M KIO_3 solution. The excess I_3^- was titrated with 50.00 mL of a 0.02000 M Na_2 S_2 O_3 solution What was the mass percent of arsenic trichloride in the original sample?

Answer 1

The first reaction is IO₃^- + 8I^- + 6H⁺   3I₃^- + 3 H₂O.

Moles of KI = 1.830g * 1mol/166.00277g = 0.01102 mol
Moles of KIO₃ = 0.050L * 0.00976 M = 0.000488 mol

First we have to find the amount reacted by subtracting the moles of I3 by the excess. So

(0.05)*(0.00889 M KIO3)*(3 mol I3 / 1 mol IO3) = 1.33*10^-3 moles I3

Now we find the excess.

((0.02)(0.05))/2 = 5 x 10^-4 moles I3

Now we find the amount reacted.

1.33*10^-3 - 5 x10^-4 = 8.3 x 10^-4 moles I3 reacted

So..

8.3 x 10^-4 AsCl3 * (181.28 g / 1 mol) = 0.1504 g AsCl3

Finally, to find the mass percentage of AsCl3 in the original sample...

0.1504/2.358 *100 = 6.37%