Question

A 2.960 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 1.520 grams of KI and 50.00 mL of a 0.00820 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?

Answer 1

According to the question

As we know that

moles S2O3^2- used = 0.02 M *50 ml

= 1 mmol

excess I2 moles present = 1/2 = 0.5 mmol

moles I- added = 1.52g * 1000/166 g/mol = 9.1566 mmol

moles IO3- = 0.00820 M * 50 ml = 0.41 mmol

limiting reagent IO3-

moles I2 formed by reaction of I- and IO3- = 3 * 0.41

= 1.23 mmol

So,

moles I2 reacted with As3+ = 1.23 - 0.5 = 0.73 mmol

mass AsCl3 present = 0.73 mmol * 181.28 g/mol/1000

= 0.132 g

Then,

mass % AsCl3 in the original sample = 0.132 g *100/2.960 g

= 4.459%