Question

A 1.199 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 1.810 grams of KI and 50.00 mL of a 0.00834 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

relate:

mol of KIO3 = MV = 0.00834*50 = 0.417 mmol

then

1 mol of IO3- = 3 mol of I2

0.417 mol of IO3- = 3*0.417 = 1.251 mmol of I2 present

therefore

initial I2 = 1.251 mmol

mmol of Na2S2O3 used = MV = 0.02*50 = 1 mmol

ratio is

2 mmol of Na2S2O3 = 1 mmol of I2

mmol of I2 excess = 1.251

mmol of I2 reacted with AsCl3 = (1.251 - 1) = 0.251 mmol

now..

1 mmol of AsCl3 = 1 mmol of I2

then

mmol of AScl3 = 0.251 mmol

mmol of AsCl3 reacted = 0.251

mass = mmol*MW = (0.251)(181.28) = 45.5012 mg

% mass of AsCl3 = (45.5012*10^-3)/(1.199) * 100 = 3.7949 % approx