Question

A 0.7538 g sample contains an unknown amount of As2O3. The sample was treated with HCL and a reducing agent, resulting in the formation of AsCl3(g), which was distilled into a beaker of water. Following the hydrolysis reaction AsCl3 + 2H2O -> HAsO2 + 3H+ + 3Cl- The amount of HAsO2 was determined by titration with 0.05264 M I2, requiring 33.64 mL to reach the equivalence point. The redox products were H3AsO4 and I-. What was the wt% As2O3 in the sample?

Answer 1

change in oxidation state of I = 1 (0 to -1)
change in oxidation state of As= 2 (+3 to +5)

Number of moles of I2 = M*V = 0.05264 * 33.64 = 1.771 mmol
Number of moles of I = 2*1.771 = 3.542 mmol

Now 1 mol of AS will require 2 moles of I for titration
So, number of moles of As = (number of moles of I) /2
= 3.542/2
= 1.771 mmol

2 AS are present in 1 As2O3
So,
number of moles of AS2O3 = 1.771/2 = 0.8855 mmol
Molar mass of As2O3=197.8 g/mol
mass of As2O3 = number of moles * molar mass
= 0.8855 mmol *197.8 g/mol
= 8.855*10^-4 * 197.8
=0.1752 g

mass % of As2O3 = mass of As2O3 * 100 / total mass
= 0.1752 * 100 / 0.7583
=23.1 %
Answer: 23.1 %