Question

A quality control technician needs to determine the percentage of arsenic (As) in a particular pesticide. The pesticide is dissolved and all of the arsenic present is converted to arsenate ions (AsO43-). Then the amount of AsO43- is determined by titrating with a solution containing silver ions (Ag+). The silver reacts with the arsenate according to the following net ionic equation: 3 Ag+(aq) + AsO43-(aq) --> Ag3AsO4(s) When a 1.0435 g sample of pesticide was analyzed this way, it required 35.0 mL of 0.175 M Ag+ solution to precipitate all of the AsO43-. What was the mass percentage of arsenic in the pesticide?

Answer 1

3 Ag⁺(aq) + AsO₄^3-(aq) --> Ag₃AsO₄(s)

Moles of Ag+ reacted = Volume of Ag+ consumed X Molarity = 35.0mL X 0.175M = 6.125 mmol

As per stoichiometry of the reaction 3 moles of Ag+ react with one mole of AsO₄^3-

1 mole of Ag+ will react with 1/3 = 0.33 moles of AsO₄^3-

6.125 mmol of Ag+ will react 0.33 X 6.125 mmol = 2.021mmol

Thus amount of AsO₄^3- = 2.021 mmol

1 mol of AsO₄^3- contain one mole of Ag+ (From the formula of AsO₄^3- )

Rherefor moles of Ag+ is 2.021 mmol of AsO₄^3- = 2.021 mmol

Mass of Ag+ in 2.021mmol of AsO₄^3- = No. Of moles of Ag+ X Gram Atomic mass of Ag

Mass of Ag+ in 2.021mmol of AsO₄^3- = 2.021 X 10^-3 X 107.86 = 0.218g

Thus mass of Ag in 1.0435g sample = 0.218g

Percentage of Ag = (Mass of Ag / Mass of sample ) X 100 = (0.218 / 1.0435) X 100 = 20.89%