Question

The arsenic in a 1.27 g sample of a pesticide was converted to AsO3−4 by suitable chemical treatment. It was then titrated using Ag+ to form Ag3AsO4 as a precipitate.

A- What is the oxidation state of  As in AsO3−4?

B- Name Ag3AsO4 by analogy to the corresponding compound containing phosphorus in place of arsenic.

C- If it took 21.0 mL of 0.106 M Ag+ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Answer 1

The arsenic in a 1.27 g sample of a pesticide was converted to AsO3−4 by suitable chemical treatment. It was then titrated using Ag+ to form Ag3AsO4 as a precipitate.

A- What is the oxidation state of  As in AsO3−4?

Assume that the oxidation state of  As in AsO3−4 is x

AsO3−4 = -3

X+ (4*-2)=-3

X-8=-3

X= +5

B- Name Ag3AsO4 by analogy to the corresponding compound containing phosphorus in place of arsenic.

Silver Arsenate Ag3AsO4

Silver Phosphate Ag3PO4

C- If it took 21.0 mL of 0.106 M Ag+ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

The chemical reaction between Ag+ and AsO4^3- is as follows:

3Ag+ + AsO4^3- = Ag3AsO4

Moles of Ag+= molarity * volume in L

= 0.106 M *0.021L

= 2.226*10^-3 Moles

Moles of Ag3AsO4

= 2.226*10^-3 Moles Ag+ * 1 moles AsO43- / 3 moles Ag+

= 7.42*10^-4 Moles AsO4^3-

1 mole of AsO4^3- = 1 mole As

7.42*10^-4 Moles AsO4^3- will be = 7.42*10^-4 Moles As

Amount of As = number of moles * molar mass

= 7.42*10^-4 Moles As * 74.92 g/ mole

= 0.055 g As

sample of a pesticide = 1.27 g

% of As in sample = [0.055/ 1.27]*100

= 4.38%