Question

A 55.00 mL solution of 0.4550 M AgNO3 was added to a solution of AsO43–. Ag3AsO4 precipitated and was filtered off. Fe3 was added and the solution was titrated with 0.2950 M KSCN. After 33.30 mL of KSCN solution had been added, the solution turned red. What mass of AsO43– was in the original solution?

Answer 1

A 55.00 mL solution of 0.4550 M AgNO3 was added to solution that means 0.45*0.05 = 0.0225 moles of AgNO3

For 1 mole of AgNO3 required Ag = 107.9 g

For 0.023 moles required Ag = 2.4817 g

So the amount of Ag added is 2.4817 g

Let us calculate the amount of Ag that reacts with KSCN

AgNO3 + KSCN --> AgSCN + KNO3

No. of moles of AgNO3 reacted with KSCN to get the end point =0.2950 * 0.033 = 0.00973 moles AgNO3

No. of moles of Ag = 0.8621g

The amount of Ag that reacted with AsO43- = 2.4817 - 0.8621 = 1.6196 g, i.e 0.015 moles

The balanced chemical equation

3 Ag+ (aq) + AsO43- (aq) → Ag₃AsO4 (s)

So no. of moles of AsO43- present in the solution = 0.015/3 = 0.005 moles

The amount of AsO43- originally present = 0.005 x 138.9 = 0.6949 g Answer