The arsenic in a 1.22-g sample of a pesticide was converted to As043- by suitable chemical treatment. It was then titrated using Ag+ to form Ag3As04 as a precipitate, (a) What is the oxidation state of As in As043-? (b) Name Ag3As04 by analogy to the corresponding compound containing phosphorus in place of arsenic, (c) If it took 25.0 ml of 0.102 M Ag+ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Solution: Step1: 3- a) Calculation of oxidation state of As in AsO : 4 Since, Arsenate ion is a polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, x + 4(-2) = -3 [x being the oxidation number of As] x - 8 = -3 x = -3 + 8 x = +5 Thus, the oxidation number of As in AsO 4-is +4. Step2: b) The name of Ag AsO 3 silv4 arsenate. Step3: 25.0 Given, volume of silver ion solution used = 25.0 mL = 1000 L = 0.025 L Molarity of silver ion solution = 0.102 M Therefore, number of moles of silver ion = volume of solution in litres x molarity = 0.025 L x 0.102 M = 0.00255 mol Step4: Now, the reaction between silver ion and arsenate ion takes place in the following manner: 3Ag (aq) + AsO (aq) -------> Ag AsO (aq) 4 3+ 4 From the above reaction, it is clear that 3 mol of Ag reacts with 1 mol of arsenate ion for complete neutralization. Therefore, 0.00255 mol of Ag reacts with ( x 0.00255 = 3.00085) mol of arsenate ion for complete neutralization. Step5: Now, 1 mol of arsenate ion = 1 mol of As = molar mass of As Therefore, 0.00085 mol of As = 0.00085 x molar mass of As = 0.00085 x 74.92 g [molar mass of As = 74.92 g/mol] = 0.064 g of As Step6: Mass percentage of Arsenic in the sample is given by: Mass % of As = (mass of As / total mass of sample) x 100% = (0.064 / 1.22) x 100% = 0.0524 x 100% = 5.24 % Thus, mass percentage of Arsenic in the pesticide is 5.24 %.