8. The arsenic in a 1.010 g sample of pesticide was converted to H3ASO4 by appropriate means. The acidwas neutralized and the arsenate precipitated quantitatively as Ag3AsO4 with exactly 40.00 mL of0.06222 M AgNO3.3Ag* + AsO4 → Ag3AsO4The excess Ag* was titrated with SCN", requiring 10.76 mL of 0.1000 M KSCN; the reaction being:Ag* + SCN- → AgSCN(s).Calculate the percentage of As2O3 (MM = 197.841 g/mol) in the original sample.

Molarity is defined as moles of solute upon the volume in litre Volume of AgNO3 present initially =40ml=40*10-3L 1L=1000ml Molarity of silver nitrate = 0.06222mol/L By definition of the molarity=moles/Volume So moles of silver nitrate =Molarity *volume Moles = 0.04L*0.06222mol/L Moles of silver nitrate= 2.4888*10-3 mol In 1mole of silver nitrate 1 mole of silver ion are present So in 2.4888 *10-3 mol of silver nitrate =2.888*10-3 Ag+ ion present The reaction of KSCN with Ag+ ion is shown below Ag+ + KSCN →AgSCN +K+ From reaction we see that 1mole of KSCN react with 1 mole of Ag+ ion Given Volume of KSCN =10.76ml=10.76*10-3 L Molarity of KSCN =0.1M So moles of KSCN =Molarity *volume Moles of KSCN= 10.76*10-3 L *0.1mol/L3 Moles of KSCN =10.76*10-4 mol Moles of Ag+ excess= Moles of KSCN =10.76*10-4 mol So moles of silver ion reacted with arsenic to precipitate as silver arsenals Moles of silver ion = (2.4888*10-3 -10.76*10-4 )mol Moles of Ag+ ion= 1.4128*10-3 mol As2O3 +3H2O→ 2H3AsO4 2H3AsO4 +6Ag+→2Ag3AsO4 +6H+ From the reaction we see that 1 mole of As2O3 is acidified to form 2 mole of H3AsO4 which will react with 6 mole of silver ion to form 2 mole of Ag3AsO4 So we can say 6 mole of Ag+ react with 1 mol of As2O3 1 mole of Ag+ react with 1/6 mol of As2O3 1.4128*10-3 mole of Ag+ react with (1.4128*10-3 )/6 mol of As2O3 1.4128*10-3 mole of Ag+ react with 2.35*10-4 mol of As2O3 1 mole of any substance weighs its molar mass So 1 mole of As2O3 weighs 197.841 gm So 2.35*10-4 mole of As2O3 weighs 197.841 gm/mol *2.5*10-4 mol =0.0465 gmMass of sample = 1.01gm Mass of As2O3 = 0.0465gm percentage of As2O3=mass of As2O3 *100mass of sample percentage of As2O3=0.0465 gm *1001.01 gm percentage of As2O3=4.65 1.01 %percentage of As2O3=4.6034 %

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Chemistry

The arsenic in a 1.010 g sample of pesticide was converted to H3ASO4 by appropriate means. The acid was neutralized and the arsenate precipitated quantitatively as Ag3AsO4 with exactly 40.00 mL of 0.06222 M AgNO3. 3Ag* + AsO4¯ → A33ASO4 The excess Ag* was titrated with SCN¯, requiring 10.76 mL of 0.1000 M KSCN; the reaction being: Ag* + SCN- → AgSCN(s). Calculate the percentage of AS2O3 (MM = 197.841 g/mol) in the original sample.

The arsenic in a 1.010 g sample of pesticide was converted to H3ASO4 by appropriate means. The acid was neutralized and the arsenate precipitated quantitatively as Ag3AsO4 with exactly 40.00 mL of 0.06222 M AgNO3. 3Ag* + AsO4¯ → A33ASO4 The excess Ag* was titrated with SCN¯, requiring 10.76 mL of 0.1000 M KSCN; the reaction being: Ag* + SCN- → AgSCN(s). Calculate the percentage of AS2O3 (MM = 197.841 g/mol) in the original sample.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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