Solve for all questions The titration solution in the cell below had a total volume of 50.0 ml and contained0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00s.C.E. || Zn(s) | titration solutionZn2* + 2e- Zn(s) E' = -0.762 vMgY2-: Kf = 6.2 x 108; Zny2- : Kf = 3.2 x 1016.What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: SeeExercise 15-B.)i)Total mmoles of Mg*2mmoles of EDTA added ismmoles of Mg*2 reacting with EDTAiv)mmoles of MgY2-v)Concentration of MgY2- in Mvi)mmoles of excess unreacted Mg+2vii)Concentration of Mg*2 in Mviii)Concentration of EDTA (Y4-) in Mix)Total mmoles of Zn+2{1:NUMERICAL:%100%0.500:0.15}x)mole of Zny2- in Mxi)Concentration of ZnY2- in Mxii)Concentration of Zn2+ in Mxiii)Potential of the cell in voltage, Ecell={1:NUMERICAL-1.364:0.1}13 marks

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The titration solution in the cell below had a total volume of 50.0 mL and contained 0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00 s.C.E. || Zn(s) | titration solution Zn2* + 2e- Zn(s) E' = -0.762 v Mgy2-: Kf = 6.2 x 108; Zny2-: Kf = 3.2 x 1016. What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: Se Exercise 15-B.) i) Total mmoles of Mg*2 ii) mmoles of EDTA added is iii) mmoles of Mg*+2 reacting with EDTA

The titration solution in the cell below had a total volume of 50.0 mL and contained 0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00 s.C.E. || Zn(s) | titration solution Zn2* + 2e- Zn(s) E' = -0.762 v Mgy2-: Kf = 6.2 x 108; Zny2-: Kf = 3.2 x 1016. What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: Se Exercise 15-B.) i) Total mmoles of Mg2 ii) mmoles of EDTA added is iii) mmoles of Mg+2 reacting with EDTA

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter24: Coulometry

Section: Chapter Questions

Problem 24.12QAP

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