Need solution urgently 1. Calculate the potential as a function of titrant volume when 50 mL of 0.100 M Fe* is titratedwith 20.0, 50.0, and 53.0 mlL of 0.100M Cet.Given: Keg = 1.7 x 1014,Fet + e• Fe2; E° = 0.767 VCe+ +e+ Cet; E0 = 1.70 V[Fe2*]E = 0.767V - 0.0592 log-Fes1[Ce3+1E = 1.7V – 0.0592 log[Ce*](n, E"+n,E",ea)Eeqn,+n,

Given, Titration of 50mL 0.100M Fe2+ solution is done with 0.100M Ce4+ solution. Before the…

1. Calculate the potential as a function of titrant volume when 50 mL of 0.100 M Fe* is titrated with 20.0, 50.0, and 53.0 mL of 0.100M Cet. Given: Keg = 1.7 x 1014, Fe + e • Fe2"; E° = 0.767 V Ce+ + e Ce+; E° = 1.70 V [Fe2*] E = 0.767V – 0.0592 log: Fes1 [Cest] E = 1.7V - 0.0592 log- [Ce+] (n, E"+n,E") Eea red- n,+n,

1. Calculate the potential as a function of titrant volume when 50 mL of 0.100 M Fe* is titrated with 20.0, 50.0, and 53.0 mL of 0.100M Cet. Given: Keg = 1.7 x 1014, Fe + e • Fe2"; E° = 0.767 V Ce+ + e Ce+; E° = 1.70 V [Fe2*] E = 0.767V – 0.0592 log: Fes1 [Cest] E = 1.7V - 0.0592 log- [Ce+] (n, E"+n,E") Eea red- n,+n,

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.29QAP

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