Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods5394+o 0.05916 [ U ]E = E+ +− logUO2 U4+ +2 /2 [ UO ][ H ]2 42Before adding these two equations together we must multiply the secondequation by 2 so that we can combine the log terms; thus3+4+oo[ Ce ][ U ]3E = E4+ 3+ + 2E2+ 4+−0. 05916logCe / CeUO2/ U4+ + +[ Ce ][ UO ][ H ]At the equivalence point we know that3+ 2+[ Ce ] = 2×[ UO ]4+ 4+[ Ce ] = 2×[ U ]22 42Substituting these equalities into the previous equation and rearranginggives us a general equation for the potential at the equivalence point.2+ 4+oo2[UO ][ U ]23E = E4+ 3+ + 2E2+ 4+−0. 05916logCe / CeUO2/ U4+ + +2[ U ][ UO ][ H ]EE =EE =+ 2Eoo4+ 3+ + +Ce / CeUO2 2 / U4 .3+ 2E−oo4+ 3+ + +Ce / CeUO2 2 / U4 .EE =3+ 2Eoo4+ 3+ + +Ce / CeUO2 2 / U430 059163log[+ ] 4H10 05916×4+log[ H3−0.07888pHAt a pH of 1 the equivalence point has a potential ofE eq172 . + 2×0.327=− 0. 07888× 1=0.712 V3Click here to return to the chapter.Practice Exercise 9.202 42Because we have not been provided with a balanced reaction, let’s use aconservation of electrons to deduce the stoichiometry. Oxidizing C 2 O 4 2– ,in which each carbon has a +3 oxidation state, to CO 2 , in which carbonhas an oxidation state of +4, requires one electron per carbon, or a total oftwo electrons for each mole of C 2 O 4 2– . Reducing MnO 4 – , in which eachmanganese is in the +7 oxidation state, to Mn 2+ requires five electrons.A conservation of electrons for the titration, therefore, requires that twomoles of KMnO 4 (10 moles of e - ) reacts with five moles of Na 2 C 2 O 4 (10moles of e - ).The moles of KMnO 4 used in reaching the end point is+ ]