Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods4812+ 4+2+initialmoles Fe − molesCe added M V[ Fe ] ==total volumeV( 0.100 M) (50.0 mL) −( 0.100 M)(10.0 mL)=50.0 mL + 10.0 mL4+3+molesCe addedCe[ Fe ] = = M Vtotalvolume + VV FeCeCeFe Fe Ce CeFe− M V+ V( 0.100 M)(10.0 mL)== 1.67× 10 −2M50.0 mL + 10.0 mLCe= 667 . × 10 −2Substituting these concentrations into equation 9.16 gives a potential of⎛667 . × 10E =+ 0. 767 V −0. 05916log⎝⎜167 . × 10−2−2M⎞M⎠⎟ =+0 . 731 VMAfter the equivalence point, the concentration of Ce 3+ and the concentrationof excess Ce 4+ are easy to calculate. For this reason we find thepotential using the Nernst equation for the Ce 4+ /Ce 3+ half-reaction.3+RTE = E − Ce CnF=4+ 3+log [ ]oCe/ e4+[ Ce ]+ 170 . V −0. log [ 3+Ce05916]4+[ Ce ]9.17For example, after adding 60.0 mL of titrant, the concentrations of Ce 3+and Ce 4+ are2+3+initialmoles FeFe F[ Ce ] = = M V etotalvolume V + V( 0.100 M)(50.0 mL)=50.0 mL + 60.0mLFeCe= 455 . × 10 −3MStep 3: Calculate the potential after theequivalence point by determining theconcentrations of the titrant’s oxidizedand reduced forms, and using the Nernstequation for the titrant’s reduction halfreaction.4+molesCe added−initial molesFe[ Ce ] =total volume4+ 2+M V=V( 0.100 M) (60.0 mL) −( 0.100 M)(50.0 mL)=50.0 mL + 60.0 mLCe Ce Fe FeFe− M V+ VCe= 909 . × 10 −3Substituting these concentrations into equation 9.17 gives a potential of⎛455 . × 10E =+ 170 . V −0. 05916log⎝⎜909 . × 10−2−3M⎞M⎠⎟ =+166 . VMAt the titration’s equivalence point, the potential, E eq , in equation 9.16and equation 9.17 are identical. Adding the equations together to givesStep 4: Calculate the potential at theequivalence point.