Analytical Chem istry - DePauw University

482 Analytical Chemistry 2.0Table 9.15 Data for the Titration of 50.0 mL of 0.100 MFe 2+ with 0.100 M Ce 4+Volume of Ce 4+ (mL) E (V) Volume Ce 4+ (mL) E (V)10.0 0.731 60.0 1.6620.0 0.757 70.0 1.6830.0 0.777 80.0 1.6940.0 0.803 90.0 1.6950.0 1.23 100.0 1.7022E E ooFe=3 2E Ce4 C30 05916eq + +++ +− . log [ + 3+][ Ce ]Fe / Fe/ e3+ 4+[ Fe ][ Ce ]Because [Fe 2+ ] = [Ce 4+ ] and [Ce 3+ ] = [Fe 3+ ] at the equivalence point, thelog term has a value of zero and the equivalence point’s potential isEeqooE3+ 2+ + EFe Fe Ce4+ C3+/ / e0. 767 V+170 . V==22= 123 . VAdditional results for this titration curve are shown in Table 9.15 and Figure9.36.Practice Exercise 9.17Calculate the titration curve for the titration of 50.0 mL of 0.0500 MSn 2+ with 0.100 M Tl 3+ . Both the titrand and the titrant are 1.0 M inHCl. The titration reaction is3+2+ 4+ +Sn ( aq) + Tl ( aq) → Sn ( aq) + Tl ( aq)Click here to review your answer to this exercise.1.61.4E (V)1.21.0Figure 9.36 Titration curve for the titration of 50.0 mLof 0.100 M Fe 2+ with 0.100 M Ce 4+ . The red pointscorrespond to the data in Table 9.15. The blue lineshows the complete titration curve.0.80.60 20 40 60 80 100Volume of Ce 4+ (mL)