Analytical Chem istry - DePauw University

538 Analytical Chemistry 2.0E (V)0.80.60.40.20.00 10 20 30 40 50Volume of Tl 3+ (mL)Figure 9.51 Titration curve for Practice Exercise9.18. The black dots and curve are theapproximate sketch of the titration curve.The points in red are the calculations fromPractice Exercise 9.17.3 ( 0. 100 M)(40.0 mL) −( 0.0500 M)(50.0 mL)[ Tl+ ] =50.0 mL + 40.0 mLand the potential isE =+ 077 . V −0.059162= 0.0167⎛0.0278 M⎞log⎝⎜0.0167 M⎠⎟ =+076 . VAt the titration’s equivalence point, the potential, E eq , potential is0. 139 V+077 . VE eq== 045 . V2Some additional results are shown here.Volume of Tl 3+ (mL) E (V) Volume of Tl 3+ (mL) E (V)5 0.121 30 0.7510 0.134 35 0.7515 0.144 40 0.7620 0.157 45 0.7625 0.45 50 0.76Click here to return to the chapter.Practice Exercise 9.18Figure 9.51 shows a sketch of the titration curve. The two points beforethe equivalence pointV Tl = 2.5 mL, E = +0.109 V and V Tl = 22.5 mL, E = +0.169 Vare plotted using the redox buffer for Sn 4+ /Sn 2+ , which spans a potentialrange of +0.139 ± 0.5916/2. The two points after the equivalence pointV Tl = 27.5 mL, E = +0.74 V and V EDTA = 50 mL, E = +0.77 Vare plotted using the redox buffer for Tl 3+ /Tl + , which spans the potentialrange of +0.139 ± 0.5916/2.Click here to return to the chapter.MPractice Exercise 9.19The two half reactions areCe( aq) + e →Ce( aq)4+ − 3+4+2+U ( aq) + 2H O→ UO ( aq)+ 4H + ( aq)+2e2for which the Nernst equations are3+o 0.05916 CeE = E4+ 3+− log [ ]Ce / Ce4+1 [ Ce ]2−