Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods485Next, we draw a straight line through each pair of points, extending theline through the vertical line representing the equivalence point’s volume(Figure 9.37d). Finally, we complete our sketch by drawing a smooth curvethat connects the three straight-line segments (Figure 9.37e). A comparisonof our sketch to the exact titration curve (Figure 9.37f) shows that they arein close agreement.Practice Exercise 9.18Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn 4+with 0.100 M Tl + . Both the titrand and the titrant are 1.0 M in HCl.The titration reaction is3+2+ 4+ +Sn ( aq) + Tl ( aq) → Sn ( aq) + Tl ( aq)Compare your sketch to your calculated titration curve from PracticeExercise 9.17.Click here to review your answer to this exercise.9D.2 Selecting and Evaluating the End pointA redox titration’s equivalence point occurs when we react stoichiometricallyequivalent amounts of titrand and titrant. As is the case with acid–baseand complexation titrations, we estimate the equivalence point of a complexationtitration using an experimental end point. A variety of methodsare available for locating the end point, including indicators and sensorsthat respond to a change in the solution conditions.Wh e r e is t h e Eq u i v a l e n c e Po i nt ?For an acid–base titration or a complexometric titration the equivalencepoint is almost identical to the inflection point on the steeping rising part ofthe titration curve. If you look back at Figure 9.7 and Figure 9.28, you willsee that the inflection point is in the middle of this steep rise in the titrationcurve, which makes it relatively easy to find the equivalence point when yousketch these titration curves. We call this a symmetric equivalence point.If the stoichiometry of a redox titration is symmetric—one mole of titrantreacts with each mole of titrand—then the equivalence point is symmetric.If the titration reaction’s stoichiometry is not 1:1, then the equivalencepoint is closer to the top or to bottom of the titration curve’s sharp rise. Inthis case we have an asymmetric equivalence point.Example 9.10Derive a general equation for the equivalence point’s potential when titratingFe 2+ with MnO 4 – .2+ − + 3+ 2+5Fe ( aq) + MnO ( aq) + 8H ( aq) → 5Fe ( aq) + Mn ( aq) + 4H O42We often use H + instead of H 3 O + whenwriting a redox reaction.