Analytical Chem istry - DePauw University

466 Analytical Chemistry 2.0Practice Exercise 9.12Calculate titration curves for the titration of 50.0 mL of 5.0010 –3M Cd 2+ with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7.Neither titration includes an auxiliary complexing agent. Compare yourresults with Figure 9.28 and comment on the effect of pH and of NH 3on the titration of Cd 2+ with EDTA.Click here to review your answer to this exercise.Sk e t c h i n g a n EDTA Ti t r a t i o n Cu r v eThis is the same example that we used indeveloping the calculations for a complexationtitration curve. You can review theresults of that calculation in Table 9.13and Figure 9.28.See Table 9.13 for the values.See the side bar comment on the previouspage for an explanation of why we areignoring the effect of NH 3 on the concentrationof Cd 2+ .To evaluate the relationship between a titration’s equivalence point and itsend point, we need to construct only a reasonable approximation of theexact titration curve. In this section we demonstrate a simple method forsketching a complexation titration curve. Our goal is to sketch the titrationcurve quickly, using as few calculations as possible. Let’s use the titration of50.0 mL of 5.0010 –3 M Cd 2+ with 0.0100 M EDTA in the presence of0.0100 M NH 3 to illustrate our approach.We begin by calculating the titration’s equivalence point volume, which,as we determined earlier, is 25.0 mL. Next, we draw our axes, placing pCdon the y-axis and the titrant’s volume on the x-axis. To indicate the equivalencepoint’s volume, we draw a vertical line corresponding to 25.0 mL ofEDTA. Figure 9.29a shows the result of the first step in our sketch.Before the equivalence point, Cd 2+ is present in excess and pCd isdetermined by the concentration of unreacted Cd 2+ . Because not all theunreacted Cd 2+ is free—some is complexed with NH 3 —we must accountfor the presence of NH 3 . The calculations are straightforward, as we sawearlier. Figure 9.29b shows the pCd after adding 5.00 mL and 10.0 mL ofEDTA.The third step in sketching our titration curve is to add two points afterthe equivalence point. Here the concentration of Cd 2+ is controlled by thedissociation of the Cd 2+ –EDTA complex. Beginning with the conditionalformation constantK′ 2−= [ CdY ]KY fCd C= α × = ×4−( 037 . )( 2.9 10 ) = 1.1×10[ ]f 2+EDTAwe take the log of each side and rearrange, arriving atlog log[ ] log [ 2−CdYK ′ =− Cd +]2+fC EDTA16 16C EDTA2pCd = logK′ + logf[−CdY ]