Analytical Chem istry - DePauw University

534 Analytical Chemistry 2.0−32+( 500 . × 10 M)(50.0 mL) −(0.0100 M)(5. 00 mL)[ Cd ] =50.0 mL + 500 . mL= 364 . × 10 −3 Mwhich gives a pCd of 2.43.At the equivalence point all the Cd 2+ initially in the titrand is now presentas CdY 2– . The concentration of Cd 2+ , therefore, is determined by thedissociation of the CdY 2– complex. First, we calculate the concentrationof CdY 2– .−32−( 500 . × 10 M)(50.0 mL)[ CdY ] =50. 0 mL + 250 . mL= 333 . × 10 −3Next, we solve for the concentration of Cd 2+ in equilibrium with CdY 2– .Kf2−−3′[ CdY ] 333 . × 10 − x= =+= 11 . × 10 162[ Cd ] C ( x)( x)EDTAMSolving gives [Cd 2+ ] as 5.50 × 10 –10 M, or a pCd of 9.26 at the equivalencepoint.After the equivalence point, EDTA is in excess and the concentration ofCd 2+ is determined by the dissociation of the CdY 2– complex. First, wecalculate the concentrations of CdY 2– and of unreacted EDTA. For example,after adding 30.0 mL of EDTA−32−( 500 . × 10 M)(50.0 mL)[ CdY ] =50. 0 mL + 300 . mL= 313 . × 10 −3( 0.0100 M)(30.0 mL) − (5.00× 10 −3M)(50. 0mL)C EDTA=50.0 mL + 30.0 mL= 625 . × 10 −4MSubstituting into the equation for the conditional formation constant andsolving for [Cd 2+ ] gives−3313 . × 10 M= 11 . × 102+ −4[ Cd ]( 625 . × 10 M)[Cd 2+ ] as 4.55× 10 –16 M, or a pCd of 15.34.The calculations at a pH of 7 are identical, except the conditional formationconstant for CdY 2– is 1.5 × 10 13 instead of 1.1 × 10 16 . The followingtable summarizes results for these two titrations as well as the results fromTable 9.13 for the titration of Cd 2+ at a pH of 10 in the presence of 0.0100M NH 3 as an auxiliary complexing agent.16M