Analytical Chem istry - DePauw University

464 Analytical Chemistry 2.0initialmoles Cd − molesEDTAadded M V − M V=total volumeV + V2+C Cd=( 500 . × 10 −3=M)(50.0 mL) −(0.0100 M)(5.0 mL)50.0 mL + 50 . mLCd Cd EDTA EDTACdEDTA= 364 . × 10 −3MTo calculate the concentration of free Cd 2+ we use equation 9.132+ −4[ Cd ] = α × C = ( 0. 0881)( 364 . × 10 M) = 321 . × 10 −4M2+CdCdwhich gives a pCd of2+ −4pCd=− log[ Cd ] =− log( 321 . × 10 ) = 3.49Step 4: Calculate pM at the equivalencepoint using the conditional formationconstant.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– .2+initialmoles Cd[ CdY2− Cd] = = M V Cdtotalvolume V + V× −CdEDTA3( 500 . 10 M)(50.0 mL)== 333 . × 10 −3M50.0 mL + 25.0 mLNext, we solve for the concentration of Cd 2+ in equilibrium with CdY 2– .At the equivalence point the initial molesof Cd 2+ and the moles of EDTA addedare equal. The total concentrations ofCd 2+ , C Cd , and the total concentrationof EDTA, C EDTA , are equal.Kf2−−3′′[ CdY ] 333 . × 10 − x= == 95 . × 10 14C C ( x)( x)CdEDTAx = C = 19 . × 10 −9MCdOnce again, to find the concentration of uncomplexed Cd 2+ we must accountfor the presence of NH 3 ; thus2+ −9[ Cd ] = α × C = ( 0. 0881)( 19 . × 10 M) = 170 . × 10 −102+CdCdMStep 5: Calculate pM after the equivalencepoint using the conditional formationconstant.and pCd is 9.77 at the equivalence point.After the equivalence point, EDTA is in excess and the concentration ofCd 2+ is determined by the dissociation of the CdY 2– complex. First, we calculatethe concentrations of CdY 2– and of unreacted EDTA. For example,after adding 30.0 mL of EDTA2+initialmoles Cd[ CdY2− Cd] = = M V Cdtotalvolume V + V× −CdEDTA3( 500 . 10 M)(50.0 mL)== 3.13× 10 −350.0 mL + 30.0mLM