Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods5330.200molNaOHL1 molC HNO6 5 3× 0.02876 L×mol NaOH139.11 gC HNO6 5 3× = 0 800 gC HNOmolC HNO6 5 3.6 5 30.800 gC HNO6 5 3× 100 = 38.%8 w/w CHNO2.006 gsampleClick here to return to the chapter.Practice Exercise 9.106 5 3The first of the two visible end points is approximately 37 mL of NaOH.The analyte’s equivalent weight, therefore, is0.1032molNaOHL1 equivalent× 0.037 L×molNaOH0.5000 gEW == 13 . × 10−338 . × 10 equivalentsClick here to return to the chapter.Practice Exercise 9.11= 38 . × 10 −32g/equivalentequivalentsAt ½V eq , or approximately 18.5 mL, the pH is approximately 2.2; thus,we estimate that the analyte’s pK a is 2.2.Click here to return to the chapter.Practice Exercise 9.12Let’s begin with the calculations at a pH of 10. Ata pH of 10 some of theEDTA is present in forms other than Y 4– . To evaluate the titration curve,therefore, we need the conditional formation constant for CdY 2– , which,from Table 9.11 is K f´ = 1.1 × 10 16 . Note that the conditional formationconstant is larger in the absence of an auxiliary complexing agent.The titration’s equivalence point requiresVeqof EDTA.M VCd Cd( 500 . × 10 −3M)(50.0 mL)= V = =EDTAM0.0100 MEDTA= 25.0Before the equivalence point, Cd 2+ is present in excess and pCd is determinedby the concentration of unreacted Cd 2+ . For example, after adding5.00 mL of EDTA, the total concentration of Cd 2+ ismL