Analytical Chem istry - DePauw University

352 Analytical Chemistry 2.0Fe 2+p(phen)13logβ 3 = 6.9Fe 3+Fe(phen)2+ 313logβ 3 = 4.6Fe(phen) 33+Figure 7.34 Ladder diagram forPractice Exercise 7.7.3Fe + 3( aq) + 3phen( aq) Fe( phen)+ ( aq)β = 6×10where phen is an abbreviation for 1,10-phenanthroline. Because b 3 islarger for the complex with Fe 2+ than it is for the complex with Fe 3+ ,1,10-phenanthroline will bind Fe 2+ before it binds Fe 3+ . A ladder diagramfor this system (Figure 6.34) suggests that an equilibrium p(phen) between5.6 and 5.9 will fully complex Fe 2+ without any significant formationof the Fe(phen) 3 3+ complex. Adding a stoichiometrically equivalentamount of 1,10-phenanthroline to a solution of Fe 2+ will be sufficient tomask Fe 2+ in the presence of Fe 3+ . Adding a large excess of 1,10-phenanthroline,however, will decrease p(phen) and allow the formation of bothmetal–ligand complexes.Click here to return to the chapter.Practice Exercise 7.8(a) The solute’s distribution ratio between water and toluene is1mol10.889 g× ×[ S ]org117.3 g 0.00500 LD = =[ S ]1mol1aq(. 1 235 g− 0.889 g) × ×117.3 g 0.01000 L3313= 514 .(b) The fraction of solute remaining in the aqueous phase after one extractionis( q )aq1Vaq=DV + VorgThe extraction efficiency, therefore, is 72.0%.(c) To extract 99.9% of the solute requiresaq20.00 mL== 0.280( 514 . )( 1000 . mL) + 20.00 mL⎛ 20.00 mL ⎞( Q aq) n= 0.001=⎝⎜( 514 . )( 1000 . mL) + 2000 . mL⎠⎟ = 0.280nna minimum of six extractions.Click here to return to the chapter.Practice Exercise 7.9log( 0. 001) = nlog( 0. 280)n = 54 .Because the weak base exists in two forms, only one of which extracts intothe organic phase, the partition coefficient, K D , and the distribution ratio,D, are not identical.