Analytical Chem istry - DePauw University

314 Analytical Chemistry 2.0leaves us withK × CA,I IE = ( R − 1)+A7.18( C )A more useful equation is obtained by solving equation 7.14 for C I andsubstituting into equation 7.18.⎡ K × ( C ) ⎤E = R − +A,I I o( 1)× RAI⎣⎢C7.19( )A o ⎦⎥The first term of equation 7.19 accounts for the analyte’s incomplete recovery,and the second term accounts for failing to remove all the interferent.Example 7.11Following the separation outlined in Example 7.10, an analysis is carriedout to determine the concentration of Cu in an industrial plating bath.Analysis of standard solutions containing either Cu or Zn give the followinglinear calibrations.SCu−1= 1250 ppm × CAoCuSZn−1= 2310 ppm × C(a) What is the relative error if we analyze samples without removing theZn? Assume that the initial concentration ratio, Cu:Zn, is 7:1. (b) What isthe relative error if we first complete the separation, obtaining the recoveriesfrom Example 7.10? (c) What is the maximum acceptable recovery forZn if the recovery for Cu is 1.00 and the error due to the separation mustbe no greater than 0.10%?So l u t i o n(a) If we complete the analysis without separating Cu and Zn, then R Cuand R Zn are exactly 1 and equation 7.19 simplifies toKE =Cu,Zn× ( C )( C )CuoZn oUsing equation 7.11, we find that the selectivity coefficient isKCu,ZnZn−1= kZnppmk= 23101250 ppm= 185 .−1 CuGiven the initial concentration ratio, Cu:Zn, of 7:1, the relative errorif we do not attempt a separation of Cu and Zn is185 . × 1E = = 0. 264, or 26.4%7