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196 Solutions Manual for Analytical Chemistry 2.1
 II = 0.17 – 0.00 = 0.17 ppb
 III = 0.00 ppb
 IV = 0.33 – 0.26 – 0.24 + 0.17 + 0.10 – 0.00 = 0.10 ppb
 V = 0.415 – 0.33 – 0.10 + 0.00 = –0.015 ppb
 VI = 0.24 – 0.17 – 0.10 + 0.00 = –0.03 ppb
 VII = 0.10 – 0.00 = 0.10 ppb
 Several of the concentrations have negative values, which, of course, 
is not possible; these values, which range from –0.03 to –0.01 suggest 
that concentrations of ±0.03 are the result of random error in the 
measurement process. 
 Based on our results, it appears that Cd2+ is present primarily as 
strong, labile organic complexes or labile metals absorbed on organic 
solids (Group II); that Pb2+ is present primarily as free metal ions and 
weak, labile organic and inorganic complexes (Group I), as strong, 
labile organic complexes or labile metals absorbed on organic solids 
(Group II), and as strong nonlabile inorganic complexes or as non-la-
bile metals absorbed on inorganic solids (Group VII); and that Cu2+ 
is present primarily as free metal ions and weak, labile organic and 
inorganic complexes (Group I), as strong, labile organic complexes or 
labile metals absorbed on organic solids (Group II), as weaker nonla-
bile organic complexes (Group IV), and as strong nonlabile inorganic 
complexes or as nonlabile metals absorbed on inorganic solids (Group 
VII).
32. Letting CCu represent the concentration of copper in seawater, we 
have the following two equations for the sample and the standard 
addition
. .k C26 1 25 0
20
0 mL
.00 mL
Cu #= & 0
. . . .k C38 4 25 0
20 5 00 25 000 mL
.00 mL ( ) mL
0.10 mLµMCu # #= +& 0
 Solving both equations for k and setting them equal to each other
.
.
.
. .
.
C C25 0
20
26 1
25 0
20 0 0 0200
38 4
0 mL
.00 mL
0 mL
0 mL µMCu Cu# #
=
+
. .C20 88 0 522 µM 30.72CCu Cu+ =
. .C9 84 0 522 µMCu=
 gives the concentration of copper as 0.0530 µM. he concentration 
of Cu2+ in mg/L, therefore, is
L
0.053 10 mol
mol
63.546 g
g
10 µg
3.37 µg/L
6 6
#
# # =
-