Page 220 - Modern Analytical Chemistry
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Chapter 7 Obtaining and Preparing Samples for Analysis 203
where C I is the concentration of interferent remaining after the separation, and
(C I ) o is the interferent’s initial concentration. The degree of separation is given by a
separation factor, S I,A , which is the change in the ratio of interferent to analyte separation factor
caused by the separation. 11 A measure of the effectiveness of a
separation at separating an analyte from
I /
CC A R I an interferent (S I,A ).
S I,A = =
( C Io ) /( C A o ) R A
7
EXAMPLE .10
An analysis to determine the concentration of Cu in an industrial plating bath
uses a procedure for which Zn is an interferent. When a sample containing
128.6 ppm Cu is carried through a separation to remove Zn, the concentration
of Cu remaining is 127.2 ppm. When a 134.9-ppm solution of Zn is carried
through the separation, a concentration of 4.3 ppm remains. Calculate the
recoveries for Cu and Zn and the separation factor.
SOLUTION
The recoveries for the analyte and interferent are
127 2 ppm
.
R Cu = =0 9891, or 98 91%
.
.
.
128 6 ppm
and
43 ppm
.
R Zn = =0 032, or 3 2%
.
.
134 9 ppm
.
The separation factor is
.
R Zn 0 032
S Zn,Cu = = =0 032
.
.
R Cu 0 9891
In an ideal separation R A = 1, R I = 0, and S I,A = 0. In general, the separation factor
–7
should be approximately 10 for the quantitative analysis of a trace analyte in the
–3
presence of a macro interferent, and 10 when the analyte and interferent are pres-
ent in approximately equal amounts.
Recoveries and separation factors are useful ways to evaluate the effectiveness
of a separation. They do not, however, give a direct indication of the relative error
introduced by failing to remove all interferents or failing to recover all the analyte.
The relative error introduced by the separation, E, is defined as
- *
S samp S samp
E = 7.14
*
S samp
*
where S samp is the expected signal for an ideal separation when all the analyte is re-
covered.
*
S samp = k A(C A) o 7.15
Substituting equations 7.12 and 7.15 into 7.14 gives
kC A o )
kC + K A,I ´ C I ) - A (
A (
A
E =
A (
kC A o )