Page 58 - Modern Analytical Chemistry

1400-CH03 9/8/99 3:51 PM Page 41

Chapter 3 The Language of Analytical Chemistry 41

SOLUTION
Since only relative concentrations are reported, we can arbitrarily assign
absolute concentrations. To make the calculations easy, let C Ca = 100 (arbitrary
units) and C Zn = 1. A relative error of +0.5% means that the signal in the
presence of Zn 2+ is 0.5% greater than the signal in the absence of zinc. Again,
we can assign values to make the calculation easier. If the signal in the absence
of zinc is 100 (arbitrary units), then the signal in the presence of zinc is 100.5.
The value of k Ca is determined using equation 3.2
100
S Ca
k Ca = = =1
C Ca 100
In the presence of zinc the signal is
S samp = 100.5 = k Ca C Ca + k Zn C Zn = (1)(100) + k Zn (1)

Solving for k Zn gives a value of 0.5. The selectivity coefficient, therefore, is
. 05
k Zn
K Ca Zn/ = = = 0.5
k Ca 1

Knowing the selectivity coefficient provides a useful way to evaluate an inter-
ferent’s potential effect on an analysis. An interferent will not pose a problem as
long as the term K A,I ´n I in equation 3.7 is significantly smaller than n A , or K A,I ´C I
in equation 3.8 is significantly smaller than C A.

3
EXAMPLE .2
5
Barnett and colleagues developed a new method for determining the
concentration of codeine during its extraction from poppy plants. As part of
their study they determined the method’s response to codeine relative to that
for several potential interferents. For example, the authors found that the
method’s signal for 6-methoxycodeine was 6 (arbitrary units) when that for an
equimolar solution of codeine was 40.
(a) What is the value for the selectivity coefficient K A,I when
6-methoxycodeine is the interferent and codeine is the
analyte?
(b) If the concentration of codeine is to be determined with an
accuracy of ±0.50%, what is the maximum relative concentration
of 6-methoxycodeine (i.e., [6-methoxycodeine]/[codeine]) that
can be present?
SOLUTION
(a) The signals due to the analyte, S A, and the interferent, S I , are

S A = k A C A S I = k I C I
Solving these two expressions for k A and k I and substituting into equation
3.6 gives
I /
SC I
K AI, =
A /
SC A