Page 63 - Modern Analytical Chemistry

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46 Modern Analytical Chemistry

Solving either equation 3.11 or 3.12 for the amount of analyte can be accomplished
by separating the analyte and interferent before the analysis, thus eliminating the
term for the interferent. Methods for effecting this separation are discussed in
Chapter 7.
Alternatively, equations 3.11 or 3.12 can be solved for the amounts of both the
analyte and the interferent. To do so, however, we must obtain two independent
values for S meas . Using a concentration method as an example, gives two equations
S meas,1 = k A,1 C A + k I,1 C I + S reag,1

S meas,2 = k A,2 C A + k I,2 C I + S reag,2
that can be solved simultaneously for C A and C I . This treatment is general. The
composition of a solution with a total of n analytes and interferents can be deter-
mined by measuring n independent signals, and solving n independent simultane-
ous equations of the general form of equation 3.11 or 3.12.

3
EXAMPLE .3
A sample was analyzed for the concentration of two analytes, A and B, under
two sets of conditions. Under condition 1, the calibration sensitivities are
k A,1 = 76 ppm –1 k B,1 = 186 ppm –1
and for condition 2
k A,2 = 33 ppm –1 k B,2 = 243 ppm –1

The signals under the two sets of conditions are
S meas,1 = 33.4 S meas,2 = 29.7
Determine the concentration of A and B. You may assume that S reag is zero
under both conditions.
SOLUTION
Using equation 3.12, we write the following simultaneous equations
–1 –1
33.4 = (76 ppm )C A + (186 ppm )C B
–1 –1
29.7 = (33 ppm )C A + (243 ppm )C B
Multiplying the first equation by the ratio 33/76 gives the two equations as
–1 –1
14.5 = (33 ppm )C A + (80.8 ppm )C B
–1 –1
29.7 = (33 ppm )C A + (243 ppm )C B
Subtracting the first equation from the second gives
–1
15.2 = (162.2 ppm )C B
Solving for C B gives the concentration of B as 0.094 ppm. Substituting this
concentration back into either of the two original equations gives the
concentration of A, C A , as 0.21 ppm.