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116 Modern Analytical Chemistry

however, their concentrations and signals will no longer be equal. In this case, an
external standardization or standard addition results in a determinate error.
A standardization is still possible if the analyte’s signal is referenced to a signal
generated by another species that has been added at a fixed concentration to all
samples and standards. The added species, which must be different from the ana-
internal standard lyte, is called an internal standard.
A standard, whose identity is different Since the analyte and internal standard in any sample or standard receive the
from the analyte’s, that is added to all
same treatment, the ratio of their signals will be unaffected by any lack of repro-
samples and standards containing the
ducibility in the procedure. If a solution contains an analyte of concentration C A ,
analyte.
and an internal standard of concentration, C IS , then the signals due to the analyte,
S A , and the internal standard, S IS , are
S A = k A C A

S IS = k ISC IS
where k A and k IS are the sensitivities for the analyte and internal standard, respec-
tively. Taking the ratio of the two signals gives
S A k A C A C A
= ´ = K ´ 5.10
S IS k IS C IS C IS
Because equation 5.10 is defined in terms of a ratio, K, of the analyte’s sensitivity
and the internal standard’s sensitivity, it is not necessary to independently deter-
mine values for either k A or k IS .
In a single-point internal standardization, a single standard is prepared, and K
is determined by solving equation 5.10
æ C öæ S ö
IS
A
K = ç ÷ç ÷ 5.11
C øè IS
è A S ø stand
Once the method is standardized, the analyte’s concentration is given by
æ C IS öæ S A ö
C A = ç ÷ç ÷
è K øè S IS ø
samp
5
EXAMPLE .8
A sixth spectrophotometric method for the quantitative determination of Pb 2+
levels in blood uses Cu 2+ as an internal standard. A standard containing 1.75
ppb Pb 2+ and 2.25 ppb Cu 2+ yields a ratio of S A /S IS of 2.37. A sample of blood is
2+
spiked with the same concentration of Cu , giving a signal ratio of 1.80.
2+
Determine the concentration of Pb in the sample of blood.
SOLUTION
Equation 5.11 allows us to calculate the value of K using the data for the
standard
æ C öæ S ö 225
.
IS
A
K = ç ÷ç ÷ = ´237 =3 05
.
.
.
C øè IS
è A S ø stand 175
2+
The concentration of Pb , therefore, is
æ C IS öæ S A ö 225
.
.
C A = ç ÷ç ÷ = ´180 =1 33 ppb Pb 2+
.
.
è K øè S IS ø 305
samp