# 7.4: Separating the Analyte from Interferents

When an analytical method is selective for the analyte, analyzing samples is a relatively simple task. For example, a quantitative analysis for glucose in honey is relatively easy to accomplish if the method is selective for glucose, even in the presence of other reducing sugars, such as fructose. Unfortunately, few analytical methods are selective toward a single species.

In the absence of interferents, the relationship between the sample’s signal, Ssamp, and the concentration of analyte, CA, is

$S_\textrm{samp}=k_\textrm AC_\textrm A\label{7.9}$

where kA is the analyte’s sensitivity. If an interferent, is present, then Equation $$\ref{7.9}$$ becomes

$S_\textrm{samp}=k_\textrm AC_\textrm A+k_\textrm IC_\textrm I\label{7.10}$

where kI and CI are, respectively, the interferent’s sensitivity and concentration. A method’s selectivity for the analyte is determined by the relative difference in its sensitivity toward the analyte and the interferent. If kA is greater than kI, then the method is more selective for the analyte. The method is more selective for the interferent if kI is greater than kA.

Note

In Equation $$\ref{7.9}$$, and the equations that follow, you can replace the analyte’s concentration, CA, with the moles of analyte, nA when working with methods, such as gravimetry, that respond to the absolute amount of analyte in a sample. In this case the interferent also is expressed in terms of moles.

Even if a method is more selective for an interferent, we can use it to determine CA if the interferent’s contribution to Ssamp is insignificant. The selectivity coefficient, KA,I, which we introduced in Chapter 3, provides a way to characterize a method’s selectivity.

$K_\textrm{A,I}=\dfrac{k_\textrm I}{k_\textrm A}\label{7.11}$

Solving Equation $$\ref{7.11}$$ for kI, substituting into Equation $$\ref{7.10}$$, and simplifying, gives

$S_\textrm{samp}=k_\textrm A(C_\textrm A+K_\textrm{A,I}\times C_\textrm I)\label{7.12}$

An interferent, therefore, does not pose a problem as long as the product of its concentration and its selectivity coefficient is significantly smaller than the analyte’s concentration.

$K_\textrm{A,I}\times C_\textrm I \ll C_\textrm A$

If we cannot ignore an interferent’s contribution to the signal, then we must begin our analysis by separating the analyte and the interferent.