# 7.5: General Theory of Separation Effiiciency


The goal of an analytical separation is to remove either the analyte or the interferent from the sample’s matrix. To achieve this separation we must identify at least one significant difference between the analyte’s and the interferent’s chemical or physical properties. A significant difference in properties, however, is not sufficient to effect a separation if the conditions that favor the extraction of interferent from the sample also removes a small amount of analyte.

Two factors limit a separation’s efficiency: failing to recover all the analyte and failing to remove all the interferent. We define the analyte’s recovery, RA, as

$R_{A}=\frac{C_{A}}{\left(C_{A}\right)_{\mathrm{o}}} \label{7.1}$

where CA is the concentration of analyte that remains after the separation, and (CA)o is the analyte’s initial concentration. A recovery of 1.00 means that no analyte is lost during the separation. The interferent’s recovery, RI, is defined in the same manner

$R_{I}=\frac{C_{I}}{\left(C_{I}\right)_{o}} \label{7.2}$

where CI is the concentration of interferent that remains after the separation, and (CI)o is the interferent’s initial concentration. We define the extent of the separation using a separation factor, SI,A [(a) Sandell, E. B. Colorimetric Determination of Trace Metals, Interscience Publishers: New York, 1950, pp. 19–20; (b) Sandell, E. B. Anal. Chem. 1968, 40, 834–835].

$S_{I, A}=\frac{R_{I}}{R_{A}} \label{7.3}$

In general, an SI,A of approximately 10–7 is needed for the quantitative analysis of a trace analyte in the presence of a macro interferent, and 10–3 when the analyte and interferent are present in approximately equal amounts.

The meaning of trace and macro, as well as other terms for describing the concentrations of analytes and interferents, is presented in Chapter 2.

Example $$\PageIndex{1}$$

An analytical method for determining Cu in an industrial plating bath gives poor results in the presence of Zn. To evaluate a method for separating the analyte from the interferent, samples with known concentrations of Cu or Zn were prepared and analyzed. When a sample of 128.6 ppm Cu was taken through the separation, the concentration of Cu that remained was 127.2 ppm. Taking a 134.9 ppm solution of Zn through the separation left behind a concentration of 4.3 ppm Zn. Calculate the recoveries for Cu and Zn, and the separation factor.

Solution

Using equation \ref{7.1} and equation \ref{7.2}, the recoveries for the analyte and interferent are

$R_{\mathrm{Cu}}=\frac{127.2 \ \mathrm{ppm}}{128.6 \ \mathrm{ppm}}=0.9891 \text { or } 98.91 \% \nonumber$

$R_{\mathrm{zn}}=\frac{4.3 \ \mathrm{ppm}}{134.9 \ \mathrm{ppm}}=0.032 \text { or } 3.2 \% \nonumber$

and the separation factor is

$S_{\mathrm{Zn}, \mathrm{Cu}}=\frac{R_{\mathrm{Zn}}}{R_{\mathrm{Cu}}}=\frac{0.032}{0.9891}=0.032 \nonumber$

Recoveries and separation factors are useful tools for evaluating a separation’s potential effectiveness; they do not, however, give a direct indication of the error that results from failing to remove all the interferent or from failing to completely recover the analyte. The relative error due to the separation, E, is

$E=\frac{S_{s a m p}-S_{s a m p}^*}{S_{samp}} \label{7.4}$

where $$S_{samp}^*$$ is the sample’s signal for an ideal separation in which we completely recover the analyte.

$S_{samp}^{*}=k_{A}\left(C_{A}\right)_{\mathrm{o}} \label{7.5}$

Substituting equation 7.4.4 and equation \ref{7.5} into equation \ref{7.4}, and rearranging

$E=\frac{k_{A}\left(C_{A}+K_{A, l} \times C_{I}\right)-k_{A}\left(C_{A}\right)_{o}}{k_{A}\left(C_{A}\right)_{o}} \nonumber$

$E=\frac{C_{A}+K_{A, I} \times C_{I}-\left(C_{A}\right)_{\circ}}{\left(C_{A}\right)_{\circ}} \nonumber$

$E=\frac{C_{A}}{\left(C_{A}\right)_{\mathrm{o}}}-\frac{\left(C_{A}\right)_{o}}{\left(C_{A}\right)_{o}}+\frac{K_{A, I} \times C_{I}}{\left(C_{A}\right)_{o}} \nonumber$

leaves us with

$E=\left(R_{A}-1\right)+\frac{K_{A, I} \times C_{I}}{\left(C_{A}\right)_{o}} \label{7.6}$

A more useful equation is obtained by solving equation \ref{7.2} for CI and substituting into equation \ref{7.6}.

$E=\left(R_{A}-1\right)+\frac{K_{A, I} \times\left(C_{I}\right)_{o}}{\left(C_{A}\right)_{o}} \times R_{I} \label{7.7}$

The first term of equation \ref{7.7} accounts for the analyte’s incomplete recovery and the second term accounts for a failure to remove all the interferent.

Example $$\PageIndex{2}$$

Following the separation outlined in Example $$\PageIndex{1}$$, an analysis is carried out to determine the concentration of Cu in an industrial plating bath. Analysis of standard solutions that contain either Cu or Zn give the following linear calibrations.

$S_{\mathrm{Cu}}=1250 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cu}} \text { and } S_{\mathrm{Zn}}=2310 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Zn}} \nonumber$

(a) What is the relative error if we analyze a sample without removing the Zn? Assume the initial concentration ratio, Cu:Zn, is 7:1. (b) What is the relative error if we first complete the separation with the recoveries determined in Example $$\PageIndex{1}$$? (c) What is the maximum acceptable recovery for Zn if the recovery for Cu is 1.00 and if the error due to the separation must be no greater than 0.10%?

Solution

(a) If we complete the analysis without separating Cu and Zn, then RCu and RZn are exactly 1 and equation \ref{7.7} simplifies to

$E=\frac{K_{\mathrm{Cu}, \mathrm{Zn}} \times\left(C_{\mathrm{Zn}}\right)_{\mathrm{o}}}{\left(C_{\mathrm{Cu}}\right)_{\mathrm{o}}} \nonumber$

Using equation 7.4.3, we find that the selectivity coefficient is

$K_{\mathrm{Cu}, \mathrm{Zn}}=\frac{k_{\mathrm{Zn}}}{k_{\mathrm{Cu}}}=\frac{2310 \ \mathrm{ppm}^{-1}}{1250 \ \mathrm{ppm}^{-1}}=1.85 \nonumber$

Given the initial concentration ratio of 7:1 for Cu and Zn, the relative error without the separation is

$E=\frac{1.85 \times 1}{7}=0.264 \text { or } 26.4 \% \nonumber$

(b)  To calculate the relative error we substitute the recoveries from Example $$\PageIndex{1}$$ into equation \ref{7.7}, obtaining

$E=(0.9891-1)+\frac{1.85 \times 1}{7} \times 0.032= -0.0109+0.085=-0.0024 \nonumber$

or –0.24%. Note that the negative determinate error from failing to recover all the analyte is offset partially by the positive determinate error from failing to remove all the interferent.

(c) To determine the maximum recovery for Zn, we make appropriate substitutions into equation \ref{7.7}

$E=0.0010=(1-1)+\frac{1.85 \times 1}{7} \times R_{\mathrm{Zn}} \nonumber$

and solve for RZn, obtaining a recovery of 0.0038, or 0.38%. Thus, we must remove at least

$100.00 \%-0.38 \%=99.62 \% \nonumber$

of the Zn to obtain an error of 0.10% when RCu is exactly 1.

7.5: General Theory of Separation Effiiciency is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Harvey.