7.5: General Theory of Separation Efficiency

[ "article:topic", "Author tag:Harvey", "concentration", "analyte", "authorname:harveyd", "showtoc:no", "analytical separation", "Separation", "recovery", "separation factor" ]

The goal of an analytical separation is to remove either the analyte or the interferent from the sample’s matrix. To achieve this separation there must be at least one significant difference between the analyte’s and the interferent’s chemical or physical properties. A significant difference in properties, however, may not be enough. A separation that completely removes the interferent may also remove a small amount of analyte. Altering the separation to minimize the analyte’s loss may prevent us from completely removing the interferent.

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

$R_\textrm A=\dfrac{C_\textrm A}{(C_\textrm A)_\textrm o}\tag{7.13}$

where CA is the concentration of analyte remaining 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

$R_\textrm I=\dfrac{C_\textrm I}{(C_\textrm I)_\textrm o}\tag{7.14}$

where CI is the concentration of interferent remaining 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.12

$S_\textrm{I,A}=\dfrac{R_\textrm I}{R_\textrm A}\tag{7.15}$

In general, SI,A should be approximately 10–7 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.

Note

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 7.10

An analytical method for determining Cu in industrial plating baths gives poor results in the presence of Zn. To evaluate a method for separating the analyte from the interferent, samples containing known concentrations of Cu or Zn are prepared and analyzed. When a sample containing 128.6 ppm Cu is taken through the separation, the concentration of Cu remaining is 127.2 ppm. Taking a 134.9 ppm solution of Zn through the separation leaves behind a concentration of 4.3 ppm Zn. Calculate the recoveries for Cu and Zn, and the separation factor.

Solution

Using equation 7.13 and equation 7.14, the recoveries for the analyte and interferent are

$R_\textrm{Cu}=\mathrm{\dfrac{127.2\;ppm}{128.6\;ppm}=0.9891,\,or\;98.91\%}$

$R_\textrm{Zn}=\mathrm{\dfrac{4.3\;ppm}{134.9\;ppm}=0.032,\,or\;3.2\%}$

and the separation factor is

$S_\mathrm{Zn,Cu}=\dfrac{R_\textrm{Zn}}{R_\textrm{Cu}}=\dfrac{0.032}{0.9891}=0.032$

Recoveries and separation factors are useful for evaluating a separation’s potential effectiveness. They do not, however, provide 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

$E=\dfrac{S_\textrm{samp}-S_\textrm{samp}^\ast}{S_\textrm{samp}^\ast}\tag{7.16}$

where Ssamp is the sample’s signal for an ideal separation in which we completely recover the analyte.

$S_\textrm{samp}^\ast=k_\textrm A(C_\textrm A)_\textrm o\tag{7.17}$

Substituting equation 7.12 and equation 7.17 into equation 7.16, and rearranging

$E=\dfrac{k_\textrm A(C_\textrm A+K_\textrm{A,I}\times C_\textrm I)-k_\textrm A(C_\textrm A)_\textrm o}{k_\textrm A(C_\textrm A)_\textrm o}$

$E=\dfrac{C_\textrm A+K_\textrm{A,I}\times C_\textrm I-(C_\textrm A)_\textrm o}{(C_\textrm A)_\textrm o}$

$E=\dfrac{C_\textrm A}{(C_\textrm A)_\textrm o}-\dfrac{(C_\textrm A)_\textrm o}{(C_\textrm A)_\textrm o}+\dfrac{K_\textrm{A,I}\times C_\textrm I}{(C_\textrm A)_\textrm o}$

leaves us with

$E=(R_\textrm A-1)+\dfrac{K_\textrm{A,I}\times C_\textrm I}{(C_\textrm A)_\textrm o}\tag{7.18}$

A more useful equation is obtained by solving equation 7.14 for CI and substituting into equation 7.18.

$E=(R_\textrm A-1)+\dfrac{K_\textrm{A,I}\times(C_\textrm I)_\textrm o}{(C_\textrm A)_\textrm o}\times R_\textrm I\tag{7.19}$

The first term of equation 7.19 accounts for the analyte’s incomplete recovery, and the second term accounts for failing to remove all the interferent.

Example 7.11

Following the separation outlined in Example 7.10, an analysis is carried out to determine the concentration of Cu in an industrial plating bath. Analysis of standard solutions containing either Cu or Zn give the following linear calibrations.

$S_\textrm{Cu}=1250\textrm{ ppm}^{-1}\times C_\textrm{Cu}$

$S_\textrm{Zn}=2310\textrm{ ppm}^{-1}\times C_\textrm{Zn}$

(a) What is the relative error if we analyze samples without removing the Zn? Assume that the initial concentration ratio, Cu:Zn, is 7:1. (b) What is the relative error if we first complete the separation, obtaining the recoveries from Example 7.10? (c) What is the maximum acceptable recovery for Zn if the recovery for Cu is 1.00 and 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 7.19 simplifies to

$E=\dfrac{K_\textrm{Cu,Zn}\times(C_\textrm{Zn})_\textrm o}{(C_\textrm{Cu})_\textrm o}$

Using equation 7.11, we find that the selectivity coefficient is

$K_\textrm{Cu,Zn}=\dfrac{k_\textrm{Zn}}{k_\textrm{Cu}}=\mathrm{\dfrac{2310\;ppm^{-1}}{1250\;ppm^{-1}}=1.85}$

Given the initial concentration ratio, Cu:Zn, of 7:1, the relative error if we do not attempt a separation of Cu and Zn is

$E=\dfrac{1.85\times 1}{7}=0.264,\;\textrm{or}\;26.4\%$

(b) To calculate the relative error we substitute the recoveries from Example 7.10 into equation 7.19, obtaining

$E=(0.9891-1)+\left (\dfrac{1.85\times1}{7} \right )\times0.032$

$E=-0.0109+0.0085=-0.0024$

or –0.24%. Note that the negative determinate error due to the analyte’s incomplete recovery is partially offset by a positive determinate error from failing to remove all the interferent.

(c) To determine the maximum recovery for Zn, we make appropriate substitutions into equation 7.19

$E=0.0010=(1-1)+\dfrac{1.85\times1}{7}\times R_\textrm{Zn}$

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

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

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