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# liquid chromatography column - Scott R.P.W.

Scott R.P.W. liquid chromatography column - John Wiley & Sons, 2001. - 144 p. Previous << 1 .. 7 8 9 10 11 12 < 13 > 14 15 16 17 18 19 .. 80 >> Next x , ,= ......................... do)
m(p) n!
THE SOLUTION OF THE DIFFERENTIAL EQUATION .
dx_
™t>)_ dv
First consider the conditions of the above equation when an initial charge of concentration X0(m) has been placed on the first plate of the column, but chromatographic development has not commenced.; that is v=0.
Then, Xm(p) = X0(m) when p=0 (i.e. the first plate)
and Xm(p) = 0 when p>0 (i.e. for any other plate in the column)
The first condition merely states that before the chromatographic development commences, the concentration in plate p=0 is that resulting from the injection of the sample on the column.The second condition states that the remainder of the column is free of solute.
Thus for plate p=0, and as there is no plate (p-1),
5^-x
dv m<°> dXmM _^) = dv
Am(0)
Integrating,
logeXo(m) = -v + constant
20
Now, when v = 0, X0(m) =¦ X0(m). Consequently, the constant = loge X0. logeXo(m) = -v + logeX0 or, X0(m) = Xoe^ ..............................................(A)
For Plate 1,
dv
— Xmro)-Xm(l) ......................................(B)
Substituting for Xm(0) from (A) in(B),
dv
Multiplying throughout by ev> dXn,o,
dXm(1)_Y „-v Y
———x0e
^Xm(t) ó _ V
+ xm(l)_xOe
e?+>U>v=^
dv
Now this equation can be recognized as the differential of a product, Hence,
dV2=x
dv
Integrating, ^m("e ~^°v+*<
Now, when v = 0, Őů( 1) = 0 , thus, ę = 0. Furthermore,
Xm(l)-X0e'vv
In a similar way it can be shown that,
Xm(2)—Xo
e'V
for Plate (2) m 12
21
for Plate (3)
Thus, for the nth Plate,
Equation (10) is the equation of prime importance that arises from the Plate Theory From this equation, it can be shown how the characteristics of the chromatogram can be used to determine the basic chromatographic properties of the column and phase system and how various other important chromatographic requirements can be calculated Equation (10) is a Poisson function, but it will be shown later that, if (n) is large, the function approximates very closely to a normal Error function or Gaussian function. Thus, since in all practical LC systems, (n) is always greater than 100, it would be expected from the conditions assumed in the derivation of the equation, that all chromatographic peaks will be Gaussian or nearly Gaussian in shape.
Elution Curves from Columns having 4, 9 and 16 theoretical
Plates
Figure 2
0.200 č
• n=4 n*9 « n=16
0.000
0
to
20
30
v (Plate Vol.)
22
The elution curves, calculated from equation (10), for a solute eluted from three columns having A, 9, and 15 plates respectively are shown in figure 3. With the exception of having a different number of theoretical plates all three columns have identical physical properties .
It is seen that as the number of plates increase (that is the column becomes longer) the peaks become more symmetrical and, as a result of greater dilution of the solute, the peaks are also reduced in height. It should be pointed out, however, that although the peaks from the column with the greater number of plates appears the broadest it will be seen later that the resolving power of a column depends on the ratio of the retention distance to the peak width and thus the column with the larger number of plates will provide the greater resolving power
The Retention Volume of a Solute
A chromatogram containing two peaks is shown in figure (4). The different dimensions along the axis of the chromatogram are described as functions of the volume of moving and stationary phases per plate, the distribution coefficients of the two solutes and the efficiency n. These functions will now be derived from equation (10) and it is appropriate to start with the simplest derivation, the retention volume Vr. The chromatogram shown in figure (4) will be employed again, later in the extension of the Plate Theory. Other characteristics will be discussed and the appropriate functions derived in order to label other dimensions on the chromatogram.
Figure 3
A Chromatogram Showing the Separation of Two Solutes
The retention volume of a solute is that volume of mobile phase that passes through the column between the injection point and the peak maximum It is therefore, possible to determine that volume by differentiating equation (10) and equating to zero and solving for (v),
Restating equation (10),
e-yvn
^m(n) \>’ p|
dXmtn) -e"vrve"nvM) dv 0 n!
-p-vv(n-l)
= Xo-^T-(n-v)
Equating to zero, n - v = 0 ,
or, v = n
This means that at the peak maximum, (n) plate volumes of mobile phase has passed through the column (remembering that the volume flow is measured in ‘plate volumes' and not ml). Thus, the volume passed through the column in ml will be obtained by multiplying by the 'plate volume', (vm + Kvs)
Thus, the retention volume (Vr > is given byr
Vr = n(vm 4 Kvs)
= nVm +nKVs
Now the total volume of mobile phase in the column, (vm), will be the volume of mobile phase per plate multiplied by the number of plates i.e. (nvm). In a similar manner the total volume of stationary phase in the column, (Vs) will be the volume of stationary phase per plate multiplied by the total number of plates, i.e. (nv)s, Previous << 1 .. 7 8 9 10 11 12 < 13 > 14 15 16 17 18 19 .. 80 >> Next 