Water and Solute Permeability of Plant Cuticles: Measurement and ...

46 2 Quantitative Description of Mass Transfer
When working with weak electrolytes, Creceiver may be maintained at zero by using
a buffer as receiver in which the solute is fully ionised. Lipophilic solutes may be
scavenged and trapped in phospholipid vesicles, which maintains the concentration
of the aqueous phase of the receiver at practically zero. Rearranging (2.25) leads to
ln(Cdonor/C0)
t
= −PA
= −k. (2.26)
Vdonor
When ln (Cdonor/C0) is plotted vs time, a straight line is obtained with slope −k.
Equations (2.4)–(2.26) treat mass transfer as a first-order process (model 3), and k
is the first-order rate constant. Permeance (P) can be calculated when k, A and Vdonor
are known. Equation (2.26) states that Cdonor/C0 decays exponentially with time
Cdonor
C0
= e −kt , (2.27)
and once k has been determined experimentally, the donor concentration can be
calculated for any time interval. The time required for the donor concentration to
decrease to one half is
ln0.5 = −kt (2.28)
and the half time (t 1/2) is 0.693/k. A simple experiment will help to demonstrate
the use of the above equations.
2.5.1 The Experiment
Again, we want to study diffusion of 2,4-D across an isolated pepper fruit CM. It
is inserted into our transport apparatus (Fig. 2.1). As donor solution, a buffer (1 ml)
having pH 2.73 is used, and as receiver (10 ml) we use borax buffer with pH 9.2. At
this pH, 2,4-D is fully ionised, and the concentration of the non-ionised species is
always zero. The membrane area exposed to solutions is 1cm2 . The experiment is
started by adding 2,4-D to the donor solution at a total concentration of 2molm −3 .
At 24-h intervals the receiver solution is quantitatively withdrawn, replaced by fresh
buffer and analysed for 2,4-D. The solute fraction (Mt/M0) of 2,4-D which penetrated
into the receiver and was sampled from it is calculated and plotted against
time (Fig. 2.9a). It increases in a non-linear fashion (circles), while the fraction
lost from the donor decreases with time (squares). It took nearly 3 days for half of
the 2,4-D to penetrate into the receiver. Plotting ln (Cdonor/C0) vs time results in
a straight line with a slope of −0.25day−1 (Fig. 2.9b). This is evidence that our
mass transport of 2,4-D was in fact a first-order process. Next we calculate t1/2 as
ln 0.5/ −0.25day −1 , which is 2.77 days. Permeance is obtained using (2.26) and SI
units
P = kVdonor
A
=
�
2.89 × 10−6s−1 �� 1 × 10−6 m3� 1 × 10−4 m2 = 2.89 × 10 −8 ms −1 . (2.29)