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

52 2 Quantitative Description of Mass Transfer
2. We use (2.9) and obtain the steady state fluxes of urea and sucrose across the cell
wall as 1.38 × 10 −3 and 5.23 × 10 −4 molm −2 s −1 respectively.
3. We solve this problem in four steps. (1) The amount of non-ionised 2,4-D in
the donor is the product of mass (0.1 l = 0.1 kg) and concentration of donor
(1 × 10 −3 moll −1 ), which is 1 × 10 −4 mol. (2) The concentration of 2,4-D in the
cuticle can be calculated using (2.12). The concentration of 2,4-D in the cuticle
is the product of the donor concentration (1 × 10 −3 molkg −1 ) and partition coefficient
(600), which is 0.6molkg −1 . (3) The area of cuticle exposed to the donor
was 1cm 2 and the thickness was 10µm, which results in a volume of cuticle of
1 × 10 −9 m 3 . With a specific weight of cuticle of 1,000kgm −3 , the mass of the
cuticle exposed to the donor is 1 × 10 −6 kg. The amount of 2,4-D in the cuticle
is the product of mass of cuticle times 2,4-D concentration in cuticle, which
amounts to 6 × 10 −7 mol. (4) According to Fig. 2.7 the solute concentration in
the cuticle during steady state is only half the maximum concentration. Hence,
in the steady state 1 × 10 −4 mol2,4-D are in the donor, while only 3 × 10 −7 mol
are sorbed in the cuticle. This is a negligible amount, and we can safely use the
donor concentration of 1 × 10 −3 moll −1 in calculating permeance. However, it
should be clear from the calculation that sorption in the cuticle would not have
been negligible if the donor volume had been only 1 ml or less.
4. From Fig. 2.9a we read the fractions of 2,4-D that would have been collected
from the receiver, had we taken only one sample after the second (0.39) or the
third day (0.53) respectively. Using (2.30) and these figures, we calculate the flow
(F) in 2 days or 3 days as 2.25 × 10 −11 and 2.04 × 10 −11 mols −1 respectively.
Before we can calculate P using (2.31), we must decide which donor concentration
to use. We are ignorant, and use the initial concentration of 1molm −3 and
obtain a P of 2.25 × 10 −7 ms −1 if we terminated the experiment after 2 days, and
2.04 ×10 −7 ms −1 if the experiment lasted 3 days. The true P calculated from the
rate constant k is 2.89 × 10 −7 ms −1 ; hence, we underestimated P by factors of
1.28 and 1.42 respectively. This may not seem a lot, but we must remember that
the 2,4-D concentration of the receiver was zero throughout, because we used a
pH of 9.2 and all 2,4-D molecules reaching the receiver were ionised. Without
this trick, ∆C across the membrane would have been much smaller and the error
in P much larger, the magnitude depending on the volume of the receiver.
5. We use (2.34) to solve this problem, and obtain 55 h and 2,273 days respectively.