Some Complications. Above it is implicitly assumed that the
diffusion coefficient in the Stokes–Einstein relation (5.16) equals that in
Fickian diffusion. This is an oversimplification:
- In Eq. (5.17) and derived equations, activities rather than
concentrationsðcÞshould be used. If the activity coefficientsðgÞ
differ significantly from unity, this can be accounted for by
multiplying the diffusion coefficient by a factorð 1 þdlng=dlncÞ. - Individual ions (say, Ca^2 þ) cannot diffuse independently since
they must be accompanied by counterions (say, SO^24 or 2Cl), to
keep the solution electroneutral. (The distance between an ion and
its counterion will be on the order of the Debye length 1=k: see
Section 6.3.2.) Moreover, neutral species (say, CaSO 4 ) will also be
transported if they show a concentration gradient. The diffusion
coefficient of an ionizable component thus is a kind of average of
those of the species involved. - Equation (5.15) concerns self-diffusion, and mass transport
involves mutual diffusion: if the solute diffuses in one direction,
then the solvent does so in the opposite one. A kind of average
diffusion coefficient must be taken, and only for low solute
concentrations is it about equal to the self-diffusion coefficient of
the solute, taking the viscosity of the solvent. This is primarily
because solute concentration will generally affect the viscosity of
the solution; in most cases it is higher than that of the solvent. - Mutual diffusion may go along with a change in volume, since
many solute–solvent mixtures have a different volume (mostly a
smaller volume) than the sum of that of both components. This
implies that the frame of reference moves; for instance, the
original interface between two layers of different concentration
moves, and this means that some transport by flow occurs also.
Question 1
A way to study diffusion rate in liquids not hindered by convection is to separate the
liquids into two large stirred compartments by means of a disk of sintered glass.
Assume that the one component contains a 10%sucrose solution and the other
initially is water. Both compartments are stirred. The sintered glass disk is 3 cm in
radius and 3 mm in thickness; the void volume (i.e., available for the liquid) fraction
in the disk is 0.25. How much time would it roughly take for 1 gram of sucrose to
diffuse through the disk?