Sedimentation Equilibrium of Mixtures of Charged Colloids

with Q(x) = ρ ˜ + (x) − ρ ˜ − (x) − ∑ N∫ i=1 Z i ˜ρ i (x) is the dimensionless total chargedensity, and φ(x) = −2πλ B dx ′ Q(x ′ )|x − x ′ | the dimensionless potential,equivalent to (B.1)-(B.4). This functional will be the equilibrium grand potentialfor the equilibrium distribution ρ i,α (x) that satisfyδΩ[ ρ˜i,α (x)]∣= 0 (4.2)δ ˜ρ i,α (x)∣∣˜ρi,α (x)=ρ i,α (x)with ρ i (x) and ρ α (x) the equilibrium densities of the components, micro-ionsrespectively.The first two terms on the right hand side of equation 4.1 are ideal gascontributions, for the colloids and the salt ions respectively. The third termrepresents the external potential of the earth’s gravity field, interacting withthe colloids, but not with the much lighter co-ions. The fourth term accountsfor the coulomb interactions between all charges. For simplicity a mean fieldapproximation is made. In this way the particles behave like an ideal gas insome self-consistent electric potential φ(x).By applying condition(4.2) one finds the following equations for the unknownρ ± and ρ i (x):ln ρ ± (x)Λ 3 ± − βµ ± ± φ(x) = 0,with µ ± = kT ln ρ s Λ 3 sln ρ i (x)Λ 3 i − βµ i − Z i φ(x) + x L i= 0, with µ i = kT ln a i Λ 3 i ,with a i ≡ 1 Λ 3 ie βµ iis called the activity, of fugacity of colloidal component i.Thanks to the salt reservoir, we can fix the chemical potential of the saltµ ± = µ s , and the equilibrium density profiles are found to satisfy:ρ i (x) = a i exp ( − x L i+ Z i φ(x) ) (4.3)ρ ± (x) = ρ s exp ( ∓ φ(x) ) (4.4)Now, by observing that d 2 φ(x)/dx 2 = −4πλ B Q(x), and using the solutions(4.3) and (4.4) we arrive at the Poisson-Boltzmann-equation:d 2dx 2 φ(x) = κ2 sinh φ(x) + 4πλ Bn∑Z i ρ i (x) (4.5)with κ −1 the Debye screening length of the reservoir. The suspension isglobally charge neutral, which is provided by the applied Neumann conditionsφ ′ (x) = 0 at the boundaries of the suspension (x = 0 and x = H). From (4.3)and (4.5) we can solve φ(x) and ρ i (x) numerically on an x-grid, i.e. by using17i=1