Case13a

# Model - Coacervation

To understand the physics behind the coacervation method requires a brief introduction into the theory of polymer solutions. Consider a solution of positively and negatively charged polymers (polycations and polyanions respectively) of equal valence $Z$, and that no salts are present in the solution. It will be implicit that throughout all that follows, temperature $T$ and pressure $p$ of the solution are fixed. For simplicity, we assume that we have a symmetric solution, which means that the volume fraction of the polycations $\phi_+$ equals the volume fraction of the polyanions $\phi_-$, so that $$\phi_+=\phi_-=\frac{\phi}{2}$$

Note that this does not imply that the polymers are of equal length. What we will now do is make use of the so-called “lattice model” to represent our solution. In this type of model, the solution is represented as an infinitely large grid (we do not care what happens at the boundary), in which each square can be occupied by part of a polycation, polyanion, or by a single solvent molecule of volume $v$. This $v$ will be our basic unit of space in the lattice, and we define the effective chain length $N$ of the polymer to be its volume in units of $v$: $$N=\frac{V_{polymer}}{v}$$ In the graphical representation of our lattice model below, the sites occupied by polycations and polyanions are represented with + and - respectively, and the sites occupied by solvent are left empty.

Figure 1: Lattice model. In a lattice model, a solution is represented as a lattice of which the sites are occupied by any of the components of the solution. By calculating the energies of the interactions between different sites, this leads to an expression for the free energy of the whole system.

In this lattice model, one can now define interaction energies between all different neighboring pairs of sites, and calculate the energy of any configuration of the whole system (Figure 1). We are particularly interested in the Helmholtz Free Energy, since we have assumed constant temperature, pressure and volume. The full derivation of this energy ranges beyond the scope of this description, but the interested reader is encouraged to read references Doi 2013; Perry & Sing 2015; Veis 2011.

The Helmholtz free energy per lattice site $f(\phi)$ in units of $k_B T$ for a solution containing polycations and polyanions is given by what is known as the Voorn-Overbeek model (Qin et al. 2014; Qin & De Pablo 2016), and reads: $$f(\phi) = \frac{\phi}{N} \cdot \ln\left(\frac{\phi}{2}\right)+(1-\phi)\cdot \ln(1-\phi)-\alpha \cdot (\sigma \phi)^{3/2}$$ Here $\alpha$ is the interaction strength from the Debye-Hückel theory, and $\sigma \equiv Z/N$ is the charge density of the polymer. For aqueous solutions at room temperature, $\alpha \approx 4.1$. Now what the system will do, like any system, is go to a state in that locally minimizes its free energy. What remains to be seen however, is whether the energy minimum towards which the system will go is indeed a phase separated stated, or a homogeneous solution. It will be a homogenous solution if and only if the mixing criterion is satisfied, and this mixing criterion states that $$\frac{\partial^2 f}{\partial \phi^2 }>0$$ (see ref Doi 2013). There will only be coacervation if this criterion is not satisfied for our system. What we can do is plot the curve for which $\frac{\partial^2 f}{\partial \phi^2 }=f''(\phi)=0$, which is known as the spinodal curve, and identify regions in this plot in which coacervation does (not) happen. In fact there is also a region in the parameter space where both phases coexist. This is defined by the coexistence (binodal) curve, and a full treatment of the theory would also include this. However, as long as we are near the critical point of the spinodal curve (the minimum in the graphs of Figure 2), the coexistence curve approximates the spinodal curve and the coexistence region is negligibly small.

From here it will be immediately clear why the coacervation method works. Taking the derivative of $f$ twice with respect to $\phi$ yields that $$\frac{\partial^2 f}{\partial \phi^2} = \frac{1}{N\phi}+\frac{1}{1-\phi}-\frac{3\alpha\sigma^{3/2}}{4\sqrt{\phi}}$$ This means that the condition $f''(\phi)=0$ can also be written as $$\sigma_N (\phi)=\left(\frac{4\sqrt{\phi}}{3\alpha}\left(\frac{1}{N\phi}+\frac{1}{1-\phi}\right)\right)^{2/3}$$ Check for yourself that if the charge density of the polymers $\sigma$ is larger than the right hand side of this equation, coacervation will occur, whereas if it is less, homogeneous mixing will occur. The value of $\sigma_N(\phi)$ can be interpreted as the minimum charge density that a polymer duo of length $N$ and volume fraction $\phi$ need to have in order to coacervate.

Now remember what we are trying to do here: we are trying to show that for a fixed charge density, longer polymers in solution coacervate, whereas shorter ones do not. In other words, we want to show that by chopping up one of the polymers, the system will go from a coacervate state into a homogeneously mixed state. Full mathematical treatment ranges beyond our scope, so instead we will take a graphical approach. We start by plotting $\sigma_N(\phi)$ for a solution containing polymers of effective chain length $N=400$ (Figure 2A). Say that the used polymers have charge density $\sigma=0.135$ and are present in a concentration $\phi=0.008$. Our system is thus at the point $P$ in Figure 2A, which lies in the region where coacervation occurs. Now what we do is chop up all the polymers from a length of $N=400$ to a length of $N=100$, without making any other change to the solution. The polymers still have exactly the same charge density, and the volume fraction is also unaltered. In other words, our system stays at the exact same point $P$. When we plot the same graph now for $N=100$, we see that the point $P$ suddenly lies outside the coacervate region, and in the region where homogeneous mixing occurs (Figure 2B).

Figure 2: The spinodal curve shifts upon decreasing polymer length. The spinodal curve shifts into the direction of positive σ when the polymers in coacervates are chopped up into smaller pieces. Here plots of $\sigma(\phi)$ are shown for polymer lengths $400$ (A) and $100$ (B). Where a solution at point P would be in the coacervate region for polymers of length $400$, it would not in the case those polymers would be chopped up into a length of $100$.

So there we have the coacervation method qualitatively explained. It can easily be shown that this explanation is general for any values $N$, and that we did not conveniently choose $N=400$ and $N=100$ to meet our statement. We saw in Figure 2 that the coacervate region in the $\sigma(\phi)$ plot “moved upwards” as we chopped up the polymers. First we rephrase the statement that $\sigma(\phi)$ “moved upwards” with decreasing $N$ as: for a fixed $\phi$, the point $\sigma(\phi)$ will be higher if $N$ is lower. This is equivalent to saying that for arbitrary $N>0$, $$\left(\frac{\partial \sigma}{\partial N}\right)_\phi <0$$ The subscript "$\phi$" means that $\phi$ is held constant. This reduces our proof to the exercise of proving that the derivative of $\sigma$ with respect to $N$ is always negative. After some basic calculus, we find $$\frac{\partial \sigma}{\partial N}=- \frac{2}{3}\cdot \left(\frac{4\sqrt{\phi}}{3\alpha\phi}\right)^{2/3}\cdot N^{-2}\cdot \left(\frac{1}{N}+\frac{\phi}{1-\phi}\right)^{-1/3}<0$$ Although the expression in the middle of the above inequality looks complicated, the important result is that it is the product of positive terms with a minus sign in front, and thus we conclude that any point on the curve $\sigma(\phi)$ “moves up” (that is, in the positive $\sigma$-direction) with decreasing $N$ as it did in Figure 2.

The theory presented so far has another important implication. Opposite to what you might expect, coacervates form only within a range of polymer concentrations. More polymers does, after some point, decrease the amount of coacervates. This may sound very counterintuitive, but as you can see for example in Figure 3, taking the system to be at point P and increasing the volume fraction of polymers $\phi$ will have the system move out of the coacervate region.

Figure 3: Increasing the polymer volume fraction moves the system out of the coacervate region. Imagine the system to be at point P in parameter space, and now increase the polymer volume fraction by adding polymers of the same charge density. This brings the system to point Q, which lies out of the coacervate region, meaning that the system will no longer form coacervates.

Adding more polyelectrolytes to a solution thus does not mean that the system will move further in the coacervate region, but out of it. In fact, having too many polymers will cause coacervates to disappear! We can prove this by showing that $\sigma(\phi)$ only has one single minimum in the region $0 <\phi<1$. First note that the minima of $\sigma$ are given by $$\frac{\partial \sigma}{\partial \phi}=0$$ Taking this derivative and equating it to zero gives the equation $$\phi^2+\left(\frac{N+2}{N-1}\right)\phi-\left(\frac{1}{N-1}\right)=0$$ Of which the exact analytical solutions are given by $$\phi=- \frac{1}{2} \cdot \frac{N+2\pm\sqrt{N^2+8N}}{N-1}$$ As we expect $N \gg 2$, we can say $N\approx N+2 \approx N-1$ we can approximate our result as $$\phi \approx \pm \sqrt{\frac{1}{4}+\frac{2}{N}}-\frac{1}{2}$$ Because by definition $\phi$ cannot be negative, the only physically relevant solution, we define the critical concentration $\phi_c$ to be $$\phi_c=\sqrt{\frac{1}{4}+\frac{2}{N}}-\frac{1}{2}$$ Note that $\lim_{N\rightarrow\infty}[\phi_c]=0$, meaning that as the polymer size increases, the lower the critical polymer volume fraction is required to have coacervation (for a fixed charge density). At the same time it proves that there is only one single extremum of $\sigma(\phi)$ for positive $\phi$, and we already know that this is a minimum from examining the graph.

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