# Help understanding osmotic pressure and van't Hoff equation

I am trying to understand how to interpret the van't Hoff equation for movement across a semi permeable membrane.

Let's say we have a concentration of 100 mM sucrose on the inside of a semi permeable membrane vesicle. On the outside we have 200 mM sucrose.

Now we can say there is a concentration difference of 100 mM. We can easily solve the equation now.

$$π = iCRT$$

Concentration = 200 - 100 = 100 mOsm

$$R$$ = 0.0821

$$T$$ = 298 K

$$π$$ = 100 x 0.0821 x 298 = 2.45 atm

now this all makes sense to me. What I struggle with is that during the process of osmosis the water moves from one side of the membrane to the other.

Therefore, the final concentration after water has equilibrated will be changed right? How do I calculate the final concentration of solutes either side of the membrane? Conversely, if I now measure the concentration of solutes either side of the membrane after equilibration of water can I infer what the starting concentration was?

What kind of physical parameters would I need to know to do this?

• Water molecules move both ways across the semi-permeable membrane, but there is a higher flux of water molecules from a higher water concentration to a lower water concentration. Commented Apr 1 at 15:58

Broadly, matter tends to shift—whenever possible—to equilibrate its chemical potential.

(The chemical potential is the partial molar Gibbs free energy at constant temperature and pressure. The Gibbs free energy is the relevant energy potential when regions are in thermal and mechanical contact with their surroundings.)

The chemical potential $$\mu$$ is related to the molar concentration $$x$$ by

$$\mu=\mu_0+RT\ln \gamma x,$$

with reference zero $$\mu_0$$, gas constant $$R$$, temperature $$T$$, and activity coefficient $$\gamma$$.

For the case of an ideal solution (no interaction among solute and solvent molecules, termed Raoultian behavior), the activity coefficient is 1, and the interpretation is straightforward: There's a lower percentage of water in impure water, and so water tends to flow from pure water (100% water) if possible to raise this percentage. Here, the concentration alone can be used as a surrogate for the chemical potential, which is convenient because the former is easier to measure and control.

From the definition of the Gibbs free energy, the chemical potential is also modulated by the pressure $$P$$ as

$$d\mu=v\,dP,$$

with molar volume $$v$$.

Therefore, the final concentration after water has equilibrated will be changed right? How do I calculate the final concentration of solutes either side of the membrane?

You set the chemical potential to be equal on both sides for each species that can pass through the membrane.

Consider, for instance, the case of pure water and a Raoultian aqueous solution separated by a membrane permeable to water only, all at constant temperature. Assume constant material properties. Equilibration of the chemical potentials of water on either side yields

$$\mu_0=\mu_0+RT\ln (1-c)+v\Delta P,$$

with molar solute concentration $$c$$. Thus,

$$\Delta P=-\frac{RT}{v}\ln(1-c),$$

(By a Taylor series expansion, this is approximately $$\frac{cRT}{v}$$ for small $$c$$. This is a version of the van't Hoff equation.)

It’s straightforward to derive the slightly more general relation

$$\Delta P=-\frac{RT}{v}\ln\left(\frac{1-c_\mathrm{left}}{1-c_\mathrm{right}}\right)$$

for impure water on both sides, where $$\Delta P$$ is the pressure excess on the left side.

Now, how does pressure arise? From a pressure head, for example. From the Laplace pressure when interface curvature exists, relevant for vesicles. From elastic stretching of a membrane, also relevant for vesicles. So another step is to relate the geometry—which governs the amount of water on either side of the semipermeable membrane—to the individual pressures on either side and then to the pressure difference.

In this way, the equilibrium pressure difference tells you the equilibrium solvent volumes, and since the solute doesn’t move, you know the equilibrium concentrations.

Conversely, if I now measure the concentration of solutes either side of the membrane after equilibration of water can I infer what the starting concentration was?

Not generally; that information is lost. However, if you can constrain some of the starting conditions in some way (e.g., atmospheric pressure, room temperature, 100% purity on one side), then it might be possible to back out the starting concentration from, say, mass conservation.

Please let me know if anything’s unclear.