I want to understand the forces involved in osmosis. If I have a molecule of water and one of a salt in the left side of a semi-permeable membrane, and a water molecule at the right side, what forces make the right side molecule travels left and remains there what forces make water molecules be more time at the left side of the membrane?

I read about a chemical potential difference between the initial and final states, but I can't understand why there is less energy in solution with more solvent. Why the impurities cause the chemical potential to reduce at the left side?

I understand that osmosis happen in other scenarios, like gases, therefore I discard think in the ionic solution or polarity of water molecules, etc.

On this site, I found:

The decrease in chemical potential occurs because there is a lower concentration of water in the solution than in the pure liquid. Statistically, fewer water molecules escape a solution into the vapor phase or freeze out onto the solid phase. That's why salt lowers the chemical potential of water in the solution.

But is still unclear to me why the salt has this effect on water molecules if their kinetic energy remains the same with or without the salt.

  • 2
    $\begingroup$ "what forces make the right side molecule travels left and remains there?" That's not what happens. Any individual water molecule diffuses around and may cross the membrane many, many times in any given period of time. $\endgroup$
    – DanielSank
    Oct 12, 2015 at 19:21
  • $\begingroup$ @DanielSank that is a very interesting comment! But what force makes statistically more molecules of water be at the left side of the membrane? $\endgroup$ Oct 13, 2015 at 18:41

2 Answers 2


Usually the driving force for diffusion is a concentration gradient, however that is not strictly true. It is actually a gradient in chemical potential which is the driving force for diffusional processes. For some simplifying assumptions this reduces again to concentration gradient as driving force.

For osmosis, the chemical potential is the driving force as the presence of any impurities cause the chemical potential to reduce compared to the pure side of the membrane. This is modeled by an equation of the form: $$\mu_i = \mu_{i,0} + RT\ln x_i$$ where $\mu_i$ is the chemical potential of species i, $\mu_{i,0}$ some reference state of chemical potential, $RT$ is the specific molar energy and most importantly $x_i$ is the mole fraction of solute $i$ in the solvent.

Now as $0\le x_i\le 1$, it follows that $\ln x_i<0$ and $\Delta\mu_i=\mu_i-\mu_{i,0}<0$, i.e. there is decrease in chemical potential in the solvent leading to diffusion of across the membrane.

  • $\begingroup$ Why the impurities cause the chemical potential to reduce? $\endgroup$ Oct 13, 2015 at 18:48
  • 2
    $\begingroup$ Well physically the addition of impurities increases the entropy. Since the chemical potential is related to the excess Gibbs free energy: $$dG_e=dH_e-TdS_e$$, increasing the entropy, decreases the excess Gibbs free energy. The impurities term $RT\ln x_i$ is a result of the excess entropy term. Is that intuitive enough? $\endgroup$
    – nluigi
    Oct 13, 2015 at 19:13
  • 1
    $\begingroup$ Well, not so much, because I understand the entropy increase, but not the relation between Gibbs free energy and reactions here. I'm thinking in how salt could affect the motion of water molecules, like when salt rises the boiling point of water. Is this related in some way? I see the math works, but I don't understand what Gibbs free energy or chemical potential are exactly modeling. That's why I proposed and example with only 3 molecules. $\endgroup$ Oct 13, 2015 at 20:28
  • 1
    $\begingroup$ @LeopoldoSanczyk Yes, it is slightly related, except that for BPE $\Delta T$ is proportional to $x_i$ rather than $\ln x_i$. However, that result is only valid for $x_i\rightarrow1$ and we can remedy this as $\lim_{x_{i}\rightarrow1}\ln x_{i}=x_{i}$. Back to the point; the presence of impurities has simply changed the energy in the system and the system will drive itself to minimum energy state (equilibrium). This energy change is quantified by the Gibbs' free energy and results in osmosis across the membrane. $\endgroup$
    – nluigi
    Oct 13, 2015 at 21:32
  • $\begingroup$ I understand that equilibrium is reached once osmosis occurs, but I'm still missing why was equilibrium lost by the presence of the salt molecule. The 2 water molecules were happy until I added a molecule of salt in the left side. Some interaction between the salt molecule and the water molecules take place to reduce the energy of this solution compared to pure water. Is salt molecule doing something to water molecule motions? $\endgroup$ Oct 13, 2015 at 22:24

John Grant Watterson claims in his paper "What drives osmosis?" that a fall in free energy, which is the rigorous thermodynamic criterion for a spontaneous change, cannot be the drive in osmotic processes:

Our models and theories require the introduction of a parameter that explicitly represents structure in liquids, which until now has had no place in the thermodynamic description of solutions. This lack is surprising, when one remembers that experimental results from the broad range of fields of colloid, clay and biological sciences have clearly established the marked effect of solutes on the structural properties of water, globally called ‘hydration phenomena’.

The introduction of such a parameter can help explain the direction in which energy flows during osmosis, which has been so puzzling to those of us interested in mechanism since the time of Pfeffer, more than a century ago. Further, elementary work cycles show, that changes in this parameter correspond to changes in the energy associated with solvent structure which can be used to produce useful work. The ability of osmotic systems to do work is familiar to all of us (indeed, a nuisance to many!), and is the basis of cytomechanics, i.e., the physical processes observed in the living cell. The fact that it still has no satisfactory explanation is clearly an urgent problem for us all.

In his article "Quantum Worlds", James Watson reviews osmotic theory of Watterson:

Osmosis is viewed as a direct result of the wave structure of water or, more specifically, of the structural aggregates of solvent molecules known as wave units or clusters. It is the structure wave itself, and not the solutes, that governs the molecular motions underlying osmosis. Since solvent can move through the semipermeable membrane, it can be considered as a single continuous medium pervading both phases. This means that the structure wave can pass unhindered from one phase to the other transferring structural energy in the process. Addition of solute breaks down the extent of solvent-solvent cooperative interactions because the molecules in contact with the solute can no longer rotate as freely as before. As a consequence, the wave length is shortened in the solution resulting in clusters smaller in size and energy but increased in number (concentration). In other words, the solute causes a decrease in the size of the pressure pixel. The increase in concentration of clusters in the solution phase is equal to the concentration of solute molecules.

At the membrane, there is a net energy flow from the energy-rich clusters of solvent into the smaller clusters of solution. This increases the tensile strength of the intermolecular bonds, so that the smaller clusters can pull solvent across the membrane increasing the pressure on the solute side. At equilibrium, the pressure in the solution has become high enough to counteract the pull of the smaller clusters and flow equalizes. At this point, the energy of the smaller clusters equals that of the pure solvent clusters.

Finally, Stephen Lower agrees with a quantum explanation in his textbook section "Some Applications of Entropy and Free Energy":

Dilution of a liquid creates uncountable numbers of new microstates, increasing the density of quantum states in the solution compared to that in the pure liquid. To the extent that these new states are thermally accessible, they will become populated at the expense of some of the microstates of the other phase.


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