# What happens when a semipermeable membrane allows only one solute to pass through?

I'm essentially a medical student where we deal a lot with osmosis. But when we are taught, it is done generally with only a single solute in consideration.

What if two different solutes are used on either side? The membrane being permeable only to one, not the other but also free for water to move.

Lets assume a beaker of base 10cm x 20cm, of any height. Add 2 litres of water to it, it would come to a height of 10cm. Dividing the beaker equally into two with the semipermeable membrane of the specifications, we'll be left with 1 litre on each side. Now add 1 osmole glucose on one side and 1 osmole NaCl on the other (you can't add only Na+ ions). Where glucose and water can move but NaCl wont.

I guess it'll be safe to assume that $\ x$ moles of glucose moves to the other side resulting in a change in concentration on both sides. Now the water would also move causing not only a change in concentration due to the change in volume distribution but also a change in hydrostatic pressure. Let the change in height be $\ y$ cm. Now the only constraint I can think of is that the pressure on either side should be equal at the base. By pressure i mean the sum of both hydrostatic and oncotic pressure. Oncotic pressure is given by $\ P_o = icRT$ and hydrostatic is by $\ P_h = h\rho g$.

Now I write $$\ P_{o1} + P_{h1} = P_{o2} + P_{h2}$$

That'd give $$\ (h+y)\rho g + \frac{n+x}{(h+y)A}RT = (h-y)\rho g + \frac{n-x}{(h-y)A}RT$$

Where $\ h$ is the height 10cm , $\ n$ is the no. of moles which we fixed as 1 mole (or rather, osmole) So rest all are constants and the only variables left here is $\ y$ and $\ x$ which we need. So rest all are constants and the only variables left here is $\ y$ and $\ x$ which we need. On solving, the equation comes to a cubic in $\ y$ which is actually not so desirable!! Also we need another equation, since it is an equation with two unknowns. So I showed all this since this is the way I would proceed. But I am now stuck.

P.S This was a post on the biology forum and then the physics forum too, a discussion of which I extended here.

• cross-posted physics.stackexchange.com/questions/241639/… – Mithoron Mar 5 '16 at 21:32
• Why do you think that glucose will contribute to the oncotic difference at all? As in why would it not be equilibriated in concentration on either sides, since there is no other force acting on it? – Satwik Pasani Mar 17 '16 at 8:15
• If you are willing to assume the concentration of the solution in any one partition is the same, then you cannot equate at the bottom. At the bottom, there has to be a net flux from the greqater height-lower height side, which is balanced at the topmost point of the semipermeable-contact by a flux in the opposite direction. You'll have to integrate along the height. (similar to the frank-starling capillary exchange, except that the exchange is balanced) – Satwik Pasani Mar 19 '16 at 12:40
• Now this is getting real complex! Could you post the solved answer @pasani – Polisetty Mar 22 '16 at 19:26
• I do not get your mindset. If the membrane is not permeable to the other species, then there is not need for equating chemical potentials for those species so the presence of them won't have influence at all (except by modify the density of the solution). – user1420303 Apr 4 '16 at 2:28

I am afraid you are stuck because you considered glucose as osmotically active. But you assumed that glucose can freely move across the membrane, which is the opposite of osmotic activity. So, short-short answer : ignore the glucose, it is an osmolyte, but it is not osmotically active on the membrane you are thinking of; its concentration is going to equilibrate on each side of the membrane : $$[\mathrm{C_6H_{12}O_6}] = \frac{1 \mathrm{mol}}{2 \mathrm{L}} = 0.5\, \mathrm{mol\, L^{-1}}$$ This solution of glucose is simply the solvent.
Also, you may have inverted terms in the equilibrium equation.

First we assume that the volume and density of the solution are unchanged by the solutes added (ideal solutions). We are left with $1 \,\mathrm{osmol\cdot L^{-1}}$ NaCl on one side of the membrane, and no osmolyte on the other side.
The equilibrium between hydrostatic and osmotic pressure ($i c R T$) then writes : $$2 \rho g y = i \frac{n}{(h+y)A} RT$$ Solving for $y$, we get $$2y = \sqrt{\frac{2inRT}{\rho g A}+h^2}-h$$ Using the following values : $i=2$, $n=0.5 \,\mathrm{mol}$, $R=8.3\,\mathrm{kg\cdot m^2\cdot s^{-2}\cdot K^{-1}\cdot mol^{-1}}$, $T=3\cdot 10^2\,\mathrm{K}$, $\rho=10^3\,\mathrm{kg\cdot m^{-3}}$, $g=9.8\,\mathrm{m\cdot s^{-2}}$ , $A=10^{-2}\, \mathrm{m^{-2}}$ and $h = 10^{-1} \mathrm{m}$, we obtain : $$2y \simeq 7\, \mathrm{m}$$ This value is actually unreachable given the volume of solvent and the geometry of your beaker. All the glucosed water would be drawn on one side, and equilibrium could not be reached.
The osmotic pressure of a solution containing $1\,\mathrm{osmol\cdot L^{-1}}=10^3 \,\mathrm{osmol\cdot m^{-3}}$ is approximately $\Pi = 2.5\, \mathrm{MPa} = 25 \, \mathrm{bar}$. To equilibrate such a pressure with hydrostatic pressure, you need a column of water of depth $z = \frac{\Pi}{\rho g} \simeq 260\,\mathrm{m}$.
The osmolarity of sea water is about $1\,\mathrm{osmol\cdot L^{-1}}$. In a desalination plant, sea water is purified by reverse osmosis using pressures from 50 to 80 bar. Osmotic equilibrium is of utmost importance in physiology, but this you must know better than me. Unless there is active transport, diffusion of solutes, or extraordinary high applied pressures (which may be relevant for deep-sea creatures), water is always going from the least to the most concentrated side of the membranes, so as to even out the osmolarities. Hope it helped.