# 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 struck, so I want help from here on!!

P.S This was a post on the biology forum, a discussion of which I extended here. Also, I don't know if this is the right approach. Please do correct me if it isn't :)

• This looks like a tough physical chemistry problem. You may want to talk to a chemistry professor who is teaching physical chemistry. One other comment - if you let this system go long enough to reach equilibrium, the rate of diffusion of water and glucose across the membrane will be the same rate in both directions, even though the rates will be different from each other. This may be enough for you to find one or two more equations for your problem. If you find these additional equations, use a non-linear equation solver to get your answer. – David White Mar 5 '16 at 19:10
• So you suggest I copy paste this in chemistry stack exchange? Can that be done? As in I remember being reprimanded once for cross posting across fora! – Polisetty Mar 5 '16 at 19:14
• crossposted chemistry.stackexchange.com/questions/47463/… Indeed crossposting is frowned upon. Migration is normal solution, but only if needed. – Mithoron Mar 5 '16 at 21:32
• So which forum should I delete in now! This seems more like a chemistry post. Do I delete it here? – Polisetty Mar 5 '16 at 23:14