# Osmosis with two different solutes

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1. Jan 8, 2015

### Mattteo

Hi all,

This is the situation: Solution A with 200 Osm urea, solution B with 200 Osm KCl, separated by a semipermeable membrane that is permeable to urea but impermeable to KCl. Each solution is in 1 L water. Calculate the equilibrium concentrations of urea and KCl.

The numerical answer actually isn't needed, but I was wondering about the driving forces at work here. As I understand, water movement is determined by total solute concentration. However, the concentration of urea should influence urea solute movement, which will in turn drive water movement to equalize the osmolarity. So, there seems to be two different forces at work: the solute diffusion of urea (dependent on urea concentration), and the diffusion of water (dependent on total solute concentrations). Both of these forces will affect the movement of water towards different equilibriums. Is this the right way to look at it? If it is, would the final equilibrium be a balance between these two forces, and thus hard to calculate?

Thank you
Matt

2. Jan 8, 2015

### Bystander

ab initio? Exact? Yes.
"Ideal solutions?" Falling off a log.
Data enough to apply various "non-ideal" models? Messy.
Recall that water activity depends primarily upon concentration(s) of dissolved solutes, not identities of the individual solutes.

3. Jan 9, 2015

### Mattteo

Hey thanks for the reply, but I'm still wondering if my analysis of the driving forces for equilibrium are correct. I.e. is there going to be urea solute diffusion activity that pushes the equilibrium in addition to the water activity, so the final equilibrium will be a balance between these two forces.

4. Jan 9, 2015

### Bystander

Your original problem statement is that there is.
Not as you've written it. Chemical activities of mobile components of the system are going to be identical for each component in every phase of the system accessible to them. "Forces" if you must use the term are that higher water activity in one phase will "force" water toward phases having lower water activity, and higher urea activity in a phase will "force" urea to phases with lower urea activity. Water activity and urea activity do not "balance."

5. Jan 9, 2015

### Staff: Mentor

Jut to keep things complete - add mass balances to what Bystander wrote (equality of activities) and you have described the system completely, all that is left is solving system of equations ;)

6. Jan 9, 2015

### Mattteo

Hmm... so in the final equilibrium, will total solute concentrations be equivalent on both sides, but urea concentrations will not be equivalent? (Since the two cases are incompatible).

7. Jan 9, 2015

### Bystander

That's the "ideal solution" approximation. Do you recall anything "different" to handling the KCl compared to handling urea in the calculation?

8. Jan 9, 2015

### Mattteo

Uh. Not really? The only difference I can think of is that KCl dissociates but since the units are Osm, wouldn't that be irrelevant. From what I'm getting, you're saying the movement of water is going to be driven solely by the total solute concentrations. Thus, though the urea concentrations are unequal at equilibrium, there is not going to be any net movement of urea, and thus no net movement of water either. The water is going to stay at the levels that equalize the total solute concentrations. I'm confused because you said "water activity and urea activity do not balance", but they seem to move towards different equilibriums (water towards the equilibrium to balance total solute concentrations, urea towards the equilibrium to balance urea concentrations). Thus, I feel they should balance or one "force" is more important. Does this mean that the urea diffusion is not going to significant affect water movement once the total solute concentrations are balanced?

I'm really just trying to get a picture of what is determining the equilibrium that is going to be reached because this is what I think would happen: urea will move across the membrane and drive subsequent water movement until the concentrations of urea are equal on both sides. However, at this point, the total solute concentration on each side will be different. This would then drive water movement to equalize the total solute concentration on each side. This water movement would result in unequal urea concentrations on both sides though, so urea would then diffuse to equalize its own concentrations. This would drive water movement that changes the total solute concentrations on both sides, forcing water to move to equalize the total solute concentrations, and so forth, so it never seems like one equilibrium is reached. There is never a case in which both total solute concentrations and urea concentrations are equal on both sides.

9. Jan 9, 2015

### Bystander

Oh, "Gibbs." This is for a bio-chem. course, isn't it. An incomplete problem statement that I filled in with the usual osmotic pressure demonstration apparatus. Is there any context in which the problem was presented/stated to you? Description of dialysis? Discussion of osmotic pressure? Anything? If there wasn't, we're going to have to set up a couple different approaches, and generate a couple different solutions.

10. Jan 9, 2015

### Mattteo

Yea sorry, I don't have an exact context to put this in. It's just part of the basic science/physiology review for medical school - osmotic pressure and solute transport. I don't see why we need different approaches though - shouldn't there be one conceptual explanation of what describes water and solute movement in a case like this? We can assume ideal solutions, conditions, etc, whatever. In class, we don't need to know anything beyond that fact that "urea and water will move towards solution B," but, as I noticed the apparent presence of two different things that could drive the system towards different equilibriums, I was wondering if there was an explanation of what affects solute and water movement in a case like this, with two different solutes, one permeable and one impermeable. I'm really not looking for a calculation or exact answer or anything, just a conceptual analysis.

11. Jan 9, 2015

### Bystander

Let's look at a couple different experimental arrangements: 1) U-tube with semi-permeable membrane at bottom, separating two solutions; 2) the mental picture behind the question given you to examine, which is just one solution in a bag made of a semi-permeable membrane sitting in a beaker containing a second solution.
The second situation is trivial since there is no difference in pressure, the urea solution will transport completely through the membrane to the KCl solution side of the membrane, either emptying the bag it's in (if it is the bagged solution), or filling/rupturing the bag originally containing a KCl solution, and the "equilibrium" state for such a system is 2 liters of 100 osm urea, 100 osm KCl solution.
The first situation allows the development of a pressure difference across the membrane, and is more interesting to examine. Water and urea move across the membrane into the KCl solution side to raise the urea activity on that side. Equilibrium is achieved when the urea activity (not necessarily concentration) is the same on both sides of the membrane, and when water activity is the same on both sides, keeping in mind that the water activity depends on solute concentrations and on the pressure difference across the membrane. At equilibrium, assuming ideal solution behavior, the urea activity is going to be slightly less than half it's initial activity and nearly the entire solution will have passed through the membrane into the KCl side, with a small amount of solution on the urea side that is a function of the pressure difference established across the membrane by experimental design, or by apparatus geometry (diameter and length of the U-tube). For all but very dilute solutions, or very high pressures, or very long (~ O(km)) tubes the volume of urea solution remaining on the urea side is going to be very small.

12. Jan 10, 2015

### Mattteo

Ah, thank you! That was exactly the kind of explanation I was looking for. I see what you mean by multiple approaches now. Thank you for your patience throughout this adventure haha. Just to check my understanding, let's say there was a 3rd situation, in which the two solutions are in a beaker separated by the semi-permeable membrane. Similar to the U-tube, the equilibrium will have equal water activity on both sides (meaning equal total solute concentrations) and unequal urea concentrations on each side (BUT equal urea activity). In the end, less than half of the original urea solution is going to be left on the original urea side as most of it moves towards the KCl side. There will be pressure forces at play too, but less so than in the U-tube.

13. Jan 10, 2015

### Bystander

You've got the essentials, methinks. Quantitative measurements and calculations of real, as opposed to ideal, solutions get a bit ugly but that's what chemists sign up to do.

14. Jan 10, 2015

### Mattteo

Ah, got it! Yea I'll leave the calculations to the pros haha. Thank you for all of the help! I appreciate it.