Osmotic Pressure

Developed by: Eric Burtson
© American Association of Immunologists 1995

Background
In the lesson on Color and Acid Diffusion, we used colored dyes to see that chemicals diffuse, spreading themselves equally throughout a solution. We also saw that when there is a semipermeable membrane separating two different solutions, the chemicals that are small enough to fit through the membrane pores will diffuse into the neighboring solution. In osmosis., solvent molecules diffuse from a region of high solvent concentration across a membrane into a region of low solvent concentration. Without osmosis, immunologists could not perform dialysis on their antibodies. They would have to find a way of purifying and testing their antibodies without changing the solution.

Because osmosis involves the movement of tiny molecules, we cannot see it happen particle by particle. Nevertheless, we can see and measure its large scale effects. Suppose there are two solutions, A and B separated by a membrane. Solution A contains pure water. B is the same volume as A and contains water too, but it has a high concentration of a chemical that is too big to diffuse through the membrane. Pretend you are a water molecule in solution A. As you and your neighboring water molecules diffuse through solution A, you move in all directions. Those water molecules diffusing in the direction of the membrane easily pass through to get to solution B. If solution B were pure water like your solution A, an equal number of water molecules would pass from B to A. After a time, there would still be the same number of water molecules in each solution. But B is not pure water. As a result, fewer water molecules are able to pass from B to A because the big molecules in B take up some of the space through which some of the water molecules would have passed. The total volume of B continues to grow until the concentration of big molecules in B becomes small enough that it does not interfere with the diffusion of water across the membrane.

What would happen if you started with a small volume of B and placed it in a vertical tube above A? As before, osmosis would cause water molecules to leave A and accumulate in B. But while that happens, the volume of B would increase, causing the solution to rise higher and higher in the tube. Just like mercury in a barometer, the pressure at the bottom of the column of solution B would increase with the height. Eventually, the pressure of the solution pushing down on the membrane would force as many water molecules back into A as those that diffuse from A. At this equilibrium state we could then measure the height of the column to determine the pressure at the bottom of the column. This pressure is called osmotic pressure.

What determines the osmotic pressure of B? Think about what happens to B as water molecules diffuse into it. The volume of B increases while the number of moles of the big molecules stays the same; therefore, the concentration of big molecules decreases. So as the height of the column of solution B increases, two things are happening: The pressure increases and the concentration decreases. If we started with a lower concentration of big molecules in B, what would happen to the final height (and osmotic pressure) of the solution? It would go down. In short, osmotic pressure is proportional to concentration.

In this experiment you will calculate the osmotic pressure of a solution column of polyethylene glycol in water. In addition, you will determine the final PEG concentration having that osmotic pressure.

Each lab group will prepare and test a different concentration of 1.0 ml of PEG solution. You will prepare the column for osmosis by transferring the solution into a pipette and fixing the membrane in place.

You will calculate the osmotic pressure of the solution from the length (height) of the pipette and the density of water. You will determine the final concentration of the solution from the starting volume and concentration and from the total volume of the tube, which you can calculate by measuring the length and width of the tube.

Procedure

1. Make a sketch showing how osmosis takes place between the pure water and the PEG solution. Use small circles for water molecules and larger squares for PEG molecules. Draw arrows on the molecules to depict their direction as they diffuse. Show also the membrane and its holes.
2. Measure the length of the wide part of the pipette. Do Processing the Data problem number one.
3. Measure the inside diameter of the wide end of the pipette tube. Do Processing the Data problem number two.
4. Obtain the concentration of your PEG solution and prepare 1.0 ml of it in a 100 ml beaker. Label the beaker.
5. Fill a 500 ml beaker near the top with distilled water. Label it with your group name and the concentration of your PEG solution.
6. Obtain about a 1.5 cm long piece of semipermeable membrane. Wave it around in the pure water to rinse off any residue.
7. Have a partner hold the pipette upright and cover the narrow tip with his or her finger. Use another pipette to transfer the solution from the beaker to the pipette.
8. Place the membrane squarely over the wide end of the pipette and slide the eppendorf collar over the end until about a millimeter of the tubing is showing.
9. Tape the pipette to the side of the beaker so that the end is only about a half of a centimeter under water. Record your observations. Let it sit overnight.
10. Measure and record the height of your column above the water surface.
11. Add your data to the class table, Height (cm) versus Concentration. Copy the table.
12. Take a quick look at other group's columns; then clean up.

Processing the Data

1. Find the osmotic pressure of the solution whose concentration permits the column to rise to the top of (but not over) the wide portion of the pipette tube. You do not need to know the concentration to calculate what that pressure will be. As in a barometer, the pressure at the bottom of a fluid column is equal to the density of the fluid times the gravitational field times the height of the column:

   p = dgh

   where d is the density of the fluid, 1.0 g/cm3; g is the gravitational field, 9.8 N/kg (0.0098 N/g); and h is the height, which you measure in centimeters. Give your answer in pascals, Pa. Remember that one Pa = one N/m2, so you will have to convert your units of pressure from N/cm2 to N/m2.

2. Calculate the volume of the wide portion of your pipette. Give your answer in cm3. Recall from mathematics that V of a cylinder = pr2h or V = p(dia)2h/4. You will use the volume when you do your concentration calculation later.
3. Use the class data table to graph Height (cm) versus Concentration for the column of PEG solution.
4. Read from the graph what concentration would cause the column to rise to the exact height of the pipette tube body. What is that concentration?
5. Solve for the final concentration of the PEG solution from problem number four. Recall that when you dilute a solution, you can find its new molar concentration M from the equation M1V1=M2V2
6. In this case you are not using molar volume, but a concentration of g per ml of solvent. Use the symbol C to represent g/ml. As above, C1V1=C2V2

Extra Credit
In the system you used in the lab, if you had longer glass tubes you would have been able to measure the osmotic pressure of higher concentrations of PEG. Even so, we could get still higher solution columns if we could preserve the original concentration of PEG as water diffuses in. Think about how we could design a system that would keep the concentration close to the starting value as the osmotic pressure builds to a maximum. Sketch the system.

Teacher Notes
Chemistry Concept: Diffusion

1. With PEG 1500, a 1000 MWCO membrane works well. A membrane of 500 MWCO is either much too slow or gets jammed. You need to soak membranes about three hours before use. Keep them wet. If you have limited resources, you can experiment with materials. I have gotten good results using PEG 8500 with a 12K MWCO membrane, even though the molecular weight is smaller than that of the membrane limits.
2. To prepare collars to hold the membrane onto the pipette, cut the ends off 1.5 ml capacity eppendorf tubes. You can size the opening by inserting the pipette end and marking its penetration depth with a circle around the tube. Take out the pipette and cut the tube along the circle with a razor blade.
3. PEG saturates at a concentration of about 1.8 g per ml of water. This concentration would produce a higher column than the pipette length. Consider using a long tube as a demo.
4. The pipettes used are the 5¾ inch long kind. These have an inside diameter of 0.6 cm and a fat body length of about 10.5 cm. So, they hold a volume of about 3 cm.
5. Assign the following concentrations to ten different lab groups: 0.10, 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90, and 1.0 g PEG/ml water.