### The Osmosis EquationTen Examples

The osmosis equation is:

π = iMRT

π is not equal to 3.14159 in this situation. π stands for the osmotic pressure and is usually expressed in the pressure unit of atmospheres.

The definition of osmotic pressure:

the amount of pressure required to stop the process of osmosis in your experimental set-up.

The lowercase letter "i" is called the van 't Hoff factor and it will be dealt with in the problems below. It is named for Jacobus Henricus van 't Hoff (Henry to his friends), who applied PV = nRT to solutions and figured out why "i" was needed and what it represents.

He was awarded the first Nobel Prize in Chemistry in 1901 and the ChemTeam thinks this is the official portrait selected from the many pictures taken at the photo session. Love that hair! From the late 1870's to the turn of the century, van 't Hoff was one of the premier chemists in the world.

M is molarity: good old moles per liter.

R is the gas constant and we will be using the same value as in the gas laws unit: 0.08206 L atm/mol K. Now, you may ask what a "gas" constant is doing in a discussion of solutions. Well, for one thing it's called the "gas" constant because it was discovered in the course of research on gases.

Also, van 't Hoff's insight was to see that PV = nRT applied to molecules of solute moving though the solvent. (There is an article called How the Theory of Solutions Arose, which is about his insight. It is in the Classic Papers section of the ChemTeam.) In essence, the molecules of solute are a "gas," dispersed through the "universe" of solvent molecules. If I were to move V to the right side, I would get:

P = (n / V)RT

(n / V) is moles divided by liters and that is molarity.

P = MRT

T is temperature, measured as usual in Kelvins.

Time for some example problems. A little bit more discussion will be woven in and out of the solution to some of the examples.

Example #1: What is the osmotic pressure of a 1.00 M solution of sucrose at 25°C?

When we insert into the equation, we have:

π = (i) (1.00 mol/L) (0.08206 L atm / mol K) (298 K)

However, there are two unknowns: π, which is the one we want, and i. What is i?

Once again, i is called the van 't Hoff factor.

The van 't Hoff factor is a unitless, empirical constant related to the degree of dissociation of the solute.

WHAT IN THE WORLD DID HE JUST SAY???

OK, OK. The value is unitless. That means it is just a number like 1 or 2. Empirical means we must determine it by experiment. You can predict what a theoretical value for i might be, but the real value is only found in an experiment. The explanation follow shortly as to why.

The key is "degree of dissociation." This refers to the fact that some molecules ionize in solution (they split into their positive and negative ions) and other do not. This idea was put forth by Svante Arrhenius in 1884 in his Ph.D. dissertation and it was soundly rejected. (Arrhenius on electrolytic dissociation links to an excerpt from his article which announced this concept to the world.) In 1903, he was awarded the Nobel Prize in chemistry for it. Today, it's part of the common ordinary high school chemistry curriculum.

The van 't Hoff factor for sucrose is 1, since sucrose does not ionize in solution. It remains as whole molecules.

The answer, to three sig figs, is 24.4 atm.

Example #2: What is the osmotic pressure (at 25°C) of seawater? It contains approximately 35.0 grams of NaCl per liter. (Seawater contains other stuff, but we'll ignore it.)

1) Convert grams to moles:

35.0 g/L ÷ 58.443 g/mol = 0.599 mol/L

2) Now, plug into the osmosis equation:

π = (i) (0.599 mol/L) (0.08206 L atm / mol K) (298 K)

3) There's that pesky van 't Hoff factor. What is its value for NaCl?

When NaCl ionizes in solution it produces Na+ ions and Cl¯ ions. One mole of NaCl produces 1 mole of each type of ion. So the van 't Hoff factor is, theoretically, equal to 2. However, we will use 1.9 and I'll explain that in a moment.

4) Plug in 1.9 and solve:

π = (1.9) (0.599 mol/L) (0.08206 L atm / mol K) (298 K)

π = 27.8 atm (to three sig figs, the van 't Hoff factor is usally ignored for sig figs)

Why did I use 1.9 for the van 't Hoff factor for NaCl rather than 2?

This has to do with a concept called ion pairing. In solution, a certain number of Na+ ions and Cl¯ ions will randomly come together and form NaCl ion pairs. This reduces the total number of particles in solution, hereby reducing the van 't Hoff factor.

Example #3: Determine the molarity of an aspirin solution that produces an osmotic pressure of 0.0555 atm at 25 °C (i = 1).

Solution:

π = iMRT

0.0555 atm = (1) (x) (0.08206 L atm / mol K) (298. K)

π = 0.00227 mol/L (to three sig figs)

Aspirin (acetosalicylic acid) actually ionizes a tiny bit in solution. However, not enough to move the van 't Hoff factor off of the value of 1.

Example #4: A physiological saline solution is 0.92% NaCl by mass. What is the osmotic pressure of such a solution at a body temperature of 37 °C?

Solution:

1) Preliminary assumptions:

Assume 100. g of the solution is present. This means 0.92 g of NaCl are present.

Assume the solution density to be 1.00 g/mL. This means we have 100. mL of solution present.

2) Determine the molarity of the NaCl:

MV = mass/ molar mass

(x) (0.100 L) = 0.92 g / 58.443 g/mol

x = 0.15741833 M

2) Determine osmotic pressure:

π = iMRT

π = (1.9) (0.15741833 mol/L) (0.08206 L atm / mol K) (310. K)

π = 7.61 atm (to three sig figs)

3) Note the van 't Hoff value used. Be aware that your teacher may want you to use the theoretical value of 2. An answer of 8.01 atm results with a van 't Hoff factor of 2.

Example #5: The osmotic pressure of a 12.996 mM solution of CaCl2 is 549.588 mmHg at 22.10 °C. How many ions per formula unit of CaCl2 must there be? The compound may not be completely ionic. (Look up millimolar (symbol = mM) here.)

Solution:

12.996 mM = 12.996 millimole/L = 0.012996 mol/L

549.588 mmHg / 760.000 mmHg / atm = 0.7231421 arm

22.10 + 273.15 = 295.25 K

π = i M R T

0.7231421 arm = (i) ( 0.012996 mol/L) (R = 0.082057366 L atm / mol K) (295.25 K)

i = 2.297

"The compound may not be completely ionic."

This hint leads to predicting a value for i that is less than 3.000. It's pretty much a given that you are going to get ion pairing with sbstances that ionize in solution.

Example #6: To determine the molar mass of an unknown protein, 1.00 mg was dissolved in 1.00 mL of water. At 25.0 °C the osmotic pressure of this solution was found to be 1.47 x 10¯3 atm. What is the molar mass of this protein?

Solution:

1) Change mg/mL to g/L:

(1.00 mg /1.00 mL) (1 g / 1000 mg) (1000 mL /1 L) = 1.00 g/L

2) Determine the molarity of the solution:

π = iMRT

1.47 x 10¯3 atm = (1) (M) (0.08206 L atm mol¯11) (298 K)

M = 0.00006011316 mol/L

3) Determine the molar mass:

1.00 g/L / 0.00006011316 mol/L = 16635 g/mol

To 3 sig figs, this is 16600 g/mol

Example #7: What is the value of the van 't Hoff factor for a 0.64 molar aqueous solution of acetone?

Solution:

Acetone does not ionize in solution at any given concentration.

The van 't Hoff factor of ANY solution of acetone will be 1.

Example #8: Arrange the following solutions in order of decreasing osmotic pressure:

(a) 0.10 M urea
(b) 0.06 M NaCl
(c) 0.05 M Ba(NO3)2
(d) 0.06 M sucrose

Solution:

1) The van 't Hoff factor times the molarity will give us the molarity when all particles in solution are considered.

(a) (1) (0.10 M) = 0.10 M
(b) (2) (0.06 M) = 0.12 M
(c) (3) (0.05 M) = 0.15 M
(d) (1) (0.06 M) = 0.06 M

Example #9: The colligative properties of a 1.26-molar solution of calcium chloride has the same colligative properties as a _____ M solution of sucrose.

1) The van 't Hoff factor times the analytical molarity for calcium chloride is this:

(3) (1.26 M) = 3.78 M

2) The sucrose solution (with a van 't Hoff factor of 1) will need to be 3.78 M to have the same osmotic pressure as a 1.26 M solution of CaCl2

Example #10: 2 solutions, a 0.1% (m/v) albumin solution (compartment A) and a 2% (m/v) albumin solution (compartment B), are separated by a semi-permeable membrane. (Albumins are colloidal proteins).

(a) Which compartment will have the higher osmotic pressure?
(b) Which compartment will lose water?
(c) If some NaCl is added to compartment A, will the sodium and chloride ions stay in compartment A, or will some cross the membrane to compartment B?

Solution:

(a) compartment B due to the higher albumin concentration in B.

(b) compartment A. Since it has the lower albumin concentration, it has the higher water "concentration."

(c) As long as the membrane isn't selective against Na+ and Cl¯ ions, then they will flow to B so they attain equilibrium. Why can the Na+ and Cl¯ flow through the membrane but not albumin? Albumin is much lager than the NaCl ions. That's why in part B, compartment A loses water to compartment B instead of compartment B losing albumin to compartment A.