Water Potential & Practice Problems
Name:_______________________________ Date:___________ Class:_____
7 Easy Steps to Understanding Water Potential
1) Log onto the PH School LabBench...copy and paste this link:
http://www.phschool.com/science/biology_place/labbench/
2) Click on Lab 1: Diffusion & Osmosis
3) From the left side of screen - select Concept 6: Water Potential
4) Go through Concepts 6-8, each explaining a different components of water potential
5) NOW, from the left side of screen...click on Exercise 3: Water Potential and Potato Cores
6) Complete Exercises 3 and 4...
7) TAKE THE LAB QUIZ...if you do well on it, and you can answer all of the practice questions in this
packet, you've got it. If not, see me for help!!!
NOW YOU SHOULD COMPLETE THE PRACTICE PROBLEMS FOUND IN THIS PACKET! Additional
help with this concept will be provided on an individual, AS NEEDED basis. BE AN ADVOCATE FOR
YOURSELF...if you don't get it, come see me. If you do get it, no need for any additional help. THIS
CONCEPT IS ABSOLUTELY ON THE TEST AND WILL ABSOLUTELY BE ON THE AP EXAM!
L. Carnes
Water potential (Ψ) is a measure of water’s potential to do work. In order to do work, an object must be able to apply
enough force to another object to cause displacement. In order for water to displace another object, water must be
moving. The largest water potential any volume of water can have, if only standard atmospheric pressure is being
applied to that volume of water, is defined as 0. This is the water potential for distilled water. Distilled water has the
greatest potential to move, and thus displace another object.
As solute is added to distilled water with no outside pressure being applied to it, the water potential of that solution
drops. But what does it mean to say that the water potential of a solution drops? It means that the water in that
solution is less likely to do work - in other words, it is less likely to move! Why is that? Well, as solute is added, the
chances become less and less that a concentration gradient can be set up between that solution and a second solution
that will favor the movement of water out of the initial solution.
In the example to the left, solute was added to the solution on
the left side of the membrane. This decreased the chances that
water would move out of the solution to the left of the
membrane and into a solution to the right of the membrane.
This means that the water on the left side of the membrane has
less potential to do work than the water on the right side. What
does this mean in terms of water potential? It means that the
solution to the left of the membrane has a more negative water
potential than the solution to the right of the membrane.
Therefore, water will flow from the right side of the membrane
to the left. Water always moves towards a more negative water
potential.
Let’s take a look at another example. The solute potential of a 0.1 M solution of distilled water and sucrose at 20º C at
standard atmospheric pressure is -0.23. If we continue adding sucrose to the solution until it reaches a concentration of
0.75 M at 20º C at standard atmospheric pressure, the solute potential continues to drop to a value of -1.87. Which
solution contains water that is less likely to do work? The one that has a higher concentration of solute and a lower
concentration of water! Think about it - if we separated a 0.1M solution of sucrose and a 0.75M solution of sucrose with
a selectively permeable membrane, which direction would the water move? Of course it would move from the 0.1M
solution into the 0.75M solution. In the process, it would be doing work! Remember, water always moves from an area
of higher water potential to an area of lower water potential.
Now that you think you’ve got water potential figured out, let’s complicate matters a little bit! Water potential (Ψ) is
actually determined by taking into account two factors - osmotic (or solute) potential (ΨS) and pressure potential (ΨP).
The formula for calculating water potential is Ψ = ΨS + ΨP. Osmotic potential is directly proportional to the solute
concentration. If the solute concentration of a solution increases, the potential for the water in that solution to undergo
osmosis decreases. Therefore, the more solute that is added to a solution, the more negative its osmotic (solute)
potential gets. If no physical pressure is applied to a solution, then the solute potential is equal to the water potential.
However, if physical pressure is applied to a solution, then it’s water potential (the potential for the water to move and
do work) will be affected. How it is affected depends upon the direction of the pressure.
How could pressure be applied to a solution? Let’s
look at another example! If a plant cell is placed
into distilled water, obviously water will move into
the cell because distilled water has a higher water
potential than the plant cell itself. However, when
the plant cell’s central vacuole fills with water, then
it will push back out on the water surrounding the
cell. The plant cell doesn’t burst due to this
pressure because it has a cell wall. An animal cell
in the same situation would burst. When the
pressure exerted outward on the water
surrounding the plant cell is equal to the osmotic
potential of the solution in the cell, the water
potential of the cell will be equal to zero. The
water potential of the plant cell will also be equal to the water surrounding it, and there will be no net movement of
water molecules. See page 768 through 770 in your textbook for other examples of the effects of applying physical
pressure on the water potential of solutions.
USEFUL FORMULAS
Formula for calculating water potential (Ψ)
 Ψ = ΨS + ΨP
Formula for calculating solute potential (ΨS)
 – iCRT, where:
o i = The ionization constant (two what degree will the substance produce ions in water):
 for NaCl this would be 2;
 for sucrose or glucose, this number is 1
o C = Molar concentration (experimentally determined)
 Iso-osmolar molarity: would allow you to place a potato in the solution and get no movement of
water.
 Point at which the line crosses 0 on the graph.
o R = Pressure constant = 0.0831 liter bar/mole K
o T = Temperature in degrees Kelvin = 273 + °C of solution
KEY POINTS TO REMEMBER
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Water always moves from [hi] to [low].
Water always moves from hypotonic to hypertonic.
[Solute] is related to osmotic potential.
[Pressure] is related to pressure potential.
Increased solutes decreases water potential.
Increased pressure raises water potential.
Always use ZERO for pressure potential in animal cells and open containers.
1 bar of pressure = 1 atmosphere.
PRACTICE PROBLEMS (Show ALL Work)
1. If a cell’s ΨP = 3 bars and its ΨS = -4.5 bars, what is the resulting Ψ?
2. The cell from question #1 is placed in a beaker of sugar water with ΨS = -4.0 bars. In which direction will the net
flow of water be?
3. The original cell from question # 1 is placed in a beaker of sugar water with ΨS = -0.15 MPa. We know that 1 MPa =
10 bars. In which direction will the net flow of water be?
4. The value for Ψ in root tissue was found to be -3.3 bars. If you take the root tissue and place it in a 0.1 M solution of
sucrose at 20°C in an open beaker, what is the Ψ of the solution, and in which direction would the net flow of water
be?
5. NaCl dissociates into 2 particles in water: Na+ and Cl-. If the solution in question 4 contained 0.1M NaCl instead of 0.1M
sucrose, what is the Ψ of the solution, and in which direction would the net flow of water be?
6. A plant cell with a ΨS of -7.5 bars keeps a constant volume when immersed in an open-beaker solution that has a ΨS
of -4 bars. What is the cell’s ΨP?
7. At 20°C, a cell containing 0.6M glucose is in equilibrium with its surrounding solution containing 0.5M glucose in an
open container. What is the cell’s ΨP?
8. At 20°C, a cell with ΨP of 3 bars is in equilibrium with the surrounding 0.4M solution of sucrose in an open beaker.
What is the molar concentration of sucrose in the cell?
9. What does increasing SOLUTE potential do to water potential?
10. What does increasing PRESSURE potential do to water potential?
Use this key to answer all the problems below. If you choose B or C, rewrite the statement so that it is complete and
true.
A = TRUE B = FALSE
C = NOT ENOUGH INFORMATION
PROBLEM: The initial molar concentration of the cytoplasm inside a cell is 2M and the cell is placed in a solution with a
concentration of 2.5M.
______1. Initially, free energy is greater inside the cell than outside.
______2. It is possible that this cell is already in equilibrium with its surroundings.
______3. Initially, solute concentration is greater outside the cell than inside.
______4. Water will enter the cell because solute potential is lower inside the cell than outside.
______5. The cell will become flaccid because the pressure potential is greater outside the cell than inside.
______6. The cell is already in equilibrium with its surroundings because of the combination of pressure potential and
solute potential inside and outside the cell.
______7. Initially, the cytoplasm is hypertonic to the surrounding solution.
______8. Initially, the numerical value of the solute potential is more negative inside the cell than outside.
______9. Net diffusion of water will be from inside the cell to outside the cell.
______10. At equilibrium, the molarity of the cytoplasm will have increased.
______11. At equilibrium, the pressure potential inside the cell will have increased.