1. What is Water Potential?

2. Water potential
the force responsible for movement of water in a system
Has the symbol psi
Is measured in bars or megapascals
                                  

3. It is a measure of the free energy of water which is less when it is has to surround solutes.
We are really stuck here, YUK!
Well I told you not to come here!
But, I HAVE to join the party!

4. Has two components:
Solute potential (also called osmotic potential) џs which is determined by solute concentration
Pressure potential џp which results from exertion of pressure on membranes/walls as water moves in or out; can be positive or negative

5. The water potential of pure water is given the value ZERO
Because pure water has the highest concentration of water molecules, and thus the highest water potential, the water potential of all other solutions must be lower than zero i.e. negative.

6. Pure water: = 0

7. Adding solute decreases water potential!
The more solute there is present in a solution the more negative it becomes.
So, solute potential will be a negative number if not pure water.
So hypertonic solutions have negative solute potentials.

8. water potential = solute potential + pressure potential
(s is pi on your paper)

9. Water moves from areas of higher water potential to areas of lower water potential (i.e. towards the more negative, concentrated region).  
water always "falls" from a high to a low water potential

10. This will occur until the water potential inside the cell equals the water potential outside of the cell.

11. ψp ψp
Pressure potential is important in plant cells because they are surrounded by a cell wall which, is strong and rigid.
When water enters a plant cell, its volume increases and the living part of the cell presses on the cell wall.
The cell wall gives very little and so pressure starts to build up inside the cell.
This has the tendency to stop more water entering the cell and also stops the cell from bursting.
When a plant cell is fully inflated with water, it is called turgid.
Pressure potential is called turgor pressure in plants) ψp ψp

12. pressure potential (ψp ) + solute (osmotic) potential (ψs)
Water potential (ψ ) =
pressure potential (ψp ) + solute (osmotic) potential (ψs)
Pressure potential (ψp):
In a plant cell, pressure exerted by the rigid cell wall that limits further water uptake
Solute potential (ψs):
The effect of solute concentration. Pure water at atmospheric pressure has a solute potential of zero. As solute is added, the value for solute potential becomes more negative. This causes water potential to decrease also.
*As solute is added, the water potential of a solution drops, and water will tend to move into the solution.

13. pressure potential (ψp ) + solute potential (ψs) (osmotic)
Water moves from a place of high water potential to a place of low water potential.
Water potential (ψ) =
pressure potential (ψp )
+ solute potential (ψs) (osmotic)
This is an open container, so the ψp = 0
This makes the ψ = ψs
The ψs =-0.23, so ψ is -0.23MPa, and water moves into the solution.

16. Can a solution with a molarity of 0
Can a solution with a molarity of 0.2 be in equilibrium with a solution with a molarity of 0.4?
YES!
Pressure
Two solutions will be at equilibrium when the water potential is the same in both solutions. This does not mean that their solute concentrations must be the same, because in plant cells the pressure exerted by the rigid cell wall is a significant factor in determining the net movement of water.

17. Solute (osmotic) potential (ψs )= –iCRT
The number of particles the molecule will make in water; for NaCl this would be 2; for sucrose or glucose, this number is 1
C =
Molar concentration
R =
Pressure constant = liter bar/mole K
T =
Temperature in degrees Kelvin (273 + °C) of solution
Yikes, what's that??????
Example Problem:
The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27°C degrees. Round your answer to the nearest hundredth. What is the water potential?
Answer:
-7.48
Solute potential
= -iCRT
= -(1) (0.3 mole/1) ( liter bar/mole K) (300 K)
= bar
Water potential
= , so water potential = -7.48

18. If this makes no sense whatsoever the key information to learn is:
The equation given
the water potential of pure water is zero
water moves from areas of higher water potential to areas of lower water potential (i.e. towards the more negative region)

19. 1. A solution in a beaker has NaCl dissolved in water with a solute potential of -0.5 bars. A flaccid cell is placed in the above beaker with a solute potential of -0.9 bars.
 p = 0 bars
a) What is the pressure potential of the flaccid cell before it was placed in the beaker?
 b) What is the water potential of the cell before it was placed in the beaker?
 w = p + s
X = 0 bars bars
X = -0.9 bars
c) What is the water potential in the beaker containing the sodium chloride?
X = 0 bars bars
X = -0.5 bars
d) How will the water move?
Water will enter the cell and leave the solution.

20. e) What is the pressure potential of the plant cell when it is in equilibrium with the NaCl solution outside?
Equilibrium means w of cell = w of solution.
wcell = pcell + scell
-0.9 bars = x bars
p of the cell = 0.4 bars
f) What is the cells final water potential when it is in equilibrium?
Same as the solutions w which = -0.5 bars
g) Is the cell now turgid/flaccid/plasmolysed?
turgid
h) Is the cell hypotonic or hypertonic with respect to the outside?
hypertonic
If it is hypo/hyper (choose one) tonic – this means that its water potential is higher/lower (choose one) than the outside.
Hypertonic means that the water potential is lower than the outside.

21. A solution in a beaker has sucrose dissolved in water with a solute potential of -0.9 bars. A flaccid cell is placed in the above beaker with a solute potential of -0.3 bars.
a) What is the pressure potential of the flaccid cell before it was placed in the beaker?
p = 0 bars since the cell is flaccid.
b) What is the water potential of the cell before it was placed in the beaker?
w = p + s
X = 0 bars bars
X = -0.3 bars
c) What is the water potential in the beaker containing the sucrose?
X = 0 bars bars
X = -0.9 bars
d) How will the water move?
Water will move out of the cell since it’s water potential value is smaller.

22. e) What is the pressure potential of the plant cell when it is in equilibrium with the sucrose solution outside? Think carefully – does the plant cell wall change shape?
p = 0 bars No
f) Also, what is the cell’s final water potential when it is in equilibrium?
-0.9 bars
g) Is the cell now turgid/flaccid/plasmolysed?
plasmolyzed
What is the cell’s solute potential when it is in equilibrium?
h) Is the cell hypotonic or hypertonic with respect to the outside?
Cell would be hypotonic
If it is hypo/hyper (choose one) tonic – this means that its water potential is higher/lower (choose one) than the outside.
Hypotonic means that its water potential value is higher than the outside.

23. So, what happens when a potato cell in put in pure water?
Water will move in or out until the wp of the cell will equals the wp surrounding the cell.
The pressure potential will increase to balance out the solute potential to equal to zero which is the wp of pure water.
No more net movement of water occurs

24. So how can we determine the water potential of potato cells?
We place potato cells in different molarities of sucrose. When enough solute is added outside of the potato cells to result in NO more NET movement of water, that is the molarity of the potato cells.

25. Units will cancel out to equal bars.
How do you go from molarity of a solution to the solute potential to figure the wp?
Another equation
solute potential = -iCRT
I = ionization constant (1 for sucrose)
C = molar concentration of sucrose (in this case where no net gain/loss of water occurs)
R = pressure constant ( liter/bars/mole 0K for sucrose)
T = temperature Kelvin (273 + C)
Units will cancel out to equal bars.

26. So what is the solute potential of a 0.1 M solution of sucrose at 22 C?
Solute potential = -iCRT
i (ionization constant) = 1
R = (from handbook)
T = temp K (273 + C of solution)
Ωs = - (1) (0.1) (0.0831) (295)
Ωs = bars

27. Calculate the water potential of a solution of 0. 15 M sucrose
Calculate the water potential of a solution of 0.15 M sucrose. The solution is at standard temperature (273K) and pressure (.0831 L bars/mol K).
w = p + s
X = 0 bars + [-iCRT]
X = 0 bars + [-(1) (0.15 mol/L) ( L bars/mol K) (273K)
X = 0 bars bars
w = bars

28. So you will graph the results of the change in weight of the potato cells in different molarity solutions of sucrose after overnite.
Where the line crosses the graph at the X axis, representing no gain or loss of water, will be the molarity of the potato cells.
Then substitute in the equation for solute potential (-iCRT)

29. So, how do you get the water potential?
Once you determine the solute potential, plug into the equation to determine the water potential. The pressure potential will be zero since water is at equilibrium (no net movement in or out.