1. Water Potential

2. Water potential??
Potential energy of water per unit area compared to pure water
Allows us to figure out where water will flow
Measured in Psi
Remember Posidon, Greek god of ocean had trident
Formula

3. Water potential What would happen? Pull water out of slug
Why
Outside of cell
Cl-
Na+
Cell Membrane
Inside of cell
NaCl immediately separates
Hydration shell formed around Na and Cl
Opens up areas and decreased H2O potential outside slug
Open area where H2O from slug can move into

4. We can measure H2O potential on either side of membrane
H2O will flow from an area of high water concentration (water potential) to and area of low concentration (water potential)
= -40 bar
= -5 bar
Water always moves from high to low water potential

5. Water always moves from high to low water potential
Water flow into roots vie osmosis
Lower water potential in steams
Lower yet in leaves
Lower more in atm
Water flows down water potential gradient

6. Equation = -5 bars Due to osmosis Pressure Potential Solute Potential
Water potential build on two things
Pressure potential
Physical squeezing
Solute potential
Water flowing through osmosis
= -5 bars
Due to osmosis
Solute potential is going to drop as I increase the # of solutes in the area
If I add two little bits of NaCl to it what will it do? Why?

7. Equation = 2 bars = -5 bars -3 bar Plant cell
Pressure potential and solute potential will determine if water will flow into or out of a cell
Pressure Potential
Solute Potential
= 2 bars
+ number; pressure is pushing water out of area
= -5 bars
Plant cell
Water continues to flow in and expand cell but because of cell wall cell wont explode
Cell wall will exert a pressure to the inside
Creates a pressure potential (measured in bar)

8. Solute potential equation
= -iCRT
i= ionization constant (factor between 1 and 2)
C= concentration (moles/ liter)
R= pressure constant ( liter/bar/mole K)
T = temperature in Kalvin (273 + oC)

9. CYU
The molar concentration of a sugar solution in an open beaker has been determined to be 0.2M. Calculate the solute potential at 22 degrees Celsius.
= - (1)(.2moles/liter)(.0831 liter/bars/mole K)(295K)
= - (1)(.2)(.0831 bars)(295)
= - (1)(.2)(.0831)(295) bars
= bars
= - 5 bars

10. What is the overall water potential?
We know that = -5 bars
In the problem it is given that our solution is in an open container
So that makes our what?
Zero
bar

11. AND DONE!