1. Water Potential

2. Ψ = Ψp + Ψs Water potential = pressure potential + solute potential
*Bozeman science video

3. Water Potential
Water Potential - In osmosis, the tendency for a system (a cell or solution) to take up water from pure water, through a differentially permeable membrane.
Water will ALWAYS move from a region of HIGH water potential to an area of LOW water potential.

4. Ψp (PRESSURE POTENTIAL)
An increase in pressure INCREASES water potential
Normal atmospheric pressure:
Ψp = 0

5. Ψs (SOLUTE POTENTIAL) Relative concentration of solutes
Addition of solutes LOWERS water potential *Make MORE NEGATIVE!

6. Ψp and Ψs are inversely related
Water Potential
Ψp and Ψs are inversely related
As pressure increases, Ψp increases
As solutes increase, Ψs decreases
(value become more NEGATIVE)

8. Finding Ψs Solute potential = –iCRT
i = The number of particles the molecule will make in water; for sucrose or glucose, this number is 1
(NaCl= Na+1 and Cl-1 so = 2 particles)
C = Molar concentration (from your experimental data)
R = Pressure constant = liter bar/mole K
T = Temperature in degrees Kelvin = °C of solution

9. Finding Ψs
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.

10. Problem #1
The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27 degrees. Round your answer to the nearest hundredth.
Ψs = -(1)(.3 mole/liter)(0.0831)(273+27)
Ψs = -7.48

11. -2.43
bars
Problem #2
-3.3 bars
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?
Ψ s= -iCRT
-(1)(0.1)(0.0831)(293)= bars Yp=0, so Y=-2.43.
The movement will be into the cell. Higher to lower.

12. Problem #3
-4.86
bars
-3.3 bars
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?
-(2)(0.1)(0.0831)(293) + 0= Ψ= -4.86
Into the environment