11/15/2012
Water Potential
Water Potential (Ψ)
• The potential of water to do work
– The more free water molecules there are in a
solution, the more work can be done by water
– Adding solute to a solution decreases the amount
of free water so it decreases the water potential
– Hypotonic solutions have greater water potential
than hypertonic solutions
– Water flows from a solution with high water
potential to a solution with low water potential
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Water Potential (Ψ)
• Measured in megapascals (MPa)
– Unit of pressure
– 1 MPa = 10 bar = 10 atmosphere
• Pure water in an open container has a Ψ = 0
– Any solute added to a solution causes its Ψ to
become more negative
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Calculating Water Potential
• Ψ = ΨS + ΨP
– ΨS = solute potential (osmotic potential)
– ΨP = pressure potential
• Physical pressure applied to the solution
• Solution in an open container has a ΨP = 0
• As water enters a plant cell ΨP increases
Calculating Solute Potential
• ΨS = -iRCT
i=
The number of particles the molecule will
make in water; for NaCl i=2; for sucrose or
glucose, i=1
C=
Molar concentration (from experimental
data)
R=
Pressure constant = 0.0831 liter bar/mole
K
T=
Temperature in degrees Kelvin = 273 + °C
of solution
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Sample Problems
1a. 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.
-7.48 bars = -0.748 MPa
1b. What is the water potential for this example? (In an open
beaker, Ψp = 0.) Round your answer to the nearest hundredth.
-7.48 bars = -0.748 MPa
Sample Problems
2a. Calculate the water potential of a solution of 0.15M sucrose
in a beaker at 20oC. (In an open beaker, Ψp = 0.)
-3.65 bars = -0.365 MPa
**Omit 2b & 2c
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Sample Problems
3a. In beaker B, what is the water potential of the distilled water in the
beaker, and of the beet core?
Distilled Water = 0
Beet Core = -0.2
3b. Which of the following statements is true for the diagrams above?
a. The beet core in beaker A is at equilibrium with the surrounding water.
b. The beet core in beaker B will lose water to the surrounding environment.
c. The beet core in beaker B would be more turgid (rigid) than the beet core in
beaker A.
d. The beet core in beaker A is likely to gain so much water that its cells will rupture
e. The cells in beet core B are likely to undergo plasmolysis.
Sample Problems
4. You measure the total water potential of a cell and find it to be
-0.24 MPa. If the pressure potential of the same cell is 0.46 MPa,
what is the solute potential of that cell?
-0.70 MPa
5. If a cell having a solute potential of -0.35 MPa is placed in a
solution of pure water (Ψ = 0), what will be its pressure potential
at equilibrium?
+.035 MPa
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Sample Problems
• 6a . A hypertonic solution has a (circle one:) HIGH / LOW
water potential compared to the cell. Why?
Low – more solute present, less water
available to do work
• 6b. According to water potential rules, which way will water
move in this system?
Water moves out of the cell
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