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 1 11/15/2012 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 2 11/15/2012 Calculating Water Potential • Ψ = Ψ S + ΨP – ΨS = solute potential (osmotic potential) – ΨP = pressure potential • Physical pressure applied to the solution • Solution in an open container has a Ψ P = 0 • As water enters a plant cell Ψ P 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 3 11/15/2012 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 o in a beaker at 20 C. (In an open beaker, Ψp = 0.) -3.65 bars = -0.365 MPa **Omit 2b & 2c 4 11/15/2012 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 5 11/15/2012 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 6.