Download Water Potential (Ψ)

Water Potential (Ψ) Water potential (Ψ) is a measure of water’s potential to do work. Water potential is a measure of the tendency of water to move from high free energy to lower free energy … a measure of the tendency of water to leave one place in favor of another place. The largest water potential any volume of water can have, if only standard atmospheric pressure is being applied to that volume of water, is defined as Ψ = 0 bars. This is the water potential for distilled water. A bar is a metric measure of pressure, measured with a barometer (about the same as 1 atmosphere). Another measure of pressure is the megapascal (MPa). 1 MPa = 10 bars. As solute is added to distilled water with no outside pressure being applied to it, the water potential of that solution drops. But what does it mean to say that the water potential of a solution drops? It means that the water in that solution is less likely to do work … in other words; it is less likely to move! Why is that? Well, as solute is added, the chances become less and less that a concentration gradient can be set up between that solution and a second solution that will favor the movement of water out of the initial solution. In the example to the left, solute was added to the solution on the left side of the membrane. This decreased the chances that water would move out of the left solution and into the right solution. This means that the water on the left side of the membrane has less potential to do work than the water on the right side.
What does this mean in terms of water potential? It means that the solution to the left of the membrane has a more negative water potential than the solution to the right of the membrane. Therefore, water will flow from the right side of the membrane to the left. Water always moves towards a more negative water potential. Let’s take a look at another example: The solute potential of a 0.1 M sucrose solution at 20 °C at standard atmospheric pressure is -­‐0.23 bars. If we continue adding sucrose to the solution until it reaches a concentration of 0.75 M at 20 °C at standard atmospheric pressure, the solute potential continues to drop to a value of -­‐1.87 bars. Which solution contains water that is less likely to do work? The one that has a higher concentration of solute and a lower concentration of water! Think about it … if we separated a 0.1M solution of sucrose and a 0.75M solution of sucrose with a selectively permeable membrane, which direction would the water move? It would move from the 0.1M solution into the 0.75M solution. In the process, it would be doing work! Water always moves from an area of higher water potential (higher free energy; more water molecules) to an area of lower water potential (lower free energy; fewer water molecules). Think of water diffusing “down” a water potential gradient! Now that you think you’ve got water potential figured out, let’s complicate matters a little bit! Water potential (Ψ) is actually determined by taking into account two factors: pressure potential (ΨP) and osmotic (or solute) potential (ΨS). The formula for calculating water potential is: Ψ = ΨP + ΨS. The water potential value can be positive, zero, or negative. Osmotic (or solute) potential is directly proportional to the solute concentration. If solute concentration increases, the potential for the water in that solution to undergo osmosis decreases. Solute potential is always negative since pure water has a solute potential of zero. The more solute that is added to a solution, the more negative its osmotic (solute) potential gets. If no physical pressure is applied to a solution, then the solute potential is equal to the water potential. However, if physical pressure is applied to a solution, then its water potential (the potential for the water to move and do work) will be affected. How it is affected depends upon the direction of the pressure. For example, pressing on the plunger of a water-­‐filled syringe causes the water to exit via any opening. Let’s look at another example! If a plant cell is placed into distilled water (a hypotonic solution), water will move into the cell because distilled water has a higher water potential than the plant cell itself. However, when the plant cell’s central vacuole fills with water, the vacuole will push back out on the water surrounding the cell. The plant cell does not burst because a plant cell has a cell wall (turgor pressure). An animal cell in the same situation would burst! Pressure potential is usually positive in living plant cells; in dead xylem elements it is often negative. When the pressure exerted outward on the water surrounding the plant cell is equal to the osmotic potential of the solution in the cell, the water potential of the cell will be equal to zero. The water potential of the plant cell will also be equal to the water surrounding it, and there will be no net movement of water molecules. If plant cells are placed in a hypertonic solution (more solute outside of the cells), water will leave the plant cells, moving from an area of higher water potential to an area of lower water potential. The loss of water from the plant cells will cause the cells to lose turgor. A continued loss of water will eventually cause the cell membrane to shrink away from the cell wall (plasmolysis). Once you know the solute concentration, you can calculate solute potential using the following formula: ΨS = -­‐ iCRT The number of particles the molecule will make in water. For NaCl this would be 2; for sucrose i = or glucose, this number is 1. C = Molar concentration (from your experimental data) R = Pressure constant = 0.0831 liter bar/mole K T = Temperature in degrees Kelvin = 273 + °C of solution. Room temperature is about 22 °C. Solve the following problems (Show ALL work!) 1. If a cell’s Ψp = 2 bars and its Ψs = -­‐3.5 bars, what is the resulting Ψ? Ψ = ΨP + ΨS Ψ = 2 bars + (-­‐3.5 bars) = -­‐1.5 bars 2. A dialysis bag containing 0.1% sucrose is placed in a beaker containing 0.4% sucrose. The beaker is open to the atmosphere. a. What is the pressure potential Ψp of the system? Zero b. What is the water potential of this dialysis bag? -­‐ 1 bars c. Water will move out of of the dialysis bag. 3. The molar concentration of a sugar solution in an open beaker is 0.3M. Calculate the solute potential and water potential at 27 degrees Celsius. Round your answer to the nearest hundredth. To solve for solute potential: ΨS = -­‐ iCRT ΨS = -­‐ (1) (0.3 moles/liter) (0.0831 liter bar/mole K) (27 + 273 K) ΨS = -­‐ 7.48 bars The pressure potential of a solution open to the air is zero, so … Ψ = ΨP + ΨS. Ψ = 0 + (-­‐ 7.48 bars) = -­‐ 7.48 bars 4. Calculate the solute potential of a 1.0 M sucrose solution at 22 °C under standard atmospheric conditions. Round your answer to the nearest hundredth. ΨS = -­‐ iCRT ΨS = -­‐ (1) (1.0 moles/liter) (0.0831 liter bar/mole K) (295 K) ΨS = -­‐ 24.51 bars 5. Suppose that a student calculates that the water potential of a solution inside a bag is -­‐6.25 bars and the water potential of a solution surrounding the bag is -­‐3.25 bars. In which direction will the water flow? EXPLAIN. Water will flow INTO the bag. This occurs because there are more solute molecules inside the bag (a value further away from zero) than outside in the solution. 6. If a potato were allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why? Potato water potential would decrease as water leaves the cells due to dehydration. 7. A plant cell has a solute potential of –2.0 bars and a pressure potential of 0.0 bars. What is its water potential? It is placed in a solution with a water potential of –1.0 bars. What will happen to this plant cell? Ψ = ΨP + ΨS Ψ = 0 + (-­‐ 2.0 bars) = -­‐ 2.0 bars Water potential = -­‐2.0 bars and the cell will gain water or swell 8. If a plant cell has a lower water potential than its surrounding environment and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain or lose water? EXPLAIN. The cell is hypertonic to its environment and will gain water. Water always moves from high water potential to low water potential. 9. Refer to the figure to the RIGHT. a. The beaker is open to the atmosphere. What is the pressure potential (ΨP)? 0 bars b. Is the greater water potential in the beaker or the dialysis bag? dialysis bag c. Water will diffuse out of the bag. Why? There is greater water potential in the bag and net water movement will be from high to low water potential. Beaker Contents: 0.4 % sucrose solution ΨS = (-­‐) 4 bars Dialysis Bag: 0.1 % sucrose solution ΨS = (-­‐) 1 bars; ΨP = 0