Notes on Water Potential

What is water potential?

water potential
The physical property predicting the direction in which water will flow, governed by solute concentration and applied pressure.


Calculating Water Potential
Water potential is calculated using the following formula:

Water potential () = pressure potential () + solute potential ()

Pressure potential (): In a plant cell, pressure exerted by the rigid cell wall that limits further water uptake.
Solute potential (): The effect of solute concentration. Pure water at atmospheric pressure has a solute potential of zero. As solute is added, the value for solute potential becomes more negative. This causes water potential to decrease also.

In sum, as solute is added, the water potential of a solution drops, and water will tend to move into the solution.

In this laboratory we use bars as the unit of measure for water potential; 1 bar = approximately 1 atmosphere.


Water Potential
The water potential of pure water in an open container is zero because there is no solute and the pressure in the container is zero. Adding solute lowers the water potential. When a solution is enclosed by a rigid cell wall, the movement of water into the cell will exert pressure on the cell wall. This increase in pressure within the cell will raise the water potential.

Look again at the equation for water potential:

Water potential () = pressure potential () + solute potential ()



So that you might better understand the procedure for calculating water potential, here is a practice problem.

Once you know the solute concentration, you can calculate solute potential using the following formula:

Solute potential () = –iCRT


i = The number of particles the molecule will make in water; for NaCl this would be 2; for sucrose 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.  Assume room temperature is = to 22 oC.

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.
The pressure potential of a solution open to the air is zero. Since you know the solute potential of the solution, you can now calculate the water potential.

Notes to help answer questions.

1.  Draw a straight line (or use the regression button on your logger pro) and determine where the line crosses the x axis.  The point at which the line crosses the x axis represents the molar concentration of sucrose with a water potential that is equal to the potato tissue water potential.  At this concentration there is no net gain or loss of water from the potato cells.

2. By graphing the data of the percent change in mass of the potato cores, it can be determined that the water potential of potato cells is equal to the water potential of an unknown sugar solution molarity.  Determine that molarity.

3.  Using the explanation above the water potential of the potato cells can be determined.

4.  If the potato is allowed to dehydrate by sitting out in the open air the water potential would decrease (be more negative) because the concentration of solutes within the cells would increase as potato cells dehydrate.  Therefore, the osmotic pressure and water potential both decrease.