1. Water Potential Problems/Solutions
The solutions can also be found in the screencast found at:

2. Water Potential (Represented by the Greek letter psi (Ψ).)
Is a measure of water’s potential to move.
Water always moves from areas of high water potential to areas of low water potential.
Water Potential=Pressure Potential + Solute Potential

3. Water Potential
If we only accounted for the presence of solutes, we would deduce that water would move into the plant cell (in the diagram to the right) continuously.
That doesn’t happen because of pressure.
Example:
Let’s assume that the total water potential of the water in the beaker is 0 bars.
Let’s also assume that the initial solute potential inside the plant cell is bars and that initial pressure potential inside the cell is zero.
Distilled Water
Plant Cell

4. Water Potential
The value of looking at water potential is that it takes into account the effect of pressure on the movement of water instead of simply looking at the concentrations of solutes.
The water potential of pure water in an open container is 0.
This is because no solutes are present and because the only pressure acting on the system is atmospheric pressure.
The pressure potential of the system is typically 0 unless you are provided with a pressure value in the problem. As more water enters a cell, the pressure increases.
The solute potential can be calculated using:

5. Water Potential The more solutes that are present, the lower
the water potential of the solution.
To calculate the ionization constant (i), use the following rules:
For all sugars, the value for i is (1). Sugars don’t ionize when dissolved into water.
For salts, the ionization constant (i) is equal to the number of ions formed per formula unit of the ionic compound.
Examples:
C—is the molar concentration. This quantity is typically measure in Molarity (moles of solute/Liter of solution)
R—is the pressure constant. The value for this constant is always: (Liter bars)/(moles K)
T—is the temperature. Be sure to use the Kelvin or absolute temperature.

6. Sample Problem 1
The molar concentration of a NaCl solution in an open container is 0.20M. The pressure potential of the solution is zero. What is the total water potential at a
temperature of 32 degrees Celsius?
Ψ=Ψp + Ψs
Since the solution is in an open contain Ψp is equal to 0. We must calculate Ψs using Ψs=-iCRT. Since the solution is a salt which dissociates into two ions per formula unit (NaClNa+ + Cl-) the “I” for the problem is 2. The concentration “C” is the molarity (0.20M); and the Kelvin temperature is 305K ( ). R is equal to ( (L X bars)/(moles X K) .
Ψs=-(2) (0.20moles/L)( (L X bars)/(moles X K)) (305 K)
This yields: bars for Ψs. Since Ψp =0, the total water potential of the solution is bars.
Since the water potential of the solution is bars and the water potential of the cell is bars, water moves out of the cell into the solution from higher water potential to lower water potential. The cell will plasmolyze.
Distilled Water
Plant Cell
If a plant cell with a total water potential of bars is placed into a beaker of the solution described above, will water move into or out of the cell? Assume that the cell’s membrane is impermeable to the solute. Justify your answer.
NaCl Solution

7. Sample Problem 2
The molar concentration of a sucrose solution in an open container is 0.30M. The pressure potential of the solution is zero. What is the total water potential at a
temperature of 23 degrees Celsius?
Ψ=Ψp + Ψs
Since the solution is in an open contain Ψp is equal to 0. We must calculate Ψs using Ψs=-iCRT. Since the solution is a sugar, it doesn’t ionize in water. The “I” for the problem is 1. The concentration “C” is the molarity (0.30M); and the Kelvin temperature is 296K ( ). R is equal to ( (L X bars)/(moles X K))
Ψs=-(1) (0.30 moles/L) ( (L X bars)/(moles X K)) (296 K)
(This yields: bars for Ψs. Since Ψp =0, the total water potential of the solution is bars.
Since the water potential of the solution is bars and the water potential of the cell is bars, water moves into the cell into the solution from higher water potential to lower water potential. The turgor pressure in the cell will increase.
If a plant cell with a total internal water potential of bars is placed into a beaker filled with the solution described above, what will happen to the cell’s total internal water potential as time passes? Assume that the cell’s membrane is impermeable to the solute. Justify your answer.
Distilled Water
Plant Cell
Sucrose Solution

8. Sample Problem 3
The molar concentration of a CaCl2 solution in an open container is 0.60M. The pressure potential of the solution is zero. What is the total water potential at a
temperature of 32 degrees Celsius?
Ψ=Ψp + Ψs
Since the solution is in an open contain Ψp is equal to 0. We must calculate Ψs using Ψs=-iCRT. Since the solution is a salt which dissociates into three ions per formula unit (CaCl2Ca+2 + 2Cl-) the “I” for the problem is 3. The concentration “C” is the molarity (0.60M); and the Kelvin temperature is 305K ( ). R is equal to ( (L X bars)/(moles X K) .
Ψs=-(3) (0.60moles/L)( (L X bars)/(moles X K)) (305 K)
This yields: bars for Ψs. Since Ψp =0, the total water potential of the solution is bars.
Since the water potential of the solution is bars and the water potential of the cell is bars, water moves out of the cell into the solution from higher water potential to lower water potential. The cell will plasmolyze.
Distilled Water
Plant Cell
If a plant cell with a total water potential of bars is placed into a beaker of the solution described above, will water move into or out of the cell? Assume that the cell’s membrane is impermeable to the solute. Justify your answer.
CaCl2 Solution

9. Sample Problem 4
1. Water will flow from Side B to Side A, from higher water potential to lower water potential.
Side A=-1.7 bars;
Side B=-0.6 bars
Side A. It has a higher solute concentration than side B.
Side B. It has a lower solute concentration than side A.
The volume of side A will increase, while the volume of side B will decrease. Water moves from side B to side A.
There are more solutes present on side A. Higher solute concentrations cause lower water potentials.
Side A S
=-1.7 bars P
=0 bars
=-0.6 bars
Side B
1. Sides A & B are separated by a semipermeable membrane. Assume that the cell’s membrane is impermeable to the solute. The triangles represent solutes that are dissolved in water. If the volumes of the two sides are initially equal and the membrane is impermeable to the solute, in which direction will water flow?
2. What is the total water potential on each side of the membrane?
3. Which side is hypertonic?
4. Which side is hypotonic?
5. What will happen to the volume of side A? Side B? Justify your answer.
6. Why is the initial solute potential on side A lower than the initial solute potential on side B?

10. Water Potential’s Relationship to Transpiration