Water Potential Problems
Water Potential ()
Water potential of pure water in an open
container is 0.
Water will flow from an area of higher
water potential to an area of lower water
potential.


Question: Which way will the water
flow? Into the cell or out of the cell?
Answer: Water will flow into the cell.
Explanation: Water will continue to flow
in both directions. However, more water
will be entering the cell. So the NET
MOVEMENT of water will be INTO the
cell. Water will flow from an area of
higher water potential to an area of lower
water potential (from  = -0.5 to  =
-2.5). This will continue until dynamic
equilibrium is reached.
 = -2.5
 = -0.5
Addition of solutes = a more negative
solute potential.
S < 0
In an open container or in an animal
cell, the pressure potential will be 0.
P = 0
In turgid plant cells, the cell wall can
exert positive pressure.
P > 0
Solute Potential (S)
i = ionization constant
• For sucrose, i = 1 because sucrose does not
ionize in water.
• For NaCl, i = 2 because NaCl ionizes when it
dissolves and becomes one Na+ and one Clion in an aqueous solution.
1 ion + 1 ion = 2 ions
Question: For MgCl2, what would i be?
Answer: i = 3
Explanation: Mg would ionize into 3 ions if
dissolved in water: one Mg2+ ion and two Clions.
C is the molar concentration of the
solution.
Moles Solute
Molarity =
Volume of Solution
M =
moles
L
R is the pressure constant.
• Always will be 0.0831 L bars / mole K
T = temperature in Kelvin
• Kelvin is the temperature in degrees Celsius +
273.
Question: Room temperature is about 20°C.
What would this be in Kelvin?
Answer: 20 + 273 = 293K
Example: If a cell’s P = 3 bars and
S = -4.5 bars, what is the resulting ?
 = P + S
Answer:
= 3 bars + (-4.5 bars)
 = -1.5 bars
Question: A cell with a  = -1.5 bars is
placed in a beaker with a solution of  = -4
bars. Will water flow into or out of the
cell?
Answer: Water will flow out of the cell from an
area of higher water potential (-1.5 bars) to an
area of lower water potential (-4 bars).
Question: What is the  of a 0.1 M
solution of sucrose in at open container at
20°C?
Answer: -2.4 bars
Explanation: S = -iCRT
L bar
S = -(1)(0.1 mole
)(0.0831
L
mole· K )(293K)
S = -2.4 bars
= P + S
 = 0 + (-2.4 bars) = -2.4 bars
= P + S
S = -iCRT
=?