Calculating water potential
Ψ = Ψp + Ψs
The combined effects of these two factors:
1. Solute concentration
2. Pressure
are incorporated into a single measurement
called Water Potential or
Ψ
Water will move across a membrance
from the solution with higher water
potential to the solution with the
lower water potential.
Water Potential
– Water potential is determined by solute potential
and pressure potential.
Ψ = Ψp + Ψs
– Water moves from regions of high water potential to
regions of low water potential.
Ψs
Solute (osmotic) potential
Once you know the solute concentration,
you can calculate solute potential using
the following formula:
• Solute potential (ΨS ) = –iCRT
Ψs
Solute (osmotic) potential
• Pure water has a solute potential (Ψs) of zero.
Solute potential can never be positive.
• Adding more solute is a negative experience;
the solute potential becomes negative.
Ψs = − iCRT
Solute potential
• i = The ionization constant
– for NaCl this would be 2;
– for sucrose or glucose, this number is 1
• C = Molar concentration (from your experimental data)
– Iso-osmolar molarity: would allow you to place a potato in
the solution and get no movement of water.
– Point at which the line crosses 0 on the graph.
• R = Pressure constant = 0.0831 liter bar/mole K
• T = Temperature in degrees Kelvin = 273 + °C of
solution
Pressure Potential – the sum of all
pressure on water.
In units of pressure: MPa or Bars
• Turgor pressure – forced caused by cell
membrane pushing against cell wall.
• Wall pressure – an equal and opposite
force exerted by cell wall. Counteracts the
movement of water due to osmosis.
• Other pressures – tension, cohesion,
atmospheric, root, etc.
Calculating Water Potential
• Water potential is calculated using the following
formula:
• Ψ = ΨP + ΨS
• Pressure potential (ΨP ): In a plant cell, pressure
exerted by the rigid cell wall that limits further
water uptake.
• Solute potential (ΨS ): The effect of solute
concentration.
– Pure water at atmospheric pressure has a
solute potential of zero.
Water potential
Chapter 32
Practice Problem
• 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.
• What is the water potential for this
example? Round your answer to the
nearest hundredth.