Download What is Water Potential?

What is Water Potential?
Water potential
• the force responsible for movement
of water in a system
• Has the symbol psi
• Is measured in bars or megapascals
It is a measure of the free
energy of water which is less
when it is has to surround
solutes.
We are
really stuck
here, YUK!
Well I told
you not to
come here!
But, I
HAVE to
join the
party!
Has two components:
• Solute potential (also called osmotic
potential) џs which is determined by
solute concentration
• Pressure potential џp which results
from exertion of pressure on
membranes/walls as water moves in
or out; can be positive or negative
The water potential of
pure water is given the
value ZERO
• Because pure water has the highest
concentration of water molecules, and
thus the highest water potential, the
water potential of all other solutions
must be lower than zero i.e. negative.
Pure water:
=0
Adding solute decreases
water potential!
• The more solute there is present in a
solution the more negative it
becomes.
• So, solute potential will be a
negative number if not pure water.
So hypertonic solutions have negative solute potentials.
water potential = solute potential +
pressure potential
(s is pi on your paper)
Water moves from areas
of higher water potential
to areas of lower water
potential (i.e. towards the
more negative,
concentrated region).
water always "falls"
from a high to a low
water potential
This will occur until the
water potential inside the
cell equals the water
potential outside of the
cell.
ψp
Pressure potential is important in plant cells
because they are surrounded by a cell wall
which, is strong and rigid.
ψp
When water enters a plant cell, its volume
increases and the living part of the cell
presses on the cell wall.
The cell wall gives very little and so
pressure starts to build up inside the cell.
ψp
This has the tendency to stop more water
entering the cell and also stops the cell from
bursting.
When a plant cell is fully inflated with
water, it is called turgid.
Pressure potential is called turgor pressure in plants)
ψp
Water potential (ψ ) =
pressure potential (ψp ) + solute (osmotic) potential (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. As solute is
added, the value for solute potential
becomes more negative. This causes
water potential to decrease also.
*As solute is added, the water
potential of a solution drops, and
water will tend to move into the
solution.
Water moves from a place of high water potential to a place of low water potential.
Water potential (ψ) =
pressure potential (ψp )
+ solute potential (ψs)
(osmotic)
This is an open
container, so the ψp = 0
This makes the ψ = ψs
The ψs =-0.23, so ψ is
-0.23MPa, and water
moves into the
solution.
Can a solution with a molarity of 0.2 be in equilibrium
with a solution with a molarity of 0.4?
YES!
Pressure
Two solutions will be at equilibrium when the water
potential is the same in both solutions. This does not
mean that their solute concentrations must be the
same, because in plant cells the pressure exerted by
the rigid cell wall is a significant factor in determining
the net movement of water.
Solute (osmotic) potential (ψs )= –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 =
R =
Molar concentration Yikes, what's
Pressure constant = 0.0831
liter bar/mole K
that??????
T =
Temperature in degrees Kelvin (273 + °C) of solution
Example 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°C degrees. Round your answer to the nearest hundredth.
What is the water potential?
Answer:
-7.48
Solute potential = -iCRT
= -(1) (0.3 mole/1) (0.0831 liter bar/mole K) (300
K)
= -7.48 bar
Water potential = -7.48 + 0, so water potential = -7.48
If this makes no sense
whatsoever the key
information to learn is:
• The equation given
• the water potential of pure water is
zero
• water moves from areas of higher
water potential to areas of lower
water potential (i.e. towards the
more negative region)
1. A solution in a beaker has NaCl dissolved in water with a solute
potential of -0.5 bars. A flaccid cell is placed in the above beaker with
a solute potential of -0.9 bars.
p = 0 bars
a) What is the pressure potential of the flaccid cell before it was
placed in the beaker?
p = 0 bars
b) What is the water potential of the cell before it was placed in the
beaker?
w = p + s
X = 0 bars + -0.9 bars
X = -0.9 bars
c) What is the water potential in the beaker containing the sodium
chloride?
w = p + s
X = 0 bars + -0.5 bars
X = -0.5 bars
d) How will the water move?
Water will enter the cell and leave the solution.
e) What is the pressure potential of the plant cell when it is in
equilibrium with the NaCl solution outside?
Equilibrium means w of cell = w of solution.
wcell = pcell + scell
-0.9 bars = x + -0.5 bars
p of the cell = 0.4 bars
f) What is the cells final water potential when it is in equilibrium?
Same as the solutions w which = -0.5 bars
g) Is the cell now turgid/flaccid/plasmolysed?
turgid
h) Is the cell hypotonic or hypertonic with respect to the outside?
hypertonic
i) If it is hypo/hyper (choose one) tonic – this means that its water
potential is higher/lower (choose one) than the outside.
Hypertonic means that the water potential is lower than the outside.
A solution in a beaker has sucrose dissolved in water with a solute
potential of -0.9 bars. A flaccid cell is placed in the above beaker with a
solute potential of -0.3 bars.
a) What is the pressure potential of the flaccid cell before it was placed in
the beaker?
p = 0 bars since the cell is flaccid.
b) What is the water potential of the cell before it was placed in the
beaker?
w = p + s
X = 0 bars + -0.3 bars
X = -0.3 bars
c) What is the water potential in the beaker containing the sucrose?
w = p + s
X = 0 bars + -0.9 bars
X = -0.9 bars
d) How will the water move?
Water will move out of the cell since it’s water potential value is smaller.
e) What is the pressure potential of the plant cell when it is in
equilibrium with the sucrose solution outside? Think carefully – does
the plant cell wall change shape?
p = 0 bars
No
f) Also, what is the cell’s final water potential when it is in equilibrium?
-0.9 bars
g) Is the cell now turgid/flaccid/plasmolysed?
plasmolyzed
What is the cell’s solute potential when it is in equilibrium?
-0.9 bars
h) Is the cell hypotonic or hypertonic with respect to the outside?
Cell would be hypotonic
i) If it is hypo/hyper (choose one) tonic – this means that its water
potential is higher/lower (choose one) than the outside.
Hypotonic means that its water potential value is higher than the
outside.
So, what happens when a potato cell
in put in pure water?
• Water will move in or out until the wp of
the cell will equals the wp surrounding
the cell.
• The pressure potential will increase to
balance out the solute potential to equal
to zero which is the wp of pure water.
• No more net movement of water occurs
So how can we determine the
water potential of potato cells?
We place potato cells in different molarities
of sucrose. When enough solute is added
outside of the potato cells to result in NO
more NET movement of water, that is the
molarity of the potato cells.
How do you go from molarity of a
solution to the solute potential to
figure the wp?
• Another equation
solute potential = -iCRT
I = ionization constant (1 for sucrose)
C = molar concentration of sucrose (in this case where
no net gain/loss of water occurs)
R = pressure constant (0.0831 liter/bars/mole 0K for
sucrose)
T = temperature Kelvin (273 + C)
Units will cancel out to equal bars.
So what is the solute potential of a 0.1
M solution of sucrose at 22 C?
•
•
•
•
Solute potential = -iCRT
i (ionization constant) = 1
R = 0.0831 (from handbook)
T = temp K (273 + C of solution)
Ωs
=
Ωs =
- (1) (0.1) (0.0831) (295)
- 2.45 bars
• Calculate the water potential of a solution of
0.15 M sucrose. The solution is at standard
temperature (273K) and pressure (.0831 L
bars/mol K).
w = p + s
X = 0 bars + [-iCRT]
X = 0 bars + [-(1) (0.15 mol/L) (0.0831 L
bars/mol K) (273K)
X = 0 bars + -3.40 bars
w = -3.40 bars
So you will graph the results of the
change in weight of the potato cells
in different molarity solutions of
sucrose after overnite.
• Where the line crosses the graph at the X
axis, representing no gain or loss of water,
will be the molarity of the potato cells.
• Then substitute in the equation for solute
potential (-iCRT)
So, how do you get the water potential?
Once you determine the solute potential, plug into
the equation to determine the water potential. The
pressure potential will be zero since water is at
equilibrium (no net movement in or out.