### Lab 4: Diffusion and Osmosis

```Lab 4: Diffusion and Osmosis
First you must review Water Potential
on the next however many slides…
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
PROBLEM – What about physical pressure?
physical pressure.
It is easy to say that water will move from
high concentration to low concentration…but
look at the cell to the right…can the water
keep going in?
NO, as water enters and the cytoplasm begins to put pressure on the cell wall, the cell wall will
begin push back (3rd law of motion) – physical pressure.
The combined effects of solute concentrations and physical pressure are given in a
single measurement called the WATER POTENTIAL (Ψ; psi) for a given solution (EACH
SOLUTION IS ASSIGNED A WATER POTENTIAL).
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
WATER POTENTIAL
How is water potential calculated?
Ψ = Ψs + Ψp
Ψ = water potential of the given solution
1. The potential pressure, in Mpa, that the water will exert
on another solution.
2. Water always move from solutions of higher water
potential (higher pressure) to solutions of lower water (lower
pressure) of course.
3. If two solutions are at equilibrium then their water
potential is the same…
physical pressure.
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
WATER POTENTIAL
physical pressure.
How is water potential calculated?
Ψ = Ψs + Ψp
Ψ = water potential of the given solution
Ψs = solute potential (osmotic potential)
1. Ψs = defined as 0 MPa for pure water; it becomes more and more negative as more and
more solute is added since the potential pressure the water would exert on a neighboring
solution should less if the water concentration is falling.
2. Always either zero or negative
3. Makes sense since the more solute, the more likely you are to bring water to you and
therefore the lower the potential pressure exerted by the water.
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
WATER POTENTIAL
physical pressure.
How is water potential calculated?
Ψ = Ψs + Ψp
Ψ = water potential of the given solution
Ψs = solute potential (osmotic potential)
Ψp = [physical] pressure potential
1. Ψp = potential pressure the water will apply on a neighboring solution due to physical
forces
2. Ψp can be positive or negative since the physical pressure can be greater than atmospheric
pressure (positive; i.e. in a turgid plant cell) or less than atmospheric pressure (negative; i.e.
xylem cells when water is flowing through by transpiration).
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
Quantitative analysis of
WATER POTENTIAL
Look at the four conditions on the right and
explain what you are observing in terms of
water potential.
Water is moving from higher water potential to lower
water potential whose values are dependent on solute
potential and pressure potential.
More solute, more negative Ψs
More pressure increases Ψp
Less pressure decreases Ψp
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
Quantitative analysis of
WATER POTENTIAL
Without looking, determine values that
make sense for solute potential and
pressure potential.
Cytosol = 0.2M
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
Quantitative analysis of
WATER POTENTIAL
Cytosol = 0.2M
Notice that Ψs of the cell doesn’t really
change since it doesn’t take much water
to enter, pressurize the cell resulting in the
cell wall pushing back with the equivalent
potential of 0.7 MPa.
Chapter 36 – Plant form and function
AIM: Describe the process of passive transport (diffusion across the membrane).
Quantitative analysis of WATER POTENTIAL
1. A solution in a beaker has sucrose dissolved in water with a solute potential of
-0.5MPa. A flaccid cell is placed in the above beaker with a solute potential of -0.9MPa.
a) What is the pressure potential of the flaccid cell before it was placed in the beaker?
0 MPa since the cell wall applied no pressure if flaccid
b) What is the water potential of the cell before it was placed in the beaker?
Ψs + Ψp = Ψ
-0.9 + 0 = -0.9MPa
c) What is the water potential in the beaker solution containing the sucrose?
Ψs + Ψp = Ψ
-0.5 + 0 = -0.5MPa
d) How will the water move?
From high to low water potential (from beaker to cell)
e) What is the pressure potential of the plant cell when it is in equilibrium with the sucrose solution outside? Also, what is its
final water potential when it is in equilibrium?
Equilibrium tells you that the water potential of both solutions must be the same. If the outer solution is -0.5MPa then the cytoplasm must
also be -0.5MPa.
Ψ = Ψs + Ψp
-0.5MPa = -0.9 MPa + Ψp
Ψp = 0.4, a positive number makes sense since the cytoplasm’s potential to apply pressure is increased if
the cell wall is pushing on it.
f) Is the cell now turgid/flaccid/plasmolysed?
turgid
g) Is the cell hypotonic or hypertonic with respect to the outside?
It is still hypertonic since not much water would have entered
before pressurizing the cell causing the membrane to push
back.
Lab 4: Diffusion and Osmosis
How can one calculate the solute
potential (ψs) for a given solution?
ψs = -iCRT
i = ionization constant (number of particles/ions formed when dissolved in
water – ex. Glucose is 1, NaCl 2, CaCl2 3, etc…)
C = Molar concentration
R = pressure const. = .0831 Liter-bars/mole-K
T = temperature (in Kelvin = 273 + °C)
Ex. What is the solute potential for a 0.15M solution of sucrose
at atmospheric pressure and a temperature of 25°C?
= -(1)(.15)(.0831)(25+273)
= -3.7 bars
***“bar” is a unit of pressure like “atmosphere”, “Torr” and “Pascal” (Pascal is the SI Unit).
***1bar = 100,000Pa = 100KPa = ~1atmosphere (760 Torr).
Lab 4: Diffusion and Osmosis
How can one calculate the solute
potential (ψs) for a given solution?
ψs = -iCRT
i = ionization constant (number of particles/ions formed when dissolved in
water – ex. Glucose is 1, NaCl 2, CaCl2 3, etc…)
C = Molar concentration
R = pressure const. = .0831 Liter-bars/mole-K
T = temperature (in Kelvin = 273 + °C)
Ex. What is the solute potential for a 0.15M solution of NaCl at
atmospheric pressure and a temperature of 25°C?
= -(2)(.15)(.0831)(25+273)
= -7.4 bars
***“bar” is a unit of pressure like “atmosphere”, “Torr” and “Pascal” (Pascal is the SI Unit).
***1bar = 100,000Pa = 100KPa = ~1atmosphere (760 Torr).
Lab 4: Diffusion and Osmosis
Explain this:
Lab 4: Diffusion and Osmosis
Explain this:
Lab 4: Diffusion and Osmosis
Water
and theBasis
fitness
the environment
Chapter 23 -– The
Chemical
of ofLife
AIM: How does one determine the pH of a solution?
Recall phenolphthalein:
PINK
COLORLESS
The structure of the phenolphthalein molecule changes in different pH values. Above pH 8,
it has a structure that reflects pink. Below 8 the structure changes and does not absorb
light. Structure determines function!!
Lab 4: Diffusion and Osmosis
1. Obviously, .1M HCl is the acid as it will donate protons to solution and drop the pH
2. The base is the .1M NaOH because it will donate -OH (hydroxide ions) to solution, which
will combine with H+ (protons) to form water thereby lower the [H+] and increasing the pH.
3. I think this was supposed to say what color is the phenolphthalein (the “dye”) in the acid.
Adding HCl will drop the pH below 8 and therefore the solution will be clear.
4. Adding NaOH will bring the phenolphthalein solution above pH 8 and therefore it will turn
pink.
Lab 4: Diffusion and Osmosis
The one with the highest surface area to volume ratio of course…(the small one)
SA:V ratio
3:1
6:1
15:1
Agar Blocks (its just like jell-o)
Design an experiment to test the hypothesis we came up with in question
Lab 4: Diffusion and Osmosis
Experimental Design
Independent Variable:The surface area of the cubes
The efficiency of the agar blocks as determined by the amount of solution that
Dependent Variable:
diffused into the block over a given amount of time.
Procedure:
1. Make agar blocks containing phenolphthalein of assorted sizes:
-2 cm3
-1 cm3
-0.5 cm3
2. The cubes start off with a pH below 8 and are therefore a white color as
shown above right. Place the cubes in NaOH solution for 10 minutes.
3. The solution should begin to diffuse into the
cubes turning the cubes pink staring on the
outside and working its way in as it diffuses.
Lab 4: Diffusion and Osmosis
Experimental Design
Procedure:
4. Remove the cubes and cut them in half:
5. Measure the size of the portion that the NaOH solution did not reach (the
white part or so-called “final volume”) using a standard mm ruler:
6. Calculate the percent
Volume of the cube that the
solution diffused into…
Lab 4: Diffusion and Osmosis
Experimental Design
Procedure:
4. Remove the cubes and cut them in half:
5. Measure the size of the portion that the NaOH solution did not reach (the
white part or so-called “final volume”) using a standard mm ruler:
6. Calculate the percent
Volume of the cube that the
solution diffused into…
Lab 4: Diffusion and Osmosis
Experimental Design
Procedure:
7. Obviously good science does not consist of one trial…Explain
what you should do and what should be done with the data.
Many trials should be performed for each cube. In addition, the mean, standard
deviation and standard errors should be calculated for each cube as well and
then a bar graph should be generated with standard errors as the error bars:
Trial #
Percent Volume
of diffusion 2cm3
block (%)
Percent Volume
of diffusion 1cm3
block (%)
Percent Volume
of diffusion
0.5cm3 block (%)
1
36
50
100
2
38
55
98
3
41
51
94
4
34
54
100
5
36
60
99
Mean +/- SD
37 +/- 2.6
58 +/- 3.9
98 +/- 2.5
Standard Error
1.2
1.8
1.1
100
90
80
70
60
50
40
30
1
2
3
How else can we display this data?
Lab 4: Diffusion and Osmosis
Experimental Design
Procedure:
8. Another excellent representation of the data (X-Y Scatter Plat):
Percent diffused
100
90
80
70
60
50
40
30
0
0.5
1
1.5
Surface Area (cm3)
2
2.5
Lab 4: Diffusion and Osmosis
Experimental Design
What must be controlled?
1. Time the cube is inside the NaOH solution
2. Temperature of the solutions (remember that the higher the
temp, the greater the diffusion rate)
3. Same amount of NaOH solution in each beaker
4. The size of the agar blocks, etc…
Lab 4: Diffusion and Osmosis
1. It must be isotonic to the cytoplasm of your cells. You can also say it must have the
same water potential as your cells. Since our cells are flaccid, Ψp=0 and therefore Ψs or
the IV solution must match Ψs of the cytosol of your cells.
2. Pure water has a Ψs=0 and therefore a Ψ=0 since Ψp=0. Therefore the water potential is
higher than that of the cytosol of your cells and water will net flow in causing lysis. Your
blood will become hypotonic relative to the cytoplasm of your cells.
3. Determine the osmolarity (solute concentration) of their blood and match it.
Lab 4: Diffusion and Osmosis
Lab 4: Diffusion and Osmosis
Lab 4: Diffusion and Osmosis
Lab 4: Diffusion and Osmosis
Pair 1
Pair 2
Pair 3
Pair 4
Control
Inside
Cell
1M
Sucrose
1M NaCl
1M
glucose
1M NaCl
Water
Outside
Cell
1M
Glucose
5%
Ovalbumi
n
1M NaCl
1M
sucrose
Water
Lab 4: Diffusion and Osmosis
Inside
Cell
Pair 1
Pair 2
Pair 3
Pair 4
Control
1M
Sucrose
1M NaCl
1M
glucose
1M NaCl
Water
1M
Glucose
5%
Ovalbumi
n
1M NaCl
1M
sucrose
Water
Initial
Mass (g)
Outside
Cell
Final Mass
after 30
Lab 4: Diffusion and Osmosis
Inside
Cell
Pair 1
Pair 2
Pair 3
Pair 4
1M
Sucrose
1M NaCl
1M glucose 1M NaCl
1M
Glucose
5%
Ovalbumin
1M NaCl
Control
Water
Initial Mass
(g)
Outside
Cell
Final Mass (g)
after 30 min
Percent
Change
1M sucrose Water
Lab 4: Diffusion and Osmosis
Pair 1
Pair 2
Pair 3
Pair 4
Control
Inside
Cell
1M
Sucrose
1M NaCl
1M
glucose
1M NaCl
Water
Outside
Cell
1M
Glucose
5%
Ovalbumi
n
1M NaCl
1M
sucrose
Water
Lab 4: Diffusion and Osmosis
Lab 4: Diffusion and Osmosis
Lab 4: Diffusion and Osmosis
Elodea
Aquatic Plant (not algae)
Lab 4: Diffusion and Osmosis
Plasmolysis
Lab 4: Diffusion and Osmosis
Design an experiment to determine the water potential of potato
cells
```