Erika Veidis
Diffusion and Osmosis Lab
I.
INTRODUCTION:
The random movement of molecules from an area of higher concentration to an
area of lower concentration (known as moving down the concentration gradient) is
called diffusion. The substance will diffuse in an attempt to reach a dynamic
equilibrium, which is when the same concentration of solute exists throughout the
area or on either side of a membrane.
Osmosis is a type of diffusion that deals with the diffusion of water. It is the
diffusion of water through a selectively membrane. The water molecules will move
from an area of higher water potential to lower water potential in an attempt to reach
equilibrium. The net (total, final) movement of water will always be from an area of
higher water potential to an area with lower water potential. Water potential is the
measure of the tendency for water to move from one area to another. It is affected by
solute concentration and pressure within the cell. Water potential is equal to pressure
potential and solute potential. The water potential of pure water at 1 atmosphere is
zero. Water potential is abbreviated by Ψ (the Greek letter psi).
In animal cells, water diffuses through the membrane depending on the solute
concentrations. In an isotonic solution, there is an equal amount of solute
concentration on either side of the membrane, so the net water movement will be
zero. In a hypertonic solution, the surrounding solution has a higher solute
concentration. The water will rush out of the cell to try and reach equilibrium, and as
a result, the cell will shrivel. Contrarily, in a hypotonic solution, there is a higher
solute concentration inside the cell, and the water rushes in through osmosis, which
will cause the cell to swell and possibly lyse. In a plant cell, the rigid cell walls
prevent the cell from lysing, so instead, as water enters the cell, the pressure inside it
increases. This pressure affects the flow of water. When the pressure inside the cell
becomes large enough, then no additional water can flow into it, even if there still is a
higher concentration inside the cell. Also, in a hypertonic solution, the water
potential outside the plant cell drops below that inside the cell, which will cause the
cell to lose water. Instead of resulting in the shriveling of the cell, plasmolysis, which
is the separation of the cell membrane and the cell wall, may occur, resulting in cell
death.
A cell membrane is selectively permeable because some things cannot pass
through it. Small, hydrophobic lipid molecules can pass easily across the membrane.
Large hydrophilic polar molecules cannot pass easily, because they get repelled by
the non-polar tails of the phospholipids (which make up the membrane). This
permeability also depends on what proteins are incorporated in the cell membrane, as
they are specific to which molecules they can transfer across the membrane.
The purpose for section 1A was to observe the diffusion of solutes in the water by
using dialysis tubing. The selective permeability of the membrane, represented by the
dialysis tubing, was observed. In section 1B, the purpose was to measure the rate of
osmosis under different conditions of solute concentration. In section 1C, the purpose
was to measure the water potential of potato cells.
In section 1A, the dialysis tubing represented the cell membrane. Its tiny holes
made the tubing selectively permeable so that it most closely resembled a cell
membrane. Only some small substances could pass through it. A dialysis tube was
filled with a solution of glucose and starch and then placed in a cup of distilled water.
After a 30-minute period of time, the solutions in the bag and in the cup were tested
with an indicator strip and with an IKI solution for the presence of glucose and starch,
respectively, to see which molecules could pass through the dialysis tubing.
In section 1B, dialysis tubing was used to study the relationship between the solute
concentration and the movement of water through a selectively permeable membrane.
This process is known as osmosis. In this section, 6 dialysis tubes were filled with
different concentrations of sucrose, weighed, and then placed in separate beakers of
distilled water. After 30 minutes, the bags were removed from the water and
reweighed. The change in mass indicated the rate of osmosis – the bags with the
highest increase in mass had the highest rate of osmosis.
In section 1C, cores of potato tissue were placed in different concentrations of
sucrose solutions and then the net movement of water was measured. Four potato
cylinders were weighed and placed in 6 solutions of varying sucrose concentrations.
After they sat overnight, the potato cylinders were removed and weighed. The
change in mass in the potato cylinders illustrated the rate of diffusion. The ones with
the highest change in mass were the ones in the lowest concentrations of sucrose,
because the water flowed into the potato cells to try and reach equilibrium, which
caused them to swell and gain mass. In the higher concentrations of sucrose, the
potato cylinders decreased in mass, because the concentration of solute was higher in
the solution than in the cells, so water flowed out of the cells. This caused the potato
cells to shrink, decreasing in mass.
1A: Glucose will be present in both the dialysis bag and the cup of distilled water,
because it is a small molecule that will be able to pass through the tiny pores in the
dialysis tubing, flowing across the membrane. Starch will be present in the bag and
not in the cup of water, because starch molecules are much larger than glucose
molecules, and will not be able to fit through the minute pores.
1B: As the concentration of sucrose increases, the rate of osmosis will increase as
well. These higher concentrations of solute (hypotonic solutions) will cause the water
to rush into the bag, making it swell and increase in mass. Conversely, the lower
concentrations of sucrose will result in a lower rate of osmosis. A 0 concentration of
sucrose will be an isotonic solution.
II.
III.
1C: If the concentration of sucrose is higher, then water potential outside the cells
will drop below the water potential inside the cells, and there will be a net loss of
water from the cells. If the concentration of sucrose is lower, then the water potential
outside the cell will be higher than the water potential inside, and there will be a net
movement of water into the potato cells.
MATERIALS:
- Distilled water
-Beakers
-IKI solution
-Glucose indicator strips
-15% glucose/ 1% starch solution
-Dialysis tubing
-Funnel
-String
-Scale
-0.0, 0.2, 0.4, 0.6, 0.8, and 1.0 M concentrations of sucrose solutions
-Potato
-Cork borer
PROCEDURE:
1A:
A cup was filled with distilled water and about 4 mL of IKI solution. This solution
was tested with a glucose indicator strip to determine the initial presence of glucose.
A dialysis bag was filled with a 15% glucose/ 1% starch solution using a funnel. A
glucose indicator strip was used to test for the presence of glucose inside the bag
initially. The bag was tied off at both ends and placed in a beaker of distilled water.
After about 30 minutes, a glucose indicator strip was used to test for the presence of
glucose in the cup and in the bag. The IKI solution changes from a gold color to a
purple color when it comes into contact with starch. At the end of the 30 minutes, the
color of the solution inside the bag and the color in the cup indicated the presence of
starch in each.
1B:
Six dialysis bags were filled with solutions of varying sucrose concentrations by
using a funnel. The sucrose concentrations were 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M,
and 1.0 M. A little extra space was left in each bag to leave room for the diffusion of
water. The bags were tied off tightly at both ends. The mass of each bag was
measured and recorded. Then each bag was placed in a separate beaker filled with
distilled water, and each beaker was labeled with which dialysis bag it contained.
After about 30 minutes, each bag was removed and weighed again. The change of
mass was recorded and used to determine the rate of osmosis.
IV.
1C:
Six beakers were filled with solutions of different sucrose concentrations, (the same
as in section 1B). The beakers were labeled. A cork borer was used to obtain 4
potato cylinders for each beaker. The mass of each set of 4 potato cylinders was
recorded before being placed in a beaker. Each beaker was covered with plastic wrap
to prevent evaporation and sat overnight. The next day, the potato cylinders were
removed and the mass of each set was recorded.
RESULTS:
1A:
Solution Color
Initial
Final
Bag
Beaker
Initial
contents
15%
Purple
glucose/ 1%
starch
H20 + IKI
Gold
Glucose Test Results
Initial
Final
Purple
Gold
+
+
-
-
1B:
Contents in
dialysis bag
0.0 M sucrose
0.2 M sucrose
0.4 M sucrose
0.6 M sucrose
0.8 M sucrose
1.0 M sucrose
Initial mass (g)
Final mass (g)
Change in mass
26.0
26.2
26.1
26.4
26.3
26.4
26.3
27.0
28.0
29.3
30.2
31.2
0.3
0.8
1.9
2.9
3.9
4.8
Percent change
in mass
+1.2
+3.1
+7.3
+11.0
+14.8
+18.2
1C:
Contents in Temperature Initial
beaker
mass (g)
Final mass Change in
(g)
mass (g)
% change
in mass
0.0 M
sucrose
0.2 M
sucrose
0.4 M
sucrose
0.6 M
sucrose
0.8 M
sucrose
1.0 M
sucrose
2.0
2.3
+ 0.3
15%
% change
in mass
(class
average)
21.4%
2.0
2.0
0
0%
6.9%
2.1
1.8
-0.3
-14%
-4.5%
2.2
1.4
-0.8
-36%
-12.8%
2.0
1.4
-0.6
-30%
-23.0%
2.2
1.7
-0.5
-22.7%
-23.5%
V.
Room
temperature
ANALYSIS:
In section 1A, the solution was purple in the bag initially, because starch was
present, and the IKI solution turns from gold to purple when it reacts with starch. It
was gold in the beaker, because starch was not present in the beginning of the
experiment. Initially, the glucose indicator strips tested positive in the bag, because
glucose was present, and tested negative in the beaker, because glucose was absent.
At the end of the experiment, the IKI solution was still purple in the bag, because
starch was still present. It was still gold in the beaker, because the starch molecules
were too large to fit through the pores in the dialysis tubing. The membrane’s
selective permeability kept starch from diffusing into beaker. There was still a higher
concentration of starch in the bag than in the beaker for this reason – it couldn’t reach
equilibrium because the membrane would not permit its transfer. The glucose
indicator strips tested positive in both the bag and the beaker. This shows that the
glucose molecules were small enough to fit through the pores in the membrane. The
glucose diffused from its higher concentration in the bag to the lower concentration in
the beaker until it reached equilibrium – when there were equal amounts of glucose in
the bag and in the beaker.
In section 1B, the change in mass increased as the concentrations of sucrose
increased. The beaker with containing the bag with the 0.0 M concentration of
sucrose contained an isotonic solution – there were equal concentrations of solute in
the bag and in the beaker (0 g). This bag had a 0.3 change in mass. This small
VI.
change in mass could be explained by the natural movement of water or by the
various sources of error. This change in mass was significantly lower than the
changes in mass of the bags containing a higher sucrose concentration. The bags with
the 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M concentrations had mass changes of 0.8 g,
1.9 g, 2.9 g, 3.9 g, and 4.8 g respectively. It is evident that the bags with the higher
concentrations had higher changes in mass, because the water (a hypotonic solution)
flowed into them in an attempt to reach equilibrium, making the bags swell and
increase in mass.
In section 1C, the potato cylinders in the beaker with the 0.0 M concentration of
sucrose had the highest percent change in mass (21.4%). This was because there was
a higher concentration of solute inside the potato cells, which caused the water in the
beaker to flow into the cells in an attempt to reach equilibrium. This caused the
potato cylinders to gain in mass. The potato cylinders in the beaker containing the
0.2 M concentration solution also gained in mass, but they gained only 6.9%.
Though there was still more solute in the potato cells than in the surrounding solution,
the difference wasn’t as great, so there was less water movement into the cells before
equilibrium was reached. When the concentrations of sucrose were lower, then the
water potential outside the cell was higher than the water potential inside, and there
was a net movement of water into the cells. In the 0.4 M solution, the potato cells
decreased in mass by 4.5%. This trend continued with each solution increasing in
mass. The 0.6 M, 0.8 M, and 1.0 M solutions had -12.8%, -23.0%, and -23.5%
changes in mass respectively. This was because there was more solute in the solution
than in the potato cells, which caused water to flow out of the cells, making them
decrease in mass. The potatoes in the 1.0 M solution had the largest decrease in
mass, because they lost the most water in an attempt to reach equilibrium. When the
concentration of sucrose was higher, the water potential outside the cells dropped
below the water potential inside the cells, and there was a net loss of water from the
cells.
In the experiments, there were various possible sources of error. In section 1B, if
the bags were tied too tightly, if not enough extra room was left, then the water
wouldn’t have enough space to diffuse into the bags. Another problem would be if
the bags leaked – then some of the sucrose solution could’ve leaked into the beaker,
altering the results.
CONCLUSION:
This experiment explored diffusion and osmosis in varying concentrations of
sucrose. The dialysis tubing represented the cell membrane, which is selectively
permeable due to the phospholipid bi-layer and its proteins. The phospholipid bilayer only allows small, hydrophobic, lipid molecules to pass through, but not large,
hydrophilic, non-polar molecules. The integral proteins only allow the passage of
specific molecules, because their active sites (dictated by the sequence of amino
acids) can only bind to a certain molecule. This is part of the lock-and-key theory.
This selective permeability allows the cell to regulate what enters and leaves it. The
dialysis tubing is selectively permeable because of its small holes – only small
substances, such as glucose, can pass through – large molecules, like starch, cannot.
This experiment also proved that placing a cell in a hypotonic solution will cause
water to diffuse into it, making it swell and gain in mass. If an excessive amount of
water is transferred, then the cell can lyse. The amount of swelling and change in
mass will increase as the concentration of solute within the cell (in this case, the
dialysis bag) increases. Contrarily, if a cell is placed in a hypertonic solution, then
water will diffuse out of the bag in an attempt to reach equilibrium, causing the cell to
shrink and decrease in mass.
If a plant cell is placed in a hypertonic solution, then the water potential outside
the cell will be less than the water potential inside. This causes water to flow out of
the cell and decrease in mass (the cell becomes flaccid). If this happens in an
excessive amount, then plasmolysis is possible. If the cell is placed in a hypotonic
solution, then the water potential outside the cell will be higher than the water
potential inside the cell. Water will move into the cells until it the pressure inside the
cell becomes too large, and no additional water can diffuse in (the plant cell becomes
turgid). The rigid plant cell wall protects the cell from lysing, but it will still increase
in mass until that point.
1B:
1. The graph of the class averages is more accurate than our group’s graph. It is closer to a
straight line, which shows that it had a constant rate of diffusion. Due to sources of error,
such as leaky bags and bags tied too tightly, our group’s data wasn’t accurate. For
example, our group found that the bag containing the 0.8 M concentration of sucrose lost
.3 g in mass. This is inaccurate, as the changes in mass should steadily increase as the
concentrations of sucrose increase.
2. As the molarity of the sucrose inside the dialysis bags increases, the change in mass
increases as well. This is because as the concentration of solute increases, more water
must diffuse into the bag to reach equilibrium, which causes a gain in mass.
3. 0.0 M bag – hypertonic solution – decrease in mass as water diffuses out of the bag to try
and reach equilibrium
0.2 M bag – hypertonic solution – decrease in mass (but on a smaller scale than the 0.0 M
bag)
0.4 M bag – isotonic solution – no change in mass
0.6 M bag – hypotonic solution – increase in mass as water diffuses into the bag
0.8 M bag – hypotonic solution – increase in mass as water diffuses into the bag (on a
larger scale than the 0.6 M bag)
4. The percent change in mass will stay the same for any amount of mass. If change in
mass were graphed, that graph would be unique to the experiment and its specific masses.
However, the percent change in mass can apply for the same experiment with any mass.
1C:
1. Molar concentration of sucrose at equilibrium = 0.32 M
2.
3. Water potential of the solution at equilibrium =
Water potential of the potato cells at equilibrium =
4. The water potential will be lower because the water evaporates, which decreases the
amount of water in the potato cells. The cells will decrease in mass and become flaccid.