Khalid Tawil
AP Bio, 6th
11/27/13
The Effects of Surface Area and Solute
Concentrations on Diffusion and Cells
Abstract
This lab investigation serves to study the effects of various conditions on the rate of diffusion
across a membrane. The central question studied was what caused plants to wilt if they are not
watered. To study this, three experiments were designed. The first dealt with the effects of the
surface area to volume ratio on diffusion, the second dealt with diffusion among hypertonic
and hypotonic solutions, and the third experiment dealt with diffusion when solute
concentration varied. Plants wilt because without water diffusion into the plant cells, the plant
cells shrivel up and are unable to support the plant.
Introduction
The central idea of the lab is to study diffusion across a membrane as well as to find out why
plants wilt if they are not watered. To solve this problem, it is necessary to find out how
watering a plant affects the plant on a cellular level. These cells are already in aqueous
environments, and water is the solvent that carries material. Because of water’s properties, it is
the universal solvent.
In order to maintain homeostasis, cells need to transport materials through their membranes.
To do this, cell membranes are selectively permeable. They allow certain particles in or out. The
process by which most material travels across a membrane is known as diffusion. The cell
membrane is comprised primarily of phospholipids with hydrophilic heads that face outward
and hydrophobic tails on the inside of the membrane. This change in polarity on the inside and
outside of the membrane prevents ions from diffusing across the cellular membrane. As a
result, only nonpolar particles can readily diffuse across. Some substances, like proteins, are
also too large to diffuse across the cell membrane even if it is nonpolar. In order for a cell to
regulate its concentration of ions and larger molecules, certain proteins embedded in the cell
membrane act as channels that control the intake and outtake of ions and larger molecules.
The various methods a cell employs to regulate the flow of materials also requires an input of
energy. There are two types of transport: passive and active transport. Passive transport does
not require an input of energy from the cell and this type of transport always operates when
materials move from regions of high concentration to regions of low concentration. Simple
diffusion through a membrane and facilitated diffusion – diffusion with the help of a protein
carrier or channel – are examples of this. Active transport requires an input of energy from the
Tawil, 2
cell directly – usually in the form of ATP. Because of an input of energy, cells can use active
transport to move molecules against their concentration gradient.
The movement of water across a membrane is known as osmosis. Water goes down its
concentration gradient but goes against the concentration gradient of solutes. Specifically,
water will travel from areas of low solute concentrations to areas of high solute concentrations
in an attempt to equalize the concentrations of solutes. A solution with a higher solute
concentration than its surroundings is hypertonic for that solute and a solution with a lower
solute concentration than its surrounding is hypotonic for that solute. A solution with an equal
solute concentration to its surroundings is isotonic. If a cell is hypertonic, water will move into
the cell and cause the cell to swell up. In plant cells with a cell wall, the pressure the cell
membrane exerts on the cell wall as it swells up is known as turgor pressure. If a cell is
hypotonic, water will move out of the cell and the cell will shrivel up.
The overall movement of water can be predicted by the area’s water potential. Water potential
is a parameter that determines the direction water will go in. Water potential is the sum of the
solute potential, which depends on the solute concentration, and pressure potential, the
exertion of pressure on a solution. The equation for water potential is given as = + ,
and solute potential
can be calculated using the equation
, where i is the
ionization constant, C the molar concentration of a solute, R the constant 0.0831 liter
bars/mole-K, and T the temperature in K (273 + oC). Water always travels from an area of high
water potential to low water potential.
To test diffusion and osmosis, three experiments were performed. First, if 3 agar cubes
dissolved with NaOH are placed in acetic solution, the acetic acid solution will diffuse most
rapidly into the agar cube with the greatest surface area to volume ratio. This is because a large
surface area means more molecules can diffuse at once but a large volume needs to be
supported by all the incoming molecules. To test this, three agar cubes dissolved in NaOH were
placed in a 1 M acetic acid solution and the rate of diffusion was analyzed as the solution
diffused into the agar.
Second, if dialysis tubing is filled up with a solution and placed in water, the mass of the dialysis
tubing will increase. This is because water will move into the dialysis tubing, where the solute
concentration is greater. For this experiment, dialysis tubing was filled with various solutions
and then placed in a beaker filled with water. The mass of the tubing was recorded before and
after a certain period of time.
Finally, if potato cores are placed in a sucrose solution and their mass increases, then the
sucrose solution has a larger water potential than another sucrose solution where the mass of
the potato cores decreased. Water potential dictates that water moves from an area of high
Tawil, 3
water potential to low water potential. Given the equation of solute potential,
,
the larger the value for C is, the lower the solute potential. The lower the solute potential is, the
lower the water potential is as well. In short, higher concentrations of solute should decrease
water potential, causing water to flow into the given solution. For this experiment, a pair of
potato cores was placed in a sucrose solution with an unknown molarity and the masses were
recorded before and after the cores were in the solution for 24 hours.
Materials and Procedures
Safety Procedures
Wear safety glasses or goggles, gloves, aprons, at all times when working with acids and other reactive
chemicals. Have spill-proof trays or pans for solutions with these chemicals. Wearing gloves is especially
important when cutting agar, be careful not to cut a finger or hand.
Procedure 1
Materials





Serrated knife
Metric ruler
250 mL of 1 M Acetic Acid solution
1000 mL beaker
Timer or clock
A cube of agar with a 1 cm side length, a cube of agar with a 1.5 cm side length, and a cube of
agar with a 2cm side length were all cut out from a 2% agar sample containing NaOH and
phenolphthalein using a serrated knife and a ruler. The surface areas and volumes of each cube
were recorded in Table 1 using the metric ruler. Then, 250 mL of 1 M acetic acid solution was
poured into a 1000 mL beaker. All three cubes were placed into the 1 M acetic acid solution
simultaneously for 7 minutes. Afterwards, a metric ruler was used to measure the surface area
and volume of the part of the agar cube that was still diffused with only NaOH. This was done
for each of the agar cubes and the data was recorded in Table 1.
Procedure 2
Materials





Five pieces of dialysis tubing
Pipette
Tap water
Metric scale
Five 250 mL beakers
Tawil, 4





200 mL of sucrose
200 mL of NaCl
200 mL of albumin
200 mL of glucose
Timer or clock
Five 20 cm pieces of dialysis tubing were each tied up on one end. A pipette was used to fill up
each piece with different solutions – sucrose, NaCl, albumin, glucose, and tap water
respectively. After being filled up with a solution the other end of each piece of dialysis tubing
was tied as well. Each piece of dialysis tubing was then weighed using a metric scale and the
mass in grams of each piece of tubing was recorded in Table 2. Five 250 mL beakers were then
filled up with 200 mL of tap water each. The five pieces of dialysis tubing were then placed in
different beakers simultaneously for 20 minutes. Afterwards, each piece of dialysis tubing was
weighed using a metric scale and the mass in grams was recorded in Table 2.
Procedure 3
Materials





Potato
Cork borer
Metric scale
Four 250 mL beakers
Prepared sucrose solutions
First, a cork borer was used to obtain 8 potato cores from a potato. The potato cores were
grouped in pairs of two and were weighed using a metric scale. The mean mass in grams of
every pair of potato cores was recorded in Table 3. 50 mL of sucrose from the prepared red
solution was poured in a 250 mL beaker. 50 mL of sucrose from the yellow, green, and blue
solutions were poured into separate 250 mL beakers as well. Each pair of potato cores was then
placed in a different colored solution for 24 hours. Afterwards, each pair of potato cores was
weighed using a metric scale and the new mean mass in grams of each pair was recorded in
Table 3.
Results/Data Collection/Analysis
Procedure 1
In short, the data from the experiment demonstrated that the rate of diffusion of a solution
across a membrane varies with the surface area of the membrane as well as the volume of the
membrane-bound object.
Tawil, 5
Table 1: Changes in Volume of Agar Cubes when Acetic Acid Diffuses into Agar
Side Length of Agar Cube (cm) Surface Areas (cm2) Volume Before (cm3) Volume After (cm3)
1.0
6.0
1.000
0.224
1.5
13.5
3.375
2.548
2.0
24.0
8.000
4.860
Table 1 above shows that smaller cubes lead to higher rates of diffusion. There are 3 agar cubes
in Table 1 of varying side lengths and the "Volumes After" column refers to the volume of the
agar cube diffused with NaOH after 7 minutes in a 1 M acetic acid solution. Larger surface areas
appear to increase the rate of diffusion, but higher volumes inhibit it.
Figure 1: Change in Volume of Agar Diffused with NaOH After 7 Minutes
Volume (cm^3)
Change in Volume of Agar Diffused with NaOH
9
8
7
6
5
4
3
2
1
0
Before
After
6
13.5
24
Surface Area of Agar Cube (cm^2)
In Figure 1, the larger the surface area of an agar cube, the larger the change in volume of the
agar that was diffused with NaOH. This does not refer to the change in volume as a percentage
of the original volume, however, merely the amount of volume that the acetic acid solution
diffused into.
Tawil, 6
Procedure 2
This experiment demonstrated that water diffuses into a region that is hypertonic to the
surrounding area.
Table 2: Rate of Diffusion through Dialysis Tubing
Solution in
Mass Before (grams) Mass After (grams)
Tubing
Sucrose
39.7
40.4
NaCl
38.2
39.2
Albumin
37.3
37.3
Glucose
39.4
41.2
Tap Water
36.8
36.5
In Table 2, the mass of almost every dialysis tubing changed after being submerged in tap
water. The mass of the tubing containing albumin did not change, however. Masses were taken
of dialysis tubes before and after they were submerged for 20 minutes in a tap water.
Figure 2: Changes in Mass of Dialysis Tubing in Tap Water
Changes in Mass of Dialysis Tubing in Tap Water
42
41
Mass (grams)
40
39
38
Before
37
After
36
35
34
Sucrose
NaCl
Albumin
Glucose
Solution in Dialysis Tubing
Tap Water
Tawil, 7
In Figure 2, the mass of the dialysis tubing containing glucose increased the most. The mass of
the control, the dialysis tubing containing tap water, decreased. Besides the dialysis tubing
containing albumin and tap water, the mass of each dialysis tubing increased.
Procedure 3
This experiment demonstrates that the rate of diffusion varies with the difference in molar
concentration.
Table 3: Diffusion of Sucrose through Potato Cores
Red
Yellow
Green
Blue
Solution Solution Solution Solution
Mass Before (grams)
3.3
3.5
3.3
3.3
Mass After (grams)
2.9
3.1
3.7
3.4
In Table 3, the greatest percent mass change occurred when the pair of potato cores were
submerged in the red-colored sucrose solution. The colored solutions all contained sucrose at
unknown molar concentrations, but the molar concentration can be extrapolated by examining
how much the mass of a specific potato core changed. The greater the mass change is, the
greater the difference in concentrations is.
Figure 3: Change in Mean Mass of Pairs of Potato Cores after 24 Hours in Sucrose Solutions
Change in Mean Mass of Pairs of Potato Cores
after 24 Hours in Sucrose Solutions
4
3.5
Mass (grams)
3
2.5
2
Before
1.5
After
1
0.5
0
Red
Yellow
Green
Color of Sucrose Solution
Blue
Tawil, 8
In Figure 3, the mass of the potato cores increased in the green and blue sucrose solutions but
decreased in the red and yellow sucrose solutions. In addition, the mass of the potato cores in
the blue sucrose solution changed the least.
Discussion
The original question of the experiment was to decide why plants that are not watered wilt. The
experiments below all deal with diffusion and osmosis, and in short, plants that are not watered
do not have the turgor pressure to support their weight. Watering plants causes water to
diffuse into the cells and produce larger cells that keep a plant upright.
Procedure 1
The hypothesis that the acetic acid solution will diffuse most rapidly into the agar cube with the
greatest surface area to volume ratio was supported by the data. This is due to the fact that a
large percentage of the cube’s volume was diffused with acetic acid after 7 minutes.
Table 4: Comparison of Surface Area and Percentage of Agar Cube Diffused with Acetic Acid
Side Length of Surface Area to % of Agar Diffused
Agar Cube (cm) Volume Ratio
with Acid
1.0
6:1
77.6
1.5
4:1
24.5
2.0
3:1
39.3
In the table above, the cube with the 1.0 cm side length had the largest surface area to volume
ratio, 6:1. In addition, 77.6% of its volume was diffused with acetic acid by the end of the
experiment. The cube that had the least amount of its volume diffused with acetic acid was the
agar cube with a 1.5 cm side length. Mathematically, the volume of a cube is given by the
equation,
, where s is the length of a side. The surface area of a cube is given by the
equation,
, where s is the length of one side of a cube as well. Since the volume
equation is of a higher degree, volume increases faster than surface area, causing a diminishing
trend of surface area to volume ratios.
This result, however, is most likely an error because the agar cube with the 2.0 cm side length
should have the smallest percentage of its volume diffused with acetic acid. Most likely, the
error involved in this experiment was probably due to incorrect measurement of the agar that
was still diffused with NaOH only. To improve the results of this experiment, the time should be
recorded when the cube completely diffuses with acetic acid in order to remove the human
error of subjectively measuring the amount of the agar that still contained NaOH.
Tawil, 9
Procedure 2
In this experiment, the hypothesis that if dialysis tubing is filled up with a solution and placed in
water and that the mass of the dialysis tubing will increase, was supported by the data as well.
In each beaker that contained dialysis tubing, the mass of the dialysis tubing increased after
being submerged in tap water for 20 minutes, with the exceptions of the tubings that contained
albumin and tap water.
Albumin is a protein that could not diffuse across the dialysis membrane, but water should have
entered the cell because there was a higher solute concentration inside the tubing than
outside. However, considering that the mass of the control, the tubing containing tap water,
decreased, this raises the possibility that some other lurking variable may be causing water to
exit the cell. In this scenario, the cell containing albumin might have originally gained mass but
water exiting due to some lurking variable might have negated the effect. To improve the
results of this experiment, distilled water should be used instead of tap water, since the solutes
in tap water might have created another concentration gradient by chance.
Procedure 3
The hypothesis that if potato cores are placed in a sucrose solution and their mass increases,
then the sucrose solution has a larger water potential than another sucrose solution where the
mass of the potato cores decreased, was also supported by the data. For each solution, the
percent mass change was found by the equation
. By calculating mass change
and knowing that the molarity of the various colored sucrose solutions varied in molarity from
0.2, 0.4, 0.6, to 0.8, the molarity of the sucrose solution could be determined. The greatest
mass increase correlates with the lowest solute concentration, since water would diffuse into
the potato cells from the low concentration solution.
Table 5: Further Analysis of Sucrose Diffusion through Potato Cores
Red
Yellow
Green
Blue
Solution Solution Solution Solution
% Mass Change
-12.12
-11.43
12.12
3.03
Molar Concentration
0.8 M
0.6 M
0.2 M
0.4 M
Water Potential (bars) -19.55
-14.66
-4.89
-9.77
After determining the molar concentration of a solution, this value was used as C in the equation for
. R was the constant 0.0831,T was 294 K, and i was 1. Since no
pressure potential acted upon the potato cells, the solute potential was equal to the water
potential. The table illustrates that the green sucrose solution had the potato cores with
greatest mass increase as well as the largest water potential.
solute potential,
Tawil, 10
One possible sources of error in the experiment include only incorrectly measuring the mass of
the potato cores by not factoring in the water that drips off the core once it is removed from its
solution. Another source of error could have been using potato cores that had small holes in
them, increasing surface area. For future experiments, the potato cores should be homogenous
as well as free of holes.
Conclusion
In summation, the rate of diffusion relies on the surface area to volume ratio as well as the
concentration gradient of a solute. The higher the surface area to volume ratio is, the higher
the rate of diffusion across a membrane is. In addition, a large concentration of a solute outside
a cell causes water to leave the cell and vice versa. This lab proved that water will diffuse
against the concentration gradient of solutes but down its own concentration gradient.
The first experiment demonstrated that larger surface areas allow more solute or water to
diffuse into a cell, but a smaller percentage of the cells total volume becomes diffused with the
solution because larger cells tend to have smaller surface area to volume ratios. The second
experiment proved that water will diffuse into a hypertonic solution because of a larger
concentration of solutes present. The third experiment demonstrated that water diffuses faster
into cells with lower solute potentials.
This investigation aided in not only developing an eye for details and trends but also helped
teach experimenters to better communicate ideas and better explain data and findings. In
addition, performing this investigation also greatly helped improve organization skills as well as
emphasize aspects like hypotheses and sources of error that should be considered more in
detail in future investigations.
Questions






When does net osmosis rate equal to zero?
How does a cell membrane affect the diffusion of different substances?
When do solutes diffuse across a membrane?
What are the effects of pressure potential on water potential?
How do multiple solute concentration gradients affect osmosis?
How does too much water affect plants?
Tawil, 11
References
Buddies, S. (2010, February 23). A strong hypothesis. Retrieved from
http://www.sciencebuddies.org/blog/2010/02/a-strong-hypothesis.php
Diffusion and osmosis. (2004, May 04). Retrieved from
http://kvhs.nbed.nb.ca/gallant/biology/osmolab.html
Lye, B. (2002, April 26). Passive transport versus active transport. Retrieved from
http://biology.kenyon.edu/HHMI/Biol113/passive_vs_active.htm
Lye, B. (2002, April 26). The movement of water in plants. Retrieved from
http://biology.kenyon.edu/HHMI/Biol113/movemet of water in plants.htm