# Determining the Water Potential of a Potato Chip

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Determining the Water Potential of a Potato Chip

Aim: To determine the water potential of a potato chip

Background knowledge

What is water potential?

Water potential is the measure of the tendency of water to move from
one place to another. Like the movement of water from potato chip to
surrounding solution or vice versa. Water always moves from an area of
high water potential to a region of lower water potential.

The addition of solutes decreases water potential making it more
negative. Therefore it is less likely to move from one place to
another. Pure water has a water potential of zero, by adding solutes
you make the water potential more negative. In plants water potential
depend on two things, Osmotic potential (due to presence or absence of
solutes) and pressure potential (due to turgor pressure).

To work out water potential you use this equation

Water potential = solute potential + pressure potential

Î¨ = Î¨s + Î¨p

In my experiment the water potential equals the solute potential
(solute potential is always negative), as the pressure potential is
negligible. However pressure potential is pressure exerted when water
enters a cell and makes it turgid. Increasing pressure increases the
tendency of water to move, therefore increases water potential.
Pressure potential is always positive therefore making water potential
less negative. In my experiment I am not going to do anything to
affect the pressure, as this will disallow me to find the water
potential using the method I have selected.

What is osmosis?

Osmosis is the movement of water molecules from a high water potential
to a low water potential through a selectively permeable membrane.
This continues to happen to till dynamic equilibrium is reached and
the water potential on both sides of the partially permeable membrane
is the same.

Plant cells always have a strong cell wall surrounding them, which is
the selectively permeable membrane. When water moves in to the cell by
osmosis it starts to swell, but the cell wall prevents it from

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### Popular Essays

bursting. Plant cells become "turgid" when they are put in dilute
solutions. This is when the cell becomes fully inflated. Turgid means
swollen and hard. The pressure inside the cell rises; eventually the
internal pressure of the cell is so high that no more water can enter
the cell. This liquid or hydrostatic pressure works against osmosis.

Kinetic energy is the energy available to the atoms in a substance to
move. The more kinetic energy there is the more movement there is as
molecules move around and collide more providing more energy.

What is plasmolysis?

When plant tissue is placed in concentrated sucrose solution the cells
lose water by osmosis and so become flaccid: this is the opposite of
turgid meaning it is loose. If you put plant cells into concentrated
sugar solutions and look at them under a microscope you would see that
the contents of the cells (protoplast) have shrunk and pulled away
from the cell wall: they are said to be plasmolysed. This only happens
when cells are placed in a solution with lower water potential then
the actual cells.

When plant cells are placed in a solution, which has exactly the same
osmotic strength as the cells they are in a state between turgidity
and flaccidity. We call this incipient plasmolysis.

Transpiration pathways

The movement of water through a plant is due to a transpiration pull.
There are three ways in which water moves through plants.

1) Apoplast pathway- where water only moves directly from cell wall to
cell wall of neighbouring cells

2) Symplast pathway is where water moves through cytoplasm and
partially permeable membrane only. It may move from cell to cell
through plasmodesmata or through adjacent plasma membranes.

3) Vacuolar pathway is through the sap of the vacuoles through
tonoplast. And through cell wall and cell membranes of adjacent cells.

Hypothesis / Prediction
=======================

Firstly I have to predict at what concentration there will be no
change in mass in the potato chip. From my preliminary results I
predict that the point of no mass change in the potato chip will be
between the concentrations of 0.1mol dm Ö¾Â³ and 0.3mol dm Ö¾Â³.

The movement of water molecules in and out of the potato cells depends
on various factors.

* The amount of water present in the cytoplasm of the potato chip
cells and in the surrounding sucrose solution.

* Also the concentration of sucrose on either side of the cells
surface membrane

* In plants pressure could also play a factor; the pressure exerted
on the cell contents by cell wall affects the movement of water in
and the cell. However this is negligible in this experiment.

With this experiment there are three possibilities:

1) The chip will either gain mass

2) Or lose mass

3) Or stay the same mass

If the sucrose solution surrounding the chip has a higher water
concentration than inside the cells of the chips, then the cells will
gain water by osmosis. This means the chip will increase in mass. This
is because water molecules move from the high concentration (diluted
sucrose solution) to a lower concentration, which is the potato chip
in this instance. As water molecules enter the cell the e pressure
inside the cell increases, therefore increasing the turgidity pressure
against the cell wall, however this is negligible.

[IMAGE]

If the sucrose solution has a lower water concentration than inside
the cells of the chip, then the chip will lose mass. This is because
the water molecules move from inside the chip to the surroundings
solution due to osmosis. As the cell loses mass through water the
protoplast of the cell gradually shrinks exerting less and less
pressure on the cell wall until there is no pressure. Therefore at
this point the pressure potential is zero. This means the water
potential will equal the solute potential.

The protoplast continues to shrink and pulls away from the cell wall
this is called plasmolysis.

[IMAGE]

If the solution is the same concentration as the concentration inside
the potato chip then the potato chips mass will remain the same. This
is because there will be no movement of water molecules as the water
potential on both sides is in equilibrium.

[IMAGE]

The partially permeable membrane has been the cell membrane in each
case. The cell membrane doesnÂ’t allow sucrose molecules through. So we
experiment is to find the water potential of the potato tuber. The
sucrose molecules wonÂ’t move inside the potato chip and affect the
water potential. So my method is still viable and sufficient to find
the water potential of the potato tube.

Kinetic theory is another factor, which affects the results. When the
sucrose is dissolved is water, the water molecules are attracted to
the sucrose molecules. The attraction of water molecules to sucrose
limits their ability to move freely. The more concentrated a solution
is the less the water molecules will move reducing the kinetic energy
in the solution. This in turn reduces the water potential as the
potential for water to move is reduced

For this experiment I am predicting that as the more concentrated a
solution becomes, meaning the lower the water potential the chips will
lose more mass. The most concentrated solution (0.4m) will lose the
most mass and the least concentrated (highest water potential) will
gain the most mass. The chip, which is the same water potential as the
solution, will stay the same mass.

Equipment / Apparatus
=====================

Equipment

Measurements

Reason for choice over alternatives

5 boiling test tubes

Large

Boiling tubes are ample enough for three potato chips to be placed in.
Beakers are too large and take up too much space

Cork borer

Medium

The small cork borer cuts them to thin so the chips tend to snap. The
big one cuts them too thick so they donÂ’t fit in the boiling tube. The
medium cork borer cuts them suitably.

Cooking Knife

Small

I could have used a scalpel but these are contaminated with all sorts
of chemicals in the lab also they are too blunt to cut potatoes.

Pipette

Pipette was used to ensure the precision of the measurement of the
liquids.

2 Measuring cylinders

(10ml) (100ml)

10ml-measuring cylinder is used to measure the sucrose, as this is
ample enough. The 100 ml measuring cylinder is used to measure the
water.

Digital scales

(To 2.d.p.)

2 decimal places is a suitable degree of accuracy for this experiment.

Test tube rack

With at least seven slots

Keeps the whole experiment in one place so there is less chance of
parts of the experiment going missing.

OHP Pen

-

To write the concentration of each solution on the test tubes so there
is no confusion as to which concentration is which.

Cling Film

Used to cover the test tube so no solution is lost through
evaporation. I could have used foil but this disallowed me to see
inside the experiment as the cling film allowed.

Lab tweezers

Large

Has to be large in order to pull the chips out of the large (deep)
boiling tubes.

1 White potato

Large

White potatoes are soft enough to place a cork borer in comfortably.
Red potatoes are too hard.

Glass rod

-

To stir the solution a spoon is too large and wonÂ’t fit in boiling
tube

Paper towels

-

Absorbs water well tissue paper would be too weak

White tile

-

To cut potato on so the table doesnÂ’t get scratched

Concentration

(Mol dm Ö¾Â³)

Mass Before (grams)

Mass After (grams)

Change in mass

% Change in mass

1

0.10m

Black

0.69

0.73

+0.04

+5.80

2

0.10m

Blue

0.65

0.71

+0.06

+9.23

3

0.10m

None

0.67

0.76

+0.11

+16.42

4

0.15m

Black

0.66

0.67

+0.01

+1.51

5

0.15m

Blue

0.62

0.62

0.00

0.00

6

0.15m

None

0.62

0.67

+0.05

+8.06

7

0.20m

Black

0.68

0.51

-0.17

-25.00

8

0.20m

Blue

0.74

0.57

-0.17

-23.00

9

0.20m

None

0.72

0.55

-0.17

-23.61

10

0.25m

Black

0.69

0.60

-0.09

-13.04

11

0.25m

Blue

0.77

0.70

-0.07

-9.09

12

0.25m

None

0.70

0.64

-0.06

-8.57

13

0.30m

Black

0.72

0.39

-0.33

-45.83

14

0.30m

Blue

0.71

0.47

-0.24

-33.80

15

0.30m

None

0.61

0.34

-0.27

-44.26

16

0.35m

Black

0.65

0.34

-0.31

-47.70

17

0.35m

Blue

0.71

0.51

-0.20

-28.17

18

0.35m

None

0.66

0.51

-0.15

-22.73

19

0.40m

Black

0.65

0.41

-0.24

-36.92

20

0.40m

Blue

0.62

0.42

-0.20

-32.26

21

0.40m

None

0.65

0.44

-0.21

-32.31

Actual results

Concentration mol (mol dm Ö¾Â³)

Average change in mass (grams) (2.d.p)

Average % change in mass (2.d.p)
================================

1

0.10

+0.07

+10.48

2

0.15

+0.02

+3.19

3

0.20

+0.17

-23.87

4

0.25

-0.07

-10.23

5

0.30

-0.28

-41.30

6

0.35

-0.22

-32.87

7

0.40

-0.22

-33.83

Analysis of results
-------------------

To test the reliability of my results I have to assess how far apart
the repeats are from the mean. If they are very far apart then that
means that the results obtained are inaccurate. If the repeats are
close together then that means the results obtained are fairly
accurate depending on how low the figure is. The lower the figure the
is the more accurate the results are. I am going to use standard
deviation to prove this.

Concentration
=============

(Mol dm Ö¾Â³)

Standard deviation of repeats

0.10

5.42

1

0.15

4.28

2

0.20

1.02

3

0.25

2.08

4

0.30

6.54

5

0.35

13.13

6

0.40

2.68

7

(1) The Chips in 0.1m solution gained mass by 10.48% as the solution
had a higher water potential then the potato chip. The standard
deviation of the results is fairly good (5.42).

(2) The chips in 0.15m solution also gained mass but to a lesser
extent by 3.19%, therefore the sucrose solution still had higher water
potential still. However it proves that the water potential was lower
than the 0.1m solution as it is more concentrated. The standard
deviation is fairly reliable (4.28).

(3) The best standard deviation of mass is the chips in the solution
0.2 mol (1.02)

This is the most reliable batch of results, because the repeats do not
vary from the mean a lot at all. It also fits the trend on the graph
and is the closest to the line best fit.

(4) 0.25moldm

One of the anomalies, which I identified on the graph, was the chips
in the concentration of 0.25 mol dmÖ¾Â³. I think this is an anomaly
because it is way out of the line of best fit also, it doesnÂ’t fit
with the general trend Â– that as the solution becomes more
concentrated the chip loses more mass. This is because the more
concentrated a solution is the lower the water potential. Therefore
due to osmosis the water will move from higher water potential (which
is the potato chip in this case) to a low water potential, which is
the solution.

However in despite of the positioning on the graph the standard
deviation of the repeats of the 0.25 mol dm Ö¾Â³ - is quite good (2.08).
Meaning the repeats donÂ’t vary a lot from the mean. This means this
anomaly could have been caused because the dilution of the
concentration is wrong. This is most likely as all the repeats were
placed in the same solution. There fore all the results of the repeats
in 0.25 mol dmÖ¾Â³ sucrose solution were wrong due to the dilution most
probably being incorrect

(5) 0.3moldm

This is an anomaly as it far from the line of best fit. Also it has
the second worst standard deviation of repeats. This again could be
due to chips being dried to different extents.

(6) 0.35moldm

From standard deviation I have found that the chips in the solution
0.35moldmÖ¾Â³concentration definitely came up with results, which are
anomalies. The repeats at this concentration are very far from the
mean therefore the standard deviation number is really high (13.13).
So the average from this concentration is definitely an anomaly. These
anomalies could have arised because the chips werenÂ’t dried in the
same way. Meaning some may have been dried better, squeezing some of
the water out of the chip, whilst others were dried lightly meaning
there was still excess water on the outside, which wasnÂ’t picked up
from the paper towel

(7) 0.40moldm

The standard deviation for these repeats is the second best and so
indicates that these repeats are reliable results also it is close to
the line of best fit on the graph.

Conclusion

The general trend, which has been represented by the graph, is that
the more concentrated the solution is the more mass is lost from the
potato chip. This proves my original prediction be true

From graph one I have concluded using the line of best fit that the
concentration at which there is 0% mass change is 0.13m

Then from graph two I allocated this point and the solute potential at
this concentration is 360kpa

Therefore the water potential of the potato chip is 360kpa (as it
equals the solute potential )

Evaluation
==========

Suitability

The method used may have been done in different ways. However the
method I used was most suitable for a school laboratory experiment. I
was able to get fairly decent results, which generally followed the
trend that I predicted. Which was the more concentrated the solution
is the more mass the chip will lose. This is due to osmosis and the
kinetic theory.

Limitations/Sources of error
----------------------------

* Drying method not done in the same manner enough

* Reliability of scales- not being set on zero

* Careless and vigorous handling- too much contact with hands

* Incorrect dilution due to carelessness

* The test tubes may have been tampered with overnight

* Cling film was unnecessary as this made the test tube more humid
for no reason.

Many improvements could have been made to the method such as:

* As well as measuring the mass changes I should have measured the
volume changes. I could have done this by using the diameter and
length of the chip and seeing how this changed after being in the
solution. This may have been a better indicator as to what the
water potential is. By examining the point in which there is no
volume change, due to incipient plamolysis.

* Instead of doing all the repeats in the same boiling tube. I
should have made a fresh solution for each chip in a separate
boiling tube. This would have reduced the chance of human error.
So if the dilution was done incorrectly it wouldnÂ’t affect all the
repeats as it did in my experiment. Also I wouldnÂ’t have to use
thread to identify which potato is which as they are in a separate
boiling tube. Thread could have affected my results as it damages
the surface cells and also disallows osmosis to occur in the small
region it covers.

* I should have used a fair drying method to remove the water
outside the surface of the potato. If I allowed them all to dry
for five minutes and then measured the mass, maybe my results
would have been more accurate. Other methods such as using a
hairdryer etc would incur more human error.

* Also by getting more people involved in the experiment I could
take the chips out at exactly the same time making it an even
fairer test. Also there would be fewer delays in the method. So
this makes the experiment quicker saves time allowing more time
for more repeats.

* I could have used a narrower range of dilution to find the point
of no mass change more precisely for example 0.1m, 0.125m, 0.15m,
0.175m, 0.2m, 0.225m, 0.25m, 0.275m, 0.3m, 0.325m, 0.35m, 0.375m,
0.4m.

* I could have used a microscope to count the amount o cells, which
are plasmolysed

* I should have used distilled water instead of tap water. Tap water
may have solutes, which would affect my dilution, which would
further affect my results. With distilled water there are no
solutes so less chance of the dilutions being the wrong
concentrations.

* I should have implemented a way of controlling the temperature.
Using a water bath could have done this.

* The cling film could have caused humidity, which could have
affected the results. So instead of using cling film I could have
used light proof sheets. Which would have also stopped variations
in light intensity affecting the results.

Bibliography cc.

Â· "Knowledge from Biology 1" (Cambridge 2001) - hypothesis

Â· "Encarta Encyclopaedia" (C-D rom 2000) - planning

Cambridge advanced sciences biology 1 (OCR)- hypothesis

http://www.phschool.com/science/biology_place/labbench/lab1/watcalc.html
(conclusion)