The aim of this experiment is to find out what effect altering the concentration of salt solution will have on osmosis in a potato chip.

Biology Coursework – Plan – Osmosis in a Potato Chip

Aim:

Background Information:

Osmosis is the diffusion of water molecules, down a water potential gradient, from an area of high concentration to an area of low concentration, through a partially or selectively permeable membrane.

The partially permeable membrane only allows the movement of small molecules or particles (often water) through it, but larger molecules such as glucose are not able to move across the membrane, as the molecules are too large. The particles will continue to diffuse until a state of equilibrium is reached. This means that the molecules or particles are evenly distributed through out both solutions and there is not a solution with a higher or lower concentration than the other.

Osmosis is usually associated with the movement of water particles in living cells. However, this can be simulated in visking tubing as well. When the vacuole in a plant cell takes in a high concentration of water molecules, it pushes the cytoplasm against the cell wall. As there is a lot of outward pressure, the cell is then able to keep its shape. The cell is then in a turgid state. The opposite of this is when there is low concentration of water molecules in the vacuole. The cell then is not full of water and collapses, which causes wilting. This is known as flaccidity.

However if the flaccidity becomes very bad and a lot of water diffuses out of the cell, the vacuole and cytoplasm continue to shrink. As the cell wall is too strong to shrink, eventually, as the cytoplasm continues to shrink, the cell membrane tears away from the cell wall. This is known as plasmolysis, and usually kills the cell.

Therefore, osmosis is very important to plant cells, as it allows them to take in water through their root hair cells, as there is normally a low concentration of water molecules in the soil. However in cases where the concentration of water in the soil is higher than the concentration in the root hair cell, active transport occurs. This is where the water is transported against the water potential gradient using energy.

A potato is good for this experiment because when water enters the cells, the concentration of water in the cell is higher than those cells surrounding it and the water diffuses into the neighbouring cells, this is then repeated until all the cells have a higher water concentration. If one cell swells up, so do all of the other cells, therefore the potato is a good object to use in this investigation.

Preliminary Experiment:

The apparatus that was used in the preliminary experiment was:

6 petri dishes – to separate the solutions

A white tile – to cut the chips on safely without marking the desktop

Distilled water

Sodium chloride solutions

Potato chips

Razor blade – to cut the chips to size

Ruler – to measure the length of the chips

Marker pen – to label the solution strengths on the petri dishes

Firstly we 18 potato chips were cut to a length of 5 cm on the white tile, using a razor blade. Then 3 of these chips were put into the petri dishes, which were then labelled, depending on the strength of the solutions that we used, these were: 0.2M, 0.4M 0.6M, 0.8M and 1.0M sodium chloride solutions. The solutions were then put into the petri dishes and left for a day. The changes in length of the chips were recorded in a table.

After analysing the results I got in the preliminary experiment, the experiment was changed. Instead of measuring the length of the chips, we would measure the mass of the chips, as it would be more accurate, because the solutions changed the shape of the chip, when they became turgid or flaccid. Not all the chips increased in length, some increased in width and became irregular, therefore not giving an accurate result for the change of length.

Pre cut potato chips will also be used, so that we don’t have to cut them to equal lengths, as it was difficult to get all of the chips exactly the same length. Each chip will be cut into 4 different parts because then I will have more results per chip and a more accurate average. Also, as the chips moved around when the solutions were poured in, and as the chips’ masses will not be the same, so the petri dishes will be labelled to show which chip is which.

The results table is maybe not as ...

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The results table is maybe not as accurate as it could have been, but it was decided that solutions to the strength of 0.5M would be used, as after this point the results of the average change in length were the same. The water threshold is reached at 0.5M so there is no need to carry on with the experiment after this point.

The drying of the chips may have affected the results because if there were excess fluids left on the chips and they weren’t dried thoroughly with tissue paper, this would have affected the mass readings of the tissues and make them heavier. Therefore to make the readings more accurate, the chips will be dried thoroughly with tissue paper before they are weighed.

Factors that affecting the Rate of Osmosis:

Temperature could affect the rate of osmosis as, if the temperature were too high, the particles would have lots of kinetic energy and increase the rate of osmosis. This will be monitored to see whether it changes dramatically through out the experiment, as it cannot be controlled. The experiment was carried out at room temperature.

Concentration is a big factor, as different concentrations of solution will have different concentrations of water molecules. Therefore if there are many water molecules in one and hardly any in the other then diffusion will take a longer time than if the concentrations of water molecules in each solution were similar and only a little movement of particles were needed to give equilibrium.

An important factor is pressure, as it can speed up or slow down the rate of osmosis. If the pressure is low, then the number of particles per unit volume is low, therefore only a small amount of particles can transfer across the partially permeable membrane. Where as if it is high, there are many more particles of sugar and water per unit volume and therefore the particles will move across the semi-permeable membrane quicker, therefore increasing the reaction rate. The experiment will be carried out at atmospheric pressure to keep the experiment fair.

The time given to the solutions to complete osmosis is also another factor because the rate of diffusion may not have reached its full potential or the state of equilibrium may have not been reached if too less time is given to the solutions.

The mass and size of the potato is another factor because if the sizes of the potato chips are not the same, then it will take different times for osmosis to be completed in each chip as a the areas that need to be filled are different. Another factor affecting the rate of osmosis is the surface area, because if there is a larger surface area the particles will be able to move more freely between the solutions, as there will be more area for osmosis to take place. However if there is a small surface area then the rate of reaction will be slower as there will be more particles moving between a smaller area and therefore less space for the particles to move freely. If the surface area: volume ratio is not the same, than the times taken for the osmosis to complete will be different.

The type of potato will also affect the rate of osmosis as different potatoes will have different structures and therefore will have different rates of osmosis. Therefore the same potato will be used to cut the chips from, this will be a maris piper potato.

Prediction:

I believe that the mass of the chips will decrease as the strength of the sugar solutions increase, because as the strength of the sugar solutions increases the concentration of the water particles in them decreases. Therefore there will be less water in the solution than in the chip and the water inside the chip will move down the water potential gradient until equilibrium is reached and past it, thus decreasing the mass of the chip.

Apparatus:

Pre-cut potato chips
6 levels of molar solution – 0.0M solution to 0.5M
6 Petri dishes – to carry out the experiments for each of the solutions in
Razors Blade - for cutting potato samples
Tile – to cut the chips on without damaging the worktop
10 cm³ Beakers - for measuring solution
Tissue - for drying potato samples
Pen - for marking the strengths of solutions
Ruler - for measuring the lengths of the chips
50 cm³ Measuring cylinder – to measure how much of the solutions we would need accurately

Method:

I will use four different strengths of salt solutions and distilled water. The strengths of the salt solutions are:

0.1M
0.2M
0.4M
0.5M

A mole is a measure of the strength of a solution. The greater the molar concentration, the stronger the solution. For example, 0.9M salt solution is stronger than 0.3M salt solution. Distilled water will also be used and has a concentration of 0.0M sugars.

1. Fill 5 petridishes with 40cm³ of a different solution in each one, do this by measuring 40cm³ of the solutions in a 50 cm³ measuring cylinder, and label which solutions are in which using the marker pen.

2. Take 5 pre cut chips and using a razor blade cut each chip into 4 equal sized parts. Measure the chips to make sure they are the same length.

3. Dry each of the 4 parts of a chip and record the initial mass of each part in a table.

4. Place the 4 parts of each chip into a petri dish and put on the lid. Do this with the 4 other chips and remember which part of each chip is which.

5. Leave the chips in the solutions in for 30 minutes. After this time, remove the chips from the solutions.

6. Dry each chip thoroughly using tissue paper and record the mass values of each part of each chip in a table. Find the % change for each of the 4 parts to each chip, by using the following formula:

% Change in mass = Change in Mass / Original Mass * 100

Fair Test:

To make sure the temperature will not affect the experiment, the experiments will be carried out at room temperature.

The same amounts of solutions and distilled water will be measured out and put into each petridish to make sure that there the reaction times are as accurate as possible.

The experiment will be left for 30 minutes until we measure the change in the mass. This is because the osmosis will be fully complete and therefore there will be no results that are unfair due to there not being enough time for the osmosis to fully complete. The chips will be left in each dish for the same amount of time.

Repetitions of this experiment will be done if there is enough time. The experiment will be repeated again so that there is a second result for the size of each chip. This is to see whether there were any anomalies due to the wrong procedure being carried out the first time.

I will use potato chips cut from the same potato because they will then have the same structure and attributes which may not necessarily be the case if there were to different potatoes with different structures. The variation is greater between two chips each from different types of potato than those from the same type of potato.

Each time I use the measuring cylinder, I will rinse it out to make sure that there are not any other liquids or substances left over that may compromise the experiment and affect the concentration of the solution being measured. The meniscus will also be measured from the bottom of the curve in order to give the exact amount of each solution needed, and no more or less.

Finally, each chip will be cut into 4 equal pieces before put into the dish so all of the chip reacts and it is a faster reaction.

Safety:

1. Be careful of the sharp edges to the razor blades and keep them well away of others when not in use.

2. Hair will be tied back, safety spectacles will be worn and all other items that are not needed for the experiment will be removed form the workstation.

3. Do not create marks on the table top while working with the blades and use the tile to stop this.

4. Do not leave any of the measuring cylinders or petridishes close to the edge so that they can be knocked over easily.

5. All general lab rules were followed.

Results

The results in the table will show the original mass, new mass, change in length and the % change in length of each of the 4 parts of each chip. The average change in mass and the average % change in mass of each whole chip will be shown. The % change in mass is calculated using the formula:

Change in Mass / Original Mass * 100

The anomalous result is the table have been highlighted in blue.

Analysis

From the results table, I can see that as the salt concentration of the solutions increased, the mass of the chip decreased. This is because as the salt concentration increased the number of water molecules in the solutions decreased and the number of salt molecules increased. Therefore as the chip is full with water and there is less water in the solution surrounding it, the water moves down the water potential gradient and as the strength of the solution increases and the number of water molecules in them decreases, more and more water moves from the chip to the surrounding solution. Therefore, this results in the mass of the chip decreasing from its original mass.

The line graph clearly shows that as the solution steadily increases in strength by 0.1M the average change in mass of the chip decreases. When the chip was put in distilled water, the positive change in mass of the chip was at its highest of 0.9425 grams, a change of 27.64%. This is because the there are no salt molecules and there are more water molecules in the solution surrounding the chip, therefore the water moves from the higher concentration, in the solution to the chip, a lower concentration, so the mass of the chip increases. The distilled water has the highest water potential, as there are no salt molecules, therefore it had the largest percentage change in mass out of all the chips, making the chip turgid.

There is a dramatic fall in the average percentage change in mass between the 0.0M solution and the 0.1M solution. This may be because in the distilled water there were no salt molecules and then in the next solution there are many salt molecules, therefore there are less water molecules. The concentration of water on the outside of the chip is less than that of the first 0.0M solution, so the water potential is less, and therefore it takes the chip less time to reach equilibrium, because there are less water molecules to move from a higher water concentration to a lower water concentration through the selectively permeable membrane of the chip.

The graph also shows that when the solution was at strength of 0.2M, the average change in mass was 0%. This shows that there was a state of equilibrium and because the concentrations of water in the solution and in the chip are both the same, there is no movement of water. Therefore the chip does not gain or lose and water i.e. the water potential has been reached. Therefore it is an isotonic solution.

As the strengths of the solutions gradually increase after the equilibrium at 0.2M the average change in mass becomes negative and the chips begin to decrease in size. This is because there were less water molecules in the salt solutions after the 0.2M solution, so the water molecules moved from a higher concentration, in the chip to a lower concentration, the solution, as there are less water molecules in the solutions because there are more salt molecules. As the graph shows that 0.3M, 0.4M and the 0.5M solutions decreased the mass of the chips, so the potato chips were becoming flaccid and some of the cells would have plasmolysed as they were deprived of the water they need to survive. Therefore theses are hypertonic solutions.

However there is an anomalous result in my graph and results table. When the salt solution is 0.4M, there is an abnormal decrease in mass of the chip. When the solution was 0.3M the decrease was by 0.60% and when the solution was 0.5M the decrease was by 0.95% but when the solution was 0.4M, the % decrease in mass of the chip was 4.55%. This may have been down to a human error, such as not taking down the mass of the chips original mass well.

My prediction ‘I believe that the mass of the chips will decrease as the strength of the sugar solutions increase…’ proved to be correct. The trend line on the graph shows this.

As osmosis controls the movement of water molecules in and out of the cells of plants, it indirectly controls the transpiration of the plant, which is the loss of water vapour from a plant through evaporation

Evaluation

I think that my experiment went reasonably well, however in my results table, there were six anomalous results. Therefore changes could have been made to the experiment to improve the accuracy of my results.

The six anomalous results are those highlighted in the results table.

These anomalous results could have occurred for different reasons:

The scales may not have been zeroed before each chip was weighed.
The person taking the readings may not have recorded the initial mass or the final mass readings accurately.
The chips may not have been dried thoroughly enough before the final mass reading was taken.
There may have been a mix up between which of the four parts of each were which and the wrong final mass readings may have been taken for each part of the chip.

In order to stop numbers 1 and 2 above from happening, the experiment could have been done in pairs with one person measuring the initial, final and changes in masses of the chips. To stop number 3 from happening, instead of using paper towel to dry the chips, the chips could have been dried using paper towel the first time then gently again by blotting paper, to make sure all the excess fluid on the chip has been removed. This is because blotting paper is more absorbent than the tissue paper. To stop there being a mix up of the chips, they could have been marked with different colours before being put into the solutions, so that they would be easily identifiable to the person writing the what the final mass readings for each part of the chip are.

Other variables that may not have been accounted for before the experiment took place were the age of the potato. The rate of osmosis would have been affected by the age of the potato, as when the potatoes are older, the permeability of the cells changes, and therefore the rate of osmosis would change, giving different results. Also, it would be better to have a new, fresh potato, as on a old potato, the likelihood of the cells being damaged is higher and therefore the rate of osmosis is slower.

Also, the experiment could have been conducted in a water bath, to make sure that there was no change in temperature during the experiment that may affect the rate of osmosis.

The times that the potatoes were left in the solution could have been varied, to find whether there was any change in the rate of osmosis after a certain time period, and then the best results from the different time period could have been used.

Another improvement could have been to use a precise cutter, that would cut the chips into exactly equal parts, and cut the chips into exactly the same dimensions every time. Although, the experiment was carried out under atmospheric pressure, the experiment could have been conducted under many pressures to find which gave the best results.

Vernier Callipers could also have been used to calculate the exact dimensions of each of the chips before and after they went into each solution to make sure that the changes in mass calculated were exact and very precise.

Another improvement could have been doing more repetitions for each solution, to get a more accurate average, or a larger range of salt concentrations between 0.0M and 0.5M could have been used to make the readings very accurate and then the anomalous results would not have as big an impact on the average changes.

Finally, before the experiment began every piece if equipment could have been checked to make sure that it worked correctly, or make sure that there were no impurities in the measuring cylinders that could have affected the experiment and make sure the scales were zeroed.