Investigate the effect of the strength of a solution on how much water moves in to or out of a potato chip.

By Victoria Wright, 11T.

In this investigation, I am going to investigate the effect of different strength solutions on how much water moves in to or out of a potato chip.

The variables involved in this investigation are:

Strength of the solution
Volume of the solution
Mass of potato chip
Length and width of potato chip

My input variable in this investigation will be the strength of the solution, in which the potato chips will be placed. By changing this variable, I will see the effect of different strength solutions on how much water goes in to or out of a potato chip, through the process of osmosis. Osmosis is the diffusion of water particles from a region of high water concentration or a dilute solution to a region of low water concentration, a concentrated solution through a selectively permeable membrane. Osmosis occurs because a selectively permeable membrane is a structure perforated with minute pores large enough to allow the passage of small molecules such as water, but to small to allow the passage of larger molecules such as salts and sugars. Therefore, it acts like a molecular sieve. The diagram below shows what happens during osmosis.

Prediction.

Due to what happens during the process of osmosis I predict that the higher the water concentration, the weaker the solution; therfore, the larger the voloume of water that will from the beaker into the vacuoles of the chip’s cells. So, consequently the stronger the soloution the larger the volume of water that will move out of the chip’s cells into the beaker.

Osmosis is the movement of water molecules from an area of high water concentration to an area of low water concentration through a semi permeable membrane, and osmosis is a vital device in the transport of fluids in living organisms, for example, in the transport of water from the soil to the roots in plants.

Therefore the scientific reason for my prediction is that, if I was to place a potato chip in the weakest soloution possible, pure distilled water there would there would be a extremely high concentration of water molecules in the beaker. Osmosis is the diffusion of water molecules from an area of high water concentration to an area of low water concentration through semi-permeable membrane. Therefore, as there is a very high concentration of water molecules in a zero molar solution, there will be more water molecules in the beacker to move through the semi-permable membrance (in this case the cell membranes of the cells in the potato chip), resulting in a large volume of water moving into the potao chip. So the chips will swell up causing the mass in,crease and as the water nolecules have moved into the potato chip the volume of solution in the boiling tube will fall.

Consquently, If I was to place my chips in a very strong one molar solution the mass of the would fall greatly. This would be because the concentration of water molecules in the chip would be far higher that the concentration of water molecules in the solution in the beaker outside the chip. Therefore, as osmosis is the movement of water molecules from an area of high water concentration to an area of low water concentration, the molecules would move from the high concentration of water molecules in the cells of the potato chip to the low concentration of water molecules inside the beaker causing the mass of the chip to decrease greatly. This is because the direction of the movement of the water molecules is always from an area of high water concentration to an area of low water concentration, the net movement of water molecules, this is shown in the diagram on the previous page. In osmosis the water, molecules will continue to diffuse until the area in which the molecules are found reaches a state of at which the concentration of the solution and potato are balanced meaning that the molecules are randomly distributed in both the potato and the solution, with no area having a higher or lower concentration than any other. Due to this, the amount that the mass of the chip increases or decreases at a certain concentration depends on the concentration of water molecules inside the cells of the potato, and different potatoes have different concentrations of water molecules in their cells, this is why it will be important for me to use the same type of potato throughout my investigation. If I was to place a chip in a solution and there was no change in the mass of the chip after being in the solution for a couple of hours, this would suggest to me that the concentration of water molecules in the solution are equal to the amount of water molecules in the potato, therefore equal concentration. However, in this situation the process of osmosis is still occuring, but we do not see a change in mass of the potato as the number of water molecules moving into the chip is equal to the number of water molecules moving out of the chip.

There will be a point, when the potato chip is place in a solution of high water concentration, where the chip will show no indication that it is increasing further in size or mass. This is because the potato cells are fully turgid and no more water can enter. Plant cells have two semi-permeable membranes, one usually pressed tightly against the cellulose cell wall and one surrounding the cell vacuole, however they act as one. As plants have a cellulose cell wall, the effect of osmosis on plant cells is different to that in animal cells. When plant cells take up water by osmosis, the cells to swell, but the cellulose cell wall prevents them from bursting unlike animal cells. Plant cells become "turgid" when they are put in dilute solutions. The pressure inside the cell rises and eventually the internal pressure of the cell is so high that no more water can enter the cell, at this point, the cell is fully turgid, and the pressure against the cellulose cell wall is called “turgor pressure”. Turgor pressure is of great importance in maintaining the shape of plant structure, it is responsible for keeping the shape of the leaf. Young seedlings, which have not yet developed much xylem, are kept upright almost entirely on turgor pressure. It gives plants their support.

Cellulose cell wall Cytoplasm Water enters by osmosis. ...

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Cellulose cell wall Cytoplasm Water enters by osmosis.

Vacuole

Vacuole enlarges and pushes cytoplasm against cellulose cell wall.

Fully Turgid Cell

Cellulose cell wall forces back until pressure in both directions is equal and no more water can enter the cell, as it is now fully turgid.

There will also be a point when the chip is place in the strong solution of low water concentration, where there is no indication that the potato chip is decreasing any further in mass or size. This is because the cell is fully plasmolysed and no more water can leave the cell. When plant cells are placed, in concentrated sugar solutions, the process is reversed. The cell loses water by the process of osmosis, causing the vacuole and cytoplasms to shrink away from the cellulose cell wall making it limp. The content of the potato cells shrinks and pulls away from the cell wall. These cells are said to be plasmolysed.

Concentrated solution

Cellulose Cell Cytoplasm shrinks Wall away from cell wall.

Outer cell membrane becomes visible.

Water leaves by osmosis.

Plasmolysed cell.

Limp cellulose cell wall

Shrunken Vacuole

Outer cell membrane

This process is called plasmolysis and the cell is said to be plasmolysed. By the cell walls becoming limp, the cell loses its turgidity. If this occurs in many plant cells, the entire plant may wilt.

To create a fair test certain areas of the experiment will have to be kept the same whilst one variable is changed. I have chosen to vary the concentration of the solution. If any of the non-variables below are not kept constant, it would mean it would not be a fair test.

In this investigation the variables I am going to keep the same to make it a fair test will be:

The volume of the solution.
The type of potato.
The size of the potato chips.
The temperature in which the solutions are kept – room temperature.
The amount of time the chips are left in the solutions.

The reason why I will keep the volume of solution the same is so that all the chips will be kept in the same conditions, as the others, the volume of the solution that the potato chips are kept in must be fair. The must be
totally covered in the solution, and the amount of solution will be kept the same because all the potato chips are the same size, so they all have equal volumes of solution this also applies to the temperature in which the solutions with the chips placed in are kept, this will be room temperature which is around 21oc so that all the chips are kept in the same conditions not to affect the reliability of the results. I will keep the size the same as the masses at the start will be different, however, this isn’t really important as the percentage change in the mass will be the same as it calculates the change in mass and not just how much the chip has lost as percentage change takes into account the mass of the chip before and after osmosis has taken place. I will keep all the chips in the different strength solutions for the same amount of time because the longer they are kept in the solutions the more likely it is for a greater increase or decrease in mass. Lastly, I will use the same type if potato because different types of potatoes have different concentration, so the point at which the solutions and the potatoes concentration are equal is different in all potatoes so if different potatoes are used, it will affect the reliability and accuracy of results. To keep the water potential of the potato initially will be kept the same by using the same type of potato, which have been treated in the same way, e.g. have all been cut using the same potato corer. I am also going to use the same balance to weigh my potato chips. This is because the measurements may slightly vary between scales.

Preliminary Work.

Before my actual investigation, I carried out the investigation for preliminary work to find the amount of time I needed to leave the chips in the solution to get a large enough change in mass and how to measure the outcome correctly and accurately for reliable results. For my preliminary investigation, I investigated two different variables; the change in mass of the chips, and the change in the volume of solution.

Table investigating the change in mass.

Table investigating the change in volume.

As this was a preliminary investigation, I only investigated what happened to the potato chips at the two extreme solutions, one molar sugar solution, and zero molar (pure distilled water). As it can be seen in both preliminary investigations there was little change in mass or volume, this could be because I only left the chips in the solutions for twenty minutes. This showed me that when for my investigation I needed to leave the potato chips for a lot longer to see a definite difference in mass.

Plan

From my preliminary work for my investigation I have decided to measure the change in mass as my outcome variable as even though neither of my preliminary results gave me a big change in mass or volume to see which was a better variable to investigate I found when I measured the volume of solution I couldn’t measure it accurately enough. I found it hard to read the measuring cylinder for a correct value as the change was only a millilitre, where as it is much more accurate to read a digital weighing scale. I decided to leave my potatoes in the solution for one hour and twenty minutes, as this was the length of my lesson so it was the longest time I could possibly leave the potato chips in the different solutions and from my preliminary results I could see that I needed to leave the chips in the solution for a longer amount of time. I decided twenty millilitres of solution was sufficient as it completely covered the chips in the boiling tubes. Therefore as the volume I was using was twenty millilitres I could now calculated the ratio of distilled water and solution I would need to make my solutions. At one molar, it was obviously twenty millilitres of solution and zero molar twenty millilitres of distilled water. Therefore at 0.5 molar it was half distilled water and half solution so ten millilitres of each. A 0.75 molar solution would be three quarters of the one molar solution and one quarter of distilled water. So a quarter of twenty is five so it was fifteen millilitres of solution and five millilitres of distilled water. Consequently, a 0.25 molar solution was the opposite of 0.75, this being fifteen millilitres of distilled water and five millilitres of solution.

1 Molar solution = 20ml of solution

0.75 Molar solution = 15ml of one molar solution and 5ml of distilled water

0.5 Molar solution = 10ml of one molar solution and 10ml of distilled water

0.25 Molar solution = 15ml of distilled water and 5ml of one molar solution

0 Molar solution = 20ml of distilled water

Therefore, from my preliminary results I can now say in this investigation the input variable I will be changing is the strength of the solutions in which the potato chips are placed. I will be using a 10ml-measuring cylinder to measure out the different solutions using my calculations above. My weakest solution will be zero molar, therefore just pure distilled water, and my strongest will be a one molar solution. In all I will take five different weights for the change in mass of the chips at, 1 molar, 0.75 molar, 0.5 molar, 0.25 molar and 0 molar. Hence, It can be seen that each solution increases or decreases in strength by 25%. I will do this investigation with three potato chips in each solution, so it can be seen if the same thing happens to each chip at that strength solution.

The outcome variable in this investigation is the change in mass of the potato chips, which I will measure using a digital weighing scale correct to one decimal place to give me accurate results.

The apparatus I will be using in this experiment are:

5 Boiling tubes
A test tube rack
Weighing Scale correct to one decimal place.
One molar sugar solution, which I will change to different strength by adding, distilled water.
A potato
Watch
10ml measuring cylinder
Scalpel
A potato corer to cut cylinders out of the potato, therefore all the chips will be the same width and would just need to be cut to length and also the potato will not have to be pealed as the corer takes the potato from the inside.

Safety and Precautions.

To keep all of the different variables that I am not changing or measuring the same, to make sure that none of them affect the results of the experiment in any way.
Whilst cutting the potato, extreme care and precaution has to be taken with the scalpel as it is very sharp and could easily cause a serious wound.
The measurements for the solutions had to be exact ratio of solution and distilled water as not to affect the outcome of the experiment.

Step-by-Step Plan.

Collect all the apparatus needed for this investigation and place them on clean surface.
Measure out the solutions using the calculation on the previous page and place each solution in boiling tubes, then place the five boiling tubes in a rack and label the boiling tubes 1m, 0.75m etc. so the strength of the solution in the boiling tube can be identified.
Use the potato corer to cut cylinders from the potatoes and cut all fifteen to the similar lengths using a scalpel and ruler to around 2cm long, however, it doesn’t matter if they are a little bit longer than each other. Measure the weights of chips using the scale, record the weights of three chips for each of the solution, and place three chips in each of the solution they are recorded for, doing this for all five solutions.
Record the time in which the chips were all placed in the solutions, leave them there for one hour and twenty minutes, and clear away unnecessary apparatus.
After the hour and twenty minutes, first remove the chips from the first solution and weigh all three chips and record the results, repeating this for all the other solutions.
After the starting and finishing masses have been recorded, the increase or decrease in mass needs to be calculated and recorded for each chip.
As the chips may have been different masses to start with the change in mass does not tell us whether the mass of the chips for that solution has actually increases or decrease so a percentage mass needs to be calculated using the formula:

Percentage change in mass = change in mass x 100

Starting mass

After the percentage, change for each chip has been calculated the average percentage change needs to be calculated by adding together all three values for the chips and dividing them by three. This giving us the percentage change in the mass of potato chips for that solution, showing whether the particular solution caused the mass of the chips to increase or decrease.

If my results were to be perfect and very accurate, I would expect to have a graph like the one below.

My Results.

Graph.

Analysis.

From my results and graph that, I have gathered that the stronger the solution in which the potato was placed, the greater the mass of the chip decreased. From my drawn graph on the previous page, it can be seen that as the strength of the solution was increased the mass of the chips decreased. This is due to osmosis. Osmosis is the process of the diffusion of water molecules from a region of high water concentration (a weak solution) to a reign of low water concentration (a strong solution) through a semi-permeable or selectively permeable membrane, like a cell wall, resulting in a change of the mass in the potato chip. The graph show the stronger the solution the more the mass decreased, this was because the concentration of water molecules was higher in the cells of the potato chip than in the solution in the beaker, so the water molecules diffused from the area of high concentration in the chip to the region of low concentration in the boiling tube through the semi-permeable cell wall in the potato, so the mass of the potato chips decreased. In the weak solutions, the concentration of water molecules was higher in the boiling tubes than in it was in the potato cells, therefore, the water molecules diffused from the area of high water concentration in the boiling tubes into the area of low water concentration in the potato cells, hence the mass increased.

On my graph on the previous page the point, which is circled on the “x-axis”, which is the point at which the concentration of the potato and solution would be equal. As this is the point on the line, of best fit, which is equal to zero, increase or decrease in mass. The point is equal to 0.175 molar, so if I was to make solution equal to 0.175 molar and leave a potato chip in it for a few hours I would see no change in mass because the concentration of water molecules would be balanced in the solution in the boiling tube with the concentration of water molecules in the cells of the potato chip, with no area having a higher or lower concentration of water molecules than the other. Therefore, there would be the same number of water molecules diffusing into the cells of the potato chip as there would be diffusing out of the cells in the potato chip, allowing no increase or decrease in mass.

This shows that my results agree with my prediction, the stronger the solution the more the mass will decrease and the weaker the solution the more the mass will increase, my graph gave me the conclusion that the stronger the solution the chip were placed in, the more the mass decreased due to the process of osmosis. Osmosis being the diffusion of water molecules for a region of high water concentration to a region of low water concentration, through a semi-permeable membrane. When I placed the potato chip in the weak zero molar solution, which therefore had a high concentration of water molecules, the water molecules diffused from the boiling tube into the cells of the potato chip increasing the mass of the potato chip. This is shown in the diagram below:

When I placed the chip in the strong one molar solution, which had a low concentration of water molecules; the molecules diffused from the high concentration of water molecules in the cells of the chip into the low concentration of water molecules in the boiling tube. Thus, causing the mass of the potato chip to decrease. This being shown in the below diagram:

On my graph, the point at which the potato cells would have become fully turgid and fully plamolysed cannot be seen. If I had left the chips in the in the weak solution longer and checked the weight of the chip every few hours I would find the chip would continue to increase, but after around twenty-four hours or more the mass of the chip would increase less and less until I would find the chip would not increase anymore as it would be fully turgid, so the potato cells wouldn’t be able to take in any more water.

The opposite would have happened if I had left the chip in a strong solution for longer and checked its weight every few hours, I would find that the weight would decrease until after around twenty-four hours or more where it would decrease less and less until the mass of the chip wouldn’t decrease any more. This because the chip would be fully plasmolysed, the cells loose water by osmosis and the vacuole and cytoplasm shrink away from the cellulose cell wall which becomes limp, the process being called plasmolysis where the cell loses its turgidity.

Evaluation

In this experiment, I think I measured the water going into or out or the potato fairy accurately by measuring the change in mass. However, from my graph it can be seen that my results are not very reliable as they are not on the line of best fit. There are many reasons why this is. Firstly, my results may have not been as accurate as they could have been as I only used a scale correct to one decimal place. If I had used a scale that was correct to two decimal places this would have given me more accurate results as the masses of the chips would have been to two decimal places instead of just one. Another reason for my results not been as accurate as they could have been, could be due to the fact that when I removed my chips from the solutions I did not dry them before I weighed them, which could have affected the reliability and accuracy of my results. The fact that the chips were still wet when I weighed them could have increased the masses of the chips, so if I had dried the chips I would have got a more accurate weight, however I don’t think it would have made a difference on the scale I was using as it was only correct to one decimal point if I had used a scale correct to two decimal points it would have picked up the change in weights if the chips were wet or dry.

Looking at my result table at my repeat results they were all very similar this showing my were results were reliable in the sense that the same thing happened to each chip in each solution. However, looking at my graph all my results are off the line of best fit especially my result for the one molar solution, which instead of the mass decreasing lower than the percentage change in mass for the 0.75 molar solution it went back up higher than the percentage change for the 0.5 molar solution showing that something went very wrong in this solution. Therefore, this effects the reliability of my results as the results in the table show they are decreasing as the solutions get stronger and more concentrated, but when it comes to the bottom of the table at one molar, the percentage mass has still decreased but not as much as the solution before it showing that something went wrong in the experiment making it less reliable, this could have been due to all the reason above for affecting the accuracy and reliability of my results put together. Another factor that could have affected the reliability of my results is the state of the potato I use, by this I mean if the potato had gone soft or if it had started to rot this could have effected the rate of osmosis giving me the odd reading in my results. In this investigation keeping it, a fair test was fairly easy but also fairly difficult, it was hard to keep the volume of solution the same only using a 10ml measuring cylinder which is usually fairly easy to keep accurate but in this experiment it was difficult as the solutions had to be made up accurately not exceeding 20ml when they were put together. What could have effected the reliability of my results is that I didn’t use two separate measuring cylinders, one for distilled water and one for solution I just used one. For example for the 0.5 molar solution I poured 10 ml of solution and poured it into the boiling tube then measured 10ml of distilled water in the same measuring cylinder. This could be another factor that affected the reliability of my results. However, by looking at my preliminary results it can be seen that because I left the potato chips in the solution an hour longer I got a greater increase and decrease in mass showing my results are quite reliable. I also found after looking at the results and graphs of other pupils in my class that their results were also of the line of best fit showing that my investigation was not the only one that gave unreliable and inaccurate results.

Changes

There are many changes I would make if I was to repeat this investigation again, the changes I would make are:

I would weigh my potato chips using a scale that was correct to two decimal places, so it would give me results that are more accurate, because of the change in mass would be more accurate, as a change of + 0.16 grams is more accurate than a change of + 0.1 grams.
I would also, after taking my chips out of the solutions dry them with a paper towel, so that the water wouldn’t cause the mass of the chips to increase making the weights inaccurate.
I would use bigger potato chips as the ones I used were very small to begin with, so if I used bigger chips the change in masses would have been more spread out.

To Further My Enquiry

A further investigation I that I would carry out would be similar to this one, I would investigate the effect of temperature on osmosis.

To carry out this investigation I would firstly do the investigation I did with different strength solutions and measuring the change in mass but this time only with one molar and zero molar, so I could have the two extreme solutions. In the investigation I just carried out the solutions were at 21oc. I would then measure out the solutions again but raise the temperature of the solutions by 10oc, I would do this another twice more and lower the temperature of one solution, so I would see the effect of temperature on osmosis at five different temperatures, 11, 21, 31, 41, 51oc.

I would increase the temperature by putting the boiling tubes with the solutions in, in a water bath and decrease the temperature by placing the boiling tubes either in ice or a water bath set at 10oc.

My prediction for this investigation would be that as I increase the temperature the less osmosis will occur, resulting little if any change in mass. This is because the heat will kill the cells in the chip and break down the semi-permeable membranes.

I could also do an investigation where I would use five different potatoes and see the effect they have on osmosis. I would set up the investigation like in this investigation and repeat it five times each time using different potatoes. I predict that the cells in different potatoes would have different concentrations of water molecules in vacuoles of their cells this would affect the change in mass at different strength solutions in the investigation.