Joshua's Water Potential and Molarity Mix Up Arguments

Joshua Everett

AP Biology

Mr. Hammer

November 21, 2014

Water Potential and Molarity Mix Up

Argument 1:

In activity one of the experiment my lab partner and I were able to conclude, through the results we received, which molarity concentration each solution was that was used in the experiment. There were six different solutions that were represented by food coloring, each a different color. Going up in increments of 0.2M, the possible molarities of these solutions ranged from 0.0M to 1.0M . After interpreting the results and drawing general conclusions, the order of molarity of the solutions were orange, dark green, light green, blue, yellow and red in respect for the range of possible molarities from least molarity to greatest molarity (0.0M to 1.0M increasing by 0.2M). In the experiment there were two processes that were done, one with a dialysis tubing cell and the other with squash and parsnip. After fully analyzing the results and making calculations, we decided to use the vegetable experiment to support our order of molarity because there were two vegetables that displayed the same order while the one dialysis tubing cell had different results.

In the vegetable experiment, we cut up each vegetable into six pieces that were  all relatively the same size.  We weighed each of the pieces for both vegetables and recorded them in a data table. We also measured out 75 mL of each solution twice into separate cups. We placed one of the squash pieces into each of the different solutions. The same process was done for the parsnip pieces as well. Once they were submerged into the solutions, we covered each of the cups with a piece of foil and we let them sit overnight. The next day we took out each piece of vegetable from the solutions and weighed them again. We recorded the weights into a data table and we started developing patterns between the two vegetables.  With the two sets of weights for each piece of vegetable we were able to calculate the percent change between the original weight and the final weight. To calculate the percent change, you subtract the final mass minus the initial mass all divided by the initial mass. Then you multiply the answer by 100 to get a percent.  For the squash, the percent change for solution red was -45.68%, orange being -6.94%, yellow being -43.04%, dark green being -4.55, light green being 20.37%, and blue being -31.43%.  For parsnip the percent change for red was -25.32%, orange was 47.62%, yellow was -20.96, dark green was 10.64%, light green was 32.09% and blue was -13.24%. We arranged the percent changes for each of the vegetables into separate lists  in order from positive percent change (+) to negative percent change (+) to see a better comparison between them.  The negative percent changes represent that the cell shrunk and the positive percent changes represents the cell swelling. When a cell shrinks it means that the cell is in a hypertonic solution in which there was a high concentration of water inside the cell and a low concentration of water outside of the cell. Through the process of diffusion, water would move across the concentration gradient and move from high to low concentration. If there is not a lot of water outside of the cell, where the solution is, that means that most of the solution contained sucrose solutes. This would mean that the solution with the vegetable that had the most negative percent change would have the greatest molarity. However, when a cell swells, it is in a hypotonic solution in which  there is a higher concentration of water on the outside of the cell where the solution is located and a lower water concentration on the inside of the cell. The water will move from outside the cell to inside the cell, making the cell swell but also making a positive percent change. If there is more water on the outside of the cell then that suggests that there is not a lot of solute of sucrose present in the water but rather more of the sucrose will be present inside the cell where less water is. This would mean the pieces of vegetables with a solution that had the most positive percent change was have the smallest molarity. The order of most positive percent change to most negative percent change for the squash was the dark green, light green, orange, blue, yellow, and red. For parsnip it was orange, dark green, light green, blue, yellow, and red. Next, we made a combined list where the placement of the solutions was determined by their similar placement from their separate lists. The blue, yellow, and red solutions were the same in placement for both vegetables. The orange and both greens moved in the lists so we decided that since orange was at the top once it would stay there even though dark green was once at the top too. But we thought that since dark green and light green are always other, it would make sense that orange starts off and the dark and light green will just follow. The list starts off with the most positive percent change and decreases into the most negative percent change as the list goes down. The order of the molarity would be orange being 0.0M, dark green being 0.2M, light green being 0.4M, blue being 0.6M, yellow being 0.8M, and red being 1.0M.

For the dialysis tubing part, we made six different cells by tying off the ends of the tubing and placing 30 mL of a  different solution in each cell. We weighed each cell and recorded it in a data table. We submerged the each cell in a different plastic cup that that had 250 mL of water. We covered each cup with foil and let them sit overnight. The next day, we removed the cells from each cup, weighed them, and recorded the weight in a data table. We then calculated the percent change of each cell and there was only one negative number and the rest were positive. The percent change for red was -6.3%, orange was 19.94%, yellow was 9.94%, dark green was 29.7%, light green was 0.97%, and blue was 30.37%. The negative percent change represents that the cell was possibly in a hypertonic solution where there would be a high concentration of water inside the cell than outside the cell. The water will move from inside the cell to outside the cell because of the process of osmosis which accounts for the decrease in weight and the negative percent change for the red solution. Because the solution is now in the cell and not in the environment, like in the vegetable experiment, the cell will only shrink if there was more water in the cell and a small amount of sucrose is present. The red solution will only have the smallest concentration of sucrose because of the water movement and negative percentage change. The positive percent changes depicted that the cell was in a hypotonic solution and that there is a  greater amount of solutes in the cell. With the greater amount of solutes, there is less water in the cell which will cause water to flow inside. As the positive percent change increases, that will determine the molarity of each solution will increase from red continuing on. For dialysis tubing the molarities would be red being 0.0M, dark green being 0.2M, light green being 0.4M, yellow being 0.6M, orange being 0.8M, and blue being 1.0M. Because red is considered 0.0M in this lab, the solution in the cell would have contained no solutes in the cell. Technically, the cell losing water from the cell would have came from the cell being isotonic to its environment. A human error that can account for this hypertonic situation could be the use of tap water and not distilled water. Tap water could contain chemicals that could have caused water to move from low concentration to high concentration which would explain why osmosis occurred.  Overall, this is not my claim because this order is not confirmed by any other experiment. I used the claim for the vegetable experiment because it was supported by both squash and parsnip and seemed more accurate.  

Argument 2:

The water potential for squash is -3.69 bars and the water potential for parsnip is -12.29 bars. My lab partner and I calculated the percent change in mass using the recorded weights of the vegetable pieces before they were each put into 75 mL of a different sucrose solution and after. Afterwards, we proceeded to make a graph comparing the percent change in both squash and parsnip to the concentration of each solution. The percent change in mass was on the y axis and the molarity was on the x axis. The purpose for this graph was to determine where a cell would be isotonic where there is zero percent change in mass. This will only occur then there is an equal amount of water entering and leaving the cell at the same time. In this isotonic stage this is no pressure being exerted on the cell, so we said there's no pressure on the cell. Next we located the point on the graph where the line crosses the x axis. This point will tell us the molarity of the solution that the cell will have at an isotonic state. For squash the molarity is 0.15M and for parsnip it is 0.50M.  We used these numbers, 0 for the pressure and 0.15M and 0.50M for the molarities, to plug into an equation to determine the water potential of the cell. The water potential equation is Ψ=Ψp+Ψs which means that the water potential equals the pressure potential plus the solute potential. The pressure potential is zero but to determine the solute potential we had to use this solute potential equation, Ψs=-iCRT. This equation means the solute potential equals to the product of negative number of particles the molecule will make in water, molar concentration, the pressure constant, and the temperature in Kelvin. The number of particles sucrose makes in water is 1, so -i equals to -1.The pressure constant is 0.0831 liter bars/mole K and the temperature is 23 degrees celsius plus 273 in order to convert it to Kelvin. After multiplying all the numbers, not forgetting the 0.15 molar concentration, I got -3.69 bars for squash. To figure out the solute potential for parsnip, I followed the same process but instead I substituted the molar concentration from 0.15M to 0.50M. The solute potential for parsnip is -12.29 bars. In order to find the water potential I had to plug the solute potential for the parsnip and squash back into the Ψ=Ψp+Ψs. The pressure is 0 for each vegetable so I just plugged in the solute potential into the equations and I got the water potential. The water potential for the squash came out to be -3.69 bars and the water potential for the parsnip was -12.29 bars. 

The title is The Water Potential of the Squash. The x axis is the molarity of the solutions and the y axis is the percent change in mass.

The title is The Water Potential of the Parsnip. The x axis is the molarity of the solutions and the y axis is the percent change in mass.