Water Potential and Molarity Mix Up Argumentation

Nida Ali

Water Potential and Molarity Mix Up Argumentation  

For the first activity of the Experimental Determination of Unknown Molarities, my group and I concluded that the 0.0M solution is the dark green solution, 0.2M solution is the light green colored solution, 0.4M is the blue solution, 0.6M is the orange solution, 0.8M solution is the red solution and lastly the 1.0M solution is the yellow solution. We interpreted our data not only based off of the first activity, but also with the second activity because our group mistakenly created a second light green dialysis tube instead of a dark green one. The piece of turnip and artichoke that was placed in the dark green solution was the softest, therefore more water must have flown into the tube resulting in a massive change in the mass. Based off of this, the green solution is the 0.0M solution. Likewise, the lighter green solution was also very soft and had a large difference in the mass of dialysis tubing of 5.2g, but also in the vegetables with a difference of 1.1g in turnips, and a .2g difference in artichoke.  A positive percent change in mass was seen in all of the colored “cells”.  Red has a percentage change of 63.7%, orange had 26.8%, yellow had 47.8%, light green had 10.1%, dark green had 18.2% and blue had 27.2% change in mass. The difference in mass of the colored “cells” of red was 18.8g, orange 9.1g, yellow 14.4g, light green 5.2g, dark green 3.1g and blue had 7.7g. The positive percent change in mass and difference in mass states that the distilled water went into the cell and displays passive transport of osmosis. Since the distilled water has a high concentration and the cell has a low concentration of water , the water will move from the higher concentration to lower concentration, and since water is hypotonic, the solution is hypertonic. We determined the molarity of the solution with our understanding of water potential and toxicity. The firmer the vegetable was after remaining in the colored solution overnight, the higher it was in molarity since more water was absorbed. Like for example, my group members and I came to the conclusion that the yellow colored solution made the turnip very firm in comparison to the dark green solution, where it was very soft. Since the turnip in the  yellow colored solution was the firmest out of all of the other solutions, we came to the conclusion that it must be the 1.0M solution. Just to confirm our results, we then checked the “cells” we created by taking them out of the beaker of water and placing them onto a paper towel. Then, we began to feel the firmness of the solutions in the dialysis tubings. The firmer the cell was, the more water it absorbed. Our results, of which molar solution was of which colored solution was proved by feeling the firmness of each tube and match those that we received after carefully feeling the intensity of the firmness.

For the second activity, the water potential of turnips is -21.2 bars and artichoke is -15.0 bars. As seen in the attached graph, the points that are plotted orange are where the artichoke and turnip water potential meet the line of molarity on past the x axis. Based off of this information, we were able to calculate the solute potential. The value used for molarity for turnip was 0.86M and for artichoke was 0.61M. These points indicate that the vegetables are at equilibrium with the solution and no pressure is present. After obtaining these values, we calculated the water potential by using it’s formula, water potential () = pressure potential () + solute potential (). By using the solute potential formula, Solute potential () = –iCRT, we calculated the solute potential first. The solute potential of turnip is  () = – i(1) x c(0.86) x r(0.0831) x t(296)= -21.2 bars. After placing the number of particles the molecule will make in water, the molar concentration, constant, and room temperature of 23 degrees Celsius converted to 296 Kelvin into the formula, we added that to the pressure potential, which was zero due to the beaker having no potential, and calculated the water potential for turnip. The same steps applied for calculating the water potential for artichoke but we only replaced the value for the molar concentration, () = –i(1) x C(0.61) x R(0.0831) x T(296) = -15.0 bars. The process of finding the water potential highlights a scientific principle of osmosis due to the inflow of water from a high concentration to a low concentration. The positive percent change in turnip were of the orange colored solution of 0.06%, light green solution 14.7%, dark green solution 28.7% and blue solution of 28.7%. These positive changes indicate that the hypotonic distilled water went into the hypertonic colored solution, and this is performing passive transport for turnips. Similarly, in the artichoke, the percent change seen in the orange colored solution was 10%, red 11.1%, light green 20%, dark green 25%, and blue of 11.1%. These positive changes in mass also displays the passive transport of osmosis of the inflow of the hypotonic distilled water into the hypertonic solution. In comparison, the negative percent changes in turnips of orange -9.8%, and in yellow of -6.8%, as well as in the artichoke in only the yellow of -22.2%, which also is passive transport. The negative difference states that water has flown out of the solution, and in this case water is hypertonic.