Jazmean's Water Potential and Molarity Mixup Lab Argumentation

Water Potential and Molarity Mixup Lab

This week, my group and I worked on the Water Potential and Molarity Mixup lab to determine the molarity of six different solutions and the water potential of two vegetables. From the data gathered from this lab, there are three possible claims to make for the molarity of the six different solutions. My group believed that even though the first task only required the data from Activity One, it would better to consider all of the data from both activities as the same solutions were used each task and more information could be gathered. Claim One is solely based off of the data gathered from Activity One. For this claim, the order of the solutions from 0.0M to 1.0M is as follows: Dark Green, Red, Orange, Light Green, Blue, and Yellow. Claim Two is based off of the data gathered from Activity Two. This claim focuses on the data from the eggplant. The order of the solutions is as follows: Dark Green, Blue, Light Green, Orange, Yellow, and Red. Claim Three is also based off of Activity Two, however this claim focuses on the data from the okra vegetable. The order of the solutions is as follows: Light Green, Orange, Dark Green, Yellow, Blue, and Red.  The claim that is the more correct claim to make is Claim Two, which focuses on the data from the eggplant and this is due to there being less errors present during the experimentation process of the eggplant. For task two, my group used eggplant and okra as sample vegetables to determine their water potential. The water potential for eggplant was -8.240196 and for okra it was -28.43974512.

Polarity

In Activity One, my group and I filled six dialysis tubings with the six different solution. Each solution was then weighed on the electric balance and placed in a plastic container filled with tap water. It was covered with aluminum foil to prevent evaporation and left on the counter overnight.

The next day, the aluminum foil was uncovered and the six dialysis tubings were weighed again and the percent change was calculated. Pictured below are the results in a chart.

In an ideal scenario, all of the numbers for the percent change should be positive and one should be zero. This is because five out of the six solutions in the dialysis tubing have a higher concentration of solutes and a lower concentration of water. One of them has no solutes and is just water with food coloring in it. Because the five solutions have a lower concentration of water, water is going to move into the tubings because the environment (the tap water in the plastic container) has a higher concentration of water. When water undergoes diffusion (also called osmosis), it travels from a high concentration to a low concentration. If water is going to be moving into the dialysis tubing, there will be an increase of mass. This means that its percent change will be a positive number. In our experiment, there are 5 positive numbers and one negative number. The 5 positive numbers were expected, but the negative number was not. The negative number means that water was loss from the tubing and that there was a higher concentration of water in the tubing than in the water in the plastic container. However, this makes sense because there is a possibility that distilled water was used for the 0.0M solution. If it was, then the distilled water would have a higher concentration of water than the tap water used as tap water as the “environment” for the tubing as it has impurities in it which can affect its concentration of water.

The tubings that had a higher percent change of mass also have a higher molarity. The high molarity solutions would have a higher percent change of mass because there is a higher concentration of solutes and a lower concentration of water, and molarity is the calculation of the concentration of a substance in a solution. The more solutes present, the higher the concentration, and the higher the molarity.

Even though this data supports part of Claim Two (it supports that Dark Green was the 0.0M solution), it was not solely used to make a claim.  As mentioned before, the experimentation process had some errors. This is because there is a possibility that one of the tubings were not tied tight and some of the solution could have leaked. Also, tap water was used as the “environment” for the tubings instead of distilled water which could of skewed the amount of water that would of been taken in by the dialysis tubings. The food coloring used to dye the solution was able to pass through the tubing which may have affected the amount of water diffused into the tubings too.

In Activity Two, my group and I used the eggplant and okra as sample vegetables to put in the solutions. We cut six similar shaped pieces of eggplant and six similar shaped pieces of okra.

We weighed each of them on the electric balance and then placed each of the six pieces of the vegetables into the six different solutions. We covered them in aluminum foil to prevent evaporation and placed them off to side overnight. The next day, we uncovered them, weighed them, and calculated their percent change of pass. Picture below are the results.

Okra

Eggplant

The pieces of the vegetables that had a negative percent change of lost water and the pieces that had a positive change of mass gained water. The negative percent changes indicate that the plant had a higher concentration of water than the solution. This means that the water in the solution diffused from the plant to the environment or from the high concentrated plant to the low concentrated solution. The positive percent changes indicate that the plant had a lower concentration of water than the solution. This means that water from the solution diffused into the plant or from the high concentrated environment to the low concentrated plant.

Both of the data sets from the plants support that the Red solution had 1.0M. However, the eggplant data was used to make the claim in the beginning. The okra was difficult to weigh because it had slime oozing from the it and the slime may have added to the weight. Therefore, the results may not be as accurate as they can be. The eggplant did not have that problem. The only error that could have been made during the experimentation process of the eggplant was the final weighing. When the eggplant was taken out of the solution, it was dripping. Some of the solution may have gotten on the electric balance. However, the extra solution that dripped on the balance may have played such a minuscule role in increasing the weight of the eggplant. Because of this, we as group believe that the eggplant data was the most reliable to use to make the claim.

Water Potential

The eggplant and okra data was used again to determine their water potential. Water potential is the measure of potential energy to move in and out of the cell of water. In order to determine their water potential, we graphed the percent change of mass.

The x-axis consisted of the different molarities and the y-axis was the percent change.

The graphs helped me figure out the molarity of the solutes in the eggplant which is needed to calculate the water potential. The molarity of the solutes in the eggplant is found when the line crosses the x-axis because this is when the molar concentration of the solution is equal to the molar concentration of the eggplant. Both of the environments are the same and therefore isotonic. For the eggplant, the line crossed the x-axis at 0.335. The okra line never crosses the x-axis so I had to use my graphing calculator to find when it does. I plugged the points into a STATPLOT and found the line of best fit. The line was -768.4837963x^3+1457.364087x^2-836.2846561x+206.5039683. The line crossed the x-axis at 1.1562. After finding the molarity of the vegetables, I then had to calculate the solute potential. The solute potential is -1 x (number of particles the molecule will make in water) x (molar concentration) x (pressure constant) x (temperature in degrees Kelvin) or -iCRT. For the eggplant it was -1(1 (sucrose only makes 1 particle in water))(0.335M)(0.0831 liter bar/mole K)(296 K) which equals -8.240196 bars. For the okra it was -1(1)(1.1562M)(0.0831 liter bar/mole K)(296 K) which equals -28.43974512 bars. After all of this, the water potential can be calculated. The water potential is equal to pressure potential + solute potential. The pressure potential for the eggplant and okra is 0 because both the solution and vegetable are in an isotonic state which means there is no pressure being exerted. This means the water potential for the eggplant is -8.240196 and the water potential for the okra is -28.43974512.

The water potential for the okra may not be as accurate as it can be because the slime may have affected its percent change of mass and therefore affect its water potential.