Introduction
Molecules are constantly in motion due to thermal energy. Therefore, the molecules will move around and bump into each other. Because of this, molecules tend to spread out evenly into available space in a process called diffusion. Diffusion is the random process of molecules moving from areas of high concentration to low concentration, thus requiring no energy. Many of the substances that enter or leave the cell do so through diffusion. Osmosis, a type of diffusion, is the movement of water across a selectively permeable membrane. In osmosis, water diffuses across the area of lower solute concentration to that of higher solute concentration until the solute concentrations of the environment and the cell are equal. Tonicity, which is the ability of a solution to gain or lose water due to osmosis, results in an environment that is isotonic, hypertonic, or hypotonic. A hypotonic solution has a lower solute concentration than the cell so water moves into the cell causing the cell to swell. A hypertonic solution has a higher solute concentration than the cell so water moves out of the cell and into the solution causing the cell to shrink. In an isotonic solution, concentration of solutes is equal inside and outside the cell so water moves across the membrane in both directions retaining the same cell size. In all three environments, water will move until it reaches equilibrium.
In this experiment, the solution in which the dialysis tube was placed in is determined by the mass in grams of the dialysis tube before put in the solution compared to the mass after. If the mass increases, the dialysis tube was in a hypotonic solution since water went in the dialysis tube. If the mass decreases, the dialysis tube was in a hypertonic solution since water left the cell. If the mass stays the same, then the dialysis tube was in an isotonic solution since water went in and out of the dialysis tube at equal rates. The polarity of water allows water to dissolve many ionic compounds such as sodium chloride. Water dissolves salt because water’s partially positive area attracts the negative chloride ions and water’s partially negative area attracts the positive sodium ions. Therefore, salt is able to dissolve in water. In this solution, salt is the solute and water is the solvent. Water potential determines the direction in which water will move. Water potential is affected by two different factors, solute potential and pressure potential. Solute potential lowers the capacity of water to do work because the solutes bind to the water molecules which lowers the number of free water molecules. Pressure potential is the physical pressure on the solution and can have a positive or negative impact on water potential. Water potential is figured out through the equation ψ (water potential) =ψs (solute potential) + ψp (pressure potential). Solute potential is figured out using the equation –iCRT. Solute potential is always negative, “i” represents the number of ions, “C” represents the molar concentration, “R” is a pressure constant and will always be 0.0831 liter x bar / mole x K, and “T” represents the temperature in Kelvin. In this experiment, dialysis tube takes the place of an actual cell. The dialysis tube acts as a cell with a permeable membrane and allows for water to move in and out of the cell but not the solute.
Question: What is the water potential of the solution in the dialysis tube? Purpose: The purpose of this lab is to figure out if the water potential of the environment affects the dialysis tube with an unknown solution. Since the dialysis tube acts as a cell because of its permeable membrane, this experiment can determine how osmosis works in an actual cell membrane. Hypothesis: If osmosis occurs between solutions with different concentrations until there is equilibrium, then the weight of the dialysis tube after put in the environment will be affected only by the molarity of the environment thus the dialysis tube that doesn’t change weight will have water potential equal to the environment it was placed in. Procedure
1) Create a dialysis tube of around 6 inches and soak it in water. Tie one end of the dialysis tube and open the other end. Insert 10 ml of the unknown solution into the dialysis tube and tie the open end but leave a fair amount of air. 4) Repeat steps 1-3 until there are 5 dialysis tubes. Then weigh each dialysis tube in terms of grams. 5) Have 5 beakers with 100 mL of tap water each. Add 0.2 moles of sugar for 100 mL of water in the first beaker, 0.4 moles of sugar in the second beaker, 0.6 moles of sugar in the third beaker, 0.8 moles of sugar in the fourth beaker, and 1.0 moles of sugar in the last beaker. Then place dialysis tube into each of the beakers. 6) Let the dialysis tubes for around half an hour then take them out and weigh them again. Data
Ψ of unknown solution
Molarity of environmentWater potential of environmentWeight before (grams)Weight after (grams)Percent Change 0.2 M-4.89 bars10.5411.064.9% ↑
0.4 M-9.77 bars10.6410.511.2% ↓
0.6 M-14.66 bars10.759.987.2% ↓
0.8 M-19.54 bars10.979.5212.7% ↓
1.0 M-24.43 bars10.568.9914.9% ↓
Data Analysis and Conclusion
Conclusion: According to the data, the molarity of the unknown solution is about 0.38M thus the water potential is -9.28 according the equation ψs= -iCRT. The data reveals that the dialysis tube was in a hypotonic environment when the environment was greater than -9.77 bars and was in a hypertonic environment when the environment was less than -9.77. As the molarity of sugar increases, the water potential of the environment decreases and more water from the dialysis tube leaves the cell. As the molarity of sugar decreases, the water potential of the environment increases and more water will enter the dialysis tube. This supports the hypothesis that the different water potentials of the environment, whether isotonic, hypertonic, or hypotonic, will result in different mass changes for each of the dialysis tube. Ultimately, the experiment proves that osmosis occurs between the cell’s environment and the inside of the cell until there is equilibrium and the concentrations of the two solutions are equal.
