Osmosis and Diffusion Lab
Pre-Lab
To begin the lab, a demonstration was given. The demonstration showed how adding an acid to a base solution they can mix and become neutralized. To demonstrate, an NaOH solution was used as the strong base along with Phenolphthalein, a base detector, in order to show there was base present. The two solution were poured together in a beaker. Once mixed, the base and base indicator turned a very dark pink. The next step was to introduce an acid. For the demonstration Vinager was added to the already existing solution. Within seconds of the acid being added, the solution went from dark pink to light pink then a clear solution.
For the group demo, a dish of "Party Gel" was pre made and dyed pink. The gel was composed of NaOH(base), Phenolphtatlein (base indicator), and agarose (gel mixture). To begin, a pumkin and an "L" shape were cut out and placed in another dish where vinegar is added to slowly diffuse and change the gel to clear. The last picture is after 24 hours of diffusing. By allowing the pieces to be cut out, the surface area was increased allowing for quicker diffusion.
Lab Procedure
Materials
Procedure
For the lab, their are several variables. The independent variable is the solutions( A-E ) and the dependent variable is the change in weight of the potato cores.
- weigh scale
- beakers
- potato core cutter
- potatoes (enough for 5 cores)
- solutions A-E
- graduated cylinder (5)
- timer
Procedure
- gather materials
- cut five potato cores of approximately equal size and weight
- weight each potato core and record
- pour solutions A, B, C, D, and E into five separate graduated cylinders up to 30 mL
- place one potato core in each separate cylinder
- time results for 24 hours
- remove the potato core from the cylinders
- reweigh the potato core and record
- analyze and calculate the increase or decrease of weight per potato core, the more the increase in weight the less concentrated the solution is
For the lab, their are several variables. The independent variable is the solutions( A-E ) and the dependent variable is the change in weight of the potato cores.
Hypothesis
The environment with the least solution density will have the greatest rate of osmosis, thus the potato will have the greatest weight increase and the additional four other solutions they will together produce results that can be used to determine the solution content.
Data
Osmosis and Diffusion of Potato Cores
Conclusion
In conclusion, the solutions can be determined most concentrated to least concentrated by the loss or gain of weight in the potato. The most concentrated was A then C and D followed up with a tie between B and E for least concentrated. In the group's findings, no potato matched the solutions concentration but D was the closest to doing so, thus human error can be assumed.
Each solute had a solute potential at 22 degrees C as follows:
formula= -iCRT
A: -1 x (.8) x (.0831) x 295
= -19.6116
B: -1 x (.2) x (.0831) x 295
= -4.9029
C: -1 x (.6) x (.0831) x 295
= -14.7087
D: -1 x (.4) x (.0831) x 295
= -9.8058
E: -1 x (1.0) x (.0831) x 295
= -24.5145
In the lab, our findings were in a different order then the actual calculated ones, thus our lists does not correspond.
Since the experiment used plant cells, if the potato cores had been left in for a longer period of time they would have continued to gain or lose water, but since plant cells have a cell wall the would not have burst in a hypotonic solution.
If the experiment was replicated with other animal cells or tissues over time the cells would burst if placed in a hypotonic solution the cells would eventually burst. They would burst because water molecules are small enough to travel in and out of the cell without being able to control it. Also since the there is no cell wall, as in plant cells, the cell will not be retained and eventually burst.
Each solute had a solute potential at 22 degrees C as follows:
formula= -iCRT
A: -1 x (.8) x (.0831) x 295
= -19.6116
B: -1 x (.2) x (.0831) x 295
= -4.9029
C: -1 x (.6) x (.0831) x 295
= -14.7087
D: -1 x (.4) x (.0831) x 295
= -9.8058
E: -1 x (1.0) x (.0831) x 295
= -24.5145
In the lab, our findings were in a different order then the actual calculated ones, thus our lists does not correspond.
Since the experiment used plant cells, if the potato cores had been left in for a longer period of time they would have continued to gain or lose water, but since plant cells have a cell wall the would not have burst in a hypotonic solution.
If the experiment was replicated with other animal cells or tissues over time the cells would burst if placed in a hypotonic solution the cells would eventually burst. They would burst because water molecules are small enough to travel in and out of the cell without being able to control it. Also since the there is no cell wall, as in plant cells, the cell will not be retained and eventually burst.
Sources of Error
As the experiment included several separate tests, errors could easily occur. One error is that the potato cores were not all the same weight to begin with. Another source of error is that the cores were not completely covered by the solution due to a larger beaker. Finally, the time for the experiment to occur was not a complete 24 hours possibly skewing the results slightly.