Plastic Bag Starch Lab

Andrew Tsai

10/28/18

Plastic Bag Starch Lab

Purpose: To determine if a plastic lunch bag is permeable to iodine or not.

Introduction: This investigation, using iodine and starch solutions, decisively portrayed the plastic bag as a cell membrane. Iodine was kept outside as starch stayed inside the bag. Based on the results of the lab, we concluded whether the plastic bag was semi-permeable.

Materials: Plastic lunch bag

100-mL beaker

Starch solution

Graduated cylinder

Rubber bands

2 test tubes

Lugol’s iodine solution

Procedure: First, we filled the plastic bag with 40 mL of starch solution, carefully sealing it with the rubber bands. Next, we mixed 10 mL Lugol’s iodine solution with 40mL of water in a beaker, isolating approximately 5 mL in a test tube for further observation. We also recorded observations for the color of the starch solution. Submerging the plastic bag in the iodine, we waited 15 minutes, then observed the color of the starch solution again. Isolating another 5 mL of the iodine, we compared colors before and after the experiment.

Data and Results:

Before

/

After

Color of starch solution inside cell / Translucent, white/clear
Grey / Translucent blue with small bits of starch floating in solution
Color of iodine solution and water outside the cell / Burgundy- black. Clear. / No distinguishable change.

When we observed the starch solution in the bag before the experiment, we saw it was grey and translucent white. The iodine, a darker color, was clear though nearly black. After 15 minutes, the starch had turned a blue color from the indicator (iodine). The iodine itself had no color change.

Conclusion: The starch turned blue, indicating there had been iodine seeping into the bag. Because the bag was impermeable to starch molecules and water molecules, it was testing one variable: the permeability of the bag in relation to the iodine. If the iodine could seep through, the starch in the bag would turn blue, which it did. If the starch could seep through, the iodine would have been black outside the bag, which it wasn’t.

Questions:

1. Cell membrane
2. No, the starch stayed inside the cell. The iodine indicated the starch in the bag by turning it blue. The iodine outside the bag did not indicate any starch outside.
3. The iodine moved inside the bag because the starch inside turned blue. When iodine comes in contact with starch, it turns the starch blue.
4. The plastic bag is semi-permeable. The starch could not pass through (no indication of starch outside by iodine) but the iodine could, by turning the starch in the bag blue. Semi-permeability is defined as being permeable to only certain substances.
5. Diffusion, or osmosis, the diffusion of water, is the name for substances moving from high to low concentrations following the concentration gradient.
6. Yes, they did. The iodine sought to equalize the concentration of iodine in the bag with the concentration in the beaker. Since it could pass through, it cost no energy and diffused inside, causing the color change in the starch.
7. a. The pore size was big enough to allow iodine molecules through. The iodine molecules were smaller than the pores.

b. The starch molecules were too big to fit through the pores. Thus, they remained in the bag.

1. No, we do not. If we followed the expected results, however, we would have had some. As the iodine diffused into the bag, the water in the bag diffused out to balance the water concentrations on both sides of the plastic bag. When the water outside the bag gained in concentration, the remaining iodine should have been diluted, and so look lighter after the experiment, in the iodine column of the table.

Andrew Tsai

10/28/18

Factors Influencing Diffusion

Purpose: To determine the effects of time and concentration of a substance in relation to diffusion.

Introduction: By submerging potato cubes in a potassium permanganate solution, we observed diffusion in the plant cells. As we altered the concentration and the time of immersion, we obtained differing results, explaining properties of diffusion.

Materials: Potato
Scalpel
4 small Styrofoam plates
Clock
5% potassium permanganate solution
1% potassium permanganate solution
0.1% potassium permanganate solution
Tweezers
Metric ruler
Sharpie marker

Procedure: First, we cut five 1 cm cubes from a potato. Setting aside one as the control, we placed the rest in a Styrofoam plate filled with 5% potassium permanganate solution. By taking one out every ten minutes, we observed the effect of time on diffusion.
Second, we cut three more 1 cm cubes. One we set into a 0.1% potassium permanganate solution, another into a 1% potassium permanganate solution, and the final one into a 5% potassium permanganate solution. After waiting 40 minutes, we used the tweezers to take them out; in this way we observed the effect of concentration on diffusion.

Data and Results:

Time in the solution / Distance of diffusion

Cube 1

/ 0 minutes / 0 mm
Cube 2 / 10 minutes / .7 mm
Cube 3 / 20 minutes / 1 mm
Cube 4 / 30 minutes / 1.2 mm
Cube 5 / 40 minutes / 1.5 mm
Concentration of chemical /

Distance of diffusion

Cube 1

/ 0.1% / 0.1 mm

Cube 2

/ 1% / 1 mm

Cube 3

/ 5% / 1.6 mm

Conclusion: As the general trend shows, as the cubes spent more time and were immersed in greater concentrations of the chemical, more diffusion occurred.
Looking at Part A of the experiment, we observe that as the time the potato cubes spent in the 5% potassium permanganate solution went up, so did the distance the chemical diffused into the potatoes. Part A’s main function, since all other conditions are kept the same, is to measure the amount of diffusion based on time. To illustrate this statement, we cite the following evidence:
From 10 minutes to 20 minutes, the distance of diffusion increased 0.3 mm, from an original 0.7 mm to 1 mm.
From 30 minutes to 40 minutes, the distance of diffusion increased 0.3mm, from an original 1.2 mm to 1.5 mm.
A general increase of about 0.3 mm per 10 minutes is given in Part A’s table. One can conclude from Part A that diffusion happens at a constant rate. As more time is invested, more diffusion is able to occur.
From Part B we obtained the results of the experiment involving differing concentrations of the potassium permanganate. Since all three potato cubes stayed in the plates for the same amount of time, the objective of Part B is to test the relation of concentration to the rate of diffusion. To support:
From 0.1% to 1%, the distance of diffusion increased 0.9 mm.
From 1% to 5%, the distance of diffusion increased 0.6 mm.
One can conclude that as a higher concentration of chemical is located outside the potato cells, the more pressure diffusion exerts to balance the concentration. As a result, a greater concentration yields greater distance diffused in the potato.

Andrew Tsai

10/28/18

What are Diffusion and Osmosis?

Purpose: To observe osmosis across a cell membrane as well as diffusion in water.

Introduction: By immersing an egg in vinegar, then syrup, then water, we observe the diffusion of water, otherwise known as osmosis, into the egg membrane. To observe diffusion in water, we drop concentrated food coloring to the bottom of a beaker of water and observe its effects.

Materials: 400 mL beaker
Water
Dropper
Food coloring
3 200 mL glass beakers with lids
Colored pencils
Wax pencil
100 mL graduated cylinder
175 mL white vinegar
150 mL clear saturated sugar mixture
Raw egg in shell

Procedure: Part A:
Allowing the water in the 400 mL beaker to settle, drop a single drop of food coloring on the surface of the liquid. Observe the effects over a period of time.
Part B:
Place 175 mL of vinegar in a 200 mL beaker, being sure to cover the egg with vinegar. Wait two days, and then observe the results. Measure and record the remaining vinegar left in the beaker. Next, place the egg in a saturated sugar mixture (150 mL) in a 200 mL beaker. Wait one day, then observe the results. Measure and record the remaining sugar solution left in the beaker; record. Finally, measure 150 mL of water in the 200 mL beaker, covering the egg completely. Wait one or two days, then remove the egg and record the results. Measure the remaining water and record.

Data and Results:

Beaker

/ Amount present when egg was put in / Amount present when egg was removed / Observations
Vinegar / 175 mL / 54 mL / Spot of calcified shell towards top of egg. Remainder of egg decalcified. Egg grew.
Syrup / 150 mL / 138 mL / Spot of calcified shell, egg has air bubble. Orange. Bigger egg. Solution much clearer.
Water / 150 mL / 143 mL / One egg burst, membrane shattered. The other egg grew bigger.

Conclusion/ Questions:
1. The food coloring stayed concentrated for the early stage. During the middle stage, it diffused somewhat. During the late stage, it diffused completely.
2. This process is called diffusion.
3. In Part B, the shell of the egg was removed by decalcification from the vinegar, an acid reacting with the basic components of the calcium carbonate, a base.
4. The acetic acid dissolved the shell, which was basic.
5.a. The egg became larger as a result of remaining in vinegar.
b. There was less liquid left in the beaker.
c. Water moved into the egg. I know because the water outside the egg was less than what we added. We added 175 mL and withdrew 54 mL. The missing mL of water and acetic acid could only have gone to the egg.
6.a. The size of the egg increased, but not much.
b. There was less liquid left in the beaker.
c. Water moved into the egg because we added 150 mL and withdrew only 138 mL. The 12 mL of sugar and water missing went into the egg through its membrane.
7.a. The size of the egg increased after remaining in water.
b. There was less liquid left in the jar.
c. Water moved in the egg because the egg was placed in a pure water solution. The amount we added (150 mL) had 7 mL diffuse into the egg, leaving 143 mL. The 7 mL was pure water.
8. The egg was larger after remaining in water. The water in the vinegar was mixed with acetic acid and other substances. When the egg was immersed in the solution, some of its internal chemicals balanced the outside. Water diffused in only to balance the substances inside and out. When the egg was immersed in pure water, however, no other substances were present outside the egg. As a result, water diffused in greater quantities through the membrane, seeking to balance the egg’s chemical balance with the outer chemical balance. The size increase and osmotic pressure was so great one egg burst. The other egg swelled more than when it was placed in vinegar.
9. This question is ambiguous. A cell could gain water if its internal concentration of water was less than the syrup’s concentration of water, and vice versa. Also, it would depend on the cell’s membrane. If the cell had a membrane permeable only to water, it could have extremely diverse effects than a membrane permeable to sugar and water. In the latter, the sugar would diffuse out to balance the sugar levels. The water would remain isotonic. In the former, a cell could lose or gain water based on sugar content inside or outside of the cell.
10. Fresh fruits and vegetables are sprinkled with water to make them crispier and prevent wilting. The plant absorbs the water, which fills out its cells to give the appearance of a healthy plant. Osmosis gives freshness.
11. The salting, combined with the water from ice, washes off to the plants. Since the salt’s concentration of water is less than the plant’s cells, the plant cells lose water. The plant wilts and could possibly die if the concentration of water outside the plant is too low and the concentration of salt is too high (a hypertonic solution).
12. Dried fruits and dried beans have no water content present in them. When cooked, they absorb the water used to cook them at a rapid pace, swelling their cells and filling out their structure. The osmosis of water seeks to establish equilibrium with 0% and 100%. As the dried fruits or beans cook, osmosis occurs.