Diffusion in Agar Cubes

Purpose: What determines the efficiency of diffusion throughout the model “cells”?

Hypothesis: The smaller the surface area and volume, the higher the rate of diffusion will be.

1 cm cube of agar

2 cm cube of agar

3 cm cube of agar

Cut agar cubes after being in base solution for 10 minutes

Close up of cut agar

Data Table

1. In terms of maximizing diffusion, what was the most effective size cube that you tested?

In terms of maximizing diffusion, the most effective cube that was tested  was the small, 1cm cube of agar.

2. Why was that size most effective at maximizing diffusion? What are the important factors that affect how materials diffuse into cells or tissues?

This size was most effective at maximizing diffusion because its surface area is proportionally larger than its volume, which can be seen through the surface area to volume ratio, 6:1. The smaller surface area and volume, the easier it is for diffusion to occur. Some factors that affect diffusion are: temperature, surface area, volume, and the size/shape of the molecule.

3. If a large surface area is helpful to cells, why do cells not grow to be very large?

A large surface area is helpful to cells as it allows for more materials to enter the cell, however, a smaller surface area is also helpful as it allows for diffusion to occur at a faster rate, and move towards the center of the cell faster. When a cell grows its volume increases at a greater rate than its surface area, causing diffusion to happen at a slower rate, thus, decreasing growth.

4. You have three cubes; A, B, and C. They have surface to volume ratios of 3:1, 5:2, and 4:1 respectively. Which of these cubes is going to be the most effective at maximizing diffusion? How do you know this?

Cube C would be the most effective at maximizing diffusion. This is due to the fact that the surface area to volume ratio (4:1) contains a surface area that is proportionately larger than the volume. There is more surface area to be diffused but, the smaller volume allows for diffusion to travel and reach the center of the cell quicker.

5. How does your body adapt surface area-to-volume ratios to help exchange gases?

Our lungs contain sacs known as Alveoli. These sacs have a large surface area to volume ratio due to their folded shape, causing diffusion to happen more quickly. Each sac contains a thin wall which increases their surface area and makes it easier for them to deal with any inflation that may occur in the lungs. This also decrease the amount of volume that diffusion needs to act upon.

6. Why can’t certain cells, like bacteria, get to be the size of a small fish?

Certain cells can’t get to the size of a small fish as it wouldn’t be practical for them. It will make it harder for cells to diffuse materials that well. The cells are small to begin with as it allows for diffusion to occur at a much greater rate. Increasing the size of the cells would only slow down the rate of diffusion for the cell.

7. What are the advantages of large organisms being multicellular?

Some of the advantages of large organisms being multicellular is that they grow. Single celled organisms can only reproduce and make more copies of themselves, they are not capable of growing. Another advantage is that due to them being multicellular, they can have multiple functions, making them more efficient.

Measuring Keq Lab

Part I:  Preparation of a standard absorption curve for FeSCN+2

Standard 0.20M Fe(NO3)3 0.0020 M KSCN 0.100M HNO3 [FeSCN+2] Absorbance
A 10.0 mL 0.0 mL 15.0 mL 0 0
B 10.0 mL 1.0 mL 14.0 mL   M 0.343
C 10.0 mL 1.5 mL 13.5 mL  M 0.518
D 10.0 mL 2.0 mL 13.0 mL  M 0.706
E 10.0 mL 2.5 mL 12.5 mL  M 0.860
F 10.0 mL 3.0 mL 12.0 mL  M 1.130

 

EQUATION:

Y : 4572.1x – 0.0068 | 

 

Part 2: Measuring Equilibrium

Test Solution 0.0020 M Fe(NO3)3 0.0020 M

KSCN

0.10 M

HNO3

Initial [Fe+3] Initial [SCN] Absorbance Equilibrium

[FeSCN+2]*

I 5.0 mL 0 5.0 mL  M 0 0 0
II 5.0 mL 1.0 mL 4.0 mL  M  M 0.174
III 5.0 mL 2.0 mL 3.0 mL  M  M 0.370
IV 5.0 mL 3.0 mL 2.0 mL  M  M 0.512
V 5.0 mL 4.0 mL 1.0 mL  M  M 0.696
VI 5.0 mL 5.0 mL 0.0 mL  M  M 0.836

 

ICE Charts

Test Solution

Keq = 260

Fe3+               +                SCN–                                FeSCN2+
I 0
C –  –  +  
E      

 

Test Solution

Keq = 280

Fe3+               +                SCN–                                FeSCN2+
I 0
C –  – 
E      

 

Test Solution

Keq = 260

Fe3+               +                SCN–                                FeSCN2+
I 0
C –  – 
E      

 

Test Solution

Keq = 290

Fe3+               +                SCN–                                FeSCN2+
I 0
C –  – 
E      

 

Test Solution

Keq = 280

Fe3+               +                SCN–                                FeSCN2+
I 0
C –  – 
E      

CONCLUSION AND EVALUATION:

  1. Comment on your Keq values.  Do your results convince you that Keq is a constant value regardless of the initial concentrations of the reactants?  Why or why not?The results of our Keq values convinced me that Keq is not a constant value. Our values jumped around between 260 and 290, leading us to the conclusion that Keq is not a constant value and will change if you alter any of the initial concentrations.
  2. Calculate the average value of Keq from your five trials.  The actual value of Keq  for this reaction at 25oC is reported as 280.  Calculate (should you use all of your values?) the percent difference of your average value from the reported value:
    Actual Value: 280
    Experimental Value: 273.91
    % difference: (273.91 – 280) / 280 = 2.17%