Diffusion Lab Report

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             Lab #1 Diffusion through Membranes

Purpose

This experiment was conducted in order to determine how diffusion rate of a solute is affected by concentration gradient and the presence of another molecule.

Introduction

     The rate at which solutes such as salt diffuse across a cell's semi-permeable membrane is affected by a variety of factors. Solutes will diffuse from where there is greater concentration of itself to where there is less concentration of itself. Salt is an electrolyte that induces conductivity of electricity in eater. In this experiment the conductivity of the solution will be measured in order to determine how much solute diffused into or out of the model cell membrane. High solute concentration difference between the interior and exterior of a cell results in diffusion. Sometimes to presence of another molecule can interfere with the movement of another. Sugar will be used as the other molecule that will be tested for interference with salt diffusion. Sugar cannot conduct an electric current and will therefore not alter the measurement of conductivity being used to measure salt diffusion.

          In this experiment higher solute concentration in the model cell should result in greater diffusion rate seen as greater conductivity. The presence of sugar with the salt solution should show decreased diffusion rate of salt as lower conductivity increase rate.

Methods and Materials

Materials:

logger pro device

Conductivity probe

three 150mm test tubes and rack

salt solution 1%, 5%, 10%

400ml beaker

dialysis tubing 2.5cm x 12cm

pippets

stirring rod

sucrose solution 5%

Procedure

       Set up logger pro device to measure electrical conductivity with probe. Prepare three dialysis tubing pouches tied at one end. Pour 15ml salt solution of 1% concentration into test tube one, 15ml of 5% salt solution into test tube three and 15ml of 10% salt solution into test tube 3. Pour 300ml of distiller water in the beaker. Begin the trials to determine conductivity rate for different solute concentration. Use pippet to place 10ml of 1% salt concentration into dialysis tubing and tie other end. Repeat this step with each test tube for subsequent trials. Place the tied tube with solution into beaker. Stir with rod or pen. Place conductivity probe into beaker and begin measurement on logger device for two minutes. Then, use device to measure slope(m) of curve to determine diffusion rate. Click analyze, select curve fit and press linear. Record the slope of the graph in the data table for diffusion rate. Replace water in beaker after each trial. Repeat this process for 5% and 10% solutions. Now fill the beaker with 5% sucrose solution and measure conductivity for two minutes. In the final trial fill another dialysis bag with 5% salt solution and place it in the sugar water beaker and measure then record the conductivity for two minutes. For more detailed instructions refer to Lab 1 Diffusion through membranes.

Results

Table 1

Set ups	Predictions	Conclusion
Solute Concentration	Higher salt concentration solutions will have greater conductivity rate	The conductivity rate increased as solute concentration increased in the tubing.
Sucrose Conductivity	Sucrose will have low or no conductivity	The sucrose solution showed very low conductivity compared to salt solutions.
Sugar Interference	Salt solution will show lower conductivity  rate due to diffusion interference by sucrose	5% salt solution showed higher conductivity rate in sucrose solution than trial in distilled water.

Table 2

Salt Concentration (%)	Rate of Diffusion (mg/L/s)
1	2.904
5	6.701
10	15.004

Table 3

Solution	Conductivity (mg/L)
Distilled water	234
Sugar water 5%	195

Table 4

Solution	Rate of Diffusion (mg/L/s)
5% Salt	6.701
5% Salt/5% Sugar	11.646

Discussion

            In Part I the three different solution concentrations showed varying rates of conductivity increase. The increase in conductivity measured is a result of diffusion of salt from the tubing into the water in the beaker. As the salt diffuses out of the bag it increases conductivity. The results of the experiment showed that as salt concentration increased the rate at which solute diffused increased. This confirms that the increased difference between solute concentration inside a semi-permeable membrane and the outside environment causes faster diffusion rate. In Part II the control group trials of distilled water and sucrose solution showed very low conductivity and no conductivity increase. The trial including 5% salt concentration in the tubing in the sucrose solution beaker tested whether or not the presence of another molecule affects diffusion. The conductivity rate increase of the 5% salt solution was greater in the sucrose beaker than the beaker with distilled water. This result shows that the presence of sucrose does not deter salt diffusion. This could mean that the sucrose induced faster movement of solute across the semi-permeable membrane.

The results of the experiment validate the hypothesis made about the salt concentration solutions, but refute the prediction made about the presence of sucrose with the salt. The hypothesis stated that there was a direct relationship between solute concentration and diffusion rate. The 5% solute concentration trial showed higher conductivity rate than the 1% and the 10% solution showed greater conductivity than the 5%. Increasing the solute concentration in the bag increased the rate at which salt diffused out. Solute diffuses through a membrane from higher concentration to lower concentration. Since there was not salt in the beaker, the difference in concentration was greater as the concentration of salt in the bag was increased.  In Part II the low conductivity of water and sucrose solution validated the hypothesis that they do not conduct electricity well. The sucrose solution showed lower conductivity that distilled water. This shows that sucrose decreases conductivity of water. Also, the diffusion rate of salt was tested with the presence of another molecule in the outside environment. The experiment refuted the hypothesis made that the presence of sucrose would decrease diffusion rate. The diffusion rate actually increased in the presence of sucrose. This was because although the beaker had sucrose in solution, the difference between the bag and beaker of salt concentration stayed unchanged. However, the greater increase in the rate of conductivity suggests that diffusion rate increased due to the presence of sucrose. A possible explanation of this result could be that the sucrose was unable to diffuse into the bag and due to osmosis more pure water moved out of  the bag as the salt moved out. Having less free water outside the bag caused more water to move from the bag into the baker. This additional water made the difference in salt concentration even greater between the inside and outside of the bag causing increased diffusion of salt.

The experiment procedure was very well set up using exact measurements and an electronic data collection tool. Possible sources of error are few, but possible sources of error could have been in the materials used. The dialysis tubing although filled with exactly 10ml of solution for each trial was not able to be tied exactly the same or sealed with no possible mistake. The presence of air or other foreign contaminants in the tubing could have affected diffusion rate. It is possible for error in the ability to seal the tubes so that only solute can move into or out of the tube allowed larger amounts of solution to move into or out of the cell. This may explain the greater conductivity measured in the 5% salt 5% sucrose trial. The tubing was also wet when being sealed thus altering the amount of water inside and outside the tubing. This might have had a slight effect on the solute concentration.

Possible solutions to improve the procedure of this experiment to minimize error would be to have dialysis tubing that is sealed at one end so that salt can diffuse, cut to have exactly 10ml of volume with one open end to pour solution into. Then the untied end can be sealed the same way as the other end. This will ensure only salt solution is inside the tube and that only salt solution can diffuse in and out.

Adendum

Questions

1.    How did you arrive at your conclusions for Part I?

 The higher salt solute concentrations showed greater increase in conductivity on the graph. The salt diffusion into the beaker solution allowed for conductivity. The slope calculated for the conductivity graph increased as solute concentration increased. Therefore there was greater diffusion rate as solute concentration increased.

2.    How did your conclusion compare to our prediction for part I? Can you account for any differences?

The conclusion of Part I validated the predictions made on the direct relationship between solute concentration and diffusion rate.

3.      If the rates in any of the three experiments varied in Part I, calculate how much faster each rate was compared to the other.

5% solution had 2.3x faster diffusion rate than the 1% solution. The 10% solution had 2.2x faster diffusion rate than the 5% and 10% was 5.17x faster than the 1%.

4.      Compare the conductivity of pure water with sugar water solution. How do you account for this?

The pure water had greater conductivity at 234mg/l than the sugar water at 195mg/l. This is due to the fact that water has slight conductivity for electricity, but the presence of sucrose molecules reduces it because it reduces the amount of free water molecules available. Because sucrose cannot form ions, there are less molecules for a current to travel on.

5.      How did your conclusion compare to your prediction in Part II? Can you account for any differences?

The conclusion of Part II invalidate the predictions that the presence of sucrose interferes and decreases the diffusion of solute. The 5% salt solution showed greater conductivity with the presence of sucrose solution. This may have been due to the fact that sucrose unable to diffuse across the membrane resulted in more free water leaving the bag. Increased movement of water across the membrane increased diffusion rate of salt in order to reach concentration equilibrium.

Extensions

1.      Make a plot of the rate of diffusion vs. the salt concentration in the dialysis bag. Estimate 3% salt solution rate of diffusion.

Based on this graph at a salt concentration of 3% the diffusion rate would be 4mg/L/s.

2.    If the results of the experiments in Part I can be extrapolated to diffusion in living systems, how would a single-celled organism respond in an oxygen rich pond compared to an oxygen poor pond?

Based on the results of this experiment a conclusion can be drawn that in an oxygen rich environment a cell would gain oxygen due to diffusion of the gas from high to low concentration. In an oxygen poor environment a cell would loose oxygen due to oxygen diffusing out of the cell from higher to lower concentration.

3.    Design an experiment to determine the effects of temperature on diffusion rate.

An experiment to determine the effects of temperature on diffusion rate would follow the same procedure as this experiment with three model cells of dialysis tubing filled with 5% salt concentration. The experimental variable would be to have three different 400ml beakers filled with distilled water heated to different temperatures: one room temperature, one near freezing and one boiling. Three trials would be observed with the conductivity probe to measure at what rate the conductivity increases.

4.    Ectotherms are organisms whoes body temperature varies with the surrounding environment. On the basis of your data from Extension Question 3, how do you expect the oxygen consumption of ectotherms to vary as the temperature varies?

As temperature increases an ectothermic organism would better absorb oxygen as the increase in internal heat would accelerate to movement of molecules through diffusion.

5.      If waste products of a single celled organism were released by the organism into the pond, how would that affect the organism’s ability to obtain oxygen as readily?

If an organism released waste product into a pond its available oxygen level would go down. This is due to the fact that waste decomposed by other organisms uses up available oxygen.

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