# OPTIMAL CONCENTRATION OF SODIUM CHLORIDE By: Melissa Kingwaya, Mervelous Malekera and Lina Kaddoura.

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OPTIMAL CONCENTRATION OF SODIUM CHLORIDE By: Melissa Kingwaya, Mervelous Malekera and Lina Kaddoura

INTRO The purpose of this lab is to find the optimal concentration of 5% NaCl The enzyme that will be studied is called catalase. Its job is to break down its substrate hydrogen peroxide (H2O2)

MATERIALS Sodium Chloride Water Hydrogen peroxide Test tubes Potato puree (enzyme) 40 ml beakers Safety Goggles Pipette Test tube rack Tweezers

PROCEDURES Find the calculations of the desired concentrations using the given informations ( C 2 C 1, V 2 1.Givens: C 1 =0.05ml C 2 =0.05 ml V 1 =? V 2 =5ml V 1 = ( C 2 )(V 2 ) / C 1 V 1 = (0.05)(5)/0.05 V 1 = 5ml Therefore, 5mL of NaCl and 0 mL of water 0ml of water

CALCULATIONS CONT. 2.Givens: C 1 =0.05ml C 2 =0.04ml V 1 =? V 2 =5ml V 1 = ( C 2 )(V 2 ) / C 1 V 1 = (0.04)(5)/0.05 V 1 = 4ml Therefore, 4mL of NaCl and 1mL of water 3. Givens: C 1 =0.05ml C 2 =0.03ml V 1 =? V 2 =5ml V 1 = ( C 2 )(V 2 ) / C 1 V 1 = (0.03)(5)/0.05 V 1 = 3ml Therefore, 3mL of NaCl and 2 ml of water

CALCULATINS CONT.. 4.Givens: C 1 =0.05ml C 2 =0.02ml V 1 =? V 2 =5ml V 1 = ( C 2 )(V 2 ) / C 1 V 1 = (0.02)(5)/0.05 V 1 = 2ml Therefore 2mL of NaCl and 3mL of water 5.Givens: C 1 =0.05ml C 2 =0.01ml V 1 =? V 2 =0.05ml V 1 = ( C 2 )(V 2 ) / C 1 V 1 = (0.01)(5)/0.05 V 1 = 1ml Therefore 1mL of NaCl and 4mL of water

PROCEDURE CONT 2. According to the calculations, add the required amount of NaCl and water in each beaker using different pipettes. For example : 5ml of NaCl, 4ml of NaCl and 1 ml of water. 3. Label each beaker with the correct name. For example: beaker # 1 : 1% NaCl with 4ml of water etc 4. Add 1ml of potato puree in each beaker using a pipette 5. Pour 10 mL of hydrogen peroxide into 5 test tubes. 6. Label them from 1-5 7. using a tweezers, pick up a paper disk and drop it into beaker # 1 for approx 5 sec

PROCEDURE CONT 8. disk inside the beaker and place it on a paper towel. Let it dry for 5 sec 9. Pick up the paper disk using the tweezers and put it in test tube # 1 10. Using your stopwatch, record the time it takes for the paper disk to rise up to the surface 11. Repeat steps # 7-10 for the rest of the % of NaCl 12. Clean up the station including the materials properly 13. After 5 sec, use the tweezers to pick up the paper

PROCEDURE OF THE CONTROL GROUP Obtain a test tube and a 40 ml beaker Add 1 mL of potato puree into the beaker Add 10 mL of hydrogen peroxide into a test tube Drop a paper disk into the beaker using the tweezers for 5 sec After 5 sec, place the paper disk on a paper towel and let it dry for 5 sec Using the tweezers, drop the paper disk in the test tube Using a stopwatch, record the time it took for the paper to rise up Clean the station including the materials used properly

OBSERVATIONS

ANALYSIS The purpose of this lab was to determined the best optimal concentration of NaCl. Due to our result, we can conclude that the paper disk in test tube # 4 rose up faster than any other paper disk. The test tube contained 10mL of H 2 O 2, but the paper disk was dipped into a beaker that contained 1ml of water, 4ml of NaCl and 1ml of the potato puree. In this case, the beaker that contained water, NaCl and potato puree acted as an enzyme solution and hydrogen peroxide as the substrate of the catalase. A "catalyst is a substance that lowers the activation energy required for a chemical reaction, and therefore increases the rate of the reaction without being used up in the process. CATALASE is an enzyme, a biological (organic) catalyst. Hydrogen peroxide is the substrate for catalase." As the catalyst was breaking down the hydrogen peroxide into water and oxygen gas, the bubbles of oxygen collected underneath the filter and made the disk rise to the surface of the hydrogen peroxide. The time it took for the filter to rise up is an indication of the rate of enzyme activity. The disk paper took long because it all depended on the concentration of the enzyme. The less concentrated the enzymes, the slower the rate of reaction. For the other disks, the enzyme might of not been as concentrated as the 4% NaCl, it might of been because the peroxide was in room temp and also the rate at which an enzyme works is influenced by many factors including pH. Comparing the 4% of NaCl to the control group, the paper disc rose faster since it only contained potato puree. The enzyme in the disk might of be more which caused the product to be formed rapidly. Overall, the procedure of this lab was done as it was suppose to, but would of be helpful if the H 2 O 2 was heated to help the reaction react rapidly.

ANALYSIS CONT

CITATIONS Source: Gen Nelson www.accessexcellence.orgwww.accessexcellence.org Adapted for Horton Biology September 2007 Mr. J. Fuller http://www.horton.ednet.ns.ca/staff/jfuller/selig/L ABS/bio12ap_adv/Catalase_paperdiscs.pdf http://www.horton.ednet.ns.ca/staff/jfuller/selig/L ABS/bio12ap_adv/Catalase_paperdiscs.pdf http://www.emsb.qc.ca/laurenhill/science/cv.html