• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3

# Question- How will temperature affect the rate of diffusion of NaOH into distilled water?

Free essay example:

Diffusion Lab                                                        Scotty Keller

Question- How will temperature affect the rate of diffusion of NaOH into distilled water?

Hypothesis- The higher the temperature the quicker the rate of diffusion because, according to Boyle’s law, the higher a substances temperature the faster the molecules move and thus the NaOH will hit the sides of the dialysis tubing more often and with more force allowing for more molecules to escape from the tubing.

Variables- The independent variable is the temperature of the water. The dependant variable is the change in the rate of diffusion. The controls are the amount of water, the amount of NaOH, the time allowed for diffusion to take place, the molarity of NaOH used, and the length of dialysis tubing used.

Materials-         6 ml of 1 M NaOH

15 inches of dialysis tubing

Three 250 ml beakers

600 ml of distilled water

1 pipette

1 PH probe

1 thermometer

Tape

String

Permanent marker

Ice

Stove top

Procedure-         1. Mix 40 grams of NaOH into 1000 ml of water. This 1M solution of NaOH is to be used for the remainder of the procedure in place of NaOH.

2. Label the beakers with the tape and permanent marker. Then, fill the beakers. Label one beaker “room temperature” and fill it with 200 ml of water. Label the second beaker “hot”, then fill it with 200 ml of water and place it on the stove top. Label the third beaker “cold”, then fill it with water and ice up to the 200 ml mark accounting for displacement. Turn on the stove.

3. Open one 5 inch length dialysis tubing.

4. Tightly tie off the bottom of the tubing with some string.

5. Fill tubing with 2 ml of NaOH.

6. Tightly tie off the top of the tubing with string.

7. Repeat steps 2-5 with the second and third 5 inch lengths of dialysis tubing.

8. Test and record the PH and temperature of the beakers using the thermometer and the PH probe. Be sure to cleanse the probe and thermometer with distlled water after each, individual test.

9. Simultaneously place a NaOH filled length of dialysis tubing in each of the water filled     bowls. Record time.

10. After 5 minutes record the PH and temperature of the beakers.

11. After 10 minutes record the PH and temperature of the beakers.

12. Clean the lab station.

Analysis-         The following formulas were using during my investigation.

#moles of solute = (mass in grams / relative atomic mass)

Ex) #moles of (NaOH) = 40 / (23 + 16 + 1)

#moles of (NaOH) = 40 / 40

#moles of (NaOH) = 1

Molarity = #moles of solute / liters of solution

Ex) Molarity = 1/1

Molarity = 1

Change in PH = PH after 10 min. – initial PH

Ex) Change in PH = 12.11 – 7.78

Change in PH = 4.33

Rate of diffusion =  change in PH / time

Ex) Rate = 4.33 / 10

Rate = .433 PH/min.

Difference in rate of diffusion = rate of diffusion – rate of diffusion (2)

Ex) Difference in rate of diffusion = .433 PH/min. - .329 PH/min.

Difference in rate of diffusion = .104 PH/min.

Conclusion-         As can be seen from the processed data NaOH diffused into the water at a higher rate when heated. Heated NaOH diffused into water at .433 PH/Minute while cooled NaOH diffused into water at .329 PH/min. The heated NaOH diffused .104 PH/Minute faster than the cooled NaOH. This proves my hypothesis which stated that the higher the temperature the quicker the rate of diffusion because, according to Boyle’s law, the higher a substances temperature the faster the molecules move and thus the NaOH will hit the sides of the dialysis tubing more often and with more force allowing for more molecules to escape from the tubing. The procedure I followed could be improved with the use of an electric thermometer for a more accurate temperature analysis. I am also afraid that a small amount of NaOH may have escaped from the tubing through the rudimentary string tie that was used. This could be improved by heat sealing (melting) the end of the tubing for a near perfect enclosure. Also, using three PH probes would better this investigation because there was a delay between the testing of the three beakers which allows more time for the NaOH to diffuse. However, all shortcomings aside, this investigation was successful in determining how temperature effects the diffusion of NaOH into distilled water.

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

Close
1. ## What factors affect the temperature change of water when heated by an electric heater?

Word count: 3435

2. ## Establish what types of soil holds the most water and to see if changing ...

Word count: 4691

3. ## Factors affecting the rate of Diffusion.

Word count: 1799

4. ## Investigating the effects of varying pH levels on the germination of cress seeds

Word count: 5088

5. ## Application of Hess's Law

Word count: 1657

# Related GCSE Aqueous Chemistry essays

1. ## Investigating the effects of varying pH levels on the germination of cress seeds

5.6 31.36 0.647933884 pH 5 ? 18.6 345.96 7.147933844 pH 6 ? 23.6 556.96 11.50743801 pH 7 ? 38.6 1489.96 30.78429752 pH 1 ? 48.4 2342.56 45.39844961 pH 2 ? 48.4 2342.56 45.39844961 pH 3 ? -10.6 112.36 2.177519379 pH 4 ? -5.6 31.36 0.607751937 pH 5 ?

2. ## Factors affecting the rate of Diffusion.

When you divide 5 by 2.5, you get 2 showing doubling. Although, this trend is not the same for other points. The considering the results, the following drawings need to be taken into consideration: In the first picture, the 10 particles in A would be moving randomly and because there

1. ## Application of Hess's Law

x2)/2 = 0.05 Moles 1 mole of Na2CO3 would produce twenty time more energy = 0.05 x 20 = 1 mole The average enthalpy change for 0.05 Moles of Na2CO3 = (104.5+209+418)/3 = 243.84 Joules For 1 mole of Na2CO3 = 243.84 Joules x 20 = 4876.7 Joules I have

2. ## Establish what types of soil holds the most water and to see if changing ...

Results Table Soil type Ph of the water added Absorbency of soil (cm3) 1st Test 2nd Test 3rd Test Average Clay Acidic (ph 1 -2) 44 47 41 44 Alkaline (ph 12 - 14) 34 32 38 34.7 Neutral (ph7)

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to