Lab 15 16 Report - CHEM 106 PDF

Preview

Summary

CHEM 106...

Description

Feratovic 1

Galjo Feratovic 11/17/20 Mina Kang CHEM 106 Lab 15/16 Report Introduction: Lab 15: When we observe a titration, we must grasp the concepts presented before us. This lab will allow us as scientists to observe the effects of the additions of NaOH into a mixture where we can see the pH of the reactions take place. This lab also informs us on the many midpoints, equivalence points (including after the addition of NaOH), buffer zones, and how these are all linked from acid dissociations in polyprotic acids. Through this, we can take into consideration how many dissociations within an acid can occur and how it’s titration can appear to be. Lab 16: After completing titration curves and pH values, we now venture into galvanic cells where we are given an anode and a cathode and through voltmeters we can observe a reaction occur and obtain a standard cell potential value. This reaction will consist of an oxidation reaction and a reduction reaction, these are half reactions that make up the full process. This lab will also be valuable for allowing us to gain an idea of how the Gibbs Free Energy formula will be implemented into this concept as well. Materials: Lab 15: -

Lab manual

-

Excel

-

Notebook

Feratovic 2

Lab 16: -

Lab manual

-

Notebook

-

Sandpaper

-

Cu electrode

-

Iron electrode

-

Zinc electrode

-

Voltmeter

-

Porous cup

-

150mL beaker

-

Thermometer

-

Solutions given

Observations and Experimental: Lab 15:

Feratovic 3

Part 1:

Volume of 0.200M NaOH

pH

1st Derivatives

0

2.42

0.39

1

2.81

0.27

2

3.08

0.2

3

3.28

0.12

4

3.4

0.1

5

3.5

0.08

6

3.58

0.09

7

3.67

0.07

8

3.74

0.07

Feratovic 4

9

3.81

0.08

10

3.89

0.06

11

3.95

0.07

12

4.02

0.07

13

4.09

0.06

14

4.15

0.05

15

4.2

0.06

16

4.26

0.07

17

4.33

0.06

18

4.39

0.06

19

4.45

0.06

20

4.51

0.05

21

4.56

0.06

22

4.62

0.05

23

4.67

0.05

24

4.72

0.09

25

4.81

0.14

26

4.95

0.16

27

5.11

0.26

28

5.37

0.17

29

5.54

0.3

29.5

5.69

0.44

Feratovic 5

30

5.91

0.76

30.25

6.1

1.56

30.5

6.49

8.12

30.75

8.52

9.92

31

11

0.65

32

11.65

0.28

33

11.93

0.26

34

12.19

0.1

35

12.29

0.12

36

12.41

0.02

37

12.43

0.01

38

12.44

0.02

39

12.46

0.02

40

12.48

0.02

41

12.5

0.01

42

12.51

0.02

43

12.53

0.01

44

12.54

0.02

45

12.56

0.01

46

12.57

0.01

47

12.58

0.01

48

12.59

0.02

Feratovic 6

49

12.61

0.01

50

12.62

0.2524

Feratovic 7

First Derivative Values: Equivalence Points: pH =9.92, 30.75mL of 0.200M NaOH Midpoint: (30.75/2) = 15.37 mL, pH = 4.2 = pKa Buffer Zone: 4.2 ± 1 Ka: 10^-4.2 = 6.3 x 10^-5 = Benzoic Acid (mm = 122.12g/n) Benzoic Acid Concentration: 1. 3g (122.12g/1n) = 0.025n / 0.1L = 0.25M 2. (0.03075 L)*(0.200M) = 0.00615n , (0.00615n)*(0.025L) = 0.25M Benzoic Acid Part 2:

Feratovic 8

Feratovic 9

Feratovic 10

Feratovic 11

0.5M NaOH added (mL)

pH

0

1.71

50

3.12

100

3.95

150

4.77

200

5.59

250

6.4

300

9.7

400

13.5

Lab 16: Part 1: cell

Exp. E (V)

Theo. E (V)

Feratovic 12

Zn(s)|Zn2+(aq,1M)||Cu2+(aq,1M)|Cu(s)

**0.700

1.1

Fe(s)|Fe2+(aq,1M)||Cu2+(sq,1M)|Cu(s)

0.687

0.78

Zn(s)|Zn2+(aq,1M)||Fe2+(aq,1M)|Fe(s)

0.354

0.32

1. 0.34 + 0.76 = 1.10 2. 0.34 + 0.44 = 0.78 3. -0.44 + 0.76 = 0.32 Part 2: cell

Exp. E (V) Theo. E (V)

A

0.998

1.10 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(.1/.1) = 1.10

B

0.976

1.10 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(1/.1) = 1.07

C

1.054

1.10 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(.1/1) = 1.13

D

0

0 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(.1/.1) = 0

E

0.021

0 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(1/.1) = 0.03

F

0.013

0 - ((8.314 J/molK)(298K)/(2)(96485 C/mol)ln(.1/1) = 0.03

Part 3:

Feratovic 13

Experimental -

Enthalpy: (-50.485)(-(2)(96485))/10^6 = 9.74 kJ/mol

-

Entropy: 2.3402(2*96485)/10^3 = 451.59 J/mol

-

Free Energy: 9.74 - (298K)(.45159) = -124.83 Theoretical

-

Entropy: (-137.7+33.2) - (-99.6 + 27.3) = -32.2 J/mol K

-

Enthalpy: -89.1 - 64.8 = -218.7 kJ/mol

-

Free Energy: -78.90 - 65.5 = -144.4 kJ mol

Feratovic 14

Experimental -

Enthalpy: (1185.1)(-(2)(96485))/10^6= -228.69 kJ/mol

-

Entropy: (-0.3165)(2)(96485) = -61.08 J/mol K

-

Free Energy: -228.69 - (298)(-0.06108) = -210.49 kJ/mol

Theoretical -

Entropy: (33.2+112.1) - (41.6 - 99.6) = 203.3 J/mol K

-

Enthalpy: -153.9 - 64.8 = -218.7 kJ/mol

-

Free Energy: -147.2 - 65.5 = -212.7 kJ/mol

Discussion and Conclusion: Lab 15: Following this lab, we can see how the data given displays how the titration occurs and how it allows us to identify a compound given to us. This lab also allowed us to see how a polyprotic acid’s titration occurs and how it contains many midpoints and equivalence points. Relying on the graphs and calculations to watch the reaction unfold before us is vital to our overall understanding of reactions in chemistry. Similar to previous labs, we can see how

Feratovic 15

additions of NaOH alters the pH of the solution and we can see many buffers occur during these reactions. Lab 16: Through the process of observing galvanic cells, we can conclude that we now understand standard cell potentials and how experimental and theoretical values can differ. Observing the graphs we can see how the two different cells differ as temperature increases and the experimental/theoretical enthalpy, entropy, and free energy values are not entirely similar but are indeed approximate. The concentrations of the half reactions also alter the cell potential as we see the largest increase when the concentrations are identical, with differing values depending on whether the cathode or anode are the larger concentration. References: “Benzoic Acid – Uses and Safety.” MSDSonline, 18 Sept. 2017, www.msdsonline.com/2015/02/16/benzoic-acid-uses-and-safety/. Smeureanu, Gabreiela. “Lab 16 part 1-3” Youtube, uploaded by Gabriela Smeureanu, 21 March 2020 https://www.youtube.com/watch?v=ygBj49Xm3gs. Smeureanu, G. & Geggier, S. (2020). General Chemistry Laboratory. New York, NY Y;, Lindquist E;Yang. “Degradation of Benzoic Acid and Its Derivatives in Subcritical Water.” Journal of Chromatography. A, U.S. National Library of Medicine, pubmed.ncbi.nlm.nih.gov/20846661/. Focus Questions: Lab 15: 1. Benzoic acid Lab 16:

Feratovic 16

Part 1: 1. 1.1, 0.78, 0.32. 2. Yes, even though it was approximate. Part 2: 1. In A, the concentrations are identical so the voltage increased.B and C are opposites of one another, but both are lower than A. A higher concentration in the anode, such as B, the voltage is lower than if it were switched where the cathode is higher, such as C. 2. No, it equates to zero. 3. In this case yes for we received slightly different values. Part 3: 1. For the first cell, we see an increase in voltage, in the second there is a decrease. These cells have opposite directions as temperature increases. 2. For the most part, the delta G values seem to be close to one another, however the enthalpy and entropy values differ, this can be due to either calculation error or basic human error. Post-Lab Assessment Questions: Lab 15: 1. We can also identify an unknown chemical by performing solubility tests, finding both freezing and boiling points, and flammability tests. If we are given a Ka/pKa value, we can compare it to values within the chart to see which chemical it aligns with. 2. C6H7O7- and C6H6O7-2 3. Yes, for it’s range is from 3.5 to 5.5, and observing the range in the first midpoint can conclude that it is within said range.

Feratovic 17

4. (at pH of 4.7, 24 mL), 0.2M = n/0.024L, (0.2M)(0.024L) = .0048n NaOH 5. Sodium azide is highly toxic and may be fatal on short term exposure. It is slightly flammable, and its instability can be serious when detonated/shocked, heated in confinement, or mixed with water. It is used in the auto industry as a chemical that detonates and sets off the airbag in a car, this is where the explosion is used to an advantage. 6. Sodium citrate and sodium hydrogen citrate. Lab 16: 1. a) Ni is the anode and Ag is the cathode, this is due to Ni being oxidized. This can be seen from its negative reduction potential value. b) Ni → Ni2+ + 2e- (anode) 2Ag+ + 2e- → Ag (cathode) c) Ni|Ni2+||Ag+|Ag d) 0.80 + 0.25 = 1.05V e) 1.05V - ((8.314)(298)/(2)(96485)) ln (0.01/.001) = 1.02 2. a) Ag(s) + Cd2+(aq) → Ag+(aq) + Cd(s) b) Pb(s) + MnO2(aq) → Pb2+(aq) + Mn2+ (aq) 3. a) 0.35V + 1.69V = 2.04V b) 2.04 - (8.314)(298)/(2)(96485) ln (⅙) = 2.06V...