Lab 15 and 16 - lab 15 and 16 my grade: 85 PDF

Preview

Summary

lab 15 and 16 my grade: 85...

Description

Cindy Jiang 16 November 2020 Professor powell Lab 15 and 16 Lab 15 “Drop the base” Introduction: This lab focuses on buffers and its tirationn levels. We are gathering our previous knowledge of acids and bases and applying it to lab 15 in order to determine an unknown. We will set up a strong acid and strong base titration, a weak acid and strong base titration, as well as analyze polyprotic acid titration curves. All data will be used to observe titration curves. Materials: - Safety glasses - Chem manual - Excel or google sheets Focus questions Part one 1. Which substance is the crystalline powder It is most Benzoic acid Part two 1. What are the major species in solution during the titration of citric acid with a strong base? Initial-C6H8O7 First midpoint- C6H8O7 and C6H7O7 are the major species First equivalence point- C6H7O7 Second midpoint- C6H7O7- and C6H6O7^2Second equivalence point- C6H6O7^2Third midpoint- C6H6O7^2- and C6H5O7^3Third equivalence point- C6H5O7^3Pass third equivalence point- OH- (NaOH) Experimentation and observation

first derivative

0.2 M NaOH added (mL)

pH

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

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

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

49

12.61

0.01

50

12.62

0.2524

Our titration curve for for 0.2 M of NaOH added

Volume at equivalence point: 30.75 Volume at midpoint: 15.4 Molarity of acid at equivalence point: 0.246 M Initial pH: =-log(0.0189)=1.72 PH at midpoint: -log[ka]=pka=pH -log(7.5E-4)=3.124 -log[ka]=pka=pH -log(1.7E-5)=4.8 -log[ka]=pka=pH -log(4.0E-7)=6.4 PH at equivalence point: PH=[3.12+4.77]/2=3.95 PH=[4.77+6.40]/2=5.58 PH=14-(4.25)=9.7 PH pass equivalence point: Volume=500mL (0.1L)(.50M)=0.05 mol of OH[OH^-1]=0.05mol/0.5L=0.1M POH= -log(0.1)=1 14-1=13. PH is 13

pH titration curve

Post lab assessment questions 1. What other methods could you have used to identify the unknown chemical? I could test the chemical’s physical traits such as the changes in the unknown chemical when reacted with lowering temperature or rising temperature. 2. What is/are the dominant species in solution at pH=4.7 H+ 3. Does the solution at pH=4.7 constitute a buffer? Explain. No. There is a steady rise in the graph where 4.7 pH is located. There is no buffer activity until pH is much more basic. 4. How many moles of strong base can be added to the solution at pH=4.7 (0.9)(0.2)=0.18 5. The fire diamond for sodium azide looks like this: What does this tell you about the hazard this compound poses? - the compound is lethal as a liquid or a gas and damages the central nervous system. Blue (4)=lethal and toxic Red (1)=fire, flammable Yellow (3)=not stable when interacts with heat or shock, can be explosive White=soluble What is this compound in the auto industry used for? -airbags, as it contains nitrogen gas

Conclusion: In lab 15, we were able to determine the concentration of an unknown acid. There were two methods: the titration curve and first derivative curve. Both ways of finding an unknown acid were effective in this lab. Based off our datas and titration curves, we are able to analyze

different parts of the graph such as the midpoint, equilibrium point, and the buffer region. By adding our base into the unknown, we were able to determine its name after experimentation and graphing of the titration curve.

Lab 16 “ LEO the lion goes GER” Introduction: This lab introduces electrical energy with experimentation with batteries. Batteries have galvanic or voltaic cells which help produce electrical energy. The cells consist of two parts, anode and cathode. If a galvanic cell is set up correctly, it can transfer the electrons to go from one electrode to another. The constant flow of the electrons will be known as an electrical current. We will set up an experiment to demonstrate the electrical current that moves through a galvanic cell. Materials: - Safety glasses - Lab coat - Lab notebook - Metallic electrodes (Cu,Fe, Zn) - Beakers - Cup - Voltmeter - Thermometer - Heating plate - Excel or google sheet Focus question Part one 1. What are the cell potentials of the three galvanic cells you obtain in the lab? Zn/Cu=1.1 Fe/Cu=0.592 Zn/Fe=0.225 2. Do your experimental/ laboratory values match the theoretical values? Our values are close but not exact. We don't expect it to be exact because of small measuring errors and calculation errors. Part two 1. How does concentration affect the cell potential? It could change the value of LnQ, which will act as a stressor on the compound, it will shift the reaction. The cell potential will either increase or decrease. 2. Do two identical half-cells constitute a galvanic cell?

Two half cells cannot constitute a galvanic cell because the cell potential would be zero and cancel out, therefore having no flow of electronegativity. 3. Do the same type of cell but different concentration constitute a galvanic cell? Based on cells d and e, the same type of cell of different concentration can constitute a galvanic cell. The concentration of anode has been decreased which increases cell potential. Part three 1. How does temperature affect the cell potential of a) the iron and copper (II) reaction and b) the zinc and copper (II) reaction? A: as the temperature increases the cell potential increases as well. For zinc and copper, as the temperature increases the cell potential decreases. 2. What are your experimental values for delta H, delta S, and delta G for these reactions? Are they in agreement with the theoretical values? Discuss any sources of experimental error. My experimental values for delta H, delta S, and delta G for the reactions is -73kj/mol, 208.5 kJ/mol, and -135.3kj/mol. The delta h and delta G are in agreement with theoretical values, however delta S is not in agreement. Experiment and observations Part 1 (determine unknowns and theoretical potentials) Determine unknowns

Unknown options

Anode

Cathode

Experiment Theoretical al (V) (V)

A = Cu2+

Cu and Cu^2+

B

A

0.62

1.1

B = Zn2+

Zn and Zn^2+

C

B

0.24

0.32

C = Fe2+

Fe and Fe^2+

C

A

0.92

0.78

Part 2 (determine theoretical cell potentials) Anode

Cathode

Experimental (V)

Theoretical (V)

Zn (0.1 M)

Cu (0.1 M)

0.914

1.1

Zn (1 M)

Cu (0.1 M)

0.893

1.07

Zn (0.1 M)

Cu (1 M)

0.923

1.13

Zn (0.1 M)

Zn (0.1 M)

0.002

0

Zn (0.1 M)

Zn (1 M)

0.03

0.0296

Cu (0.1 M)

Cu (1 M)

0.02

0.0296

Post lab questions 1. A voltaic cell is constructed with an Ag/Ag+ half cell and a Ni/Ni2+ half-cell. a. Which metal will serve as the anode and which as the cathode> justify your answer quantitatively. - nickel is the anode because it is the most negative. The cathode would be silver. b. Write a cell reaction for this cell. Write the half-reactions and label them as oxidation and reduction c. Write the line notation for this cell d. Determine the standard cell potential e. If [Ag+]=0.0010 M, determine the cell potential

2. Write a balanced equation from line notation:

3. In the lead storage battery, lead serves as the anode, and lead-coated with lead oxide serves as the cathode. Both electrodes dip into an electrolyte solution of sulfuric acid. The electrode reactions are a. Calculate the standard cell potential, E, for this cell at 25 C. The concentration of sulfuric acid in a car battery is 6 M, when the battery is fully charged. Use the nernst equation to determine the cell potential, E. remember that sulfuric acid is a strong acid (use only the first proton)

Conclusion In lab 16, we were able to determine the cell potential in 2 ways. We used the voltmeter and formulas to find our values. At standard conditions, we used E ° cell = E ° ox + E ° red to find cell potential. In non-standard conditions, the Nernst equation is used to find cell potential. Lab 16 is mainly a calculation based lab, therefore we kept in mind that percent errors will be a thing as we cannot reach exact results. Some examples include temperature mistakes, calculation mistakes, and measurement rounding. Lab 16 helped us understand the galvanic processes that occur in our everyday life and showed us the scientific side of our everyday electronic experiences....