CHM2046L Final Exam UF Spring 2021

beer-lambert law

A = εdc

A = absorbance

ε = molar absorptivity (L mol-1 cm-1)

d = distance the light travels through the solution (the cuvette) in cm

c = concentration of the solution (mol/L)

Beer's Law calibration curve:

Use the following data to generate a calibration curve of Absorbance vs Concentration at λmax = 420.6 nm.

concentration 1: 6e-5 M; absorbance 1: 1.330

concentration 2: 2e-5 M; absorbance 1: 0.451

concentration 1: 1e-5 M; absorbance 1: 0.239

concentration 1: 2e-6 M; absorbance 1: 0.042

Record the equation and analyze your data:

- do all of the data points fall on the trendline? Is your R2 value greater than 0.990?

- what does y represent?

- what does x represent?

- why should the y-intercept of the linear trendline be very close to 0?

equation: y=22066x + 0.008

yes, all points fall on the trendline or are very close. R2 = 0.9998

y = absorbance

x = solute concentration (mol/L)

a y-intercept of 0 indicates 0 absorbance at 0 M concentration of colored solute

Serial Dilution:

- Perform calculations to determine the amount of 6.00x10-5 M stock solution needed to prepare 20.00 mL of 2.00x10-5 M dye solution.

- Perform calculations to determine the amount of 2.00x10-5 M stock solution needed to prepare 20.00 mL of 1.00x10-5 M dye solution.

- Perform calculations to determine the amount of 1.00x10-5 M stock solution needed to prepare 20.00 mL of 2.00x10-6 M dye solution.

x = 6.67 ml

x = 10 ml

x = 4 ml

using this Beer's Law calibration curve equation, y=22066x + 0.008, at λmax = 420.6 nm, determine the concentrations for each of the absorbances values below:

absorbance 1: 2.224

absorbance 2: 1.558

if absorbance 1 is from a 1e-3 M stock solution and absorbance 2 is from a 1e-4 M stock solution, do all of these data points follow Beer's Law? (did the concentrations you obtained using the equation match the concentrations of the stock solutions?). If not, why? (***hint: look at the excel graph you generated and determine the maximum absorbance at λmax = 420.6 nm that falls on/very close to the trendline)

for absorbance 1:

x = 1e-4 M

for absorbance 2:

x = 7.02e-5 M

No. These concentrations obtained from the absorbances, y = 2.224 and y = 1.558, do not follow Beer's Law. The concentrations calculated do not match the stock solution concentrations, 1e-3 M and 1e-4 M. The maximum absorbance value that follows Beer's Law is 1.330 at λmax = 420.6 nm because 1.330 falls on the trendline. The absorbances, y = 2.224 and y = 1.558, do not follow the trendline because their absorbances surpass the maximum absorbance at λmax = 420.6 nm.

consider that you have a 100 mM stock solution and you need to prepare 10 ml of a 30 mM solution.

How many ml of the stock solution do you need?

How many ml of deionized water do you need?

enter each of your answers to 2 sig figs

3.0 ml stock

7.0 ml DI water

visible light wavelengths

red: 630-750 nm

orange: 590-630

yellow: 560-590

green: 480-560

blue: 430-480

violet: 400-430

if a solution appears green, approximately what wavelength of light is it absorbing?

a. 400 nm

b. 470 nm

c. 520 nm

d. 580 nm

e. 630 nm

f. 700 nm

(image of wavelengths provided when u flip this card)

f. 700 nm

if the solution is green, then it is absorbing red light (opposite color on the color wheel)

for a red dye, a calibration curve for absorbance versys concentration was plotted and it yielded a trendline with an equation of y= 31,870x + 0.0012. If an unknown sample of red dye has an absorbance of 0.807, what is the concentration of the sample?

enter your answer in units of micromolar (μM)

25.3

the concentration of dye in Solution A is 23.8 M. A serial dilution is performed to make Solutions B and C.

In the 1st dilution, 7 ml of Solution A is diluted with 12 ml of water to make Solution B.

then, 2 ml of Solution B is then diluted with 1 ml of water to make Solution C.

What is the concentration od dye in Solution C?

3 sig figs in molarity

5.85

suppose you make a calibration curve as described in the pre-lab information and get a linear equation in the form of y=mx+b. Assuming the path length is 1 cm, what is represented by the "y" in the equation?

a. concentration

b. molar absorptivity

c. absorbance

d. path length

c. absorbance

How does concentration affect rate of reaction?

Increasing reactant concentration increases reaction rate.

factors that increase the rate of reaction:

- increasing temperature

- increasing reactant concentrations

kinetics: prepare approximately 20.0 mL each of the following solutions of NaOH, using the NaCl solution to dilute the concentrated (0.300 M) solution of NaOH.

how many ml of NaOH are added? how many ml of NaCl is needed to dilute the NaOH?

0.200 M NaOH

0.150 M NaOH

0.100 M NaOH

0.050 M NaOH

0.025 M NaOH

x = 13.3 ml of NaOH added, 6.7 ml of NaCl needed

x = 15 ml of NaOH added, 5 ml of NaCl needed

x = 13.3 ml of NaOH added, 6.7 ml of NaCl needed

x = 10 ml of NaOH added, 10 ml of NaCl needed

x = 10 ml of NaOH added, 10 ml of NaCl needed

kinetics:

Use these graphs to determine the reaction order with respect to P2-

***it's the first tab at the bottom left hand corner labeled "reaction order w respect to P2"

link: https://uflorida-my.sharepoint.com/:x:/g/personal/vannahoang_ufl_edu/Ef0GT2FCnExPhHHy4xXBmMYBdraCKQEeq8anS0hQSiiKOQ?e=pcjEP7

the order of this reaction is 1st order because the graph for ln(absorbance) versus time was the most linear graph. It has an R2 = 0.9995 compared to the 1/(Absorbance) graph's R2 value of 0.9915

Use your data (graph of ln[OH-] vs k1) to determine the overall rate constant (k) of the reaction. for each concentration To do this, calculate the value of k from the equation k1 = k[OH-]^m.

***extra info from the lab: m = 1 for OH- bc k1/[NaOH] had the least variation

link to graph: ***click on the Sheet2

https://uflorida-my.sharepoint.com/:x:/g/personal/vannahoang_ufl_edu/Ef0GT2FCnExPhHHy4xXBmMYBdraCKQEeq8anS0hQSiiKOQ?e=pcjEP7

***hint: use the fact that this reaction is 1st order as u determined earlier. Use the equation of the graph.

y = 1.2235x - 4.1596

lnk = 1.2235x - 4.2586

e^(lnk) = e^(-4.1586)

answer: k = 0.0156

in this lab (kinetics) you will need to prepare solutions using dilutions. starting with the stock 0.3 M NaOH solution, how would you prepare a 0.2 M NaOH solution (using 0.3 M NaCl as the diluent)?

to prepare 24 ml of 0.2 M NaOH solution, you would add ______ ml of 0.3 M NaOH stock solution and _____ ml of 0.3 M NaCl solution

answer each to 1 digit after the decimal

16.0

8.0

Kinetics Generate graphs (pre-lab)

Concentration vs Time data for the following decomposition was collected at 300 K. The data shown below.

What is the rate constant for the reaction at this temperature (in units of min-1)?

H₂O₂ (aq) → H₂O(l) + 1/2 O₂ (g)

time (min):

0

500

1000

1500

2000

[H₂O₂] (mol/L)

0.01524

0.00473

0.00144

0.00045

0.00014

1st order (look at the units) so u need to create a ln[H₂O₂] vs time graph

rate constant for 1st order rxn:

slope = -k1

so:

slope = -0.0023 = -k

answer: k = 0.0023

my graph for reference:

https://uflorida-my.sharepoint.com/:x:/g/personal/vannahoang_ufl_edu/EUHnUeEFbjJCtVf_csC8IY4B2aB05ncB92B2k8H3rk-QAw?e=1J1UIp

the reaction A + B → C is studied similarly to our study of phenolphthalein fading kinetics.

The corresponding rate law is rate = k[A]^m[B]^m

in this particular experiment, the concentration of A is sufficiently high that the pseudo-order rate law: Rate = k1[B]^n can be written, where k1=k[A]^m.

if the study establishes tht the rxn is 1st order in B (n=1) & tht the pseudo rate constant k1 has the following values at varying concentrations of B, what is the order of the rxn in react A (what is the value of m)?

K1:

0.0034

0.0020

0.0010

[B]

0.200

0.100

0.050

m=1

a student generates a plot of ln(k) vs ln[OH-]

the results yields a straight line with the equation: y = 1.84x + 1.01

according to the data, what is the order of the reaction in OH-, to the nearest whole #?

2

slope = 1.84

nearest whole # = 2 so order of rxn is 2

you are generating a calibration curve for absorbance vs concentration for a colored solute. using only the values provided, what would be the value of the slope of the line through these points? provide your response to the nearest whole #

concentration (M):

0.001

0.003

absorbance:

0.373

0.521

answer: 74

you begin preparation of the calibration curve to measure absorbance vs concentration of FeSCN2+. To do so, u add 2.276 ml of 0.2 M Fe(NO3)3 to a cuvette and then directly add 164 μL of 0.001 M KSCN. What is the resulting concentration of FeSCN2+, assuming complete conversion of SCN- to FeSCN2+?

enter in units of mM to 4 digits after decimal

0.0672 margin of error =/- 0.05

for the iron thiocyanate system, what is the value of the equilibrium constant, Kc, if the following are the concentrations of all species present. answer to 3 digits after decimal

[FeSCN2+] = 0.459

[Fe3+] = 0.053

[SCN-] = 0.061

141.973

under certain conditions, Kc for the iron thiocyanate system has a value of 88.4. If [Fe3+] = 0.032 M & [SCN-] = 0.024 M, what must be [FeSCN2+]? Answer to 3 digits after the decimal: ___ M

0.068

0.0679 margin of error: +/- 0.01

if the initial amount of Fe(NO3)3 transferred to the cuvette is 0.045 mol, & the absorbance measurements indicate that 0.019 mol of FeSCN2+ are present @ equilibrium, what must [Fe3+] at equilibrium? the total volume of solution in the cuvette is 2.53 ml. answer to 2 digits after the decimal

10.28

addition of which of the following will increase the solubility of CaCO3 in water? Consider the equilibrium process:

CaCO₃ ⇌ Ca²⁺ + CO₃²⁻

select all that apply:

Na₂CO₃

HCl

NaHSO₄

HCl & NaHSO₄

HCl can react w/ carbonate ion so more CaCO3 can dissolve

HSO4- = weak acid. Can protonate CO32-, increasing solubility of CaCO3

Na2CO3 would add common ions and would decrease solubility

methyl orange (HMO) is an acid.base indicate. its 2 forms in soln are HMO (red) & MO- (yellow)

When HMO is added to distilled water, the soln is yellow. How would you turn the soln red?

a. add acid

b. add base

c. add DI water

a. add acid

Mg(OH)2 is slightly soluble in water (Ksp = 1.8e-11). NaOH is added to a soln of Mg(NO3)2 until a precipitate forms & persists. The soln is then heated & the precipitate dissolves. what can you conclude from these observations>

a. dissolution of Mg(NO3)2 is endothermic

b. dissolution of Mg(NO3)2 is exothermic

c. dissolution of Mg(OH)2 is endothermic

d. dissolution of Mg(NO3)2 is exothermic

a. dissolution of Mg(OH)2 is endothermic

NaOH (aq) is transferred into a test tube. CaCl2 (aq) is added to the tube. what is the formula for the precipitate (if any) tht forms?

Ca(OH)2

To K2CO3 (aq) in a test tube is added CuSO4(aq). After the precipitation process, phenolphthalein is added to the test tube. If some [OH-] remains in soln, what color should the indicator become?

a. colorless

b. pink

b. pink

Phth = pink is basic soln & colorless in acidic

if u have a 7.794 M soln of NaOH, what is the pH of the soln? Assume the soln is @ 25 C. answer to 1 digit after decimal

14.9

the Ka of a weak acid = 1.15e-5. What is the predicted pH of a 0.16 M soln of the weak acid?

2 decimal places

2.87

predict whether each of the following salts should be acidic, basic, or neutral in aq soln:

1. NaNO3

2. NH4Cl

3. Na2CO3

1. neutral

2. acidic

3. basic

suppose u want to prepare a buffer w/ a pH of 4.33 using formic acid. what ratio of [sodium formate]/[formic acid] do u need to make this buffer? formic acid's ka = 1.8e-4

3 sig figs

3.85

suppose u need to prepare 25 ml of formate buffer w/ a ratio of 4.2 of [sodium formate]/[formic acid] by mixing 0.1 M formic acid & 0.1 M sodium formate. how many mL of sodium formate do u need to make this buffer (assuming the rest is formic acid)? 1 decimal place

20.2

suppose u are performing a titration of a weak acid. which of the following indicators could be used for visualizing the equivalence point?

select all that apply

a. methyl red

b. phenolphthalein

c. crystal violet

d. bromocresol green

e. none of the above

phenolphthalein only

suppose u perform a titration of an unknown weak acid soln. u start w. 4 ml of the weak acid & find tht it takes 19.6 ml of 0.05 M NaOH to reach the equivalence point. what is the concentration of the unknown weak acid soln?

3 sig figs in M

0.245

suppose u perform a titration of a monoprotic weak acid & u determine the pH @ the half-equivalence point = 4.97. what is the pKa of the acid?

4.97

pka = pH @ half-equi. pt.

suppose u perform a titration of a diprotic acid & u determine the following info:

pH @ 1st equivalence point: 4.28

pH @ 2nd equivalence point: 8.78

pH @ 1st half-equivalence point: 2.1

pH @ 2nd half-equivalence point: 6.8

what is the pKa2 for this unknown acid?

6.8

suppose u perform a titration of a weak acid & u found tht the equivalence point occured at 14.3 ml of added NaOH. @ what volume would u use the pH to determine the pKa. of the acid?

enter to 3 sig figs

7.15

pH = pka @ half-equivalence point

total # of electrons in the 3d orbitals of Ti3+

1

in the coordination compound [Pt(NH3)2Cl2], the coordination # & oxidation # of the central atom are, respectively

4, 2

if a compound appears orange, what color of light is absorbed?

blue

which of the following colors of light corresponds to the highest energy?

a. red

b. violet

c. green

b. violet

which of the following complexes will absorb a photon with the lowest energy?

[Co(H2O)6]3+

[Co(CN)6]3-

[CoF6]3-

[CoF6]3-

assume the molar absorptivity (ε) for CoCl42- at 700 nm is 493 M-1 cm-1. If d = 1.07 cm for the cuvette, and if the measured absorbance (A) at 700 nm = 1.574, what is [CoCl42-]?

4 digits after decimal

0.003

a bottle in the lab is labeled [CoCl2•6H2O] = 0.68 M in 1.25 M HCl. If you determine [CoCl42-] to be 0.436 M @ a particular temperature using absorbance measurements, what must be the value of [Co(H2O)62+] in the soln at that temperature?

3 digits after decimal

0.244

you're provided w/ a bottle labeled [CoCl2•6H2O] = 0.68 M in 4.6 M HCl. you heat a small volume of the soln in a hot water bath to 50 C. If u determine that [CoCl42-] = 0.047 M @ 50 C at equilibrium, what must be the equilibrium concentration of [Cl-] at 50 C?

units of M 3 digits after decimal

4.506

[CoCl42-(aq)] = 0.045 M, [Co(H2O)62+ (aq)] = 0.289 M, & [Cl-(aq)] = 0.145 M at @ 5°C, what is value of ∆G at 5°C in kj/mol?

2 digits after decimal

-13.56

a plot of ∆G (kj/mol) vs. T (k) yields a straight line w/ the equation: y = -13.416x + 191.4. What is the value of ∆S (kj/mol K) for the rxn?

2 digits after decimal

13.42

Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to 1 compartment of the microwell plate. She then makes a soln of MCln by adding KCl to M(NO3)n. She adds 8.165 ml of the resulting mixture to a 2nd compartment of the microwell plate.

Sally knows n (the charge on the metal ion) = +3

She already calculated [M^n+] in the prepared MCln soln using the Nernst eqn. [M^n+] = 9.903 M

how many moles of [M^n+] are dissolved in that compartment of the microwell plate?

3 digits after decimal place

0.081

# of moles M^n+ = (9.903 mol Mn2+/L)(8.165e-3 L) = 0.081

consider the concentration cell in which the metal ion has a charge of 1+ & the soln concentrations are:

dilute soln = 0.081 M

concentrated soln = 1.377 M

what is the predicted Ecell, using the Nernst eqn?

3 digits after the decimal

0.072 V

Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to 1 compartment of the microwell plate. She then makes a soln of MCln by adding KCl to M(NO3)n. She adds 5.564 ml of the resulting mixture to a 2nd compartment of the microwell plate.

Sally knows n (the charge on the metal ion) = +3

She already calculated [M^n+] in the prepared MCln soln using the Nernst eqn. [M^n+] = 1.38 M

how many moles of [Cl-] must be dissolved in that compartment?

**hint: start by finding how many mol of [M^n+] are dissolved in tht compartment of the microwell plate

3 digits after decimal place

0.023

if E°anode = 1.696 V & E°cathode = 2.328 V, what is E°cell? ____ V

2 digits after decimal

0.63

consider the galvanic cell described by (N and M are metals):

N(s)|N²⁺(aq)||M⁺(aq)|M(s)

if E°anode = 1.641 V & E°cathode = 0.253 V, and [N²⁺(aq)] = 0.444 M and [M⁺(aq)] = 0.829 M, what is Ecell, using the Nernst eqn? ____ V

2 digits after decimal

-1.38

Nernst equation

Ecell = E˚cell - {(0.0592/n)(logQ)}

n = moles of electrons

Q = concentrations (products over reactants)

E˚cell = E˚cathode - E˚anode

in a particular electroplating process, the metal being plated has a 2+ charge & molar mass = 138.068 g/mol. If 644.82 C of charge pass through the cell, how many g of metal should be plated?

F = 96,500 C/mol e-

2 digits after decimal

0.46

in a particular electroplating process, the metal being plated has a 2+ charge. If 793.29 C of charge pass through the cell, how many moles of metal should be plated?

F = 96,500 C/mol e-

4 digits after decimal

0.0041

Jill performs the electroplating experiment & calculates the theoretical mass of copper to be plated as 1.741 g. She weighted the brass disc (which was plated in the experiment) before & after the electroplating process:

mass of disc before: 2.21 g

after: 3.701 g

what was the percent yield of Jill's electroplating process?

85.6

percent yield

(actual yield/theoretical) x 100%

percent error

(|theoretical - actual| / theoretical) x 100

in a particular electrolysis process, E°cathode = -1.31 V & E°anode = 2.068 V. What minimum voltage must be applied to drive the nonspontaneous process?

2 digits after the decimal

3.38

in the half-rxn shown, what is the oxidizing agent?

2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻ (aq); E° = -0.83 V

a. H₂O(l)

b. H₂(g)

c. OH⁻ (aq)

d. none

a. H₂O(l)

le chatelier's lab:

The indicator that was used is methyl violet, which can be represented by the formula HMV. HMV is yellow and MV- is violet in color. Begin by writing the equation for the dissociation of HMV to MV-.

Explain which form of the indicator was present when the indicator was added to deionized water. Explain how the addition of acid (1 M HCl) and base (1 M NaOH) affected the equilibrium. Reason your answer using Le Chatelier's Principle.

HMV ⇌ H⁺ + MV⁻

in water:

HMV (aq) + H₂O(l) ⇌ H₃O⁺ (aq) + MV⁻ (aq)

Adding HCl will result in more HMV being formed since HCl will react with MV-. The solution will become more acidic and it will shift left, turning green. Adding base shifts the equilibrium to the right because NaOH will react with H3O+. The solution will turn violet.

determine the pH of each of the HCl solutions:

1 M

0.1 M

0.01 M

0.001 M

0

1

2

3

acids and bases lab:

find the theoretical pH of the following:

1. 0.1 M HCl

2. 0.01 M HCl

3. 0.001 M HCl

4. 0.1 M CH3COOH (Ka = 1.8e-5)

5. 0.1 M NH3 (Kb = 1.8e-5)

6. 0.1 M NaOH

1. 1

2. 2

3. 3

4. 2.87

5. 11.13

6. 13

acids/bases lab:

determine whether they should be neutral, acidic, or basic in solution

NaCl

NaCH3COO

NH4Cl

CuCl2

Na3PO4

Na2HPO4 (amphoteric)

NaH2PO4 (amphoteric)

Ka HPO42- = 4.5e-13

Ka H2PO4- = 6.3e-8

Ka H3PO4 = 7.1 x 10-3

n= neutral, a = acidic, b = basic

1. n

2. b

3. a

4. a

5. b

6. b

7. a

titrations lab: titrating 20 ml of unknown weak acid with 1 M NaOH

From your graph,

1. determine the concentration of the unknown acid solution. Describe how you obtained this information.

2. determine the Ka value for the acid and its likely identity. Describe how you obtained this information.

3. based on the identity and concentration of the acid that you determined, determine the theoretical pH of the equivalence point. Show your calculations in your notebook.

https://drive.google.com/file/d/1fVaeC82CpKpEYgxUHyjnxy0UjfUCjlDn/view?usp=sharing

1. (M acid)(V acid)=(M base)(V base)

Macid(20 ml)=(1 M NaOH)(~7.625 ml of NaOH) =

*answer: 0.381 M*

***approximated from the graph

2. pka = pH @ half equivalence pt

7.625/2 = 3.8 ml NaOH (approx)

pH @ 3.8 ml ~ 4.4 so

*answer: pka ~ 4.4* which is closest to pka of acetic acid (pka ~ 4.75)

3. pH = 9.09 (use Kb)

From your graph, https://drive.google.com/file/d/1fVaeC82CpKpEYgxUHyjnxy0UjfUCjlDn/view?usp=sharing

(2nd graph; the one for the 20 ml of unknown diprotic acid)

1. determine the concentration of the unknown acid solution. Describe how you obtained this information.

2. determine the pKa1 and pKa2 value for the acid. Describe how you obtained this information. Determine the likely identity of your diprotic acid.

***this is approximate

as long as you are relatively close, you're good

1.

*if u use the 2nd equivalence point, u need to divide whatever u got by 2, but if u used the 1st equivalence point, u don't divide by 2

MacidVacid=MbaseVbase

(Macid)(20 ml) = (1 M NaOH)(21.25 ml)

Macid = 1.0625

1.0625/2 = 0.531 M

*answer = 0.531 M*

2.

1st equivalence point: (10.25 + 11)/2 = 10.625 ml

1st half-equivalence point: 10.625/2 = 5.3125 ml

pH = pKa1 ≈ 2.2

answer: ~ 2.2 closest to pKa of maleic acid (pKa1 = 1.94)

10.625 + 5.3125 ≈ 16 ml NaOH

pH ~ 5.91 ≈ pKa

answer: pKa2 5.91 (close to pKa2 of maleic acid = 6.22)

Gibbs free energy, ΔG

ΔG = -RTlnK

spontaneous when ΔG < 0

ΔG = 0 when at equilibrium

thermodynamics lab part 4 q 1:

Create a calibration curve of the absorbance of the 0.0020 M, 0.0010 M, and 0.00050 M CoCl2 solutions in 12 M HCl. Add a linear trendline, including an equation. Write the equation of your calibration curve

absorbance: 1.318, 0.657, 0.310, respectively

y = 670.43x - 0.0205

thermodynamics lab part 4 q 2: using the equation, y = 670.43x - 0.0205, convert absorbance into concentration of CoCl₄²⁻ for each temperature:

24°C: absorbance = 0.145

34°C: absorbance = 0.243

44°C: absorbance = 0.376

54°C: absorbance = 0.577

64°C: absorbance = 0.855

0.0002469 M

0.000393 M

0.0005914

0.0008912

0.001306

thermodynamics lab part 4 q 3:

calculate the concentrations of Co(H2O)62+ and Cl- using the initial total concentrations of Co2+ and Cl- and the concentration of CoCl42- from thermodynamics lab part 4 q 2:

***from procedures: initial [Co(H2O)62+] = 0.05 M and initial [Cl-] = 5.1 M

Co(H2O)62+(aq) + 4 Cl-(aq) ⇌ CoCl42-(aq) + 6 H2O(l)

24°C:

[Co(H2O)62+] = 0.05-(2.469e-4) = *0.04975*

[Cl-] = 5.1 - 4(2.469e-4) = *5.099*

34°C:

[Co(H2O)62+] = 0.0496

[Cl-] = 5.098

44°C:

[Co(H2O)62+] = 0.0494

[Cl-] = 5.0976

54°C:

[Co(H2O)62+] = 0.0491

[Cl-] = 5.096

64°C:

[Co(H2O)62+] = 0.0487

[Cl-] = 5.095

thermodynamics lab part 4 q 2-3:

4. calculate Kc for the reaction at each temperature using concentrations obtained in q 3-4

5. calculate ΔG∘ at each temperature, in kJ/mol. Make sure to convert your temperature to Kelvin.

4.

at 24°C: K = 7.3e-6

34: K=1.17e-5

44: 1.77e-5

54: 2.69e-5

64: 3.98e-5

5.

24°C: ∆G°=29.206 kJ/mol

34°C: ∆G°=28.999

44: ∆G°=28.852

54: ∆G°=28.623

64: 28.400

1. Create a scatter plot of ΔG∘ vs T. Make sure to plot the temperature in Kelvin. Make sure to include a descriptive title and axis titles (with units). Add a linear trendline, including the equation and R2 value.

data:

24°C: ∆G°=29.206 kJ/mol

34°C: ∆G°=28.999

44°C: ∆G°=28.852

54°C: ∆G°=28.623

64°C: ∆G°=28.400

2. Based on your graph, answer the following: What is ΔH∘ for this reaction? What is ΔS∘ for this reaction? Is this reaction spontaneous at all temperatures, high temperatures, low temperatures, or never?

graph:

https://drive.google.com/file/d/1BZgYTcUUlx9KOr5l8CMw6IdRjmyM4ONQ/view?usp=sharing

1. y = -0.0199x + 35.121; R2 = 0.995

2.

∆H° = 35.121 kj/mol

∆S° = -(-0.0199) = 0.0199

∆H & ∆S are both + so spontaneous at high temperatures

(bc ∆G = ∆H - T∆S)

galvanic/voltaic cell diagram (similar to the one we drew for the galvanic cells lab)

Use the Nernst equation to calculate the theoretical cell potential for the concentration cell and compare with your measured value. Report the % error for your measurements.

***from procedures

one cell has: 0.050 M Cu(NO3)2

one cell has: 0.50 M Cu(NO3)2

***from data:

Trial 1: cell potential = 0.135 V

Trial 2: 0.137 V

Trial 3: 0.108 V

answer:

Ecell = 0.0296 V

328.04% error

steps:

measured value: average E°cell = 0.1267 V

theoretical value:

Ecell = 0 - (0.0592/2 mol e-)•log[0.05]/[0.5] = 0.0296 V

% error = |(0.0296 - 0.1267)/0.0296| • 100% = 323.04%

Use the Nernst equation and the information you collected about the Pb/PbI2 concentration cell to complete the following calculations:

Trial 1: E°cell = 0.284 V

Trial 2: 0.329 V

Trial 3: 0.762

1. Use the cell potential for the Pb/PbI2 cell and the known [Pb2+] (3 mL of a 0.10 M Pb(C2H3O2)2 solution was used) to calculate [Pb2+] in the mixture containing PbI2

- calculate the [I-] in the solution (hint: how many mol I- dissolves when 1 mol of Pb2+ dissolves?)

- Use your data to calculate the Ksp of PbI2.

- The accepted value for Ksp of PbI2 is 9.8 x 10-9. How does your experimental Ksp compare with the accepted value?

1. [Pb2+] in PbI2 = [dilute] = 1.9e-8 M

2. [I-] = 3.8e-8 M

3. ksp = [1.9e-9][3.8e-8]²=2.71e-23

4. the ksp that I obtained was extremely inaccurate compared to the accepted value of 9.8e-9

voltaic cell with 0.1 M KI and Phenolphthalein

when a current was added, one electrode turned yellow and one electrode turned magenta. the magenta electrode bubbled. a black coating is visible on the electrode that turned yellow.

write the half and overall balanced equations for each electrode (positive and negative) and the minimum voltage necessary for each reaction

cathode/- electrode:

2H₂O (l) + 2e⁻ → 2OH⁻ + H₂ (g)

anode/+ electrode:

2I⁻ → I₂ (s) + 2e⁻

overall balanced:

2H₂O (l) + 2I⁻ → 2OH⁻ + H₂ (g) + I₂ (s)

E°cell = -0.83 - 0.56 = -1.39 V

minimum voltage = 1.39 V

voltaic cell with 0.5 M K2SO4 and bromothymol blue

when a current was added, one electrode turned yellow and one electrode turned a darker blue. both electrodes had bubbles. no precipitate forms.

write the half and overall balanced equations for each electrode (positive and negative) and the minimum voltage necessary for each reaction

cathode/- electrode:

2H₂O (l) + 2e⁻ → 2OH⁻ + H₂ (g)

anode/+ electrode:

2H₂O (l) → 4e⁻ + 4H⁺ + O₂ (g)

overall:

6H₂O (l) → 4H⁺ + O₂ (g) 2OH⁻ + H₂ (g)

E°cell = -0.83 - 1.23 = -2.06 V

minimum voltage = 2.06 V

voltaic cell with 0.5 M NaBr and Phenolphthalein

when a current was added, one electrode turned magenta and one electrode turned yellow. the magenta electrode bubbled.

write the half and overall balanced equations for each electrode (positive and negative) and the minimum voltage necessary for each reaction

cathode/- electrode:

2H₂O (l) + 2e⁻ → 2OH⁻ + H₂ (g)

anode/+ electrode:

2Br⁻ (aq) → Br₂ (l) + 2e⁻

overall:

2Br⁻ (aq) + 2H₂O (l) → 2OH⁻ + H₂ (g) + Br₂ (g)

E°cell = -0.83 - 1.07 = -1.90 V

minimum voltage = 1.90 V

electrolytic cells lab:

1. calculate the theoretical number of moles of copper that should have plated out onto the brass disc. **from data: Coulombs: 122.6

2. Use your data from measuring the mass of the electrode and the brass disc to calculate the actual number of moles of copper that plated out.**from data: initial mass of brass disc = 4.57 g; initial mass of Cu electrode = 0.86 g; final mass of brass disc = 4.61 g; final mass of Cu electrode = 0.82 g

3. Calculate the % yield of copper plated out.

4. Write the oxidation and reduction half-reactions.

1. theoretical: 6.35e-4 mol Cu

122.6 C • (1 mol e-/96500 C) • (1 mol Cu / 2 mol e-) = 6.35e-4 mol

2. actual: 6.29e-4 mol

4.61-4.57=0.04 g Cu

0.04 g Cu • (1 mol Cu/63.55 g Cu) = 6.29e-4 mol

3. % yield = (6.29e-4/6.35e-4) • 100% = 99.1%

4. redox half rxn:

ox: Cu(s) → Cu²⁺ (aq) + 2e⁻

red: Cu²⁺ (aq) + 2e⁻ → Cu(s)