Lab quizzes/pre-lab

Question 1

Determination of an Equilibrium Constant

write the equilibrium expression for a given reaction
calculate the initial concentrations of reactants from the molarities and volumes used
determine the equilibrium concentrations of reactants and products from the absorbance and calibration curve data (such as the equation of the line for the calibration curve)
calculate the equilibrium constant, K

Answer 1

K= products/reactants*ignore solids
(more details ltr)

Question 2

Chemical Equilibrium and Le Chåtelier’s Principle

be able to predict the direction in which the equilibrium will shift as a result of changing the concentration or temperature
be able to write a balanced chemical equation that describes a particular equilibrium mixture
know the three types of equilibria that were investigated: acid-base, solubility, and complex ion

Answer 2

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Question 3

pH Measurements

know the Arrhenius, Bronsted-Lowry, and Lewis definitions of acids and bases; be able to identify whether a compound is an acid or base
be able to calculate the pH of a solution from the concentration of hydrogen ion
calculate the concentration from the pH
calculate the Ka for a weak acid or base
use the color change of various acid-base indicators to estimate the pH

Answer 3

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Question 4

Acid-Base Buffers

understand how buffers work
apply the Henderson-Hasselbalch equation to calculations involving buffer solutions
be able to calculate the amounts of conjugate acid and base needed to prepare a buffer of a particular pH. Conversely, given the composition of the buffer, calculate the pH

Answer 4

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Question 5

General

significant figures in measurements and calculations (Appendix F)
volumetric measurements, titrations, and dilutions (Appendix D)
laboratory safety (p. i-ix of the syllabus)

Answer 5

more details ltr

Question 6

(1) Write the equilibrium constant expression, K, for the reaction we studied for Experiment #1 (hint: Fe and SCN)

Answer 6

K=[FeNCS 2+]/[Fe3+][SCN-]

Question 7

(1) 3.00 mL of 0.0020 M Fe(NO3)3, 4.00 mL of .0020 M KSCN, and 3.00 mL of H2O are mixed. From the absorbance and calibration curve of the equilibrium concentration of FeNCS2+ is found to be 8.75e-5 M. How many moles of Fe3+ are present at equilibrium? Show your work.

Answer 7

8.75e-5 M x .010 L = 8.74 e -7 (moles consumed and product) then intial moles .0020 M x .003 L = 6e-6.intial - consumed = 5.126 e-6

Question 8

(1) In the calibration curve measurement, you’ll mix 1.00 mL of 2.00 e-3 M SCN with 19.0 mL of 0.200 M Fe3+

Why is the concentration of Fe3+ 100x larger than SCN-? What is the concentration of FeNCS2+ in the resulting solution?

Answer 8

Part A: higher charge and moles (double check)

Part B: 1. M1V1 = M2V2, where:
-M1V1 is the moles of SCN- reacted
-V2 is the total volume of solution after mixing SCN- and Fe3+
2. Let M2 = concentration og FeNCS2+ formed
3. (1.00mL)(2.00x10^-3) = M2(1.00mL + 19.0mL)
4. M2 = 0.0001

Question 9

(prelab 1) Suppose a student mixes 4.00 mL of 2.00 x 10-3 M Fe(NO3) with 4.00 mL of 2.00 x 10-3 M KSCN and 2.00 mL of water. The student then determines the [FeNCS2+] at equilibrium to be 7.85 x 10-5 M. Find the equilibrium constant for the following reactions.
Fe3+eq + SCN-eq ↔ FeNCS2+eq

Step 1: Calculate the initial number of moles of Fe3+ and SCN- (use equation 12)

Step 2: How many moles of FeNCS2+ are present at equilibrium? What is the volume of the equilibrium mixture?

Step 3. Calculate the number of moles of Fe3+ and SCN- remaining at equilibrium. Use Equation (13) and the results of Steps 1 and 2.

Step 4. Calculate the concentrations of Fe3+, SCN-, and FeNCS2+ at equilibrium. Use Equation (14) and the results of step 3.

Step 5. What is the value of the equilibrium constant for the reaction? Use Equation (10) and the results from step 4?

Step 6: What is K?

Answer 9

1: (2.00 e-3)(.004 L)
Moles of Fe3+: 8.00 e-6
Moles of SCN-: 8.00 e -6

2. (7.85 e-5 M)(.010 L)
   Volume of equlibrium mixture (mL): 10
   moles of FeNCS2+: 7.85e-7 which is equal to moles of reactants consumed

How many moles of Fe3+ and SCN- are consumed to produce the FeNCS2+?

Moles of Fe3+ consumed: 7.85e-7
Moles of SCN- consumed: 7.85 e-7

3. Calculate the number of moles of Fe3+ and SCN- remaining at equli using equation 13 and results of steps 1 and 2

(initial moles) - (moles consumed) =
Moles of Fe3+ and SCN-: 7.22 e-6

4. Concentrations at equli
   Moles consumed/total volume (L) = Molarity
   [Fe3+] and [SCN-]: 7.22 e -4
   [FeNCS2+]: 7.85 e-5

K = 150.8 = products/reactants

Question 10

(2) explain Le châtlier principle in no more than 2 sentences

Answer 10

System -> stress -> relieve -> equilibrium

Question 11

(2) reaction of nitric oxide to form dinitrogen monoxide and nitrogen dioxide is endothermic.
Heat + reactants -> 2NO(g)

What effect will be seen if the temperature of the system at equilibrium is increased by 25 degrees?

Answer 11

Partial pressure of (NO/O2/N2) will (increase/decrease)

No and increase - more reactants

Question 12

(2) What is an acid-base indicator? What was the indicator used?

Answer 12

Changes color at certain pH
Indicator: Pheno

Question 13

(2) Write the equilibrium constant expression, Ksp, for the following reaction:

PbCl2 (s) -> PB2+ (aq) + 2Cl-(aq)

Answer 13

Ksp= [PB2+][Cl]^2

Question 14

(2) 5.00 ml of 0.0020 M Fe(NO3)2, 3.00 mL of .0020 M KSCN, and 2.00 mL of H2O are mixed. From the absorbance and calibration curve, the equli concentration of FeNCS2+ is found to be 6.63 e-5 M. What are the equilibrium concentrations of Fe3+ and SCN- in mol/L?

Answer 14

Find initial and consumed

Consumed = 6.63 e -5 x .01 L = 6.63 e-7
Initial Fe3+ = 1e-5 - (6.63e-7) x .01 = 9 e- 4
Initial SCN- = 6e-6 - (6.63e-7) x .01 = 5 e-4

Question 15

(prelab 2) You have a container in which SO2(g), O2(g), and SO3(g) are in equilibrium with each other at 120 °C. Answer the following questions given that the formation of sulfur trioxide is exothermic.

a. Balanced chemical equation which describes the equilibrium
b. Explain what influence adding O2 to the container will have on the equilibrium (explain in terms of [SO3]).
c. Explain what effect decreasing the volume of the container will have on the equilibrium (explain in terms of [SO3]).
d. Explain what effect increasing the temperature of the container will have on the equilibrium (explain in terms of [SO3]).

Answer 15

a. 2SO2(g)+O2(g)⇌2SO3(g)

b. By adding more reactant, the Le Châtelier’s Principle states that to reach equilibrium, the reaction will shift forward (right) to use up the additional reactants. This will also increase the concentration of product, SO3.

c. By decreasing volume, the Le Châtelier’s Principle states that the pressure will increase, which will also mean the reaction will shift right due to fewer gas molecules to counter it. This would mean increasing the concentration product, SO3.

d. By increasing the temperature, the exothermic reaction would make the heat a product. This would make the assumption under Le Châtelier’s Principle that we added more products and so the reaction shifts left to absorb up the excess products in the reverse direction. This would make the concentration of SO3 decrease as well.

Question 16

(3) calculate the pH of a 0.005 M Barium hydroxide [Ba(OH)2] solution

Answer 16

.005 x 2 = 0.01 = 10 e -2
- log (pOH = 10 e-2) = 2
PH = 14 - 2 = 12

Question 17

(3) what is the bronsted Lowry definition of acid and base? Which of the chemical species are which?

NaCl, NaCN, KNO3, NH4Cl

Answer 17

Acid: NH4Cl
Base: NaCN

Question 18

(3) You measure the pH of a 0.15 M solution of benzoic acid (Hbenz) and find it to be 4.05. What is the value of Ka?

Answer 18

Find concentration of H+ by 10^-pH = x
Then find Ka by (x)(x) / .15-(x) derived by ICE table

Question 19

(3) Which reagent, when added to solution of Hbenz, will shift the equilibrium in the direction of decreasing concentration of Hbenz?

Nabenz
HCl
NaOH

Answer 19

“strong base like NaOH. The base reacts with the acidic H+ ions from benzoic acid forming water and reducing the concentration of benzoic acid, thus pushing the equilibrium left”

Question 20

(prelab 3)
1. Find the pH of a solution whose hydrogen ion concentration is
a. 2.0 e -4
b. 0.050
c. 8.0

2. A solution of a weak acid was tested with the indicators used in this experiment. The colors observed were:
   Methyl violet: violet

Congo red: violet

Bromocresol green: yellow

Thymol blue: yellow

3. The pH of a 0.10 M HCN solution is 5.15.

What is the Hydrogen ion concentration? What is the CN- concentration? What is the Ka?

Answer 20

1. -log[H3O+]
   a. 3.70
   b. 1.30
   c. -.90
2. Essentially what number do they have in common = 3
3. pH = -log [H+]
   So, H+= 7.1e-6, CN-=7.1e-6, Ka= Products/reactants so 5.0e-10

Question 21

(4) You need to prepare 20.00 mL of a buffer solution having a pH of 4.50 from 0.15 M weak acid (HA) and 0.15 M conjugate base (A-). The Ka of HA is 8.2 e-5. What volume of HA is required to make the buffer?

Answer 21

pH = 4.50
Ka= 8.2 e -5

4.50 = -log (8.2 e-5) + log [A-/HA]
4.50-4.09=log [A-/HA]
2.57x=20-x

3.57x=20, x=5.6

Question 22

(4) A student measured the pH of a solution of 0.1 M Na2HPO4 and found that it was 8.85. Write the net ionic equation that explains the measured pH

Answer 22

HPO2- 4 (aq) + H2O (l) -> H2PO-4(aq) + OH- (aq)

Question 23

(4) Use the table to determine the approx pH of a solution that exhibits the following indicator colors

Answer 23

Essentially what numbers do they have in common

Question 24

(prelab 4)
1. Formic acid, HFor, has a Ka value of 1.8×10-4. You need to prepare 250 mL of a buffer having a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL of HFor and NaFor should be mixed to make the desired buffer?

Hint: If we let x = mL of HFor, then 250 - x = mL NaFor.

2. When five drops of 0.10 M NaOH were added to 20 mL of the buffer in question 1, the pH went from 3.25 to 3.31. Write a net ionic equation to explain why the pH did not go up more than this.

Answer 24

find pka=-log(1.8e-4) = 3.74
then use pH=pKa +log(A-/HA)
3.25 = 3.74 + log (a/ha)
=.324

250= 1.324x
x= 188.82
250-x = 61.18 mL

*I originally put 250-x/x = .325 which it should’ve been .324 however the .325 was correct

2. HA+OH-→A-+H2O

Question 25

(5) Write:
A. The chemical equation that describes what happens when solid PbI2 is added to water, and b, the Ksp expression for PbI2

Answer 25

A. PbI2 (s) -> PB2+ (aq) + 2I- (aq)
B. Ksp = [PB2+][I-]^2

Question 26

(5) A solution was prepared by combining 5.00 mL of 0.0120 M PB(NO3)2, 3.00 mL of 0.300 M KI, and 2.00 mL of 0.20 M KNO3. The equli concentration of iodide ions was determined spectroscopically to be 5.25 e-3 M.

A. How many moles of I- were present initially?

B. How many moles of I- were precipitated?

Answer 26

3e-3 x 0.300 M KI = 9e-5 mol initially

5.25 e-3 M x .01 L = 5.25 e -5

Initial - equli = 3.75 e -5 precipitated moles

Question 27

(5) Hbenz has a ka value of 6.3 e-5. You need to prepare a buffer having a pH of 4.15 from 0.10 M Hbenz and 0.10 M Nabenz. How many mL of the NaBenz solution should you add to 20 mL of the 0.10 M Hbenz solution to make buffer?

Answer 27

pH = pKa + log (HA/A)
pH = 4.15
pKa= 6.3 e-5

.89 = HA/A which would mean .89=x/20 = x = 17.8 mL

Question 28

(Prelab 5)

1. a. Write the equation for the solubility product for lead iodide.

b. Explain the meaning of this equation in your own words.

2. A student mixed 4.00 mL of 0.0120 M Pb(NO3)2 with 4.00 mL of 0.0300 M KI and 2.00 mL of 0.20 M KNO3 and observed the formation of a yellow precipitate.
   a. What is the molecular formula of the precipitate?
   b. How many moles of Pb2+ are present initially? (moles = M x V)
   c. How many moles of I- are present initially?
   d. The concentration of I- at equilibrium is experimentally determined to be 6.5×10-3 M. How many moles of I- are present in the solution (10 mL)?
   e. How many moles of I- precipitated?
   (You have already determined how many moles were present initially and how many remained in solution.)
   f. How many moles of Pb2+ remain in solution?
   g. What is the concentration of Pb2+ in the equilibrium solution? (The volume of the equilibrium solution is still 10 mL.) (M)
   h. Determine Ksp of PbI2 from this data.

Answer 28

1. a. Ksp= [Pb2+][I-]2

b. Since there is only solid as a product, the denominator doesn’t exist due to holding no weight in the expression for K. This leaves us with Ksp, “equilibrium constant that corresponds to the dissolution of a slightly soluble or insoluble ionic compound in water”

2.
a. PbI2
b. (4e-3)(.0120 L) = 4.8 e-5 moles
c. (4e-3)(.0300 L) = 1.2 e-4 moles
d. (6.5e-3) (.010 L) = 6.5 e-5 I- moles present in the solution
e. Precipitated I- moles= Initial - equilibrium = 5.5 e -5
f. Pb2+ moles remaining in solution = stoichiometric relationship so 5.5e-5/2 = 2.75 e-5 moles
g. Concentration of Pb2+ in the equilibrium solution: mol/L = 2.05 e-3 M
h. (I-)^2(Pb2+) = 8.66 e-8

Question 29

(6) in today’s experiment, the EDTA will be standardized by titrating it against which of the following?

A. A sample of hard water
B. CaCO3
C. 0.03 M MgCl2
D. PbI2
E. Solid Ca(No3)2

Answer 29

B

Question 30

(prelab 6)
1. A 0.492-g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask.
a. How many moles of CaCO3 are used (M.W. = 100.1 g/mol)? (mol)
b. What is the molarity of the Ca2+ in the 250 mL of solution? (M)
c. How many moles of Ca2+ are in a 25.0-mL aliquot of the soln in b.? (mol)

2. 25.00-mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg2+ requires 2.32 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 27.52 mL of the EDTA to reach the end point.
   a. How many mL of EDTA are needed to titrate the Ca2+ ion in the aliquot? (mL)
   b. How many moles of EDTA are there in the volume obtained in a.? (mol)
   c. What is the molarity of the EDTA solution? (M)
3. A 100-mL sample of hard water is titrated with the EDTA solution in problem 2. The same amount of Mg2+ is added, and the volume of EDTA required is 21.15 mL.
   a. What is the volume of EDTA used in titrating the Ca2+? (mL)
   b. How many moles of EDTA are there in that volume? (mol)
   c. How many moles of Ca2+ are there in 100 mL of water? (mol)

Answer 30

1a. 0.492 g CaCO3 x 1 mol/100.1 g CaCO3 = 0.0049
1b. mol/L = 0.0049/.250 = 0.0196
1c. 0.025 L x (1.96e-2 mol CaCO3/1L) = 4.90e-4 mol Ca2+

2a. 27.52 mL - 2.32 mL = 25.20 mL
2b. 25.00 mL Ca2+ /1000 mL x (1.96e-2 mol Ca2+/1 L) = 4.90 e-4 mol EDTA (EDTA mol = CA2+)
2c. 4.90e-4 mol EDTA/25.20 mL /1000 mL = 1.94 e -2

3a. 21.15 mL EDTA - 2.32 mL ETDA = 18.83 mL EDTA
3b. 4.05 mL EDTA /1000 mL x (1.94 e-2 M) = 7.86 e -5 mol EDTA
3c. 7.86 e-1 mol EDTA = mol Ca2+

Question 31

(7) A student constructs a voltaic cell from a 1 M Zn2+ solution and a 1 M Ni2+ solution connected by a salt bridge. There is an Zn electrode in the Zn2+ solution and a Ni electrode in the Ni2+ SOLUTION. ecell is measured to be 0.505 V and the Ni electrode is positive.

1. Write the balanced net ionic equation for the reaction that occurs in the cell
2. Which statement(s) is/are true?
   a. the Zn electrode is the cathode
   b. the Ni electrode is the anode
   c. Reduction occurs at the Ni electrode
   d. Oxidation occurs at the Zn electrode
3. Ecathode equals -0.258 V. What is the E anode?
4. Write the half-reaction that occurs at the cathode

Answer 31

1. Zn (s) + Ni2+ (aq) -> Zn2+ (aq) =Ni (s)
2. reduction occurs at Ni electrode
   oxidation occurs at Zn electrode
3. E cell = e cathode - e anode
   0.505 V = -0.258 V - Eanode
   -0.763
4. cathode = reduction -> Ni electrode
   Ni2+ (aq) + 2e- -> Ni(s)

Question 32

(7) A 75.00 mL sample of hard water (with Mg2+ added) is titrated with 0.0100 M EDTA. A blank solution with the same amount of MG2+ requires 2.50 mL of EDTA to reach its end point. A total of 32.00 mL of EDTA was used for the titration of the 75.00 mL sample of hard water (with Mg2+) M.W CaCO3 = 100.1 g/mol

Answer 32

EDTA for Ca2+: 32.00 mL - 2.50 mL (for Mg 2+) = 29.50 mL

0.01 M EDTA = x/0.0295 ; x = 2.95 e- 4 mol EDTA : mol Ca2+

2.954 e -4 mol Ca 2+ x 100.1 g/mol Ca2+ = 0.0296 g Ca2+ per 75 mL H2O

0.0296 g x 100 mg/g = 29.57 mg/0.075 L = 394.3 ppm

Question 33

(Prelab 7) A student measures the potential of a cell made up with 1 M CuSO4 in one solution and 1 M AgNO3 in the other. There is a copper electrode in the CuSO4 and a silver electrode in the AgNO3. The student finds the potential of the cell, E°cell , to be 0.47 V, and that the copper electrode is negative. Show complete calculations for the following questions.

1. at which electrode is oxidation occuring?
2. write the half reaction taking place at the anode?
3. write the half reaction taking place at the cathode
4. what is the e-anode given that the potential of the silver, silver ion electrode, E-Ag,Ag+, is taken to be 0.00 V?
5. If E°Ag,Ag+ equals 0.80 V, what is E°anode?
6. Write the net ionic equation for the reaction that occurs in the cell studied by the student
7. The student adds 6 M NH3 to the CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)42+ ion. The cell Voltage, Ecell, increases to 0.92 V and the copper electrode is still negative. Find the concentration of Cu2+ ion in the cell (use Equation (8) in your lab manual). Show your calculations.
8. In question 7 the [Cu(NH3)42+] is about 0.01 M and [NH3] is about 2 M. Given these values and [Cu2+] from question 7, determine K for the following reaction. Show your calculations.

Cu(NH3)42+(aq) ⇌ Cu2+(aq) + 4NH3(aq)

Answer 33

1. Copper
2. Cu(s)→Cu2+(aq) + 2e-
3. Ag+(aq) + e- → Ag(s)
   *E°cathode - E°anode = E°cell
4. E°anode (V)= -0.47 = 0.00 V - 0.47
5. E°anode (V) = 0.33 = 0.80-0.47
6. Cu(s)+2Ag+(aq)⇌Cu2+(aq)+Ag(s)
7. Ecell = E°cell - 0.0592/n log ([Cu2+]/[Ag]^2)
   0.92 - 0.47 x -2/0.0592 = log([Cu2+]) = -15.2 M
   [Cu2+] = 6.3 e-16 M
8. K = Reactants/Products = 1.0 e -12