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Review Answers

2000 REVIEW #1

Kc & Kp

2H2S(g) ↔ 2H2(g) + S2(g)

When heated, hydrogen sulfide gas decomposes according to the equation above.

A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container.

The sealed container is heated to 483 K and 3.72x10-2 mol of S2(g) is present at

equilibrium.

a. Write the expression for the equilibrium constant, Kc, for the decomposition

reaction represented above.

b. Calculate the equilibrium concentration, in mol L1-, of the following gases in

the container at 483 K.

i. H2(g)

ii. H2S(g)

c. Calculate the value of the equilibrium constant, Kc, for the decomposition

reaction at 483 K.

d. Calculate the partial pressure of S2(g) in the container at 483 K.

e. For the reaction H2(g) + ½S2(g) ( H2S(g) at 483 K, calculate the value of the

equilibrium constant, Kc.

a. [pic]

b. First find the molarity of the H2S and then recognize that the moles of S2 is the equilibrium moles of S2 and place both in an ICE table. The beginning moles of S2 and H2 is 0. Also be aware of the coefficients in the reaction.

[pic]

| |2H2S |↔ |2H2 |+ |S2 |

|Initial |0.0800 | |0 | |0 |

|Change |-0.0595 | |+0.0595 | |+0.0298 |

|Equilibrium |0.0205 | |0.0595 | |0.0298 |

Therefore: [H2]=0.0595M and [S2]=0.0298

c. Simply plug into the equilibrium expression the equilibrium concentrations

[pic]

d. Use PV=nRT

[pic]

e. Recognize that this is ½ of the given reaction. So the value of K must be take to the ½ power. Note that a common misconception is to cut the value of K in half. But since this is a power function it must be take to the ½ power.

0.251(1/2)=0.500

2001 REVIEW #2

Ksp

Answer the following questions relating to the solubility of the chlorides of silver

and lead.

a. At 10°C , 8.9x10-5 g of AgCl(s) will dissolve in 100. mL of water.

i. Write the equation for the dissociation of AgCl(s) in water.

ii. Calculate the solubility, in mol L1-, of AgCl(s) in water at 10°C.

iii. Calculate the value of the solubility-product constant, Ksp, for

AgCl(s) at 10°C.

b. At 25°C, the value of Ksp for PbCl2(s) is 1.6x10-5 and the value of Ksp for

AgCl(s) is 1.8x10-10.

i. If 60.0 mL of 0.0400M NaCl(aq) is added to 60.0mL of 0.0300M

Pb(NO3)2(aq), will a precipitate form? Assume that volumes are

additive. Show calculations to support your answer.

ii. Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated

PbCl2 solution to which 0.250 moles of NaCl(s) have been added.

Assume that no volume change occurs.

iii. If 0.100M NaCl(aq) is added slowly to a beaker containing both

0.120M AgNO3(aq) and 0.150M Pb(NO3)2(aq) at 25°C, which will

precipitate first, AgCl(s) or PbCl2(s)? Show calculations to support

your answer.

a. i. AgCl↔Ag+ + Cl-

ii. [pic]

iii. Ksp = [Ag+][Cl-]=(6.23 ( 10-6)(6.23 ( 10-6)=3.89 ( 10-11

b. i. Find Qsp and see if it is bigger than the Ksp.

[Cl-]=0.0200M, [Pb2+]=0.0150M (Note since the volume is doubled the concentrations are cut in half.

PbCl2 ↔ Pb2+ + 2Cl-, Qsp =[Pb2+][Cl-]2 =[0.0150][0.0200]2 =6.00 ( 10-6 1 1 point

E° is positive, so K > 1 1 point

Note: The student’s score in part (d) is based on the sign of Eocell calculated in part (b).

Note on Overall Question: If in part (a) a student incorrectly identifies Ni as being oxidized,

partial credit is earned if subsequent parts are followed through consistently.

1997 REVIEW #19

Electrochemistry

In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride of iron, producing Fe(s) and Cl2(g).

a. Write the equation for the half-reaction that occurs at the anode.

Since oxidation occurs at the anode, I am looking for the chemical that oxidizes: That would be the one that has the oxidation number (charge—sorta) go up. That would be the Cl-: So

2Cl- ( Cl2 + 2e-

b. When the cell operates for 2.00 hours, 0.521 grams of iron are deposited at one electrode. Determine the formula or the chloride of iron in the original solution.

This is a big stoich problem. There are two ways to solve this. The first was done my Mr. Bergmann and is long, the 2nd by Mr. Sams which is much more elegant. I will present the Sam’s Version first because it is easier.

Version Sams: Solve for the mass of Fe and see which comes up with 0.521g Fe. You need to know that Fe can be either Fe2+ or Fe3+

[pic]

Version Bergmann

[pic]

The Fe will either be Fe2+ or Fe3+ so…

Fex+ + xe- ( Fe

I will try guess and check…

If the reaction is Fe3+ + 3e- ( Fe than the overall reaction would be

Fe3+ + 3e- ( Fe

2Cl- ( Cl2 + 2e-

2Fe3+ + 6Cl- ( 3Cl2 + 2Fe

Then: [pic]

Note that this is NOT the same moles of Cl2 from above.

So it could also be….

Fe2+ + 2e- ( Fe

2Cl- ( Cl2 + 2e-

Fe3+ + 2Cl- ( Cl2 + Fe

[pic]

These moles of Cl2 match the moles based on the electrochem data so the formula of the Chloride is FeCl2

c. Write the balanced equation for the overall reaction that occurs in the cell.

Fe2+ + 2e- ( Fe

2Cl- ( Cl2 + 2e-

Fe3+ + 2Cl- ( Cl2 + Fe

d. How many liters of Cl2(g), measured at 25°C and 750 mmHg, are produced when the cell operates as described in part (b)?

Use PV=nRT: First we need moles Cl2

[pic]

e. Calculate the current that would produce chlorine gas from the solution at a rate of 3.00 grams per hour.

Current is C/sec

[pic]

2000 REVIEW #20

Electrochemistry

Answer the following questions that relate to electrochemical reactions.

a. Under standard conditions at 25°C , Zn(s) reacts with Co2+ (aq) to produce Co(s).

i. Write the balanced equation for the oxidation half reaction.

Use the table of reduction potentials to look up the following values.

Co2+ + 2e- ( Co -0.28V

Zn2+ + 2e- ( Zn -0.76V

The Zn will be oxidized so that the value of the reduction potential, Eo will be positive. Therefore:

Zn ( Zn2+ + 2e-

ii. Write the balanced net-ionic equation for the overall reaction.

Co2+ + 2e- ( Co -0.28V

Zn ( Zn2+ + 2e- 0.76V

Zn + Co2+ ( Zn2+ + Co

iii. Calculate the standard potential, E°, for the overall reaction at 25°C.

=0.48V

b. At 25°C , H2O2 decomposes according to the following equation.

2H2O2(aq) → 2H2O(l) + O2(g) E°=0.55 V

i. Determine the value of the standard free energy change, ΔG°, for the reaction at 25°C.

∆G=-nFEo

=-2(96485)(0.55V)=-106100J/mol

Note that I used a 2 for the electrons. When you break out the reaction there are 2 moles e- that cancel in the reactions:

2e- + 2H+ + H2O2 (2 H2O

H2O2 ( O2 + 2H+ + 2e-

ii. Determine the value of the equilibrium constant, Keq, for the reaction at 25°C.

∆G=-RTlnK

-106100J/mol=-8.3145(298K)(lnK)

-42.8 = lnK

e-42.8=elnK =K

K=2.53 ( 10-19

iii. The standard reduction potential, E°, for the half reaction O2(g) + 4H1+ (aq) +4e- → 2H2O(l)

has a value of 1.23 V. Using this information in addition to the information given above, determine the value of the standard reduction potential, E°, for the half reaction below.

O2(g) + 2H1+ (aq) + 2e- → H2O2(aq)=target

Use the value from above and the one given in the following reaction

Reaction 1: 2H2O2(aq) → 2H2O(l) + O2(g) E°=0.55 V

Reaction 2: O2(g) + 4H1+(aq) +4e- → 2H2O(l) E°=1.23 V

This is an adding reactions type of question

Flip Rxn 1 and cut it in half ½ O2 + H2O ( H2O2 -0.55V

Cut Rxn 2 in half ½ O2 + 2H+ + 2e-( H2O 1.23 V

Reactions now add to O2 + 2H+ + 2e-( H2O2

E0 = 0.68V

This question is tricky. You will be tempted to multiply the value of the reaction by ½ because you cut the reaction in ½. This is done with thermodynamic data (∆H), but not done with Eo values. Watch this…

c. In an electrolytic cell, Cu(s) is produced by the electrolysis of CuSO4(aq). Calculate the maximum mass of Cu(s) that can be deposited by a direct current of 100. amperes passed through 5.00 L of 2.00M CuSO4(aq) for a period of 1.00 hour.

Since Cu in CuSO4 has a 2+ charge: Cu2+ + 2e- ( Cu

[pic]

1998 REVIEW #21

Kinetics

Answer the following questions regarding the kinetics of chemical reactions

a. The diagram to the right shows the energy pathway for the reaction O3 + NO → NO2 + O2.

Clearly label the following directly on the diagram.

i. The activation energy (Ea) for the forward reaction.

ii. The enthalpy change (ΔH) for the reaction.

b. The reaction 2N2O5 → 4NO2 + O2 is first order with respect to N2O5.

i. Using the axes at right, complete the graph that represents the change in [N2O5] over time as the reaction proceeds.

ii. Describe how the graph in (i) could be used to find the reaction rate at a given time, t.

You could take the slope of the line at any point. This can also be done by h taking the derivative of the line.

iii. Considering the rate law and the graph in (i), describe how the value of the rate constant, k, could be determined.

Plot a graph of ln [N2O5] verses time. The slope of that line will be = the absolute value of k

iv. If more N2O5 were added to the reaction mixture at constant temperature, what would be the effect on the rate constant, k? Explain.

Since this is a first order reaction: Rate=k[N2O5]. Adding more N2O5 will increase the rate based upon the rate law

c. Data for the chemical reaction 2A → B + C were collected by measuring the concentration of A at 10-minute intervals for 80 minutes. The following graphs were generated from the analysis of the data. Use the information in the graphs above to answer the following.

[pic]

i. Write the rate-law expression for the reaction. Justify your answer.

Since a straight line occurs when you plot lnA and not 1/A vs time, this reaction is first order. Therefore: Rate = k[A]1

ii. Describe how to determine the value of the rate constant for the reaction.

Slope of the line=-k

1997 REVIEW #22

Kinetics

2 A + B → C + D

The following results were obtained when the reaction represented above was studied at 25°C.

|Experiment |Initial [A] |Initial [B] |Initial Rate of Formation of C (mol|

| | | |L-1 min-10 |

|1 |0.25 |0.75 |4.3 ( 10-4 |

|2 |0.75 |0.75 |1.3 ( 10-3 |

|3 |1.50 |1.50 |5.3 ( 10-3 |

|4 |1.75 |?? |8.0 ( 10-3 |

a) Determine the order of the reaction with respect to A and B. Justify your answer.

[pic]

Order is first order with respect to A and to B

Rate=k[A]1[B]1

b) Write the rate law for the reaction. Calculate the value of the rate constant, specifying units.

Rate=k[A]1[B]1

From Reaction 1

4.3 ( 10-4 = k(0.25)(0.75)

K=2.3 ( 10-3 M-1min-1

Be careful on how you do the rate units. I like to look at the whole picture:

[pic]

c) Determine the initial rate of change of [A] in Experiment 3.

[pic]

d) Determine the initial value of [B] in Experiment 4.

8.0 ( 10-3 = 2.3 ( 10-4[1.75M][B]

[B]=19.9M

e) Identify which of the reaction mechanisms represented below is consistent with the rate law developed in part (b). Justify your choice.

1

A + B → C + M Fast

M + A → D Slow

2

B ⇌ M Fast equilibrium

M + A → C + X Slow

A + X → D Fast

3

A + B ⇌ M Fast equilibrium

M + A → C + X Slow

X → D Fast

Mechanism 2 is the correct one:

Rate = k[M][A]

But from step 1 [M]=[B]

Substituting in for M

Rate = k[B][A]

1996 REVIEW #23

Kinetics

The reaction between NO and H2 is believed to occur in the following three-step

process.

NO + NO ⇌ N2O2 (fast)

N2O2 + H2 → N2O + H2O (slow)

N2O + H2 → N2 + H2O (fast)

a) Write a balanced equation for the overall reaction.

2NO + 2H2 ( N2 +2 H2O

(b) Identify the intermediates in the reaction. Explain your reasoning.

N2O2 and N2O are reaction intermediates. They appear as products in the steps and then cancel, therefore they are intermediates.

(c) From the mechanism represented above, a student correctly deduces that the rate law for the reaction is rate = k[NO]2[H2]. The student then concludes that (1) the reaction is third-order and (2) the mechanism involves the simultaneous collision of two NO molecules and an H2 molecule. Are conclusions (1) and (2) correct? Explain.

(1): is correct:

Rate is determined by the slow step which is step 2

Rate = k[N2O2][H2], but since N2O2 is an intermediate you must:

[N2O2]=[NO][NO] from step 1.

Then you substitute for N2O2 in the rate law and you get

Rate = k[NO]2[H2]

(2) this is incorrect. Since the rate law depends on the slow step, what you are waiting for is for the N2O2 to collide with the H2. It is third order, but only because of the odd fact that the N2O2 is a reaction intermediate and is = to [NO]2

(d) Explain why an increase in temperature increases the rate constant, k, given the rate law in (c).

As temperature is increased reaction rate increases. This is due to more collisions happening with a greater force.

1999 REVIEW #24

Kinetics

2NO(g) + Br2(g) → 2 NOBr(g)

A rate study of the reaction represented above was conducted at 25°C. The data

that were obtained are shown in the table below.

|Expt |Initial [NO] M |Initial [Br2] M |Initial Rate of appearance of NOBr |

| | | |(M/sec) |

|1 |0.0160 |0.0120 |3.24 ( 10-4 |

|2 |0.0160 |0.0240 |6.38 ( 10-4 |

|3 |0.0320 |0.0060 |6.42 ( 10-4 |

(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.

[pic]

(b) Determine the order of the reaction with respect to each reactant, Br2(g) and NO(g). In each case, explain your reasoning.

[pic]

With respect to NO order is =1

With respect to Br2 order =2

(c) For the reaction,

i. write the rate law that is consistent with the data, and

Rate = k[NO][Br2]2

ii. calculate the value of the specific rate constant, k, and specify units.

[pic]

(d) The following mechanism was proposed for the reaction:

Br2(g) + NO(g) → NOBr2(g) slow

NOBr2(g) + NO(g) → 2 NOBr(g) fast

Is this mechanism consistent with the given experimental

observations? Justify your answer.

NO: rate is determined by the slowest step. This mechanism would yield the rate: Rate=k[Br2][NO] and this is not consitent with the actual rate which is Rate =k[NO][Br2]2

1996 REVIEW #25

Bonding/Structure/IMF

Explain each of the following observations in terms of the electronic structure

and/or bonding of the compounds involved.

(a) At ordinary conditions, HF (normal boiling point = 20°C ) is a liquid, whereas HCl (normal boiling point = -114°C ) is a gas.

The key point here is that both of these molecules have a similar lewis structure (you should draw it)!! But since the H is bonded to F in the HF this exhibits hydrogen bonding and thus has stronger forces of attraction and therefore a higher boiling point

(b) Molecules of AsF3 are polar, whereas molecules of AsF5 are nonpolar.

Again: You MUST draw the Lewis Structure. AsF5 shape = trigoanal bipyrimidal and AsF3 is pyramidal. This makes AsF3 polar due to an unbalanced shape and AsF5 nonpolar due to a balanced shape: Note: A full answer must have a Lewis Structure

(c) The N–O bonds in the NO21- ion are equal in length, whereas they are unequal in HNO2.

Draw Lewis Structures of Both. HNO2 has a hydrogen sticking onto an H this causes a pull of the electrons (via electronegativity) on the oxygen that has the H attached. Note: A full answer must have a Lewis Structure

(d) For sulfur, the fluorides SF2, SF4, and SF6 are known to exist, whereas for oxygen only OF2 is known to exist.

This one is a bit tricky: There are two answers: First: S is bigger than O (though in the same chemical family-column) So it can accommodate more atoms around it. Also, especially when you make SF6, you have a hybridization of d2sp3 and this means that you must have a d orbital available. Since there is no2d orbital, oxygen cannot have that kind of hybridization.

1997 REVIEW #26

Bonding/Structure/IMF

Consider the molecules PF3 and PF5

(a) Draw the Lewis electron-dot structures for PF3 and PF5 and predict the

molecular geometry of each.

Draw them

Geometry is the same as asking about shape: PF3 = pyramidal with one lone pair of e- around the P

PF5 is trig bipyrimidal with no extra electrons around the central atom

(b) Is the PF3 molecule polar, or is it nonpolar? Explain.

Due to the unbalanced shape (geometry), PF3 is polar

(c) On the basis of bonding principles, predict whether each of the following

compounds exists. In each case, explain your prediction.

i. NF5

Doe not exist: There is no 2d orbital for the dsp3 hybridization

ii. AsF5

Does exist: Draw the lewis structure to prove it.

1999 REVIEW #27

Bonding/Structure/IMF

Answer the following questions using principles of chemical bonding and molecular structure.

(a) Consider the carbon dioxide molecule, CO2, and the carbonate ion, CO32-.

i. Draw the complete Lewis electron-dot structure for each species.

Draw them: CO has a triple bonds

CO32- has resonance structures (plural)

Account for the fact that the carbon-oxygen bond length in CO32- is greater than the carbon-oxygen bond length in CO2.

You must discuss the fact that with resonance structures the “real” structure is the average of all of the structures. This in essence makes for each bond being like a 1.33 bond. Single bonds are long and triple bonds are very short.

(b) Consider the molecules CF4 and SF4.

i. Draw the complete Lewis electron-dot structure for each molecule.

CF4 will be tetrahedral and SF4 will be See-Saw. Draw these

iii. In terms of molecular geometry, account for the fact that the CF4 molecule is nonpolar, whereas the SF4 molecule is polar.

Due to the see-saw shape, SF4 will be polar and due to the tetrahedral shape CF4 will be nonpolar. Explain this

1997 REVIEW #28

Bonding/Structure/IMF

Explain each of the following observations using principles of atomic structures and/or bonding.

(a) Potassium has a lower first-ionization energy than lithium.

These atomic theory questions all come back to size: Size matters. Potassium has a larger atomic radius (more electron shells than Li) Therefore the outermost electron (valence) is further from the nucleus and is less tightly held and easier to break.

(b) The ionic radius of N3- is larger than that of O2-.

N3- and O2- have the same number of electrons (10). But O2- has one more proton. Therefore a stronger nuclear force is holding the same number of electrons, drawing it into the nucleus and making O2- smaller.

(c) A calcium atom is larger than a zinc atom.

Ca and Zn are in the same period. (4th principal quantum number). As you go across a row the trend is that elements get smaller. This is due to the same electron shell (4th) is being added to at the same time that additional protons are being added to the nucleus. This causes the size of the electron cloud to shrink as you go across a row.

(d) Boron has a lower first-ionization energy than beryllium.

This goes against the main trend. So what is the deal: Draw the electron config with the box notation and explain about full and ½ full shells being more stable.

2000 REVIEW #29

Bonding/Structure/IMF

Answer the following question about the element selenium, Se (atomic number 34).

(a) Samples of natural selenium contain six stable isotopes. In terms of

atomic structure, explain what these isotopes have in common, and how they differ.

Isotopes have the same number of protons but different numbers of neutrons. This affects the mass. Isotopes have different masses (Mass numbers)

b) Write the complete electron configuration (e.g., 1s22s2… etc.) for a selenium atom in the ground state. Indicate the number of unpaired electrons in the ground-state atom, and explain your reasoning.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 3p4

Unpaired = 2: Here you MUST draw the box notation showing the arrows in the 3p orbital to get full credit

(c) In terms of atomic structure, explain why the first ionization energy of selenium is

i. Less than that of bromine (atomic number 35), and

Again: Box notation and explain that increasing the number of electrons it is harder to break the electron free

ii. Greater than that of tellurium (atomic number 52).

This is an exception and needs to be explained by showing the box notation of Se vs Te. Te has 3 electrons in each of 3 separate 3p orbitals. Se has 2 electrons paired up in the first 3p orbital and is less stable and thus has a lower first IE.

d) Selenium reacts with fluorine to form SeF4. Draw the complete Lewis electron-dot structure for SeF4 and sketch the molecular structure. Indicate whether the molecule is polar or nonpolar, and justify your answer.

Draw: This is a see-saw shape (dsp3) and therefore polar.

2003 REVIEW #30

Bonding/Structure/IMF

|Compound |Formula |∆Hvap (kJ/mol) |

|Propane |CH3CH2CH3 |19.0 |

|Propanone |CH3COCH3 |32.0 |

|1-propanol |CH3CH2CH2OH |47.3 |

Using the information in the table above, answer the following questions about organic compounds.

To fully get this question you should draw the Lewis Structures of each compound then it is an easy question

(a) For propanone,

i. draw the complete structural formula (showing all atoms and bonds);

[pic]

ii. predict the approximate carbon-to-carbon-to-carbon angle.

120o due to the sp2 hybridzation

(b) For each pair of compounds below, explain why they do not have the same value for their standard heat, o ΔHvap . (You must include specific information about both compounds in each pair.)

i. Propane and propanone

The intermolecular attractive forces in propane are dispersion forces only (LDF). The IMF in porpanone are dispersion AND dipole-dipole. Since the IMF of dipole is stronger than only LDF, this accounts for the higher value for the propanone. Again: This will best understood if you DRAW THE LEWIS STRUCTURES OF EACH

ii. Propanone and 1-propanol

The key is that propanol has hydrogen bonding which is stronger than dipole forces.

(c) Draw the complete structural formula for an isomer of the molecule you drew in part (a) (i).

[pic]

(d) Given the structural formula for propyne below,

[pic]

i. indicate the hybridization of the carbon atom indicated by the arrow in the structure above;

sp

ii. indicate the total number of sigma (σ) bonds and the total number of pi (π) bonds in the molecule.

Sigma=6

Pi = 2

1998 REVIEW #31

Colligative properties

An unknown compound contains only three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.

a) Determine the empirical formula of the compound.

Percent to mass, mass to mole, divide by small, times till whole

[pic]

b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze at a temperature 15.2 Celsius degrees below the normal freezing point for camphor. Determine the molar mass and apparent molecular formula of the compound. (The molal freezing point depression constant, Kf, for camphor is 40.0 kg ▪ K ▪ mol-1.)

[pic]

c) When 1.570 grams of the compound is vaporized at 300°C and 1.00 atmosphere, the gas occupies a volume of 577 milliliters. What is the molar mass of the compound based on this result?

[pic]

e) Briefly describe what occurs in solution that accounts for the difference between the results obtained in parts (b) and (c).

The substance must actually ionize into two parts thus accounting for the formula. This makes the real formula to be C7H12O2

1999 REVIEW #32

Colligative properties

[pic]

Answer the following questions, which refer to the 100 mL samples of aqueous solutions at 25°C in the stoppered flasks shown above.

a) Which solution has the lowest electrical conductivity? Explain

C2H5OH is an alcohol and therefore does not dissociate appreciably and thus has the lowest electrical conductivity

b) Which solution has the lowest freezing point? Explain.

Lowest f.p. will be the substance with the greatest ∆Tf: Since MgCl2 ionizes into three ions it has the greatest: ∆Tf = iKfm: i=3

MgCl2 ( Mg2+ + 2Cl-

c) Above which solution is the pressure of the water vapor greatest? Explain.

Greatest Water Vapor will be the one with the lowest value of i. This is the alchohol, C2H5OH

d) Which solution has the highest pH? Explain.

NaF:

F- + HOH ↔ HF + OH-

This is a base. CH3COOH is an acid (low pH), etc…

2001 REVIEW #33

Colligative properties

[pic]

Answer the questions below that relate to the five aqueous solutions at 25°C shown above.

a) Which solution has the highest boiling point? Explain.

∆T=iKb m: Greatst b.p. will be thw one that dissociates into the most ions: That is the Pb(NO3)2. i=3

b) Which solution has the highest pH? Explain.

Highest pH: Most basic: KC2H3O2

C2H2O2- + HOH ↔HC2H3O2 + OH-

c) Identify a pair of the solutions that would produce a precipitate when mixed together. Write the formula of the precipitate.

Pb(NO3)2 + 2NaCl ( PbCl2 (s) + 2NaNO3

PbCl2 = ppt (it is yellow)

d) Which solution could be used to oxidize the Cl1- (aq) ion? Identify the product of the oxidation.

The KMnO4

2Cl- ( Cl2 + 2e-

MnO4- ( Mn2+

f) Which solution would be the least effective conductor of electricity? Explain.

C2H5OH is worst since is not an ionic compound and will not dissociate into ions

1996 REVIEW #34

Acid-Base (Lab)

A 0.500-grams sample of a weak, nonvolatile acid, HA, was dissolved in sufficient water to make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH solution. Predict how the calculated molar mass of HA would be affected (too high, too low, or not affected) by the following laboratory procedures. Explain each of your answers.

a. After rinsing the buret with distilled water, the buret is filled with the standard NaOH solution; the weak acid HA is titrated to its equivalence point.

Adding distilled water to the buret will cause the concentration of the NaOH to be too small. This will necessitate using more volume of NaOH to reach the equivalcence point. More volume will increase the number of moles of acid: And then when you solve for the molar mass (grams/moles) the moles will be larger and hence the MM will be HIGHER

b. Extra water is added to the 0.500-gram sample of HA.

No effect. The water is simply the dissolving medium. Only the moles of HA (determined by the mass) will neutralize the acid

c. An indicator that changes color at pH 5 is used to signal the equivalence point.

In this reaction the pH at the equivalence point will need to be basic since HA is a weak acid and A- is the conjugate base of the weak acid. The indicator will change, but the reaction will not be at the equivalence point. So less volume of NaOH will be used and hence the moles of the acid will be lower and since MM = g/mole the value of the MM will be higher than expected.

d. An air bubble passes unnoticed through the tip of the buret during the titration.

This will increase the amount of measured NaOH that is required to reach the equivalence point so this will increase the moles of NaOH, hence the moles of HA. And more moles of HA make for a smaller MM of HA.

1997 REVIEW #35

Lab

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An experiment is to be performed to determine the mass percent of sulfate in an unknown soluble sulfate salt. The equipment shown above is available for the experiment. A drying oven is also available.

a) Briefly list the steps needed to carry out this experiment.

1) Mass Mixture(unknown sulfate salt) on balance

2) Dissolve mixture into flask

3) Add 0.20 M BaCl2 and a ppt will form. (BaSO4)

4) Mass Filter Paper

5) Filtrate the ppt through the filter paper using the funnel, and a cylinder. This make take a few times to get all of the ppt out of the solution.

6) Dry the filter paper in the drying oven and mass after it is completely dry.

b) What experimental data need to be collected to calculate the mass percent of sulfate in the unknown?

Mass of Filter paper dry

Mass of Filter paper with ppt (dry)

Mass of unknown sulfate salt

c) List the calculations necessary to determine the mass percent of sulfate in the unknown.

Mass of dry ppt = mass of filter paper with ppt- mass of filter paper alone

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d) Would 0.20-molar MgCl2 be an acceptable substitute for the BaCl2 solution provided for this experiment? Explain.

No: MgSO4 is not a ppt, so it would not work.

2000 REVIEW #36

Lab

The molar mass of an unknown solid, which is nonvolatile and a nonelectrolyte, is to be determined by the freezing point depression method. The pure solvent used in the experiment freezes at 10°C and has a known molal freezing-point

depression constant, Kf. Assume that the following materials are also available.

• Test tubes • Stirrer • Pipet • Thermometer • Balance

• Beaker • Stopwatch • Graph paper • Hot-water bath • Ice

(a) Using the two sets of axes provided below, sketch cooling curves for (i) the pure solvent and for (ii) the solution as each is cooled from 20.0°C to 0.0°C.

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(b) Information from these graphs may be used to determine the molar mass of the unknown solid.

i. Describe the measurements that must be made to determine the molar mass of the unknown solid by this method.

Mass of Solute

Mass of Solvent

Do a cooling curve of the pure solvent (it should freeze at 10 C)

Dissolve a given amount of solute into the solvent and freeze. Then determine the f.p. from the curve (flat portion that is lower than 10C)

iii. Show the setup(s) for the calculation(s) that must be performed to determine the molar mass of the unknown solid from experimental data.

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Solve for Molality and then determine the moles of the solute by:

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Now take the grams (measured at the beginning of the experiment) and divide by the moles of the solute (from the calculation above)

iii. Explain how the difference(s) between the two graphs in part (a) can be used to obtain information needed to calculate the molar mass of the unknown solid.

The flat portion of the graph (10C) is the f.p. the flat portion of the 2nd graph is at some number less than 10C. the difference between the flat portions is the change in the freezing point.

(c) Suppose that during the experiment a significant but unknown amount of solvent evaporates from the test tube. What effect would this have on the calculated value of the molar mass of the solid (i.e., too large, too small, or no effect)? Justify you answer.

This would make your concentration (Molality) higher. This would make the f.p. depression lower and thus the moles of the solute calculated higher. Since MM=g/mole the mole term will be larger and thus the MM value will be LOWER.

c) Show the setup for the calculation of the percentage error in a student’s result if the student obtains a value of 126 g mol-1 for the molar mass of the solid when the actual value is 120. g mol-1.

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1998 REVIEW #37

Lab

An approximately 0.1-molar solution of NaOH is to be standardized by titration. Assume that the following materials are available.

• Clean, dry 50 mL buret • Analytical balance

• 250 mL Erlenmeyer flask • Phenolphthalein indicator solution

• Wash bottle filled with distilled water • Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard)

(a) Briefly describe the steps you would take, using materials listed above, to standardize the NaOH solution.

1. Dissolve some solid HA into a beaker

2. Add some phenolphthalein to the beaker

3. Put the NaOH in the Buret

4. Add NaOH to the solution below until it turns pink

(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution.

NaOH + KHP ( NaKP + HOH

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__gKHP = mass KHP measured in the trial

__LNaOH is the amount measured in the titration from the buret.

(c) After the NaOH solutions has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0.0 to 35.0 mL. Clearly label the equivalence point on the curve.

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(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c).

At ½ of the equivalence point (12.5mL) the pH=pKa. Once the pKa is known:

10-pKa=Ka

(e) The graph to the right shows the results obtained by titrating a different weak acid, H2Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve.

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H2A ( HA- + H+

HA- ( H+ + A2-

A2- is present at this equvilance point

1999 REVIEW #38

Lab

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A student performs an experiment to determine the molar mass of an unknown gas. A small amount of the pure gas is released from a pressurized container and collected in a graduated tube over water at room temperature, as shown in the diagram above. The collection tube containing the gas is allowed to stand for several minutes, and its depth is adjusted until the water levels inside and outside the tube are the same. Assume that:

• the gas is not appreciably soluble in water

• the gas collected in the graduated tube and the water are in thermal equilibrium

• a barometer, a thermometer, an analytical balance, and a table of the equilibrium vapor pressure of water at various temperatures are also available.

(a) Write the equation(s) needed to calculate the molar mass of the gas

Molar Mass = g/mol

PV=nRT: Solve for moles from this equation

Ptot=PH2O + PH2

(b) List the measurements that must be made in order to calculate the molar mass of the gas.

Mass of the cylinder before

Mass of the cylinder after

Volume of the Gas

VP of the Water

Temperature of the Gas (= to Temperature of the water)

Atmospheric Pressure

(c) Explain the purpose of equalizing the water levels inside and outside the gas collection tube.

This will have the pressure’s equal inside the eudiometer and outside. Otherwise the weight of the water column will affect the value.

(d) The student determines the molar mass of the gas to be 64 g mol-1. Write the expression (set-up) for calculating the percent error in the experimental value, assuming that the unknown gas is butane (molar mass 58 g mol-1). Calculations are not required.

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d) If the student fails to use information from the table of the equilibrium vapor pressures of water in the calculation, the calculated value for the molar mass of the unknown gas will be smaller than the actual value. Explain.

The value of the PH2 will be too high. Then n=PV/RT will be too large. The moles will be too large and the MM will be to LOW.

2000 REVIEW #39

Lab

A volume of 30.0 mL of 0.10 M NH3(aq) is titrated with 0.20 M HCl(aq) . The value of the base-dissociation constant, Kb, for NH3 in water is 1.8 x 10-5 at 25 oC.

a) Write the net-ionic equation for the reaction of NH3(aq) with HCl(aq) .

NH3 + H+ ( NH41+

b) Using the axes provided below, sketch the titration curve that results when a total of 40.0 mL of 0.20 M HCl(aq) is added dropwise to the 30.0 mL volume of 0.10 M NH3(aq) .

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c) From the table below, select the most appropriate indicator for the titration. Justify your choice. Indicator pKa

Methyl Red 5.5

Bromothymol Blue 7.1

Phenolphthalein 8.7

I would use methyl: the pH at the equivalace point will be somewhere in the range of 5. It should be acidic due to the presence of the acidic NH4+ (the conjugate acid of NH3). And the pKa gives you the approximate pH that the indicator will change it’s color.

d) If equal volumes of 0.10 M NH3(aq) and 0.10 M NH4Cl(aq) are mixed, is the resulting solution acidic, neutral, or basic? Explain.

Ka of NH3 = 1.8 ( 10-5

Kb of NH4+ ≈ 1 ( 10-9

Due To the fact that Ka x Kb = Kw

Since NH3 is a stronger base than NH41+ is an acid, the solution will favor the NH3 and will therefore be BASIC

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Ea

∆H

[N2O5]

Time

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