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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.
Kc 
b.
[ H 2 ]2 [ S 2 ]
[ H 2 S ]2
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 S 2 and H2 is 0. Also be aware of the coefficients in the
reaction.
3.40 gH 2 S 1molH 2 S
1
x
x
 0.0800MH 2 S
1
34 gH 2 S 1.25LH 2 S
3.72 x10 2 molS 2
 0.0298MS 2
1.25 LS 2
Initial
Change
Equilibrium
2H2S
0.0800
-0.0595
0.0205
↔
2H2
0
+0.0595
0.0595
+
S2
0
+0.0298
0.0298
Therefore: [H2]=0.0595M and [S2]=0.0298
c.
Simply plug into the equilibrium expression the equilibrium concentrations
Kc 
d.
[ H 2 ]2 [ S2 ] (0.0595) 2 (0.0298)

 0.251
[ H 2 S ]2
(0.0205)2
Use PV=nRT
Latm
P(1.25L)  (3.72 x102 mol)(0.0821 molK
)( 483K )
P  1.18atm
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.
AgCl↔Ag+ + Cl-
i.
ii.
8.9x10 -5 g 1mol
x
 6.23x10 6 M
0.10L
143.5g
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<1.6  10-5 therefore no ppt will form.
iii.
Use common ion effect. You will know the [Cl-] and then solve for the [Pb2+] using the Ksp
equation.
Ksp=1.6  10-5=[Pb2+][0.25M]2
[Pb2+]=2.56  10-4 M
ii.
Competitive precipitation: the hardest of all Ksp problems. Solve for the of the Cl-. The lowest
answer wins.
AgCl↔Ag+ + ClPbCl2↔Pb2+ + 2Cl-5
2
Ksp= 1.8x10-10 = [0.120][s]
Ksp=1.6  10 =[0.150][2s]
[Cl-]=0.0206M
[Cl-]=1.5  10-9 M
iii.
The AgCl will ppt first due to the lower concentration of the Cl-
1998 REVIEW #3
Ksp
Solve the following problem related to the solubility equilibria of some metal
hydroxides in aqueous solution.
a. The solubility of Cu(OH)2(s) is 1.72x10-6 gram per 100 milliliters of solution at
25°C.
i. Write the balanced chemical equation for the dissociation of
Cu(OH)2(s) at 25°C.
ii. Calculate the solubility (in moles per liter) of Cu(OH)2(s) at 25°C.
iii. Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25°C.
b. The value of the solubility-product constant, Ksp, for Zn(OH)2(s) is 7.7x10-17 at 25°C.
i. Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C in a solution with a pH of 9.35.
ii. At 25°C, 50.0 milliliters of 0.100-molar Zn(NO3)2 is mixed with 50.0 milliliters of 0.300-molar NaOH. Claculate
the molar concentration of Zn2+(aq) in the resulting solution once equilibrium has been established. Assume that
volumes are
additive.
1) Cu(OH)2 ↔ Cu2+ + 2OH-
1.72 x106 g 1mol
2)
x
 1.76 x10 7 M
0.100L
97.5g
3)
4)
5)
6)
7)
Ksp =[Cu2+][OH-]2 =[s][2s]2=4s3=4(1.76  10-7)3=2.18  10-20
pOH = 14-9.35 = 4.6
10-pOH =10-4.65 = 2.24  10-5
Ksp = 7.7  10-17 = [Zn2+][2.24  10-5]2
[Zn2+]=1.53  10-7 M
b. i. Ksp = 7.7  10-17 = [Zn2+][OH-]2 =4s3 s= molar solubility = 2.68  10-6 M
Zn(NO3)2
+ 2NaOH
Zn(OH)2
+2 NaNO3

5 mmol
15mmol
0
-5
-10
+10
0
5 mmol
10mmol
Not important
5
mmol
[OH-]=
 0.050 M
100mL
Ksp = 7.7  10-17 = [Zn2+][OH-]2 = [Zn2+][0.050M]2
[Zn2+]=3.08  10-14
1998 REVIEW #4
Le Châtelier’s Principle
C(s) + H2O(g) ↔ CO(g) + H2(g) ΔH°= +131 kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), and H2(g) at equilibrium. State whether the
number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following
disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except
for the given disturbance. Explain each answer with a short statement.
a. Additional H2(g) is added to the equilibrium mixture at constant volume.
b. The temperature of the equilibrium mixture is increased at constant volume.
c. The volume of the container is decreased at constant temperature.
d. The graphite pellets are pulverized.
This is a essay question, but don’t think of it like an essay question: think rather of a short answer.
a.
b.
c.
d.
moles of CO will go down due to the reaction shifting to the left
Since this reaction is endothermic (H is positive): this causes the reaction to shift to the right making
more CO
Volume decreased makes the pressure go up. When pressure goes up the reaction shifts to the side with the
least moles of gas. Threfore this will shift to the left since the reactants have only one mole of gas and the
products have 2 moles of gas.
No effect. Solids do not affect the position of equilibrium.
1996 REVIEW #5
Acid-Base
Ka & Kb
HOCl  OCl1- + H1+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid-dissociation constant, Ka,
for the reaction represented above is 3.2 x 10 -8.
a. Calculate the [H1+] of a 0.14-molar solution of HOCl.
b. Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and
calculate the numerical value of the equilibrium constant for the reaction.
c. Calculate the pH of a solution made by combining 40.0 milliliters of 0.14- molar HOCl and 10.0 milliliters of
0.56-molar NaOH.
d. How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl to obtain a buffer
solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a negligible change in volume.
e. Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O(l) → H1+ + Cl1- + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065-molar.
a.
HOCl
H+
+
OCl↔
0.14
0
0
-x
+x
+x
0.14-x
X
X
[ H  ][OCl  ]
x2
Ka 
 3.2 x108 
[ HOCl ]
(0.14  x)
x  [ H  ]  6.69 x10 5 M
b) OCl- + HOH  HOCl + OHc) Start with stoich
OH+ HOCl
HOH

5.6 mmol
5.6 mmol
-5.6
-5.6
0
0
At equivalence point
[OCl-] = 5.6mmol/50mL = 0.112 M
+ OCl0
+5.6
5.6
OCl+ HOH
HOCl
+ OH↔
0.112
0
0
-x
+x
+x
0.112-x
X
x
This is the conjugate base so I must find Kb for this substance
Kw / Ka =1  10-14/3.2  10-8 =3.12 x10-7
3.12 x10 - 7 
x2
 1.87x10 - 4 M
(0.112  x)
 log( 1.87x10 - 4 M )  pOH  3.73 :: pH  10.27
OHx
-x
0
d) First put in and ICE table
+ HOCl
HOH
+ OCl
10 mmol
0
-x
+x
10-x
x
Now use Henderson-Hasselbach Equation
Find pKa = -log(3.2 x 10-8)=7.49.
pH = pKa + log(base/acid)
 [OCl  ] 
 x 
  7.49  log 
7.49  7.49  log 

 10  x 
 {HOCl ] 
x  5mmolNaOH  0.00500molNaOH
e. don’t overthink this question. Simply realize that the: [HOCl]=[H+]=0.065M:: pH = -log(o.o65M)=1.19
1999 REVIEW #6
Acid-Base
NH3 + H2O ↔ NH4+ + OHKa & Kb
In aqueous solution, ammonia reacts as represented above. In 0.0180M NH3(aq) at 25°C, the hydroxide ion concentration, [OH1-],
is 5.60 x 10-4M. In answering the following, assume that the temperature is constant at 25°C and that the volumes are additive.
a. Write the equilibrium-constant expression for the reaction represented above.
b. Determine the pH of 0.0180M NH3(aq).
c. Determine the value of the base ionization constant, Kb, for NH3(aq).
d. Determine the percent ionization of NH3 in 0.0180M NH3(aq).
e. In an experiment, a 20.0 mL sample of 0.0180M NH3(aq) was placed in a flask and titrated to the equivalence point and
beyond using 0.0120M HCl(aq).
i. Determine the volume of 0.0120M HCl(aq) that was added to reach the equivalence point.
ii. Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120M HCl(aq) was added.
iii. Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120M HCl(aq) was added.
a.
[ NH 4 ][OH  ]
Kb 
[ NH 3 ]
b) [OH1-], is 5.60 x 10-4M
pOH=-log(5.6  10-4)=3.25
pH=14-pOH=14-3.25=10.75
c) Use ICE table and solve for x
NH3
+ H2O
↔
NH4+
0.0180M
0
-5.6  10-4
+5.6  10-4
0.01744
5.6  10-4
+ OH0
+5.6  10-4
5.6  10-4
[ NH 4 ][OH  ] (5.6 x104 )(5.6 x104 )
Kb 

 1.8 x105
[ NH 3 ]
0.1744
d)

e)
NH3
Percent Ionization
x
5.6 x104
x100 
x100  3.1%
initialCon centration
0.0180
Reaction
+ H+

NH41+
0.0200 LNH 3 0.0180molNH 3 1molHCl
1LHCl
x
x
x
 0.030 LHCl  30.0mLHCl
1
1LNH 3
1molNH 3 0.012molHCl
NH3
36
-18
18
NH3
36
-36
0
Convert everything to mmol
+ H+
NH41+

18
0
-18
+18
0
18
This is ½ the equivalence point so pH=pKa
pKb =-log(1.8E-5) = 4.74
pKb + pKa = 14: pKb=9.26=pH
+ H+
48
-36
12

NH41+
0
+36
36
Excess acid: [H+]=12mmol/60mL=0.20M: -log(0.20)=0.70=pH
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.

Mass some amount of KHP (maybe 0.50g) and dissolve. Also add a few drops of phenolphthlein

Place 0.1 M NaOH in buret

Add NaOH unitl the solution turns pink

Do Stoich to determine the concentration of the NaOH. Stoich would look something like this:
0.50 gKHP 1molKHP 1molNaOH
1
x
x
x
 MolarityNaOH
1
MMKHP 1molKHP VolNaOH
(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution.
Done above
(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.
(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 the equivalence point (12.5mL) the pH=pKa.. Then use the equation 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.
H2Y + OH-  HY- + HOH
HY- + OH-  Y2- + HOH
Answer: Y2-
1997 REVIEW #12
Thermodynamics
For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3
and Cl2 are produced as the temperature is increased.
PCl5(g) ⇌ PCl3(g) + Cl2(g)
a. What is the sign of ΔS° for the reaction? Explain.
Since you are going from one mole of a gas to two moles of a gas this system is
getting more disordered and therefore entropy is increased. ΔS° is positive
b. What change, if any, will occur in ΔG° for the reaction as the temperature is increased?
Explain your reasoning in terms of thermodynamics principles.
Since when Temp is increased there are more molecules of PCl3 and Cl2, this
reaction is endothermic and ∆H is positive.
∆G = ∆H-T∆S: = + - T(+). At higher temperatures the T∆S term will be larger and
make the value of ∆G become more negative, thus making the reaction more
spontaneous at higher temperatures.
c. If He gas is added to the original reaction mixture at constant volume and temperature,
what will happen to the partial pressure of Cl 2? Explain.
He gas is an inert gas. It will not increase the partial pressures of any of the
gasses therefore there will be no change in the equilibrium mixture.
d)
If the volume of the reaction mixture is decreased at constant temperature to half
the original volume, what will happen to the number of moles of Cl 2 in the reaction
vessel? Explain.
When volume is decreased the pressure goes up. This favors the formation of the
reactant (less moles of gas). Therefore the number of moles of Cl2 will go down.
1999 REVIEW #13
Thermodynamics
Answer the following questions in terms of thermodynamic principles and concepts of kinetic
molecular theory.
a. Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2NH3(g) → CO(NH2)2(s) + H2O(l) ΔH°298 = -134 kJ
i.
For the reaction, indicate whether the standard entropy change, ΔS°298,
is positive, or negative, or zero. Justify your answer.
This system is going from 3 moles of gas on the reactant side to 1 mole of solid
and one mole of liquid. The system is becoming more ordered, therefore ΔS is
negative.
ii.
ii. Which factor, the change in enthalpy, ΔH°298, or the change in
entropy, ΔS°298, provide the principal driving force for reaction at 298 K.
Explain
the enthalpy will be the principle driving force. Since ΔS is negative this does not
favor a spontaneous change. The value of ΔH is negative and this does contribute
to the value of ΔG being negative and will drive the reaction to be spontaneous.
iii.
iii. For the reaction, how is the value of the standard free energy change,
ΔG°, affected by an increase in temperature? Explain.
Since ΔG= ΔH-T ΔS: and we have a (-) value of ΔH and a (-) value of ΔS
ΔG = (-) –T(-): Higher values of T will cause the reaction to be less spontaneous,
more positive.
b. Some reactions that are predicted by their signs of ΔG° to be spontaneous at room
temperature do not proceed at a measurable rate at room temperature.
i.
Account for this apparent contradiction.
This is a separate question of reaction kinetics. Just because ΔG is negative only
tells the chemist that the reaction will proceed, not how fast it will proceed.
Diamonds graphite is a spontaneous process, but the kinetics of the reaction
cause it to happen very very slowly.
ii.
A suitable catalyst increases the rate of such a reaction. What effect does the
catalyst have on ΔG° for the reaction? Explain.
No effect. The kinetics do not have an effect on the thermodynamics of a
reaction. The catalyst will speed up the process, but not make it more or less
spontaneous.
1996 REVIEW #14
Thermodynamics
C2H2(g) + 2H2(g) → C2H6(g)
Information about the substances involved in the reaction represented above is summarized in the
following tables.
Substance
S° ( J mol⋅K )
H  (kJ/mol)
f
C2H2 (g)
H2(g)
C2H6
Bond
C-C
C=C
C-H
H-H
200.9
130.7
-----
226.7
0
-84.7
Bond energy (kJ/mol)
347
611
414
436
a. If the value of the standard entropy change, ΔS°, for the reaction is - 232.7 joules per mole ▪
Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.
The trick on this one is to realize that the value -232.7 joules/mol K is the overall value for
the entire reaction. Here we want to find the individual amount for the C2H6.
S   Pr oducts   Re ac tan ts
 232.7  x  200.9  2(130.7) 
 232.7  x  462.3
x  229.6 J / molK
b. Calculate the value of the standard free-energy change, ΔG°, for the reaction. What does the sign
of ΔG° indicate about the reaction above?
ΔG=ΔH-TΔS: First though we must find ΔH. Note that when you then covert it that the value of  S must be
converted to kJ.
H   Pr oducts   Re ac tan ts
 (84.7)  (226.7)  311.4
G  H  TS
kJ
kJ
 311.4 mol
 (298 K )(0.2296 molK
)
 379.4kJ / mol
c. Calculate the value of the equilibrium constant, K, for the reaction at 298K?
 G=-RTlnK
- 379,400  -(8.3145J/molK)(298K )lnK
lnK  153.1
e lnK  e153  3.78 x10 66
d. Calculate the value of the C ≡ C bond energy in C2H2 in kilojoules per
mole.
Bonds Broken-bonds Formed = ΔH
C≡C 1(x)
C=C 1(611)
H-H 2(436)
C-H 6(414)
C-H 2(414)
Total
1700+x
3095
1700+x -3095 = -379
X=1016kJ/mol=Bond energy of C≡C
1998 REVIEW #15
Thermodynamics
C6H5OH(s) + 7O2(g) → 6CO2(g) + 3H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the
equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer
the questions that follow.

Substance
Standard Heat of Formation
Absolute Entropy S at ° C
H f at 25 C (kJ/mol)
C(graphite)
CO2(g)
H2(g)
H2O(l)
O2(g)
C6H5OH(s)
0.00
-393.5
0.00
-385.85
0.00
?
(J/mol K)
5.69
213.6
130.6
69.91
205.0
144.0
a. Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.
2.000 gC6 H 5OH 1molC6 H 5OH
x
 0.02128molC6 H 5OH
1
94 gC6 H 5OH
___ kJ
64.98kJ

 3054kJ / mol
___ mol 0.02128
 3054kJ / mol
This must be negative since this is exothermic
H 
b. Calculate the standard heat of formation, of
H f
, of phenol in kilojoules per mole at 25°C.
H   Pr oducts   Re ac tan ts
 3054  6(393.5)  3(385.85)   x 
x  464.5kJ / mol  H f
c. Calculate the value of the standard free-energy change, ΔG°, for the combustion of phenol at 25°C.
Frist find ∆S:
S   Pr oducts   Re ac tan ts
J
3(69.91)  6(213.6)  144  7(205)  87.67 molK
Now find ∆G using: ∆G=∆H-T∆S
=-3054-298(-0.08767)=-3028kJ/mol
e.
If the volume of the combustion container is 10.0 liters, calculate the final pressure in the
container when the temperature is changed to 110. °C. (Assume no oxygen remains
unreacted and that all products are gaseous.)
Here I will use PV=nRT. The trick will be finding the moles of gaseous product. So I
will do stoich to determine the moles: Volume and temperature are given in the
problem above:
2.000 gC6 H 5OH 1molC6 H 5OH
9molGases
x
x
 0.1915molGases
1
94 gC6 H 5OH 1molC6 H 5OH
PV  nRT
Latm
P(10 L)  (0.1915mol)(0.0821 molK
)(383K )
P  0.602atm
1996 REVIEW #16
Electrochemistry
Sr(s) + Mg2+ ⇌ Sr2+ + Mg(s)
Consider the reaction represented above that occurs at 25°C. All reactants and products are in their
standard states. The value of the equilibrium constant, Keq, for the reaction is 4.2 x 1017 at 25°C.
a. Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain your
prediction.
Since Keq is very large, the reaction favors the products. This means that the reaction is
spontaneous and therefore the value of E° is positive.
b. Identify the oxidizing agent for the spontaneous reaction.
Oxidizing agent is the substance that reduces. Mg2+ is losing charge and therefore is the
oxidizing agent
c. If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change?
Justify your answer.
This can be determined by using the Nernst Equation:
Ecell = E°cell –RT/nF (lnQ)
Since E°cell is (+) and the value of ln(Q) is (+) then a higher T will make the –RT/nF (lnQ)
term become larger which will
lower
the value of Ecel.
d. How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar
solution of Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2? Explain.
Again we would use the Nernst Equation but now the Q term becomes important:
RT  [ Sr 2  ] 

ln 
nF  [ Mg 2  ] 
RT  [0.1] 

ln 

nF  [1.0] 
Ecell  E cell 
Ecell  E cell
Ln(0.1)=a negative number so this will cause the –RT/nF (lnQ) term to be negative. And
when you subtract a negative number you get a (+) number. So this will increase the
value of cell potential
f. When the cell reaction in (d) reaches equilibrium, what is the cell potential?
This akin to a battery dying. So at equilibrium the value of the cell potential is =0. This
can also be proven mathematically by realizing that at equilibrium ∆G=0. And using the
equation ∆G=-nFE, E must be 0 if ∆G=0
1998 REVIEW #17
Electrochemistry
Answer the following questions regarding the electrochemical cell shown to the right.
a. Write the net-ionic equation for the spontaneous reaction that occurs as the cell
operates, and determine the cell voltage.
Ag+ +1e-  Ag
0.80 V
Cd  Cd2+ + 2e+0.40 V
2Ag+ + Cd Cd2+ +2Ag 1.20V
b. In which direction do anions flow in the salt bridge as the cell operates? Justify
your answer.
Since the electrons are leaving the Cd side(Anode) that is making the silver
side(cathode) more negative. This means that the cations are flowing
toward the Silver side and the anions are flowing through the salt bridge
towards the Cd side.
c. If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what
will happen to the cell voltage? Explain.
Adding additional AgNO3 is adding Ag+ to the reaction. This follows LeChatlier’s principle and adding more
silver ions (a reactant) will drive the reaction to the right and cause the products to be favored and the
voltage will go UP.
d. If 1.0 gram of solid NaCl is added to each half-cell, what will happen to the cell voltage? Explain.
NaCl will react with the Ag+ (Ag+ + Cl-  AgCl). This will take away Ag+ ions which will lower the [Ag+], thus
driving the reaction to the right, favoring the reactants. This will lower the value of the voltage. Cl- does not
react with the Cd2+.
Note: this could also be explained using the Nernst equation and what happens to Q with lower amounts of
Ag+
f.
If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.
This would lower the concentrations of each. This must be understood by looking at the Nernst Equation.
Ecell  Ecell 
RT  [Cd 2  ] 

ln 
nF  [ Ag1 ]2 
The trick on this one has to do with the fact that the [Ag+] is squared. This will cause
the value of Q to be larger since you are now dividing by a smaller number squared.
This will make the value of ln(Q) to be smaller and make the value of Ecell to be larger.
2001 REVIEW #18
Electrochemistry
Answer the following questions that refer to the galvanic cell shown in the diagram
the right. (A table of standard reduction potentials is printed on the green insert
and on page 4 of the booklet with the pink cover.)
a. Identify the anode of the cell and write the half-reaction that occurs
there.
b. Write the net ionic equation for the overall reaction that occurs as the
cell operates and calculate the value of the standard cell potential, o cell E .
c. Indicate how the value of Ecell will be affected if the concentration of
Ni(NO3)2(aq) was changed from 1.0M to 0.10M and the concentration of
Zn(NO3)2(aq) remained at 1.0M. Justify your answer.
d. Specify whether the value of Keq for the cell reaction is less than 1,
greater than 1, or equal to 1. Justify your answer.
(a) The anode is the electrode on the right (Zn is the anode) 1 point
Point is also earned if the student points to the Zn cell in the diagram.
The half-reaction is Zn Zn2+ + 2 e 

1 point
(b) Zn + Ni2+ Zn2+ + Ni
1 point
o
E cell = (0.25 V) (0.76 V) = 0.51 V
1 point
Some work must be shown to support the answer.
(c) Ecell would decrease
1 point
Since Ni2+ is a reactant, a decrease in its concentration decreases the 1 point
driving force for the forward reaction
or
[ Zn 2  ]
Ecell = ERTn^ln Q , where Q 

[ Ni 2  ]
Decreasing the [Ni2+] would increase the value of Q , so a larger number
would be subtracted from E, thus decreasing the value of Ecell .
(d) K > 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.
to
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 + 2eb. 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+
2.00hrs 3600 sec 0.250C 1molee 1molFe 56 gFe
x
x
x
x
 0.522 gFe
1
1hr
1sec
96485C 2mole 1molFe
2.00hrs 3600 sec 0.250C 1molee 1molFe 56 gFe
x
x
x
x
 0.348 gFe
1
1hr
1sec
96485C 3mole 1molFe
Version Bergmann
2.00hrs 3600 sec 0.250C 1molee 1molCl2
x
x
x
 0.00932molCl2
1
1hr
1sec
96485C 2mole 
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 + 2e2Fe3+ + 6Cl-  3Cl2 + 2Fe
Then: 0.521gFe x 1molFe x 3molCl2  0.0139molCl
2
1
56 gFe
2molFe
Note that this is NOT the same moles of Cl2 from above.
So it could also be….
Fe2+ + 2e-  Fe
2Cl-  Cl2 + 2eFe3+ + 2Cl-  Cl2 + Fe
0.521gFe 1molFe 3molCl2
x
x2
 0.0093molCl2
1
56 gFe
2molFe
These moles of Cl2 match the moles based on the electrochem data so the formula of the Chloride is FeCl 2
c. Write the balanced equation for the overall reaction that occurs in the cell.
Fe2+ + 2e-  Fe
2Cl-  Cl2 + 2eFe3+ + 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
0.521gFe 1molFe 1molCl2
x
x
 0.0093molCl2
1
56 gFe 1molFe
PV  nRT
Latm
(0.987atm)(V )  (0.0093molCl2 )(0.0821 molK
)( 298K )
V  0.231LCl2
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
3.00 gCl2 1molCl2 2mole 96485C
1hr
C
x
x
x
x
 2.26 sec
 2.26 A
1hr
71gCl2 1molCl2 1mole 3600 sec
2000 REVIEW #20
Electrochemistry
Answer the following questions that relate to electrochemical reactions.
a. Under standard conditions at 25°C , Zn(s) reacts with Co (aq) to produce Co(s).
2+
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+ + 2eii.
Write the balanced net-ionic equation for the overall reaction.
Co2+ + 2e-  Co
-0.28V
Zn  Zn2+ + 2e0.76V
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+ + 2eii.
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
1hr 3600 sec 100C 1mole 1molCu 63.55 gCu
x
x
x
x
x
 119 gCu
1
1hr
1sec 96485C 2mole  1molCu
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.
Ea
∆H
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
[N2O5]
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
Time
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.
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
2A+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.
Rate 2 k[0.75] x [0.75] y 1.3 x10 3


Rate1 k[0.25] x [0.75] y 4.3 x10  4
3 x  3 :: x  1
Rate3 k[1.50]1 [1.50] y 5.3 x10 3


Rate 2 k[0.75]1 [0.75] y 1.3 x10 3
(2)( 2 y )  4
2y  2
y 1
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:
M
 k ( M ) 2
min
So to cancel out you will need M -1 min 1
Rate 
c) Determine the initial rate of change of [A] in Experiment 3.
5.3x10 3 M min 1 C 2M min 1 A
x
 0.0106M min 1 A
1
1
1M min C
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.
3.24 x104 M sec 1 NOBr
1M sec 1 Br2
x
 1.61x10 4 M sec 1
1
1
2M sec NOBr
(b) Determine the order of the reaction with respect to each reactant, Br2(g) and NO(g). In
each case, explain your reasoning.
Rate2 k[0.0160]x [0.0240] y 6.38 x104


Rate1 k[0.0160]x [0.0120] y 3.24 x10 4
2 x  2 :: x  1
Rate3 k[0.0320]1[0.0060] y
6.42 x10 4


Rate2 k[0.0160]1[0.0240] y 6.38 x10 34
(2)(0.25 y )  1
0.25 y  0.5
y2
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.
3.24 x104  k[0.0160]1[0.0120]2
k  141
Rate  k[ NO][ Br2 ]2
M
 k M 3
sec
k  M  2 sec 1
k  141M  2 sec 1
(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][Br 2]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 SF 2, 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: PF 3 = 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.
NF5
i.
Doe not exist: There is no 2d orbital for the dsp3 hybridization
AsF5
ii.
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 carbonoxygen 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 CF 4 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 O 2- 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
Propane
Propanone
1-propanol
Formula
CH3CH2CH3
CH3COCH3
CH3CH2CH2OH
∆Hvap (kJ/mol)
19.0
32.0
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);
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).
(d) Given the structural formula for propyne below,
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
 65.60 gC 1molC

C :
x
 5.47 molC / 1.56  3.5 
1
12 gC




9.44 gH 1molH
H :
x
 9.44molH / 1.56  6  x 2  C7 H12O2
1
1gH




25 gO 1molO
 O :

x
 1.56molO / 1.56  1
1
16 gO


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.)
T f  iK f m
15.2C  (1)( 40.0 KgK
)( m)
mol
0.38molCompound 0.01608kgCamphor
x
 0.00611molCompound
1kgCamphor
1
grams
1.570 g
MolarMass 

 256 g / mol
moles 0.00611mol
EmpiricalM assofC7 H12O2  128 g / mol
Therefore :: C7 H12O2 x 2  C14 H 24O4
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?
PV  nRT
PV
(1.00atm)(0.577 L)
n

 0.0123mol
Latm
RT (0.0821 molK
)(573K )
MolarMass 
grams
1.570 g

 128 g / mol
moles 0.0123mol
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
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+ + 2Clc) 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 + OHThis is a base. CH3COOH is an acid (low pH), etc…
2001 REVIEW #33
Colligative properties
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 + OHc) 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 Cl 1- (aq) ion? Identify the product of the
oxidation.
The KMnO4
2Cl-  Cl2 + 2eMnO4-  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
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
__ gBaSO4 1molBaSO4 1molNa2 SO4 MMNa2 SO4
x
x
x
 massNa2 SO4
1
MMBaSO4 1molBaSO4 1molNa2 SO4
massNa2 SO4
x100  % Na2 SO4inMixture
massMixture
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.
(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.
g
gramsSolute

mol
molSolute
T f  K f m
MM 
Solve for Molality and then determine the moles of the solute by:
T f
Kf
m
__ molSolute __ kgSolvent
x
 molesSolut e
kgSolvent
1
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 2 nd 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.
126
%error 

g
 120 mol
x100
g
126 mol
g
mol
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
__ gKHP 1molKHP 1molNaOH
1
x
x
x
1
MMKHP 1molKHP __ LNaOH
__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.
(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.
H2A  HA- + H+
HA-  H+ + A2A2- is present at this equvilance point
1999 REVIEW #38
Lab
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.
PercentError 
(64  58)
x100
58
(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) .
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 NH 4+ (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