CHEM 1225
Exam III
John III. Gelder
April 8, 1999
Name ________________________
TA's Name ________________________
Lab Section _______
INSTRUCTIONS:
1. This examination consists of a total of 7 different pages.
The last two pages includes a periodic table and a solubility
table. All work should be done in this booklet.
2. PRINT your name, TA's name and your lab section
number now in the space at the top of this sheet. DO
NOT SEPARATE THESE PAGES.
3. Answer all questions that you can and whenever called
for show your work clearly. Your method of solving
problems should pattern the approach used in lecture.
You do not have to show your work for the multiple
choice (if any) or short answer questions.
4. No credit will be awarded if your work is not shown in
problems 3 – 5, 7, 8, 10 and 11.
5. Point values are shown next to the problem number.
6. Budget your time for each of the questions. Some
problems may have a low point value yet be very
challenging. If you do not recognize the solution to a
question quickly, skip it, and return to the question after
completing the easier problems.
7. Look through the exam before beginning; plan your
work; then begin.
8. R e l a x and do well.
SCORES
Page 2
Page 3
Page 4
Page 5
Page 6
TOTAL
_____
(20)
_____
(22)
_____
(22)
_____
(18)
_____
(18)
______
(100)
CHEM 1225 EXAM II
PAGE 2
(12) 1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify all
products phases as either (g)as, (l)iquid, (s)olid or (aq)ueous.
(8)
Pb(NO3)2(aq →
a)
K2CrO4(aq) +
b)
H2SO4(aq) +
2KOH (aq)
c)
NaOH(aq) +
HC2H 3O 2(aq) → NaC2H 3O 2(aq) + H2O (aq)
d)
CaCO3(s) +
HCl(aq) → CaCl2(aq) + CO2(aq) + H2O (aq)
→
PbCrO4(s) + 2KNO3(aq)
K 2SO4(aq) + 2H2O (aq)
2. Write the balanced ionic and balanced net ionic chemical equations for any two of the reactions in
Problem 1. (Remember to include the correct charges on all ions and the phase of each species.)
1a, 1b, 1c or 1d)
Ionic equation:
2K + (aq) + CrO4 2– (aq) + 2NO3 – (aq) + Pb2+ (aq) → PbCrO4 (s)
+ 2K+ (aq) + 2NO3 – (aq)
Net Ionic equation:
CrO 4 2– (aq) + Pb2+ (aq) → PbCrO4 (s)
1a, 1b or 1c)
Ionic equation:
2H + (aq) + SO4 2– (aq) + 2OH– (aq) + K+ (aq) → 2K+ (aq) + SO4 2– (aq) + 2H2 O (aq)
Net Ionic equation:
H + (aq) + OH– (aq) → H2 O (aq)
1a, 1b or 1c)
Ionic equation:
H + (aq) + C2 H 3 O 2 – (aq) + OH– (aq) + Na+ (aq) → Na+ (aq) + C2 H 3 O 2 – (aq) + H2 O (aq)
Or more correct
HC 2 H 3 O 2 – (aq) + OH– (aq) + Na+ (aq) → Na+ (aq) + C2 H 3 O 2 – (aq) + H2 O (aq)
Net Ionic equation:
H + (aq) + OH– (aq) → H2 O (aq)
Or more correct
HC 2 H 3 O 2 – (aq) + OH– (aq) → C2 H 3 O 2 – (aq) + H2 O (aq)
CHEM 1225 EXAM II
PAGE 3
(12) 3. Describe how you would prepare;
a) 500.00 mls of a 1.25 M Na2SO4 solution.
1.25 mol
0.500 L  1 Liter  = 0.625 mol Na2 S O 4
142 g
0.625 mol  1 m o l = 88.8 gm Na2 S O 4
Wearing your safety goggles, add the Na2 SO 4 solid to approximately 300 mL of
water and be sure the solid has completely dissolved. Then add enough water to
obtain 500.0 mL solution. (Note: we do not have to worry about how much
water is added, only the final volume.)
b) 1 liter of 0.733 M KMnO4 from a solution which is 1.39 M KMnO4.
M 1 V 1 = M2 V 2
M 2V 2
0.733 M·1 L
V 1 =  M 1  =  1 . 3 9 M  = 527 mLs
Add enough water to 527 mLs of 1.39 M KMnO4 to get 1 L of
0.733 M solution.
(10) 4. Some sulfuric acid is spilled on a bench top in the laboratory. Sodium hydrogen carbonate is sprinkled
on the spill. The balanced equation describing the reaction which takes place,
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + 2H2O(l)
Calculate the mass of sodium hydrogen carbonate that must be weighed out to react with 400. mLs of
6.00 M H2SO4 that was spilled.
1 L
6 . 0 0 m o l H 2 S O 4 
400.0 mLs  1 0 0 0 m L 
1 L
 = 2.40 mol H2 S O 4
2 mol NaHCO 3 8 4 g N a H C O 3
2.40 mol H2 SO 4  1 m o l H 2 S O 4   1 m o l N a H C O 3  = 403 g NaHCO3
CHEM 1225 EXAM II
PAGE 4
(10) 5. Nitric acid reacts with barium hydroxide according to the equation;
2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O(g)
Calculate the volume of 0.259 M nitric acid required to exactly neutralize 18.0 mLs of 0.185 M barium
hydroxide.
1 L
 0.185 mol Ba(OH) 2   2 m o l H N O 3  

0.0180 L 
1 L
  1 mol Ba(OH) 2   0 . 2 5 9 m o l H N O 3  = 0.0260 L HNO3
(4)
6. Write the equilibrium expression for each of the following chemical equations;
a)
K =
[NCl 3 ] 2
[N 2 ][Cl 2 ] 3
b)
[CO 2 ] 8 [H 2 O] 1 8
K=
[O 2 ] 1 7
(8) 7. The equation describing the industrial preparation of ammonia is;
Suppose that a reaction mixture at a given temperature, at equilibrium, was analyzed and found to
contain 3.45 x 10–4 M NH3, 8.17 x 10–4 M N2, and 0.580 M H2. Calculate the magnitude of the
equilibrium constant for the reaction.
[ N H 3]2
[N 2 ][H 2 ] 3
[3.45 x 10–4]2
K =
[ 8.17 x 10–4][ 0.580]3
K = 7.47 x 10– 4
K =
CHEM 1225 EXAM II
PAGE 5
(10) 8. When 0.981 moles of NO, 0.483 moles of Cl2 and 0.400 moles of NOCl are sealed in a 1.00 L flask at
220 ˚C, the following equilibrium is established,
After the mixture achieves equilibrium analysis shows the concentration of NOCl to be 0.222 M.
Calculate K for the reaction.
Initial
Change
Equilibrium
0.981
0.483
0.400
0.222
Since 0.222 moles of NOCl are formed at equilibrium, 0.178 moles of NOCl must
have reacted. So 0.178 moles is the change in NOCl. We now use stoichiometry to
calculate the change of NO and Cl2 .
2 mol NO
0.178 mol NOCl  2 m o l N O C l = 0.178 mol NO
 1 mol Cl2 
0.178 mol NOCl  2 m o l N O C l = 0.089 mol Cl2
Remember since NOCl reacted, NO and Cl2 are formed.
Now the ICE table looks like
Initial 0 . 9 8 1 0 . 4 8 3
Change + 0 . 1 7 8 + 0 . 0 8 9
Equilibrium
1.16 0.661
0.400
-0.178
0.222
[NOCl] 2
[Cl 2 ][NO] 2
[0.222] 2
K =
[0.661][1.16] 2
K = 0.055
K =
(8)
9. For the system
How will the [NH3] be effected (increase, decrease or no change) when the equilibrium is disturbed
by;
a) Removal of O2
increase
b) Addition of H2O increase
c) Increase in temperature
increase
d) Decrease in the volume of the reaction container increase
CHEM 1225 EXAM II
PAGE 6
(9) 10. Calculate the solubility of PbCO3 in pure water. Ksp = 1.0 x 10–13
I
C
E
-
0
+x
x
0
+x
x
x = solubility of PbCO3
K sp = [Pb2+ ][CO 3 2– ]
1.0 x 10–13 = (x)(x)
1.0 x 10– 1 3 = x2
3.16 x 10–7 M = x = solubility of PbCO3
(9) 11. The concentration of Mg2+ in a saturated solution of Mg(OH)2 is 1.16 x 10–4 M. Calculate the
magnitude of the equilibrium constant, Ksp, for Mg(OH)2.
I
C
E
-
0
0
+1.16 x 10– 4 2(1.16 x 10 – 4 )
1.16 x 10 – 4
2.32 x 10 – 4
K sp = [Mg2+ ][OH – ] 2
K s p = [1.16 x 10– 4 ][ 2.32 x 10– 4 ] 2
K sp = 6.24 x 10– 1 2
CHEM 1225 EXAM II
Periodic Table of the Elements
IA
1
VIIIA
1
2
H
He
IIA
1.008
3
2
PAGE 7
IIIA IVA VA VIA VIIA 4.00
4
Li Be
6.94 9.01
11
12
3
Na Mg
22.99 24.30
19
20
4
5
6
7
5
6
7
8
9
10
B
C
N
O
F
Ne
10.81 12.01 14.01 16.00 19.00 20.18
13
14
15
16
17
18
IIIB IVB VB VIB VIIB
21
22
K Ca Sc Ti
23
24
25
VIII
26
27
IB
28
Al
Si
P
S
Cl Ar
31
32
33
34
35
IIB 26.98 28.09 30.97 32.06 35.45 39.95
29
30
36
V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb Sr
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
I
Xe
85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs Ba La Hf Ta W Re Os Ir
Pt Au Hg Tl Pb Bi Po At Rn
132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)
87
88
89 104 105 106 107 108 109
Fr Ra Ac Rf Db Sg Bh Hs Mt
(223) 226.0 227.0 (261) (262) (263) (262) (265) (266)
58
Lanthanides
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 (145) 150.4 152.0 157.2 158.9 162.5 164.9 167.3 168.9 173.0 175.0
90
91
92
93
94
95
96
97
98
99 100 101 102 103
Actinides
Th Pa
U
Np Pu Am Cm Bk Cf Es Fm Md No Lr
232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
Ion
Solubility
–
NO3
ClO4
Cl
–
–
SO4
2–
2–
CO3
3–
PO4
2–
CrO4
-OH
S
2–
Na
+
NH4
+
K
+
Solubility Table
Exceptions
soluble
none
soluble
none
soluble
except Ag , Hg2 , *Pb
soluble
except Ca , Ba , Sr , Hg , Pb , Ag
insoluble
except Group IA and NH4
insoluble
except Group IA and NH4
insoluble
except Group IA, IIA and NH4
insoluble
except Group IA, *Ca , Ba , Sr
insoluble
except Group IA, IIA and NH4
soluble
none
soluble
none
soluble
none
+
2+
2+
2+
2+
2+
2+
2+
+
+
+
+
2+
2+
2+
+
*slightly soluble
CHEM 1225 EXAM II
PAGE 8