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CHEM 1515.001 Name ________________________ Exam III John III. Gelder TA's Name ________________________ November 7, 2001 Lab Section _______
INSTRUCTIONS :
- This examination consists of a total of 9 different pages. The last three pages include a periodic table, some useful mathematical equations, a solubility table and a table of equilibrium constants. All work should be done in this booklet.
- 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.
- 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 or short answer questions.
- No credit will be awarded if your work is not shown in problems 3, 5, 6b and 7.
- Point values are shown next to the problem number.
- 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.
- Look through the exam before beginning; plan your work; then begin.
8. ReRellaaxx^ and do well.
Page 2 Page 3 Page 4 Page 5 Page 6 TOTAL
SCORES _____ _____ _____ _____ _____ ______ (23) (20) (30) (9) (16) (100)
(9) 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. Soluble ionic compounds should be written in the form of their component ions.
a) HI (aq) + RbOH (aq) → H 2 O (l) + Rb + (aq) + I – (aq)
b) 3 Na 2 S (aq) + 2 Fe(NO 3 ) 3 (aq) → Fe 2 S 3 (s) + 6Na + (aq) + 6NO 3 – (aq)
c) C 6 H 5 NH 2 (aq) + HCN (aq) → C 6 H 5 NH 3 + (aq) + CN – (aq)
(8) 2. Write the ionic and net ionic chemical equations for 1a).
1a) Ionic equation:
H + (aq) + I – (aq) + Rb + (aq) + OH – (aq) → H 2 O (l) + Rb + (aq) + I – (aq)
Net Ionic equation:
H + (aq) + OH – (aq) → H 2 O (l)
(6) 3. Give the name or draw the Lewis structure for each of the following compounds.
C
H
H
H
C
H
H
C H H
C
H
C
H
C
H
H
C
H
H H
trans-3-heptene
4,5-dimethyl-2-hexyne
C
H
H
H
C C^
C
H C H
H
H
C H
C
H
H H
C
H
H
H
2,3-dimethyl-2-butene
C
H
H
H
C
H
H
H
C C
C H
H H
C
H
H
H
(30) 5. Calculate the pH for each of the following solution;
a) 0.00360 M Ba(OH) (^2)
Ba(OH) 2 → Ba 2+^ + 2OH - I 0.0036 0 ~ F 0 .0036. NaOH is a strong base and completely dissociates in water. Therefore,
[OH - ] = 0.0072 M and pOH = 2.14, pH = 11.
b) 0.750 M HC (^) 3H5O2 (propionic acid)
HC 3 H 5 O (^2) ääää Η +^ + C 3 H 5 O 2 – I .750 ~0 0 C -x +x +x x = [H C 3 H 5 O 2 ] (^) D E .750 – x x x
K (^) a =
[H +^ ][C 3 H 5 O 2 –^ ]
[HC 3 H 5 O 2 ] = 1.3 x 10^
-
1.3 x 10 -5^ =
(x)(x) (0.750 - x) Assume 0.750 – x = 0.
1.3 x 10 -5^ =
x 2 0.750 =
1.3 x 10 -5^ =
x 2 0.750 = x = 0.0031 M = [ Η + ] : pH = 2.
c) 0.218 M (CH 3 ) 3 N
(CH 3 ) 3 N + H 2 O ääää (CH 3 ) 3 NH + (aq) + OH - (aq) I 0.218 M - 0 0 C -x - +x +x let x = [ (CH 3 ) 3 N ] (^) R E 0.218 – x - x x
K (^) b =
[(CH 3 ) 3 NH +^ ][OH-]
[(CH 3 ) 3 N ]
6.4 x 10 -5^ =
(x) (x) 0.218 - x assume x << 0.
6.4 x 10 -5^ =
x 2 . 3.7 x 10 -3^ M = x = [OH -^ ] pOH = 2. pH = 11.
Short Answer: Parts a – d are worth 4 points, part e is worth 6 points and part f is worth 9 points.
(20) 6a. Identify the acid, base, conjugate acid and conjugate base in the following chemical equation,
HPO 24 – (aq) + HC2H3O2(aq) → H (^) 2PO 4 – (aq) + C 2 H 3 O 2 – (aq) base acid conjugate acid conjugate base
b) K for the reaction in 6a above has a value of 290. Identify the stronger acid and base in the reaction.
acid HC 2 H 3 O 2 base HPO2– 4
c) The pH of a 5.50 x 10-4^ M HClO 3 solution is about 3.30. Is HClO 3 a strong acid or a weak acid? Explain.
For a pH of 3.30 the [ Η + ] = 5.01 x 10 -4^. The concentration of [ Η + ] is nearly the same as the initial concentration of HClO 3. The percent ionization is;
5.01 x 10 - 5.50 x 10 -^
x 100 = 91%
So nearly all of the has HClO 3 dissociated, so HClO 3 is a strong acid.
d) Identify the chemical specie(s) in the highest concentration in 6c. (Note: excluding water.)
[ Η + ] and [ClO 3 –^ ] are at the largest concentrations in the solution.
e) Identify each of the following substances as an acid or a base and write a chemical equation that describes the acid or base character. i) (CH3)2NH 2 +(aq)
(CH 3 ) 2 NH 2 +(aq) ääää H +(aq) + CH 3 ) 2 NH (aq) acid ii) C 4 H 7 O 2 – (aq)
C 4 H 7 O 2 – (aq) + H 2 O (aq) ääää HC 4 H 7 O (^) 2(aq) + OH – (aq) base
90
91
Th Pa
92
93
94
(244)
95
(243)
96
(247)
U Np Pu Am Cm
97
(247)
98
(251)
99
(252)
100
(257)
101
(258)
103
(260)
102
(259)
Bk Cf Es Fm Md No Lr
58
59
Ce Pr
60
61
(145)
62
63
64
Nd Pm Sm Eu Gd
65
66
67
68
69
70
Tb Dy Ho Er Tm Yb
37
Rb
38
Sr
39
Y
40
Zr
41
42
43
(98)
44
45
46
47
Nb Mo Tc Ru Rh Pd Ag
48
49
50
51
52
54
53
Cd In Sn Sb Te I Xe
3
4
Li Be
11
12
Na Mg
19
K
22
20
Ca
21
Sc Ti
23
24
25
26
27
28
29
V Cr Mn Fe Co Ni Cu
30
31
32
33
34
36
35
Zn Ga Ge As Se Br Kr
5
6
B C
8
O
9
F
7
N
10
Ne
13
14
Al Si
16
S
17
Cl
15
P
18
Ar
2
He
55
Cs
56
Ba
57
La
72
Hf
73
74
Ta W
75
76
77
78
79
Re Os Ir Pt Au
80
81
82
83
84
(209)
86
(222)
85
(210)
Hg Tl Pb Bi Po At Rn
1
H
IA
IIA IIIA IVA VA VIA VIIA
VIIIA
IIIB IVB VB VIBVIIB VIII IB IIB
Lanthanides
Actinides
Periodic Table of the Elements
87
(223)
Fr
88
Ra
89
Ac
104
(261)
105
(262)
106
(263)
71
Lu
Rf Db Sg
108
(265)
109
(266)
107
(262)
Bh Hs Mt
Useful Information
Kw = 1.0 x 10-14^ Kp = Kc(RT )∆n
pH = –log[H +^ ] pH + pOH = 14
x 1 ,2 =
–b± b^2 – 4ac 2a for ax^
(^2) + bx + c = 0
6.023 x 10^23
R = 0.
L·atm mol·K = 8.^
J
mol·K
E.1 DISSOCIATION CONSTANTS FOR ACIDS AT 25 ˚C
E.2 DISSOCIATION CONSTANTS FOR BASES AT 25˚C
- Name Formula Ka1 Ka2 Ka
- Acetic HC 2 H 3 O 2 1.8 x 10– - 6 H 7 O^6 8.0 x Ascorbic HC - –
- 3 AsO^4 5.6 x Arsenic H - – 3 1.0 x 10– 7 3.0 x 10 –
- 3 AsO^3 6.0 x Arsenous H - –
- Benzoic HC 7 H 5 O 2 6.5 x 10 –
- Boric H 3 BO 3 5.8 x 10 –
- Butyric acid HC 4 H 7 O 2 1.5 x 10 –
- Carbonic H 2 CO 3 4.3 x 10 – 7 5.6 x 10 –
- Cyanic HCNO 3.5 x 10 –
- Citric H 3 C 6 H 5 O 7 7.4 x 10 – 4 1.7 x 10 – 5 4.0 x 10 –
- Formic HCHO 2 1.8 x 10–
- Hydroazoic HN 3 1.9 x 10 –
- Hydrocyanic HCN 4.9 x 10 –
- Hydrofluoric HF 7.2 x 10–
- Hydrogen chromate ion HCrO 4 – 3.0 x 10 –
- Hydrogen peroxide H 2 O 2 2.4 x 10 –
- Hydrogen selenate ion HSeO 4 – 2.2 x 10 –
- Hydrogen sulfate ion HSO 4 – 1.2 x 10 –
- Hydrogen sulfide H 2 S 5.7 x 10 – 8 1.3 x 10 –
- Hypobromous HBrO 2.0 x 10 –
- Hypochlorous HClO 3.0 x 10 –
- Hypoiodus HIO 2.0 x 10 –
- Iodic HIO 3 1.7 x 10 –
- Lactic HC 3 H 5 O 3 1.4 x 10 –
- Malonic H 2 C 3 H 2 O 4 1.5 x 10 – 3 2.0 x 10 –
- Oxalic H 2 C 2 O 4 5.9 x 10 – 2 6.4 x 10 –
- Nitrous HNO 2 4.5 x 10–
- Phenol HC 6 H 5 O 1.3 x 10 –
- Phosphoric H 3 PO 4 7.5 x 10 – 3 6.2 x 10 – 8 4.2 x 10 –
- Paraperiodic H 5 IO 6 2.8 x 10 – 2 5.3 x 10 –
- Propionic HC 3 H 5 O 2 1.3 x 10 –
- Pyrophosphoric H 4 P 2 O 3.0 x 10 – 2 4.4 x 10 –
- Selenous H 2 SeO 3 2.3 x 10 – 3 5.3 x 10 –
- Sulfuric H 2 SO 4 strong acid 1.2 x 10 –
- Sulfurous H 2 SO 3 1.7 x 10 – 2 6.4 x 10 –
- Tartaric H 2 C 4 H 4 O 6 1.0 x 10 –3 4.6 x 10 –
- Ammonia NH 3 1.8 x 10 – 5 Hydroxylamine HONH 2 1.1 x 10 – Name Formula K b Name Formula K b
- Aniline C 6 H 5 NH 2 4.3 x 10 –10 Methylamine CH 3 NH 2 4.4 x 10 –
- Dimethylamine (CH 3 ) 2 NH 5.4 x 10 –
- Ethylamine C 2 H 5 NH 2 6.4 x 10 – 4 Trimethylamine (CH 3 ) 3 N 6.4 x 10 –
- Hydrazine H 2 NNH 2 1.3 x 10 –
