CHEM 10123/10125, Quiz 4
February 29, 2012
Name_____________________
(please print)
Significant figures and correct units count, show charges as appropriate, and please box your answers!
Periodic Table of the Elements
IA
VIIIA
(1)
(18)
1
1
2
3
4
5
6
7
2
H
IIA
IIIA
IVA
VA
VIA
VIIA
He
1.0080
(2)
(13)
(14)
(15)
(16)
(17)
4.0026
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.9410
9.0122
10.811
12.011
14.007
15.999
18.998
20.179
11
12
13
14
15
16
17
18
Na
Mg
IIB
Al
Si
P
S
Cl
Ar
22.990
24.305
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
26.982
28.086
30.974
32.066
35.453
39.948
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.098
40.078
44.956
47.880
50.942
51.996
54.938
55.847
58.933
58.690
63.546
65.380
69.723
72.610
74.922
78.960
79.904
83.800
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
85.468
87.620
88.906
91.224
92.906
95.940
98.907
101.07
102.91
106.42
107.87
112.41
114.82
118.71
121.75
127.60
126.90
131.29
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
132.91
137.33
138.91
178.49
180.95
183.85
186.21
190.20
192.22
195.09
196.97
200.59
204.38
207.20
208.98
208.98
209.99
222.02
87
88
89
104
105
106
107
108
109
110
111
112
IIIB
IVB
VB
VIB
VIIB. . . . . . . . . . VIIIB . . . . . . . . . .IB
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
Ds Uuu Uub
223.02
226.03
227.03
(261)
(262)
(266)
(264)
(270)
(268)
(281)
114
Uug
58
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
140.12
140.91
144.24
145.91
150.36
151.97
157.25
158.93
162.50
164.93
167.26
168.93
173.04
174.97
95
96
103
90
91
92
93
94
Th
Pa
U
Np
Pu
232.04
231.04
238.03
237.05
244.06
Am Cm
243.06
247.07
97
98
99
100
101
102
Bk
Cf
Es
Fm
Md
No
Lr
247.07
242.06
252.08
257.10
258.10
259.10
260.11
1. (6 points) For each reaction below, a) circle the stronger acid and base, b) indicate whether
the equilibrium will lie to the left or the right, and c) indicate whether the equilibrium constant K
will be greater than or less than 1.
H2S(aq) + HSe–(aq) ⇌ H2Se(aq) + HS–(aq)
equil. to the
left or to the
right?
left
K<1
or
K>1?
K<1
H2Te(aq) + I–(aq) ⇌ HTe–(aq) + HI(aq)
left
K<1
H3AsO4(aq) + H2PO4–(aq) ⇌ H2AsO4–(aq) + H3PO4(aq)
left
K<1
2. (5 points) Use Lewis structures to diagram the reaction, and identify the Lewis acid and
Lewis base. Use arrows to show the movement of electrons.
BF3(aq) + CN–(aq)  NCBF3–(aq)
Kb Values for Selected Weak Molecular Bases at 25 °C
Name of Base
Formula
Butylamine
C4H9NH2
Methylamine
CH3NH2
Ammonia
NH3
Hydrazine
N2H4
Kb
5.9 x 10–4
4.4 x 10–4
1.8 x 10–5
1.7 x 10–6
3. (7 points) SHOW ALL WORK. A student prepared a buffer solution by dissolving 0.12 mol
CH3NH2 and 0.095 mol CH3NH3Cl in 250 mL of water. What is the pH of the buffer?
CH3NH3+(aq) ⇌ CH3NH3(aq) + H+(aq)
Kb = 4.4 x 10–4, so Ka = (1.0 x 10–14)/(4.4 x 10–4) = 2.2727 x 10–11
Ka = [H+][A–]/[HA] = [H+][CH3NH2]/[CH3NH3+]
0.12/0.250 = 0.48 M CH3NH2
0.095/0.250 = 0.38 M CH3NH3+
“HA”
“A–”
H+
I 0.38 M
0.48
0
C -x
+x
+x
E 0.38 - x M
0.48 + x
x
Assume x is small.
2.27 x 10–11 = x(0.48)/(0.38), so x = 1.8 x 10–11
(yes, x is small assumption works; it’s many orders of magnitude
smaller than conc)
–log x = 10.75 = pH
4. (4 points) You have the following reagents on hand:
Solids (pKa of acid form is given)
Benzoic acid (4.19)
Sodium acetate (4.74)
Potassium fluoride (3.14)
Ammonium chloride (9.26)
Solutions
5.0 M HCl
1.0 M Acetic acid (4.74)
2.6 M NaOH
1.0 M HOCl (7.46)
What combinations of reagents would you use to prepare buffers at the following pH values?
a. 3.0
KF + HCl
b. 7.0 HOCl + NaOH
5. (8 points) SHOW ALL WORK. Suppose 30.0 mL of 0.100 M HCl is added to a 250 mL
portion of a buffer composed of 0.250 M HIO3 (iodic acid) and 0.200 M NaIO3 (sodium iodate).
Give the new concentrations of HIO3 and IO3– ions after the addition of the strong acid.
30.0 mL HCl soln x (0.100 mol H+)/(1000 mL HCl soln) = 0.00300
mol H+
H+(aq) + IO3–(aq)  HIO3(aq)
(0.250 mL) x (0.250 mol HIO3/1000 mL HIO3 soln) = 0.0625 mol HIO3
(0.250 mL) x (0.200 mol NaIO3/1000 mL NaIO3 soln) = 0.0500 mol
NaIO3
0.0625 mol HIO3 + 0.00300 mol H+ = 0.0655 mol HIO3
0.0500 mol NaIO3 – 0.00300 mol H+ = 0.0470 mol NaIO3
New volume = 280 mL = 0.280 L
(0.0655 mol HIO3)/(0.280L) = 0.233 M HIO3
(0.0470 mol NaIO3)/(0.280L) = 0.168 M NaIO3
CHEM 10123/10125, Quiz 4
February 29, 2012
Name_____________________
(please print)
Significant figures and correct units count, show charges as appropriate, and please box your answers!
Periodic Table of the Elements
IA
VIIIA
(1)
(18)
1
1
2
3
4
5
6
7
2
H
IIA
IIIA
IVA
VA
VIA
VIIA
He
1.0080
(2)
(13)
(14)
(15)
(16)
(17)
4.0026
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.9410
9.0122
10.811
12.011
14.007
15.999
18.998
20.179
11
12
13
14
15
16
17
18
Na
Mg
IIB
Al
Si
P
S
Cl
Ar
22.990
24.305
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
26.982
28.086
30.974
32.066
35.453
39.948
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.098
40.078
44.956
47.880
50.942
51.996
54.938
55.847
58.933
58.690
63.546
65.380
69.723
72.610
74.922
78.960
79.904
83.800
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
85.468
87.620
88.906
91.224
92.906
95.940
98.907
101.07
102.91
106.42
107.87
112.41
114.82
118.71
121.75
127.60
126.90
131.29
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
132.91
137.33
138.91
178.49
180.95
183.85
186.21
190.20
192.22
195.09
196.97
200.59
204.38
207.20
208.98
208.98
209.99
222.02
87
88
89
104
105
106
107
108
109
110
111
112
IIIB
IVB
VB
VIB
VIIB. . . . . . . . . . VIIIB . . . . . . . . . .IB
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
Ds Uuu Uub
223.02
226.03
227.03
(261)
(262)
(266)
(264)
(270)
(268)
(281)
114
Uug
58
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
140.12
140.91
144.24
145.91
150.36
151.97
157.25
158.93
162.50
164.93
167.26
168.93
173.04
174.97
95
96
103
90
91
92
93
94
Th
Pa
U
Np
Pu
232.04
231.04
238.03
237.05
244.06
Am Cm
243.06
247.07
97
98
99
100
101
102
Bk
Cf
Es
Fm
Md
No
Lr
247.07
242.06
252.08
257.10
258.10
259.10
260.11
1. (6 points) For each reaction below, a) circle the stronger acid and base, b) indicate whether
the equilibrium will lie to the left or the right, and c) indicate whether the equilibrium constant K
will be greater than or less than 1.
H2S(aq) + Br–(aq) ⇌ HS–(aq) + HBr(aq)
equil. to the
left or to the
right?
left
K<1
or
K>1?
K<1
PH3(aq) + NH2–(aq) ⇌ NH3(aq) + PH2–(aq)
right
K>1
H2SeO4(aq) + HSeO3–(aq) ⇌ H2SeO3(aq) + HSeO4–(aq)
right
K>1
2. (5 points) Use Lewis structures to diagram the reaction, and identify the Lewis acid and
Lewis base. Use arrows to show the movement of electrons.
GaCl3(aq) + CH3O–(aq)  CH3OGaCl3–(aq)
3. (7 points) SHOW ALL WORK. A student prepared a buffer solution by dissolving 0.12 mol
N2H4 and 0.095 mol N2H5Cl in 250 mL of water. What is the pH of the buffer?
N2H5+(aq) ⇌ N2H4(aq) + H+(aq)
Kb = 1.7 x 10–6, so Ka = (1.0 x 10–14)/(1.7 x 10–6) = 5.882 x 10–9
Ka = [H+][A–]/[HA] = [H+][N2H4]/[N2H5+]
0.12/0.250 = 0.48 M N2H4
0.095/0.250 = 0.38 M N2H5+
“HA”
“A–”
H+
I 0.38 M
0.48
0
C -x
+x
+x
E 0.38 - x M
0.48 + x
x
Assume x is small.
5.882 x 10–9 = x(0.48)/(0.38), so x = 4.657 x 10–9
(yes, x is small assumption works; it’s many orders of magnitude
smaller than conc)
–log x = 8.33 = pH
4. (4 points) You have the following reagents on hand:
Solids (pKa of acid form is given)
Benzoic acid (4.19)
Sodium acetate (4.74)
Potassium fluoride (3.14)
Ammonium chloride (9.26)
Solutions
5.0 M HCl
1.0 M Acetic acid (4.74)
2.6 M NaOH
1.0 M HOCl (7.46)
What combinations of reagents would you use to prepare buffers at the following pH values?
(You do not need to include amounts)
a. 5.0 Sodium acetate + HCl
Or Na acetate + acetic acid
Or acetic acid + NaOH
b. 9.0 Ammonium chloride + NaOH
5. (8 points) SHOW ALL WORK. Suppose 28.0 mL of 0.100 M HCl is added to a 300 mL
portion of a buffer composed of 0.300 M HN3 (hydrazoic acid) and 0.250 M NaN3 (sodium
azide). Give the new concentrations of HN3 and N3– ions after the addition of the strong acid.
28.0 mL HCl soln x (0.100 mol H+)/(1000 mL HCl soln) = 0.00280
mol H+
H+(aq) + N3–(aq)  HN3(aq)
(0.300 mL) x (0.300 mol HN3/1000 mL HN3 soln) = 0.0900 mol HN3
(0.300 mL) x (0.250 mol NaN3/1000 mL NaN3 soln) = 0.0750 mol NaN3
0.0900 mol HN3 + 0.00280 mol H+ = 0.0928 mol HN3
0.0750 mol NaN3 – 0.00280 mol H+ = 0.0722 mol NaN3
New volume = 328 mL = 0.328 L
(0.0928 mol HN3)/(0.328L) = 0.283 M HN3
(0.0722 mol NaN3)/(0.328L) = 0.220 M NaN3