NAME:_________________________
Spring 2006
INSTRUCTIONS:
1.
Student Number:______________________
Chemistry 2000 Test #3A
____/ 50 marks
1) Please read over the test carefully before beginning. You should have
6 pages of questions, and a data/periodic table sheet (7 pages total).
2) If your work is not legible, it will be given a mark of zero.
3) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
4) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
5) You may use a calculator.
6) You have 60 minutes to complete this test.
The trichlorocadmate(II) complex exists in equilibrium with free cadmium(II) and
chloride ions:
2+
Cd(aq) + 3 Cl(aq)
CdCl3(aq)
Kf = 2.5 x 102
A solution contains 4.7 × 10-3 M Cd2+, 5.1 × 10-3 M Cl- and 6.6 × 10-5 M CdCl3-. Is this
solution at equilibrium? If not, indicate the direction in which the reaction must proceed
to reach equilibrium.
[6 marks]
Q =
[CdCl 3− ]
[Cd 2+ ][Cl − ]3
Q =
(6.6 × 10 -5 )
(4.7 × 10 -3 )(5.1 × 10 -3 ) 3
Q = 1.1 × 10 5
Q and Kf are not equal. Therefore, the system is not at equilibrium.
Q is greater than Kf. Therefore, there are more products (and/or fewer reactants) now
than there would be at equilibrium. As such, the reaction must proceed in reverse to
reach equilibrium.
NAME:_________________________
Student Number:______________________
2.
The pKa of [Co(OH2)6]3+ is 8.9.
[3 marks]
(a)
What is the conjugate base of [Co(OH2)6]3+?
[Co(OH2)5(OH)]2+
(b)
Conjugate base is formed when an acid loses H+.
As such, it has one less hydrogen atom, and the
charge is 1 less positive.
Calculate the pKb of the conjugate base.
pKa + pKb = 14
Therefore:
3.
pKb = 14 – pKa = 14 – 8.9 = 5.1
Choose any amphoteric compound. Draw its structure. Identify what feature(s) allow it
to act as an acid and what feature(s) allow it to act as a base.
[5 marks]
An amphoteric compound is one which can act as either an acid or a base. Common
correct answers included water, bicarbonate, hydrogen phosphate, etc. Note that, in all
these examples, the hydrogen atom(s) is/are bonded to the oxygen atom(s)!!!
..O..
e.g.
.. ..
O
..
-1
C
..O..
H
To be an acid, a compound had to have a hydrogen atom that could be removed as a
proton (H+). This requires that removal of H+ produce a relatively stable species (the
conjugate base). Often, this is presented as the hydrogen atom being attached to an
electronegative atom – since the electronegative atom can bear the residual negative
charge relatively easily.
To be a base, a compound has to be able to accept a hydrogen atom from an acid (by
donating an electron pair to make the base-H bond). This requires that it have a lone pair.
Note that, in bicarbonate (the example above), it’s the lone pair on the oxygen atom with
a negative charge that’s making the molecule act as a base – not the lone pair on the
oxygen atom between C and H.
NAME:_________________________
4.
Student Number:______________________
The figure below represents the titration curve of one compound in Table 1 with another
compound in Table 1. (see data sheet)
[12 marks]
14.00
12.00
10.00
Veq #2 = 40.00 mL
pH
8.00
6.00
pKa2 = 6.5
4.00
Veq #1 = 20.00 mL
2.00
pKa1 = 2.5
0.00
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
50.00
volume titrant added (mL)
(a)
Label the equivalence point volume(s) and pK value(s) on the titration curve.
(b)
Complete the following sentences. You may use either formulae or names of compounds.
Do not give generic answers like “strong acid” or “weak acid”.
i.
H2SeO3 was titrated with OH-.
ii.
Of the indicators listed in Table 2 (data sheet), thymolphthalein would be the best
choice for this titration. This is because it has the pKa closest to 10 (the pH of the
steeper equivalence point).
iii.
When 25 mL of titrant have been added, the major species in solution* is HSeO3-.
* Excluding water and spectator ions!
iv.
When 35 mL of titrant have been added, the major species in solution* is SeO32-.
* Excluding water and spectator ions!
NAME:_________________________
5.
Student Number:______________________
The pH of a saturated aqueous solution of manganese(II) hydroxide is 9.87. Calculate the
Ksp for manganese(II) hydroxide.
[8 marks]
Mn2+(aq) + 2 OH-(aq)
Mn(OH)2(s)
Ksp = [Mn2+][OH-]2
where [OH-] = 2 [Mn2+]
We know that pH = 9.87. Therefore, pOH = 14 – 9.87 = 4.13.
[OH-] = 10-pOH = 10-(4.13) = 7.4 × 10-5
[Mn2+] = ½ [OH-] = ½ (7.4 × 10-5) = 3.7 × 10-5
Ksp = (3.7 × 10-5)(7.4 × 10-5)2 = 2.0 × 10-13
NAME:_________________________
6.
Student Number:______________________
Somebody forgot to label the sample vials for a new chemistry experiment, leaving Ying
to figure out which vial contained which compound. From her prep list, she knew that
the compounds were: AgClO4, NaNO3, NH4Cl and K2S. All four of them are white salts
so, to determine which was which, she knew she’d have to use their chemical properties.
She put temporary labels on the sample vials (A, B, C, and D) then took a small amount
of each solid and dissolved it in water. They all formed clear colourless solutions.
[8 marks]
She measured the pH of each solution.
• Solutions A and C had neutral pH values
• Solution B had a pH of ~4
• Solution D had a pH of ~9
She then mixed a drop of each solution with a drop of each of the other solutions.
• Precipitates formed when the following pairs of solutions were mixed:
o B&C
○ C&D
• There was no observable reaction when the following pairs of solutions were mixed:
o A&B
○ A&D
o A&C
○ B&D
From this information, she was able to identify which salt was which and properly label
the vials. Match each of the four salts to its temporary label.
A = NaNO3
B = NH4Cl
C = AgClO4
D = K2S
Step 1: Break each salt into its ions, giving
Na+ (neutral) and NO3- (neutral)
NH4+ (acidic) and Cl- (neutral)
Ag+ (neutral or faintly acidic) and ClO4- (neutral)
2 K+ (neutral) and S2- (basic)
Step 2: Solution B is acidic (pH 4) therefore NH4Cl
Solution D is basic (pH 10) therefore K2S
Solutions A and C are neutral therefore NaNO3 and AgClO4
Step 3: Both precipitates formed have the same solution (C) as a reactant. This must be AgClO4
because Ag+ is the only cation here that isn’t on the “always soluble” list.
Solution A is the only one that forms no precipitates. This must be NaNO3 because Na+
and NO3- are both on the “always soluble” list.
NAME:_________________________
7.
Student Number:______________________
Describe how you would prepare 1.0 L of a pH 5.00 buffer. You have access to the
following reagents:
[8 marks]
• 0.50 M pyridine solution (C5H5N, 79.101 g/mol)
• solid pyridinium chloride (C5H5NHCl, 115.562 g/mol)
• concentrated hydrochloric acid (HCl, 12 M)
Clearly indicate which reagent(s) you would use and what quantity of each.
There are two basic approaches to this question. You can either mix the weak base with
its conjugate acid (Method A), or mix the weak base with HCl – generating the conjugate
acid in situ (Method B).
Step 1: Find the pKa of pyridine’s conjugate acid (Methods A and B)
The conjugate acid of pyridine (C5H5N) is pyridinium (C5H5NH+).
The Kb of pyridine is listed on Table 1 as 1.5 × 10-9.
Ka(pyridinium) × Kb(pyridine) = Kw
Ka(pyridinium) = Kw / Kb(pyridine) = (1.0 × 10-14) / (1.5 × 10-9) = 6.7 × 10-6
pKa = -logKa = -log(6.7 × 10-6) = 5.18
Check your math by ensuring that this is within one unit of the buffer’s final pH.
Step 2: Find the required ratio of conjugate base : conjugate acid (Methods A and B).
pH = pK a + log
[conj.base]
[conj.acid]
[conj.base]
= 10 pH − pK a = 105.00−5.18 = 0.67
[conj.acid]
Step 3 for Method A: Find the mass of conjugate acid (pyridinium hydrochloride) that you need
to add to the 0.50 M pyridine solution.
[conj.base]
= 0.67
[conj.acid]
[conj.base] 0.5M
[conj. acid] =
=
= 0.75M
0.67
0.67
m pyridinium chloride = 0.75
mol
g
×1.0 L × 115.562
= 87 g
L
mol
To make a pH 5.00 buffer, dissolve 87 g of solid pyridinium chloride in 0.50 M pyridine.
NAME:_________________________
Student Number:______________________
Step 3 for Method B: Find the amount of concentrated HCl that you need to add to the 0.50 M
pyridine solution to make a solution with the correct ratio of pyridine :
pyridinium.
C5H5N(aq) +
H3O+(aq)
→ C5H5NH+(aq) + H2O(l)
To get [C5H5N] : [C5H5NH+] to be 0.67 : 1, we need to add 1 mole of HCl for every 1.67 moles
of pyridine. Since we are starting with 0.50 moles of pyridine (1.0 L × 0.50 mol/L) , that means
we need 0.30 moles of HCl.
0.50 mol pyridine ×
1 mol HCl
= 0.30 mol HCl
1.67 mol pyridine
Then calculate the volume of HCl necessary to get that many moles.
VHCl =
0.30 mol HCl
= 0.025 L = 25 mL
12 mol/L
Therefore, add 25 mL of 12 M HCl to 1.0 L of 0.50 M pyridine solution. (The added volume of
HCl does not significantly affect the total volume of solution because 1.0 L + 0.025 L = 1.0 L.)
NAME:_________________________
Student Number:______________________
Some Useful Data
Constants
R = 8.3145 J · mol-1 · K-1
R = 0.082057 L · atm · mol-1 · K-1
Kw = 1.0 × 10-14 at 25 ˚C
Table 1: Dissociation Constants for Selected Acids and Bases
Acid
Ka
Base
acetic acid, CH3CO2H
Ka = 1.8 × 10-5
ammonia, NH3
Ka1 = 4.2 × 10-7
carbonic acid, H2CO3
carbonate, CO32Ka2 = 4.8 × 10-11
fluoroacetic acid, CH2FCO2H
Ka = 2.2 × 10-3
fluoride, Fformic acid, HCO2H
Ka = 1.8 × 10-4
hydroxide, HOhydrochloric acid, HCl
Ka = 107
methylamine, CH3NH2
+
hydronium, H3O
Ka = 1.0
pyridine, C5H5N
Ka1 = 7.5 × 10-3
phosphoric acid, H3PO4
Ka2 = 6.2 × 10-8
phosphate, PO43Ka3 = 3.6 × 10-13
Ka1 = 2.7 × 10-3
selenous acid, H2SeO3
selenite, SeO32Ka2 = 2.5 × 10-7
water, H2O
Ka = 1.0 × 10-14
water, H2O
Table 2: Some Common Indicators
Indicator
crystal violet
2,4-dinitrophenol
bromothymol blue
phenolphthalein
thymolphthalein
1
Kb
Kb = 1.8 × 10-5
Kb1 = 2.1 × 10-4
Kb2 = 2.4 × 10-8
Kb = 1.4 × 10-11
Kb = 1.0
Kb = 5.0 × 10-4
Kb = 1.5 × 10-9
Kb1 = 2.8 × 10-2
Kb2 = 1.6 × 10-7
Kb3 = 1.3 × 10-12
Kb1 = 4.0 × 10-8
Kb2 = 3.7 × 10-12
Kb = 1.0 × 10-14
Colour Change
yellow → blue
colourless → yellow
yellow → blue
colourless → red
colourless → blue
pKa
1
3.5
7
8.5
10
Chem 2000 Standard Periodic Table
18
4.0026
1.0079
He
H
2
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
107
Np
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré