NAME:_________________________
Spring 2006
Chemistry 2000 Test #3B
INSTRUCTIONS:
1.
Student Number:______________________
____/ 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 pH of a saturated aqueous solution of iron(II) hydroxide is 9.20. Calculate the Ksp
for iron(II) hydroxide.
[8 marks]
Fe2+(aq) + 2 OH-(aq)
Fe(OH)2(s)
Ksp = [Fe2+][OH-]2
where [OH-] = 2 [Fe2+]
We know that pH = 9.20. Therefore, pOH = 14 – 9.20 = 4.80.
[OH-] = 10-pOH = 10-(4.80) = 1.6 × 10-5
[Fe2+] = ½ [OH-] = ½ (1.6 × 10-5) = 7.9 × 10-6
Ksp = (7.9 × 10-6)(1.6 × 10-5)2 = 2.0 × 10-15
NAME:_________________________
2.
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
pKa1 = 10.4
12.00
Veq #1 = 10.00 mL
10.00
8.00
pH
pKa2 = 6.4
6.00
Veq #2 = 20.00 mL
4.00
2.00
0.00
0.00
5.00
10.00
15.00
20.00
25.00
volume of 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.
CO32- was titrated with HCl (or H3O+).
ii.
Of the indicators listed in Table 2 (data sheet), 2,4-dinitrophenol would be the
best choice for this titration. This is because it has the pKa closest to 3.2 (the pH
of the steeper equivalence point).
iii.
When 7.5 mL of titrant have been added, the major species in solution is HCO3-.
* Excluding water and spectator ions!
iv.
When 12.5 mL of titrant have been added, the major species in solution is HCO3-.
* Excluding water and spectator ions!
NAME:_________________________
3.
Student Number:______________________
The tetracyanocuprate(I) complex exists in equilibrium with free copper(I) and cyanide
ions:
3+ + 4 CN Cu(aq)
Cu(CN)4(aq)
Kf = 5.0 x 1026
(aq)
A solution contains 1.1 × 10-4 M Cu+, 2.3 × 10-7 M CN- and 2.4 × 10-2 M Cu(CN)43-. Is
this solution at equilibrium? If not, indicate the direction in which the reaction must
proceed to reach equilibrium.
[6 marks]
Q =
[Cu(CN) 34− ]
[Cu + ][CN − ]4
Q =
(2.4 × 10 -2 )
(1.1 × 10 -4 )(2.3 × 10 -7 ) 4
Q = 7.8 × 10 28
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:_________________________
4.
(a)
The pKa of [Al(OH2)6]3+ is 5.1.
What is the conjugate base of [Al(OH2)6]3+?
[Al(OH2)5(OH)]2+
(b)
Student Number:______________________
[3 marks]
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:
5.
pKb = 14 – pKa = 14 – 5.1 = 8.9
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:_________________________
6.
Student Number:______________________
Describe how you would prepare 1.0 L of a pH 2.50 buffer. You have access to the
following reagents:
[8 marks]
• 0.50 M fluoroacetic acid solution (CH2FCO2H, 78.043 g/mol)
• solid sodium fluoroacetate (CH2FCO2Na, 100.025 g/mol)
• solid sodium hydroxide (NaOH, 39.997 g/mol)
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 acid with
its conjugate base (Method A), or mix the weak acid with NaOH – generating the
conjugate base in situ (Method B).
Step 1: Find the pKa of fluoroacetic acid (Methods A and B)
The Ka of fluoroacetic acid is listed on Table 1 as 2.2 × 10-3
pKa = -logKa = -log(2.2 × 10-3) = 2.66
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 = 10 2.50− 2.66 = 0.70
[conj.acid]
Step 3 for Method A: Find the mass of conjugate base (sodium fluoroacetate) that you need to
add to the 0.50 M fluoroacetic acid solution.
[conj.base]
= 0.70
[conj.acid]
[conj. base] = 0.70 [conj. acid] = 0.70 × 0.50 M = 0.35 M
m sodium fluoroacetate = 0.35
mol
g
× 1.0 L × 100.025
= 35 g
L
mol
To make a pH 2.50 buffer, dissolve 35 g of solid sodium fluoroacetate in 0.50 M
fluoroacetic acid.
NAME:_________________________
Student Number:______________________
Step 3 for Method B: Find the amount of concentrated NaOH that you need to add to the 0.50 M
fluoroacetic acid solution to make a solution with the correct ratio of
fluoroacetate : fluoroacetic acid.
CH2FCO2H(aq) + OH-(aq) → CH2FCO2-(aq) + H2O(l)
To get [CH2FCO2-] : [CH2FCO2H] to be 0.70 : 1, we need to add 0.70 moles of NaOH for every
1.70 moles of fluoroacetic acid. Since we are starting with 0.50 moles of fluoroacetic acid (1.0 L
× 0.50 mol/L) , that means we need 0.20 moles of HCl.
0.50 mol fluoroacetic acid ×
0.70 mol NaOH
= 0.20 mol NaOH
1.70 mol fluoroacetic acid
Then calculate the mass of NaOH necessary to get that many moles.
m NaOH = 0.20molNaOH × 39.9971g / mol = 8.2 g
Therefore, add 8.2 g of solid NaOH to 1.0 L of 0.50 M fluoroacetic acid solution.
NAME:_________________________
7.
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: KClO4, Na2CO3, NH4Br, and Pb(NO3)2. 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 = KClO4
B = NH4Br
C = Pb(NO3)2
D = Na2CO3
Step 1: Break each salt into its ions, giving
K+ (neutral) and ClO4- (neutral)
NH4+ (acidic) and Br- (neutral)
Pb2+ (neutral or faintly acidic) and 2 NO3- (neutral)
2 Na+ (neutral) and CO32- (basic)
Step 2: Solution B is acidic (pH 4) therefore NH4Br
Solution D is basic (pH 10) therefore Na2CO3
Solutions A and C are neutral therefore KClO4 and Pb(NO3)2
Step 3: Both precipitates formed have the same solution (C) as a reactant. This must be
Pb(NO3)2 because Pb2+ 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 KClO4 because K+
and ClO4- are both on the “always soluble” list.
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