Eastern Mediterranean University
Department of Chemistry
Department of Chemistry
CHEM247 Analytical Chemistry I
Fall 2012-2013 Final Exam
11-01-2013
08.30-10.30
Student No:
Q. 1
Name
Q. 2
Surname:
Q. 3.
Signature:
Q. 4.
Q. 5.
Q. 6.
Q. 7.
Q. 8.
TOTAL AVAILABLE MARKS = 50
TOTAL
(Fin weight is 40%)
INSTRUCTIONS:
1.
Write your student number, name and surname, and sign the question booklet.
2.
The exam consists of 5 questions. To get full marks, answer all questions. Show all your steps.
3.
The Periodic Table provided may be necessary to answer some of the questions.
4.
Use of mobile phones, exchange of calculators or erasers is not allowed.
5.
You can see your papers in the first10 days after the announcement of the results.
Periodic Table of Elements
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
1
H
1.008
3
Li
6.94
11
Na
22.99
19
K
39.10
37
Rb
85.47
55
Cs
132.91
87
Fr
223.02
18
2
4
Be
9.01
12
Mg
24.30
20
Ca
40.08
38
Sr
87.62
56
Ba
137.33
88
Ra
226.03
5
21
Sc
44.96
39
Y
88.91
71
Lu
174.97
103
Lr
260.11
Lanthanides
22
23
24
Ti
47.88
40
Zr
91.22
72
Hf
178.49
V
50.94
41
Nb
92.91
73
Ta
180.95
Cr
52.00
42
Mo
95.94
74
W
183.85
57
La
138.91
89
Actinides
Ac
227.03
58
Ce
140.12
90
Th
232.04
59
Pr
140.91
91
Pa
231.04
25
Mn
54.94
43
Tc
98.91
75
Re
186.2
60
Nd
144.24
92
U
238.03
26
Fe
55.85
44
Ru
101.07
76
Os
190.2
61
Pm
146.92
93
Np
237.05
27
Co
58.93
45
Rh
102.91
77
Ir
192.22
62
Sm
150.36
94
Pu
244.06
28
Ni
58.69
46
Pd
106.42
78
Pt
195.08
63
Eu
151.97
95
Am
243.06
29
Cu
63.54
47
Ag
107.87
79
Au
196.97
64
Gd
157.25
96
Cm
247.07
30
Zn
65.39
48
Cd
112.41
80
Hg
200.59
65
Tb
158.93
97
Bk
247.07
B
10.81
13
Al
26.98
31
Ga
69.72
49
In
114.82
81
Tl
204.38
66
Dy
162.50
98
Cf
251.08
6
C
12.01
14
Si
28.09
32
Ge
72.61
50
Sn
118.71
82
Pb
207.2
67
Ho
164.93
99
Es
252.08
7
N
14.01
15
P
30.97
33
As
74.92
51
Sb
121.75
83
Bi
208.98
68
Er
167.26
100
Fm
257.10
8
O
16.00
16
S
32.07
34
Se
78.96
52
Te
127.6
84
Po
208.98
69
Tm
168.93
101
Md
258.10
9
F
19.00
17
Cl
35.45
35
Br
79.90
53
I
126.90
85
At
209.99
70
Yb
173.04
102
No
259.10
He
4.003
10
Ne
20.18
18
Ar
39.95
36
Kr
83.80
54
Xe
131.29
86
Rn
222.02
FALL 2012-2013
CHEM247 Midterm Exam
Page 2
Question 1
A 5.00% (w/w) Fe(NO3)3 (241.86 g/mol) solution has a density of 1.042 g/mL. Calculate
a) the molar analytical concentration of Fe(NO3)3 in this solution.
b) the molar NO3− concentration in the solution.
c) the mass in grams of Fe(NO3)3 contained in each liter of this solution?
d) the pFe of this solution
(a)
(b)
C Fe( NO3 )3 
NO 

3
(c)
(d)
5.00 g Fe( NO3 ) 3 1.042 g solution 1000 mL solution
mol Fe( NO3 ) 3



 0.215 M Fe( NO3 ) 3
100 g solution
mL solution
L solution
241.86 g Fe( NO3 ) 3

0.215 mole Fe( NO3 )3
3 mole NO3



 0.645 M NO3
L
mole Fe( NO3 )3
M Fe( NO3 )3 
0.215 mole Fe( NO3 )3 241.86 g Fe( NO3 )3 52.0 g Fe( NO3 )3


L
mole Fe( NO3 )3
L
pFe = -log[Fe3+] =−log(0.215) = 0.668
Question 2
Data are presented below for the calcium concentration in water that was analysed by titrimetric and
gravimetric methods.
ppm Ca
ppm Ca
Gravimetric Titrimetric
771
765
781
738
803
765
781
794
784
752
784
799
791
744
a) Determine the average and standard deviation for calcium concentration by each method.
b) How do the precision of the two methods compare with each other? Please explain briefly.
c) What other information is necessary in order to compare the accuracy of the two analytical methods
in this particular case? Please explain briefly.
d) Can you suggest the names of suitable reagents and their purpose for conducting titrimetric and
gravimetric analysis of calcium in aqueous solution?
e)
a)
avgCGr
= 785 ppm
sGr = 9.91 ppm
avgCTi
= 765 ppm
sTi = 23.6 ppm
b) Since sGr < sTi precision of gravimetric method is better because leff dispersion of data
c) We need to know true value of Ca concentration so as to estimate accuracy.
d) For gravimetric analysis the OXALATE anion C2O42− and
EDTA (titrant) and Erichrome black T (indicator) for titrimetric analysis.
FALL 2012-2013
CHEM247 Midterm Exam
Question 3
The solubility products for a series of iodides are:
CuI Ksp = 1×10−12
AgI
Ksp = 8.3·10−17
PbI2 Ksp = 7.1×10−9
List these four compounds in order of decreasing molar solubility in
a) water and
b) a 0.010 M solution of the solute cation.
Ksp = [Cu+][I-] = 110-12 = (S)2
Ksp = [Ag+][I-] = 8.310-17 = (S)2
Ksp = [Pb2+][I-]2 = 7.110-9 = S(2S)2 = 4S3
(a) Solubilities in water
S CuI  [Cu  ]  [I  ] S  1  10 12  1  10 6
S AgI  [Ag  ]  [I  ] S  8.3  10 17  9.1  10 9
1 
(7.1  10 9 )
3
S PbI2  [Pb ]  [I ] S 
 1.2  10 3
2
4
Therefore
PbI2  CuI  AgI in water
2
(b)Solubilities in solution containing 0.01 M of the cation
SCuI 
1  10 12
 1  10 10
0.010 M
SAgI 
8.3  10 17
 8.3  10 15
0.010 M
SP bI2 
1 7.1  10 9
 4.2  10 4
2 0.010 M
Therefore
PbI2  CuI  AgI in a 0.010 M solution of the solute cation
Question 4
Write the mass-balance expressions for each of the following solutions.
a) 0.30 M in H2SO4
b) 0.0800 M in HClO and 0.200 M in NaClO.
c) 0.25 M in NaF and saturated with CaF2 .
(a)
0.30 = [H2SO4] + [HSO4-] + [SO42-]
(b)
0.200 + 0.0800 = [ClO-] + [HClO]
0.200 = [Na+]
(c)
[F-] + [HF] = 0.25 + 2[Ca2+]
[Na+] = 0.25
Page 3
FALL 2012-2013
CHEM247 Midterm Exam
Page 4
Question 5
Aqueous NaOH is introduced into a solution that is 0.040 M in Cu2+ and 0.050 M in Mn2+.
(Ksp Cu(OH)2 = 4.810-20 and Ksp Mn(OH)2 = 210-13)
a) Which hydroxide precipitates first?
b) What OH− concentration is needed to start precipitation of the first hydroxide?
c) What is the concentration of the cation forming the less soluble hydroxide when the more soluble
hydroxide begins to form?
[Cu2+][OH-] 2 = 4.810-20
(a)
[Mn2+][OH-] 2 = 210-13
Cu(OH)2 precipitates first because Cu(OH)2 is less soluble than Mn(OH)2 because Cu begins to
precipitate when [OH  ]Cu 
(4.8  10 20 )
 1.095  10 9 M
0.04
and Mn begins to precipitate when [OH  ]Mn 
(2  10 13 )
 2.00  10 6 M
0.05
SMn(OH)2>SCu(OH)2
(b)
Cu2+ begins to precipitate when
[OH−] =
4.8  10 20 / 0.040 = 1.09510-9 M
(c)
Mn2+ begins to precipitate when
[OH−] =
2  10 13 / 0.050 = 2.0010-6 M
the concentration of Cu at this point is;
[Cu2+] = 4.810-20/(2.210-6)2 = 1.2010-8 M
Question 6(
a. Explain the difference between a colloidal and a crystalline precipitate.
A colloidal precipitate consists of solid particles with dimensions that are less than 10 -4 cm. A crystalline precipitate consists
of solid particles with dimensions that at least 10-4 cm or greater. As a consequence, crystalline precipitates settle rapidly,
whereas colloidal precipitates remain suspended in solution unless caused to agglomerate.
b. Explain the difference between peptization and coagulation of a colloid.
Coagulation, or agglomeration, is the process by which colloidal particles coalesce to form larger aggregates. Peptization
refers to the process by which a coagulated colloid reverts to its original dispersed state. Heating, stirring and adding an
electrolyte can coagulate colloidal suspensions. Washing the coagulated colloid with water often removes sufficient
electrolyte to permit the re-establishment of repulsive forces that favor return to the colloidal state.
c. Explain what is meant by digestion in gravimetric analysis and why is it carried out.
Digestion is a process in which a precipitate is heated in the presence of the solution from which it was formed (the mother
liquor). Digestion improves the purity and filterability of the precipitate.
d. Explain the difference between the equivalence point and the end point of a titration.
The equivalence point in a titration is that point at which sufficient titrant has been added so that stoichiometrically equivalent
amounts of analyte and titrant are present. The end point in a titration is the point at which an observable physical change
signals the equivalence point.
FALL 2012-2013
CHEM247 Midterm Exam
Page 5
Question 7
Acetic acid (HA) is a weak acid. Its acid dissociation constant is Ka = 1.75×10−5 M.
HA(aq)  A−(aq) + H+(aq)
Write three expressions for Ka, mass balance and charge balance. Using these equations calculate the pH of
a 0.010 M acetic acid solution. List any assumptions you make about concentrations in your calculations.
CH 3COOH


C 2 H 5 COO   H 
[CH 3COO  ][H  ]
eqn 1 K a 
 1.75  10 5 eqn 2 [H  ]  [CH3COO  ] and
[CH 3COOH]
eqn 3 [CH 3COOH]  0.010 M  [H  ]  0.010 M
Thus eqn 1 becomes
[H  ]2
 1.75  10 5
(0.010 M)
M
[H  ]  1.75  10 5 (0.010 M)  4.18  10 4 and
pH  3.378
Question 8
a) What is the pH of the solution that results when 0.153 g of Mg(OH)2 is mixed with 75.0 mL of 0.100 M
HCl?
Amount of Mg(OH)2 taken =
0.153 g Mg(OH) 2
 2.6235mmol
0.05832 g Mg(OH) 2 / mmol
Amount of HCl = (75 ×0.100) mmol = 7.5 mmol
Neutralisation is 2 HCl per one Mg(OH)2 therefore HCl is in excess
Remaining CHCl = (75.00.100 – 2.62352)/75.0 = 0.03004 M
[H3O+] = 0.01366 and pH = -log(0.01366) = 1.87
b) Treatment of a 0.3000-g sample of impure potassium chloride with an excess of AgNO3 resulted in the
formation of 0.2725 g of AgCl. Calculate the percentage of KCl in the sample
143.32 g
74.55 g
M KCl 
mole
mole
 1 mole AgCl   1 mole KCl   74.55 g KCl 
  
  
0.2725 g AgCl  

mole

 143.32 g   1 mole AgCl  
 100%  47.25%
0.3000 g impure sample
M AgCl 
c) A 0.5142-g sample of primary-standard-grade Na2CO3 required 39.55 mL of an H2SO4 solution to reach
the end point in the reaction
CO32− + 2H+  H2O + CO2(g).
What is the molarity of the H2SO4?
MNa2CO3  105.99g / mole
0.5142 g Na 2 CO3 
C H 2 SO4 
1 mole Na 2 CO3
2 mole H  1 mole H 2 SO4


105.99 g
mole Na 2 CO3
2 mole H 
 0.1227 M H 2 SO4
1L
39.55 mL 
1000 mL