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APP601S - ANALYTICAL PRINCIPLES AND PRACTICE - 1ST OPP - JUNE 2022 |
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g
NAMIBI A UNIVERSITY
OF SCIENCE AND TECHNOLOGY
FACULTY OF HEALTH, APPLIED SCIENCES AND NATURAL RESOURCES
DEPARTMENT OF NATURAL AND APPLIED SCIENCES
QUALIFICATION: BACHELOR OF SCIENCE
QUALIFICATION CODE: 07BOSC
LEVEL: 6
COURSE CODE: APP601S
COURSE NAME: ANALYTICAL PRINCIPLES AND
PRACTICE
SESSION: JUNE 2022
DURATION: 3 HOURS
PAPER: THEORY
MARKS: 100
FIRST OPPORTUNITY EXAMINATION QUESTION PAPER
EXAMINER(S) | DR JULIEN LUSILAO
MODERATOR: | DR MARIUS MUTORWA
INSTRUCTIONS
1. Answer ALL the questions in the answer book provided.
2. Write and number your answers clearly.
3. All written work MUST be done in blue or black ink.
PERMISSIBLE MATERIALS
Non-programmable Calculators
ATTACHMENTS
List of useful tables, formulas and constants
THIS QUESTION PAPER CONSISTS OF 11 PAGES (Including this front page and attachments)
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Question 1: Multiple Choice Questions
[20]
1.1 How many moles of Na* ions are in 20 mL of 0.40 M Na3POq?
(2)
(A) 0.0080
(B) 0.050
(C) 0.024
(D) 0.20
1.2 Amass of 5.4 grams of Al reacts with an excess of CuCl in solution, as shown below:
3CuCl2 + 2Al — 2AlCl3 + 3Cu
What mass of solid copper (Cu) is produced?
(2)
(A) 0.65¢
(B) 85g
(C) 13¢
(D) 19g
1.3 For the reaction
2C(s) + O2(g) — 2CO(g)
What are the signs of AH and AS?
(2)
(A) They are both negative
(B) They are both positive
(C) AH is positive and AS is negative
(D) AH is negative and AS is positive
1.4 The ion-product constant for water (Kw) at 45 °C is 4.0 x 10-44. What is the pH of
pure water at this temperature?
(2)
(A) 7.0
(B) 6.7
(C) 7.3
(D) 13.4
1.5 The amount of chloride ion in a water sample is to be determined by adding excess
silver nitrate (MW AgNOs: 169.91 g-mol).
Cl(aq) + AgNO3(aq)— AgCl(s) + NOz(aq)
If 1.0 g of silver chloride (MW AgCI: 143.25 g-mol-) is precipitated, what mass of
chloride ion is in the original sample?
(2)
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(A) 0.34 ¢
(B) 0.50g
(C) 0.25 g
(D) 0.75 g
1.6 The ionization of benzoic acid is represented by this equation.
Ce6HsCOOH(aq) <> H*(aq) + CsHsCOO (aq)
If a 0.045 M solution of benzoic acid has an [H*] = 1.7 x 10-3, what is the Kz of
benzoic acid?
(2)
(A) 7.7 x lO
(B) 6.4 x 10-5
(C) 3.8 x 107
(D) 8.4 x 107
1.7 Mg3N2(s) + 6H20(I) > 2NH3(aq) + 3Mg(OH)2(s)
If 54.0 grams of water are mixed with excess magnesium nitride, then how many
grams of ammonia are produced?
(2)
(A) 1.00
(B) 17.0
(C) 51.0
(D) 153
1.8 For the reaction
CeHsOH(aq) + CN-(aq) <* HCN(aq) + CeHsO-(aq)
The equilibrium constant for this reaction is less than 1. What is the strongest base
in this system?
(2)
(A) CeHsOH(aq)
(B) CN“(aq)
(C) CeHs0-(aq)
(D) HCN(aq)
1.9 The solubility product constant, Ksp, of Ag3POz is 1.8 x10°78. What is the molar
solubility of Ag3PO4 in water? Neglect any hydrolysis.
(2)
(A) 1.6x 1075
(B) 8.4x1077
(C) 1.3x 1079
(D) 4.5x 1079
1.10 The balanced equation for the reduction of the nitrate anion by the Fe(II) ion in
an acidic solution is
(2)
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(A) 3Fe?*(aq) + NO3 (aq) + 4H*(aq) — 3Fe?*(aq) + NO(g) + 2H20(I)
(B) Fe2*(aq) + NO3 (aq) + 8 H+(aq) — Fe**(aq) + NO(g) + 4H20(I)
(C) 2Fe?*(aq) + 2NO3 (aq) + 4H*(aq) — 2Fe?*(aq) + 2NO(g) + 4H20(I)
(D) 3Fe?*(aq) + NO(g) + 2H20(I) — 3Fe?*(aq) + NO3 (aq) + 4H+(aq)
Question 2
[15]
2.1 To test a spectrophotometer’s accuracy a solution of 60.06 ppm K2Cr207 in 5.0 mM
H2SO, is prepared and analysed. This solution has an expected absorbance of 0.640
at 350.0 nm in a 1.0-cm cell when using 5.0 mM H2SO, as a reagent blank. Several
aliquots of the solution produce the following absorbance values.
0.639
0.638
0.640 0.639
0.640
0.639
0.638
(a) Calculate the mean and standard deviation of the measured absorbance values. (2)
(b) Determine whether there is a significant difference between the experimental
mean and the expected value at a = 0.01 (i.e. P= 99%).
(6)
2.2 One way to check the accuracy of a spectrophotometer is to measure absorbencies
for a series of standard dichromate solutions obtained from the National Institute
of Standards and Technology. Absorbencies (A) are measured at 257 nm and
compared to the accepted values. The results obtained when testing a newly
purchased spectrophotometer are shown below.
Standard
1
2
3
4
5
MeasuredA
0.2872
0.5773
0.8674
1.1623
1.4559
ExpectedA
0.2871
0.5760
0.8677
1.1608
1.4565
Determine if the tested spectrophotometer is accurate at a = 0.05.
(7)
Question 3
[15]
3.1 A solution containing 3.47 mM of analyte and 1.72 mM of standard gave peak areas
of 3,473 and 10,222, respectively, in a chromatographic analysis. Then 1.00 mL of
8.47 mM standard was added to 5.00 mL of unknown solution, and the mixture was
diluted to 10.0 mL. This solution gave peak areas of 5 428 and 4 431 for the analyte
and standard, respectively.
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(a) Calculate the response factor for the analyte.
(2)
(b) Find the concentration of the standard in the 10.0 mL of mixed solution.
(2)
(c) Find the analyte concentration in the 10.0 mL of mixed solution.
(2)
(d) Find the analyte concentration in the original unknown.
(2)
3.2 The concentration of phenol in a water sample is determined by separating the
phenol from non-volatile impurities by steam distillation, followed by reacting with
4-aminoantipyrine and K3Fe(CN)¢ at pH 7.9 to form a colored antipyrine dye.
A phenol standard with a concentration of 4.00 ppm has an absorbance of 0.424.
A water sample is steam distilled and a 50.00-mL aliquot of the distillate is placed in
a 100-mL volumetric flask and diluted to volume with distilled water. The absorbance
of this solution is found to be 0.394.
(a) What is the concentration of phenol (in parts per million) inthe water sample? — (3)
(b) What calibration method has been used here?
(2)
(c) Briefly explain your choice of the calibration method.
(2)
Question 4
[20]
4.1 For the following unbalanced reaction at 25°C
Fe?* +MnOg Fe?* + Mn** (acidic medium)
(E°Fe3+/re2+ = 0.771 V; E°vnoa-/vin2+ = 1.51 V)
(a) Write the balanced oxidation and reduction half reactions as well as the overall
reaction.
(3)
(b) Calculate the standard potential, £°, of the reaction.
(1)
(c) Calculate the equilibrium constant, K, of the reaction.
(2)
(d) Calculate the potential, £, under the following conditions: [Fe2*] = 0.50 M,
[Fe?*] = 0.10 M, [MnO."] = 0.025 M, [Mn2*] = 0.015 M, and a pH of 7.00.
(2)
4.2 Calculate the pH of the solution that results from the addition of 0.040 moles of
HNOs to a buffer made by combining 0.500 L of 0.380 M HC3HsQ2 (Ka = 1.30 x 10°)
and 0.500 L of 0.380 M NaC3HsO2. Assume addition of the nitric acid has no effect
on volume.
(5)
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4.3 The following diagram shows the variation of the solubility of potassium tartrate
(K2CaHaO¢) when adding different concentrations of several substances.
0.05 |-
0.04 |—-
0.08
0.02 |-
Glucose —
0.01
0.00
a er e ee
|
0.00 0.02 0.04 0.06 0.08 0.10
Concentration of added
substance (M)
(a) Explain the trend observed with the addition of MgSOq and NaCl.
(2)
(b) Why MgSO, produces a higher solubility than NaCl?
(1)
(c) Explain the trend observed with the addition of glucose.
(2)
(d) Why does the solubility of potassium tartrate decrease with the addition of KCI? (2)
Question 5
[30]
5.1 50.0 mL of 0.0400 M formic acid (HCOOH, Ka = 1.80 x 10%) was titrated with
0.120 M NaOH.
(a) Write the balanced reaction of the titration.
(1)
(b) calculate the volume of added titrant at the equivalence point.
(2)
(c) Calculate the pH after addition of the following volumes of the titrant
(i) 0.0 mL
(4)
(ii) 10.0 mL
(4)
(iii) 20.0 mL
(4)
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5.2 25.0 mL of 0.01 M V** is titrated using 0.01 M Ce**
(£°v3+/v2e = - 0.255 V; E%ceat/ce3+ = + 1.72 V).
(a) Write the two redox half-reactions, the overall reaction and the potential (E)
expressions for both redox half-reactions.
(5)
(b) Calculate the potential of the titration after addition of
(i) 15.0 mL Ce**
(5)
(ii) 25.0 mL Ce**
(5)
TOTAL MARK = [100]
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Data Sheet
tealeutated
_
= i H VN
d
Coseutatea = Sy Jn
t calculated =
s?(N, —1)+s¢(N, -1) +...
S pooled =
—N—,*+Ny,+ ae=eee — Nsets of data
Confidence
dFergereedeosm 50%
90%
95%
99%
ee i
1.000
6.314
12.706
2
0.816
2.920
4.303
63.656
9.925
5-
00..771287
21..091453
2> .454771
43..073027
7
0.711 |1.895 | 2.365 | 3.499
8
0.706
1.860
2.306
3.355
9
0.703
1.833
2.262
3.250
10
0.700
1.812
2.228
3.169
11
0.697
1.796
2.201
3.106
12
0.695 |1.782 | 2.179 | 3.055
13
0.694
1.771
2.160
3.012
14
0.692
1.761
2.145
2.977
15
0.691
1.753
2.131
2.947
16
0.690
1.746
2.120
2.921
17
0.689 1.740 |2.110 | 2.398
18
0.688
1.734
2.101
2.878
19
0.688
1.729
2.093
2.861
20
0.687
1.725
2.086
2.845
21
0.686
1721
2.080
2.831
22
0.686
1.717
2.074
2.819
23
0.685
1.714
2.069
2.807
24
0.685
1.711
2.064
2.797
25
0.684
1.708
2.060
2.787
26
0.684
1.706
2.056
2.779
27
0.684
1.703
2.052
2.771
28
0.683
1.701
2.048
2.763
29
0.683
1.699
2.045
2.756
30
0.683
1.697
2.042
2.750
31
0.682
1.696
2.040
2.744
32
0.682
1.694
2.037
2.738
33
0.682
1.692
2.035
2.733
34
0.682
1.691
2.032
2.728
35
0.682
1.690
2.030
2.724
pax+2lhn
Geeaa ed Critical Values for the Rejection Quotient
N
Conf9i0d%ence | Conf9i5d%ence | Conf9i9d%ence
3
0.941
4
0.765
5
0.642
‘
0.560
0.970
0.829
0.710
0.625
0.994
0.926
0.821
0.740
7
0.507
8
0.468
9
0.437
10
0.412
0.568
0.526
0.493
0.466
0.680
0.634
0.598
0.568
N = number of observations
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F(0.05, conum, odenom) for a Two-Tailed F-Test
onum> 1
2
3
4
5
odenU
1
647.8 799.5 864.2 899.6 921.8
2
38.51 39.00 39.17 39.25 39.30
3
17.44 16.04 15.44 15.10 1488
4
12.22 10.65 9.979 9.605 9.364
5
10.01 8.434 7.764 7.388 7.146
6
8.813. 7.260 6.599 6.227 5.988
7
8.073 6.542 5.890 5.523 5.285
8
7.571 6.059 5.416 5.053 4.817
9
7.209 5.715 5.078 4.718 4.484
10
6.937 5.456 4.826 4.468 4.236
11
6.724 5.256 4.630 4.275 4.044
12
6.544 5.096 4.474 4.121 3.891
13
6.414 4965 4.347 3.996 3.767
14
6.298 4.857 4.242 3.892 3.663
15
6.200 4.765 4.153 3.804 3.576
16
6.115 4.687 4.077 3.729 3.502
17
6.042 4.619 4.011 3.665 3.438
18
5.978 4.560 3.954 3.608 3.382
19
§.922 4.508 3.903 3.559 3.333
20
5.871 4461 3.859 3.515 3.289
co
5.024 3.689 3.116 2.786 2.567
6
937.1
39.33
14.73
9.197
6.978
5.820
5.119
4.652
4.320
4.072
3.881
3.728
3.604
3.501
3.415
3.341
3.277
3.221
3.172
3.128
2.408
7
948.2
39.36
1462
9.074
6.853
5.695
4995
4.529
4.197
3.950
3.759
3.607
3.483
3.380
3.293
3.219
3.156
3.100
3.051
3.007
2.288
8
956.7
39.37
1454
8.980
6.757
5.600
4.899
4.433
4.102
3.855
3.644
3.512
3.388
3.285
3.199
3.125
3.061
3.005
2.956
2.913
2.192
9
963.3
39.39
14.47
8.905
6.681
5.523
4.823
4.357
4.026
3.779
3.588
3.436
3.312
3.209
3.123
3.049
2.985
2.929
2.880
2.837
2.114
10
968.6
39.40
14.42
8.444
6.619
5.461
4.761
4.259
3.964
3.717
3.526
3.374
3.250
3.147
3.060
2.986
2.922
2.866
2.817
2.774
2.048
15
9849
39.43
14.25
8.657
6.428
5.269
4.568
4.101
3.769
3.522
3.330
3.177
3.053
2.949
2.862
2.788
2.723
2.667
2.617
2.573
1.833
20
993.1
39.45
14.17
8.560
6.329
5.168
4.467
3.999
3.667
3.419
3.226
3.073
2.948
2.844
2.756
2.681
2.616
2.559
2.509
2.464
1.708
oo
1018
39.50
13.90
8.257
6.015
4.894
4.142
3.670
3.333
3.080
2.883
2.725
2.596
2.487
2.395
2.316
2.247
2.187
2.133
2.085
1.000
Physical Constants
Gas constant
R
Boltzmann constant
k
Planck constant
h
Faraday constant
F
Avogadro constant
Lor Na
Speed of light in vacuum
c
Mole volume of an ideal gas
Vin
= 22.71 L mol (at 1 bar and 273.15 K)
Elementary charge
e
Rest mass of electron
Me
Rest mass of proton
Mp
Rest mass of neutron
Mn
Permitivity of vacuum
Eo
Gravitational acceleration
g
= 8.315) K+ molt
= 8.315 kPa dm? K+ mol?
= 8.315 Pa m? K? mol?
= 8.206 x 10% L atm K+ mol?
= 1.381x 1073) K+
= 6.626 x 1074) K+
= 9.649 x 10* C molt
= 6.022 x 10*3 mol
= 2.998 x 108 ms?
= 22.41 L mol (at 1 atm and 273.15 K)
= 1.602 x 10°C
= 9.109 x 10° kg
=1.673x 10?’kg
= 1.675 x 10°’ kg
= 8.854 x 10°C? J4m? (or Fm?)
=9.807 ms?
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Conversion Factors
1W
1J
1 cal
leV
1Latm
1 atm
1 bar
1L
1 Angstrom
1 micron (j1)
1 Poise
1 ppm
5 mp =
5 ott
Oe F
Cetf Cattf
7905916
AG? = -RT InK
n
=1Js*
= 0.2390 cal=1Nm=1VC
= 1Pam?=1kgm?s?
= 4,184J
= 1.602 x 1079)
= 101.3 J
= 1.013 x 10° N m? = 1.013 x 10° Pa =
760 mmHg
=1x10°Pa
=1m0 3 =17dim?
=1x107m =0.1 nm = 100 pm
=10°m=1 um
= 0.1 Pas=0.1N sm?
= 1g g*=1mgkg"*
= 1 mg L? (dilute aqueous solutions only)
Smp _
S ike
Cs C, V+Mo Vad +C,, VetVad Moa
po po ge B= F295
md
=
nm
10
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