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APP601S - ANALYTICAL PRINCIPLES AND PRACTICE - 1ST OPP - JUNE 2023 |
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nAm I BI A un IVERSITY
OF SCIEnCE
TECHnOLOGY
FACULTYOF HEALTH,NATURAL RESOURCESAND APPLIEDSCIENCES
DEPARTMENT OF BIOLOGY, CHEMISTRY AND PHYSICS
QUALIFICATION: BACHELOR OF SCIENCE
QUALIFICATION CODE: 07BOSC
LEVEL: 6
COURSE CODE: APP601S
COURSE NAME: ANALYTICALPRINCIPLESAND
PRACTICE
SESSION: JUNE 2023
DURATION: 3 HOURS
PAPER: THEORY
MARKS: 100
FIRSTOPPORTUNITY EXAMINATION QUESTION PAPER
EXAMINER(S) DR EUODIA HESS
MODERATOR: DR MARIUS MUTORWA
ATTACHMENT
List of Useful Tables, Formulas and Constants
THIS EXAMINATION PAPER CONSISTS OF 9 PAGES (Including this front page and
attachments)
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Question 1: Multiple Choice Questions
[30]
1.1 Which of the following glassware is not recommended for accurate measurements
of volumes?
(2}
(A} A graduated cylinder
(B} A volumetric flask
(C} A volumetric pipette
(D) A measuring pipette
1.2 A chemical or physical principle that can be used to study an analyte is called
(2)
(A} A technique
(B} A procedure
(C) A protocol
(D) A method
1.3 The ability of an analytical balance to measure the smallest detectable increment
of mass is called
(2)
(A) The balance accuracy
(B) The balance precision
(C) The balance sensitivity
(D) None of the above
1.4 In statistics, the precision of repeated measurements is characterised by
(2)
(A) The standard deviation
(B} The relative standard deviation
(C) The variance
(D) All of the above
1.5 An amphoteric substance
(2)
(A) Has neither acid or base properties
(B) Turns litmus paper red and blue
(C} Is insoluble in base, but dissolves in an acid
(D) Reacts with both an acid and a base
1.6 Consider the equilibrium reaction
4NH3(g) + 302(g) 2N2(g} + 6H20(g) tiH = -1268 kJ
Which change will cause the reaction to shift to the right?
(2)
(A) Increase the temperature
(B} Decrease the volume of the container.
(C} Add a catalyst to speed up the reaction.
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(D) Remove the gaseous water by allowing it to react and be absorbed by KOH.
1.7 Sodium nitrate, heated in the presence of an excess of hydrogen, forms water
according to the two-step process
2NaN03 - 2NaN02 + 02
2H2+ 02 - 2H20
From the reactions above, how many grams of sodium nitrate are required to form
9 grams of water?
(2)
(A) 21.3
(B) 42.5
(C) 69.0
(D) 85.0
1.8 What is the molarity of the sulphate ion in a solution prepared by dissolving 17 .1 g
of aluminium sulphate, Al2(S04)3,in enough water to prepare 1.00 L of solution?
Neglect any hydrolysis.
(2)
(A) 1.67 x 10-2 M
(B) 5.00 x 10-2 M
(C) 1.50 x 10-1 M
(D) 2.50 x 10-1 M
1.9 A reaction for which tlH < 0 and ilS < 0 is most likely to have which of these
thermodynamic properties?
(2)
(A) The reaction cannot be spontaneous at any temperature.
(B) The reaction will tend to be spontaneous at low temperatures.
(C)The reaction will tend to be spontaneous at high temperatures.
(D) The spontaneity of the reaction will be independent of temperature.
1.10 Consider the equilibrium reaction
Cd2+(aq) + 4NH3(aq)
Cd(NH3)i+(aq)
The equilibrium constant of the reaction is called
(2)
(A) Overall formation constant
(B) Stepwise formation constant
(C) Cumulative formation constant
(D) Both (A) and (C)
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Question 2
[20)
A researcher at NUST investigated the quantitative determination of Cr in high-alloy
steels using a potentiometric titration of Cr(VI). Before the titration, samples of the
steel were dissolved in acid and the chromium oxidized to Cr(VI) using peroxydisulfate.
Shown here are the results (as %w/w Cr) for the analysis of a reference steel.
16.968; 16.922; 16.840; 16.883; 16.887; 16.977; 16.857; 16.728
2.1 Calculate the mean of the measurements (to 5 significant figures).
(2)
2.2 Determine the median of the measurements
(2)
2.3 Calculate the standard deviation (2 significant figures).
(2)
2.4 Calculate the relative standard deviation (in%)
(2)
2.5 Comment on the precision of the measurements
(2)
2.6 Calculate the 95% confidence interval about the mean.
(4)
2.7 What does this confidence interval mean?
(2)
2.8 One of the measured values appears to be an outlier. Identify that value and use
The Dixon Q test to confirm whether that measurement is reliable at 95%
confidence level.
(4)
Question 3
[10)
A 10.0-g sample containing an analyte is transferred to a 250-ml volumetric flask and
diluted to volume. When a 10.0 ml aliquot of the resulting solution is diluted to 25.0 ml
it gives signal of 0.235 (arbitrary units). A second 10.0-ml portion of the solution is
spiked with 10.0 ml of a 1.0-ppm standard solution of the analyte and diluted to
25.0 ml. The signal for the spiked sample is 0.502.
3.1 What type of standardisation is used in this analysis? Explain your answer.
(2)
3.2 Express ppm in gram per litre (g/l)
(1)
3.3 Calculate the analyte concentration (in g/l) in the 10.00 ml aliquot.
(3)
3.4 Calculate the gram of analyte in the 250 ml solution.
(2)
3.5 Calculate the weight percent of analyte in the original sample.
(2)
Question 4
4.1 Write equilibrium constant expressions for the following reactions.
(a) AgCl(s) + 2NH3(aq) 0 Ag(NH3)i+(aq) + c1-(aq)
[20)
(1)
(1)
4.2 Calculate the standard state potential and the equilibrium constant for the following
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redox reaction: Mno4-(aq) + H2S03(aq) Mn 2+(aq) + soi-(aq) acidic solution.
Assume that [H30+] is 1.0 M.
E0 (MnQ4-/ Mn 2+) = +1.51 V and E0 (Sol-/SO/-) = +0.172 V
(8)
4.3 Calculate the ionic strength of a 0.050 M NaCl solution.
(2)
4.4 Calculate the pH of the solution that results from the addition of 0.040 moles of
HN03 to a buffer made by combining 0.500 L of 0.380 M HC3Hs02(Ka= 1.30 x 10- 5)
and 0.500 L of 0.380 M NaC3Hs02.Assume addition of the nitric acid has no effect
on volume.
(8)
Question 5
[20]
Calculate the pH at each point listed below for the titration of 100.0 ml of 0.100 M
cocaine, symbolised as "B", (Kb= 2.6 x 10-6) with 0.200 M HN03. The points to calculate
are at added HN03 volumes of 0.0, 10.0, 25.0, 50.0 and 60.0 ml.
5.1 0.0 ml of added HN03
(4)
5.2 10.0 ml of added HN03
(4)
5.3 25.0 ml of added HN03
(4)
5.4 50.0 ml of added HN03
(4)
5.5 60.0 ml of added HN03
(4)
END
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Data Sheet
l calc.11/ated
Jn d
fcalculated = -
Sd
Spooled=
s;(Na -l)+s~(N 6 -l)+ ....... .
Na+ Nb+ ...... -Nsetsofdata
Confidence
degrees
Freedom
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
50%
1.000
0.816
0.765
0.741
0.727
0.718
0.711
0.706
0.703
0.700
0.697
0.695
0.694
0.692
0.691
0.690
0.689
0.688
0.688
0.687
0.686
0.686
0.685
0.685
0.684
0.684
0.684
0.683
0.683
0.683
0.682
0.682
0.682
0.682
0.682
90%
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.696
1.694
1.692
1.691
1.690
95%
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
2.042
2.040
2.037
2.035
2.032
2.030
99%
63.656
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.744
2.738
2.733
2.728
2.724
JJ=X- ±- ts
Critical Values for the Rejection Quotient
Q,,,;t(Reject if C2.e>xpQ,,,;tl
N
90%
Confidence
95%
Confidence
99%
Confidence
3
0.941
0.970
0.994
4
0.765
0.829
0.926
5
0.642
0.710
0.821
6
0.560
0.625
0.740
7
0.507
0.568
0.680
8
0.468
0.526
0.634
9
0.437
0.493
0.598
10
0.412
0.466
0.568
N = number of observations
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F(0.05, anum, adenom) for a Two-Tailed F-Test
1
2
3
4
5
adenV.
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 14.88
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 4.965 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
5.922 4.508 3.903 3.559 3.333
20
5.871 4.461 3.859 3.515 3.289
00
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
14.62
9.074
6.853
5.695
4.995
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
14.54
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
984.9
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
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
Vm
= 22.71 L moI- 1 (at 1 bar and 273.15 K)
Elementary charge
e
Rest mass of electron
Rest mass of proton
Rest mass of neutron
Permitivity of vacuum
Gravitationa I acceleration
g
Conversion Factors
= 8.315 J K-1 moI- 1
= 8.315 kPa dm 3 K-1 moI- 1
= 8.315 Pa m 3 K-1 moI- 1
= 8.206 x 10-2 L atm K-1 moI- 1
= 1.381 x 10-23 J K-1
= 6.626 x 10-34 J K-1
= 9.649 x 104 C moI- 1
= 6.022 x 1023 moI- 1
= 2.998 x 108 m s-1
= 22.41 L moI- 1 (at 1 atm and 273.15 K)
= 1.602 X 10-19 C
= 9.109 X 10-31 kg
= 1.673 X 10-27 kg
= 1.675 X 10-27 kg
= 8.854 x 10-12 C2 J-1 m-1 (or F m-1)
= 9.807 m s-2
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lW
1J
1 cal
1 eV
1 L atm
1 atm
1 bar
1L
1 Angstrom
1 micron(µ)
1 Poise
1 ppm
ss:unp
C Vo
AV
f
C
A
_VV_..!!.C_s+td
V
V
f
f
= 1 J s-1
= 0.2390 cal = 1 N m = 1 V C
= 1 Pa m3 =1 kg m2 s-2
= 4.184 J
= 1.602 X 10-19 J
= 101.3 J
= 1.013 x 105 N m-2 = 1.013 x 105 Pa =
760 mmHg
= 1 x 105 Pa
= = 10-3 m3 1 dm 3
= 1 x 10-10 m = 0.1 nm= 100 pm
= = 10-6 m 1 µm
= 0.1 Pas= 0.1 N sm-2
= 1 µg g-1 = 1 mg kg-1
= 1 mg L-1 (dilute aqueous solutions only)
s=P
c,\\
cAv+VvO
+c
V
stdv+5.ldv
o
>id
o
5.ld
nG0 = -RT lnK
Eo = 0.05916 logK
n
Eo=Eo -Eo
ml
ai:
= E Eo _
0.05916
n
a1~,;Q
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PERIODIC TABLE OF THE ELEMENTS
18
l
2
H
1.00794 2
13
14
15
He
16
17 4.00260
34
5 6 7 8 9 10
Li Be
B C N 0 F Ne
6.941 9.01218
10.81 12.011 14.0067 15.9994 18.9984 20.179
11 12
Na Mg
22.9898 24.305 3
4
5
6
13 14 15 16 17 18
Al Si p s Cl Ar
7
8
9
10
11
12 26.9815 28.0855 30.9738 32.06 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
39.0983 40.08 44.9559 47.88 50.9415 51.996 54.9380 55.847 58.9332 58.69 63.546 6538 69.72 7259 74.9216 78.96 79.904 83.8
-0
°""'
(1)
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
\\0
0.....
\\0
85.4678 87.62 88.9059 91.22 92.9064 95.94 (98) 101.07 102.906 106.42 107.868 112.41 114.82 118.69 121.75 127.6 126.9 131.29
55 56 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba Lu Hf Ta w Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
132.905 137.33 174.967 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204383 207.2 208.908 (209) (210) (222)
87 88 103 104 105 106 107 108 109 110 111 112
114
116
118
Fr Ra Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub
Uuq
(223) 226.025 (260) (261) (262) (263) (264) 265) '268) (269) (272) (269)
Uuh
Uuo
Lanthanides: 57 58 59 60 61 62 63 64 65 66 67 68 69 70
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
138.906 140.12 140.908 144.24 (145) 150.36 151.96 157.25 158.925 162.50 161.930 167.26 166.934 173.04
Actinides:
89 90 91 92 93 94 95 96 97 98 99 100 101 102
u Ac Th Pa
Np Pu Am Cm Bk Cf Es Fm Md No
227.028 232.038 231.036 238.029 237.048 (244) (243) (247) (247) (251) (252) (257) (258) (259)