-1 Universiti I j Malaysia PAHANG Engineering . Technology Creativity

FACULTY OF CHEMICAL & NATURAL RESOURCES ENGINEERING

FINAL EXAMINATION

COURSE : ANALYTICAL CHEMISTRY

COURSE CODE : BKF1243

LECTURER : EMAN N. ALl

FARHAN BT MOHD SAID DATE : 18 JUNE 2014

DURATION : 3 HOURS

SESSION/SEMESTER : SESSION 2013/2014 SEMESTER II

PROGRAMME CODE : BKB/BKC/BKG

INSTRUCTIONS TO CANDIDATE:

1. This question paper consists of FOUR (4) questions. Answer ALL questions. 2. All answers to a new question should start on new page. 3. All the calculations and assumptions must be clearly stated. 4. Candidates are not allowed to bring any material other than those allowed by

the invigilator into the examination room.

EXAMINATION REQUIREMENTS: 1. APPENDICES - List of Formulas and Statistic Tables

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

This examination paper consists of NINE (9) printed pages including front page.

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QUESTION 1

a) Two different laboratories were participated in an inter-laboratory study involved

the determination of the Ca level in water samples. The following results are

replicated determination of the Ca for laboratories A and B in ppm.

Table 1: Ca levels of water sample of Sungai Gebeng for Laboratories A and B

Result Ca level (ppm) Ca level (ppm)

no. Laboratory A Laboratory B

1 10.1 10.6 2 10.0 10.5 3 10.4 11.1

i) State the appropriate hypothesis of the above analysis

(2 Marks)

ii) Conduct a statistical test to determine whether the laboratories differ at the

95% confidence level?

(12 Marks)

iii) Do they differ at 99% confidence level (F 11 = 18)?

(2 Marks)

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b) Ammoniacal nitrogen can be determined by a treatment of the sample with

chioroplatinic acid. The product is slightly soluble ammonium chioroplatinate:

H2PtC16 + 2NH4 (NHPtC16 + 2H

The precipitate was then decomposed on ignition, yielding metallic platinum and

gaseous products (02, NH3, HCI).

i) Perform the complete reaction of the precipitation when the ignition process

complete.

(2 Marks)

ii) Determine the weight and percentage of ammonia in a sample, if 0.2115 g gave

rise to 0.4693 g of platinum.

(7 Marks)

QUESTION 2

a) A solution of Ba(OH) 2 was standardized against 0.1175 g of primary standard grade

benzoic acid, C61-15COOH (122.12 g!mol). The end point was observed after

additional of 45 ml of base.

i) Determine the molarity of the base.

(7 Marks)

ii) If the standard deviation of weighing was ±0.2 mg and for the volume was

±0.03 ml, determine the standard deviation/absolute uncertainty of the molarity

of the base.

(5 Marks)

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b) Titrimetric method based on silver nitrate is sometime called argentometric method.

As a new staff member in the Fleischmann's Company, you are asked to calculate

and sketch the titration curve of p1 for the titration of 50.0 ml of 0.0250M KI with

0.0500M AgNO 3 . To sketch the curve, you should cover three reactions, which are

pre-equivalence, equivalence and post- equivalence.

Given: For AgI, Ksp = 8.3x10'7

(13 Marks)

QUESTION 3

Spectroscopic analytical method is one of analytical methods which are based on

measuring the amount of radiation produced or absorbed by molecular or atomic species

of interest. The absorbance and concentration relationship should follow Beer's Law.

a) The complex formed between Cu(I) and 1,10-phenanthroline has a molar

absorptivity of 7000 L cm -1 mo1 1 at 435 nm, the wavelength of maximum

absorption. Calculate:

i) The absorbance of a 6.77 X 10 M solution of the complex when measured in

a 1.00 cm cell at 435 nm.

(4 Marks)

ii) The percent transmittance of the solution in (a).

(4 Marks)

iii) The concentration of a solution that in a 5.00 cm cell has the same absorbance

as the solution in (a).

(3 Marks)

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iv) The path length through a 3.40 X 10 M solution of the complex that is needed

for an absorbance that is the same as the solution in (a).

(3 Marks)

b) Identify factors that cause the Beer's law relationship to depart from linearity.

(5 Marks)

c) A sophisticated ultraviolet/visible/near-JR instrument has a wavelength range of

185 to 3000 nm. What are its wave number and frequency ranges?

(4 Marks)

d) Sketch the energy level diagram in case of absorption and emission.

(2 Mark)

QUESTION 4

Chromatography analytical method is the most general and common separation

technique, where separation, identification and determination of compound can be done.

a) The following data were obtained by gas-liquid chromatography on a 40 cm packed

column.

Compound tR (mm) W(min)

Air 1.9

Methylcyclohexane 10.0 0.76

Methylcyclohexane 10.9 0.82

Toluene 13.4 1.06

Calculate:

i) An average number of plates from the data.

(5 Marks)

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ii) An average plate height for the column.

(3 Marks)

iii) The resolution for Methylcyclohexane and Methylcyclohexane.

(3 Marks)

iv) The resolution for Methylcyclohexane and Toluene.

(2 Marks)

v) If the resolution 1.5 is desired in separating Methylcyclohexane and

Methylcyclohexane. How many plates are required to get the required

resolution?

(3 Marks)

vi) How long must the column be if the same packing is employed?

(3 Marks)

vii) What is the retention time for Toluene on the column in part (vi).

(3 Marks)

b) Sketch the three main parts in gas chromatography.

(3 Marks)

END OF QUESTION PAPER

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APPENDICES

Appendix A

Table 2: F-table for one-tailed test at 95% confidence level

FTabIe forOne-Tajied lest at c 005 (95% Confidence Level)

.1 2 3 4. , 5 - 6 7 8 9 10 15 20

1 161 4 199.5 215.7 224.6 230.2 - 2.34.0 236.8 238.9 240.5 2439 2459 248.0 2543 2 18.51 19.00 19.16 19.25 1930 1933 19.35 1937 19.38 19.40 19.43 19.45 19.50 -3 10.13 9.552 9.277 9.117 9013 8941 8.587 8.945 8.812 8.786 8-703 9.660 8526 4 7.709 6.941 6591 6383 6.256 6.153 60-4 6.041 5.999 5364 5.858 5303 5.628

6.608 5.786 5-409 S 192 5050 4.950 4.876 4.613 4.772 4.733 4.619 4.558 4365 5.987 5143 4.757 4.534 37 4394 4.207 4.147 4, 0-59 4.060 3933 3.874 3.669

7 5391 4337 4.347 4.120 1972 3.866 3.787 3.726 3.677 1637 3.511 3.445 3.230 8 5,318 4.459 4,066 3.838 3.687 3.581 3.500 3438 3.3SS 3.347 1210 3.150 2.928 9 5.117 4.256 1863 3.633 3482 3,374 1293 3.230 3.179 3,137 3.005 2.938 2.707

10 4.965 4.103 3.708 3.479 3326 3.217 3.135 3.072 3.020 2.978 2,845 2.774 2S38 11 4.844 3.982 3.587 3.337 3.204 3.095 3012 2.948 2.896 2854 2.719 2 646 2.404 12 4.747 3.885 3.490 3.259 3106 2.996 2.913 2.849 2.796 2.753 2.617 2.544 2.296 13 4.667 3.806 3.411 3.179 3.025 2,915 2.832 2,767 2.714 2671 2.533 2.459 2.206 14 4.600 3.739 3.344 3.112 2.958 2.848 2.764 2.699 2.646 2.602 2.463 2,388 2.131 15 4.534 3.682 3.287 3.056 2.901 2.790 2.707 2.641 2.583 2.544 2.403 2.328 2.066

16 4.494 3.634 3.239 3.007 2.852 2.741 2.657 2.591 2.538 2.494 2.352 2.276 2.010

17 4451 3.592 3.197 2.965 2810 2.699 3614 2.548 2.494 2.450 2.308 2.230 1.960

18 4.414 3.555 3.150 2.928 2.773 2661 2.577 2.510 2,456 2412 2.269 2.191 1,917

19 4.381 3.552 3.127 2.895 2.740 2.628 2,544 2477 2.423 2.378 2.234 2.155 1.878

20 4.351 3,493 3.098 2865 2.711 2.599 2.514 2.447 2.393 2.348 2.203 2.124 1.843

3.842 2.996 2.605 2.372 2.214 2.099 2.010 1.938 1,880 1.831 1,656 1.570 1.000

= degrees of freedom in numerator v = degrees of freedom in denominator

Table 3: F-table for two-tailed test at 95% confidence level

F-Table for Two-Tailed Test at a = 0.05 (95% Confidence Level)

v2 lv i a 1 2 3 4 5 6 7 8 9 10 15 20 °°

1 647.8 799.5 864.2 899.6 921.8 937.1 948.2 956.7 9613 9.68.6 984.9 993.1 1018

2 38.51 39,00 39.17 39.25 39.30 39.33 39.36 39.37 39.39 39.40 3943 3945 39.498

3 17.44 16.04 15.44 15.10 14.88 14.73 14.62 14.54 14.47 14.42 14.25 14.17 1.3.902

4 12.22 10.65 9.979 9,605 9.364 9.197 9.074 8380 8.905 8.844 8.657 8.560 8.257

5 10.01 8,434 7.764 7.388 7.146 6.978 6.853 6.757 6.681 6.619 6:428 6329 6.015

6 8.813 7.260 6.599 6.227 5.988 5.820 5.695 5.600 5.523 5.461 5.269 5.168 4.849

7 8.073 6.542 5.890 5.523 5.285 5.119 4.995 4.899 4.823 4.761 4.568 4.467 0.142

8 7.571 6.059 5.416 5.053 4.817 4.652 4.529 4,433 4.357 4.295 4.101 3.999 3.670

9 7.209 5.715 56078 4.718 4.484 4.320 4.197 4.102 4.026 3.964 3,769 3.667 3333

10 6:937 5.456 4.826 4.468 4.236 4.072 3.950 1855 3.779 3.717 3.522 1419 1080

11 6.724 5156 4630 4.275 4.044 3.881 3.759 3,664 3.588 3.526 3.330 3.226 2.883

12 6.544 5.096 4474 4.121 3.891 3.728 3.607 3.512 3.436 3.374 3.177 3.073 2.725

13 6.414 4.965 4.347 3.996 3.767 3.604 1483 3.388 3.312 3.250 3.053 2.948 2.596

14 6.298 4.857 4.242 3.892 3663 3.501 3380 3.285 1209 3,147 2149 2.844 2.487

15 6,200 4.765 4.153 3.804 3.576 3.415 3.293 3.199 3,123 1060 2.862 2.756 2.39.5

= degrees of freedom in numerator; v 2 = degrees of freedom in denominator

VA

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IA The Periodic Table of the Elements WA VA VIA VIIA 4.003 hA

- 4

LuBe

lilA 5

MT 8 9

. B 0F10.81 99,00 20.98

m ri

Na I MgVB VIB VflB VIJIB

13 1i 14 i1 15 I 05 17 is Al Si P S Cl Ar

III HR 20.98 30.08 34.97 32.07 25.45 38.95 2295 L 2431 IIIB 1V13

19 20 21 22 23 24 26 20 77 28

K Ca Sc Ti V Cr Mn Fe Co Ni 44.56 47.61 60.94 52,05 54.44 05.55 55.93 68.69

29 343 31 32 33 34 35 345

Cu Zn Ga Ge As Se Br Kr '82.00 50.39 68.72 72.61 14.92 18.90 78.90 8300

39.10

E8547

40.00

30

Sr39

Y40

Zr41

Nb47

Mo 95.54

43

Tc(50)

44

109,1

RPd 06

102,9

.16

'(00.4

17

Ag 107.0

49

Cd 112.4

49 (4 54

in Sn Sb Te I 114.3 1001 121.43 877,8 026.5 138.3

55

Cs

87.62

55

Ba

58.91

57

La

9122

72

Hf

92.91

73

Ta

74

W

75

Re 160.2

75

Os

850.2

77

Ir 192.2

75

Pt

195.1

79

Au

197.0

85

Hg 2006.

51 92 83 84 85 88

TI Pb Bi Po At Rn 204.4 227.2 005.0 (200) (210) 222)

1325

FrlIRa

137.3 130.4

59

Ac

178.5

104

Rf

150.9

106

1Db

153.5

(06

Sg

107

Bh

150

Hs

108

Mt

110 1

UunUuu Iii 112

Uub

55 59 (.0 -51 04 3 ... . --

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

1 5409 54 (145) 04 20 3 9

('0 91 0 04 ]

67

Th Pa Pu' mBkfs212.0 (231) (244) (043) (Li

(24fl (251) (252) (257) 258 (259) [i!

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Appendix B

hc- Po PT

E=hv=—=hcv 2

R = 2AZ 2[(tR)B—(tR)A]

WA + WB WA+WB - Z

= X -

kA - tR tM

tM

XThLI0

(tR)3-tM

(tR)A tM

(tR)l - (R)

N=16(t/w)2(t (R,- (R)

-;-

(R) 1 -xp_

(R)2jIN

N= L/H s=1J j1

N—i

- zSR -SA+SB

ii= X±—

I 2 2

SA/('4) +1

Fexp = R1,ftA) B) s

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