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Universiti Malaysia PAHANG Engineering Technology . Creativity FACULTY OF CHEMICAL & NATURAL RESOURCES ENGINEERING FINAL EXAMINATION COURSE : ANALYTICAL CHEMISTRY COURSE CODE : BKF1243 LECTURER : EMAN N. ALl FARHAN BINTI MOHD SAID DATE : 7 JANUARY 2015 DURATION : 3 HOURS SESSION/SEMESTER : SESSION 2014/2015 SEMESTER I 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: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO This examination paper consists of SEVEN (7) printed pages including front page.

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Page 1: BKF1243-ANALYTICAL CHEMISTRY 11415.PDF

Universiti Malaysia PAHANG Engineering Technology . Creativity

FACULTY OF CHEMICAL & NATURAL RESOURCES ENGINEERING

FINAL EXAMINATION

COURSE : ANALYTICAL CHEMISTRY

COURSE CODE : BKF1243

LECTURER : EMAN N. ALl

FARHAN BINTI MOHD SAID

DATE : 7 JANUARY 2015

DURATION : 3 HOURS

SESSION/SEMESTER : SESSION 2014/2015 SEMESTER I

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:

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

This examination paper consists of SEVEN (7) printed pages including front page.

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CONFIDENTIAL

BKB/BKCIBKG/141511BKF1243

QUESTION 1

Evaluating data of any analysis is considered an important step in analytical chemistry.

Analysis of several plant-food preparations for potassium ion yielded the following data:

Sample Percent K* 1 5.15, 5.03, 5.04, 5.18, 5.20 2 7.18, 7.17, 6.97 3 4.00, 3.93, 4.15, 3.86 4 4.68, 4.85, 4.79, 4.62 5 6.04, 6.02, 5.82, 6.06, 5.88

The preparations were randomly drawn from the same population.

a) Find the mean and standard deviation for each sample.

(12 Marks)

b) Obtain the pooled value Spooled.

(4 Marks)

c) Why the result at (b) is a better estimate of a than the standard deviation from

any one sample.

(2 Marks)

d) Check if you have any outlier value for the sample 5 at confidence level 95%.

(5 Marks)

e) What is the median value for sample 4?

(2 Marks)

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QUESTION 2

a) Steel which only contains Ag and Cu was analyzed by dissolving the steel in HNO3,

adding 10 3 in excess and weighing the mixture of A910 3 and Cu(10 3 )2 formed after

filtering process. If the steel sample weighs 0.2175 g and the weight of precipitate

formed is 0.7391 g, determine the content percentage of steel (Ag, Cu).

(12 Marks)

b) Determine the pH of a solution prepared by mixing 0.250 mol of acetic acid and

0.100 mol sodium acetate and diluted to 500 ml. Then, calculate the pH when 10.0

ml of this solution is diluted to 500 ml?

Ka for acetic acid is 1.75x 10.

(11 Marks)

c) Determine [Hf] for a solution of pH 10.25.

(2 Marks)

QUESTION 3

Measurements based on light and other forms of electromagnetic radiation are widely

used throughout analytical chemistry one of which is spectroscopy method.

a) Define spectroscopy.

(2 Marks)

b) Absorption spectroscopy based on UV & Visible and JR radiation is one of the

most useful tools available for quantitative and qualitative analysis:

i) List the important characteristics of spectrophotometric that make it very

good tool in quantitative analysis?

(5 Marks)

ii) What are the qualitative applications for UV Visible spectroscopy?

(2 Marks)

iii) What kind of compounds can be identified using IR spectroscopy?

(2 Marks)

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iv) Why the AAS is currently the most widely used of all spectroscopy

methods?

(3 Marks)

c) The molar absorptivity for the complex formed between bismuth (III) and

thiourea is 9.32 X 10 3 LIcm. mol at 470 nm, the concentration is 1.07 X 10-5M.

Calculate:

i) The absorbance of the complex at 470 nm in a 1.00 cm cell.

(2 Marks)

ii) The absorbance of the complex using 5.00 cell.

(2 Marks)

iii) The transmittance of the solutions described in (i), and (ii).

(2 Marks)

iv) What is the conclusion from results obtained in (i), (ii), and (iii)?

(2 Marks)

d) List THREE (3) main components in Atomic Absorption Spectroscopy (AAS).

(3 Mark)

QUESTION 4

The goals of an analytical separation are usually to eliminate or reduce interferences so

that quantitative analytical information can be obtained about complex mixtures. One of

the separation methods is chromatography technique.

a) Define each of the following?

i) Stationary phase.

(2 Marks)

ii) Retention Time.

(2 Marks)

iii) Selectivity factor.

(2 Marks)

b) What is the difference between gas-liquid and liquid-liquid chromatography?

(2 Marks)

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c) List the variables that lead to:

i) Band broadening.

(5 Marks)

ii) Band separation in gas-liquid chromatography.

(2 Marks)

d) Liquid chromatography with the following date is used for an analysis.

Length of Packing 24.70 cm Flow rate 0.313 mL/min

A chromatogram of a mixture of species A, B, C, and D provided the following data:

Retention Time (mm) Width of Peak Base (W), (mm)

Non retained 3.1 A 5.4 0.41 B 13.3 1.07 C 14.1 1.16 D 21.6 1.72

Calculate: i) The number of plates from each peak.

(2 Marks)

ii) The plate height for the column.

(2 Marks)

iii) The resolution for species B and C.

(1.5 Marks)

iv) The selectivity factor for species B and C.

(1.5 Marks)

v) The length of column necessary to separate the species B and C with a

resolution of 1.5.

(1.5 Marks)

vi) The time required to separate the two species on the column in part (v).

(1.5 Marks)

END OF QUESTION PAPER

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APPENDICES IV

xi =1=1

N

IN I— (x—x) 2

S = 1! 1=1

N—i

texp =S

_x2—x1 exp -

Xn -Xi

= xn - X11-1Qexp

-

A = —logT = _Iog L = log-- = abC = ebC PO PT

R = 2AZ 2[(tR)B—(tR)A]

WA + WB WA+WB

F tRtM kA -

a = (tR)B — tM

(tR)A tM

N=

N= L/H

(R)1

(R)2 1/Ic:;-

(tR)l - (R)

(tR )2 (R)

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Critia1 Vla !ucc cr t P. j. in O oterit Q

Q11 (Rtjt1 if Q

Number or Observations 90% Confidence 959fci Ciimlidence 99% Confidence

3 0.941 0.970 0994 4 0.765 0.829 0.926 S o4)

(M I() O82I

6 0-56() 0.625 0.740 7 0507 0.569 0.680

0.326 0.634 9 0.437 0.493 0.598

10 0.412 0.466

IA The Periodic Table of the Elements

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