HOOF- EN QWAQWA KAMPUS / MAIN- AND QWAQWA CAMPUS
CEM 314
DIE UNIVERSITEIT VAN DIE VRYSTAAT / UNIVERSITY OF THE FREE STATE
TOETS 2 / TEST 2 - 17 APRIL 2010
ASSESSORE / ASSESSORS:
DR. K. VON ESCHWEGE
MR. T.A. TSOTETSI
MODERATOR / MODERATOR: PROF. W. PURCELL
TYDSDUUR / DURATION: 2 URE / HOURS
PUNTE / MARKS: 85
BEANTWOORD ALLE VRAE EN TOON ALLE BEREKENINGE (TENSY ANDERS GESPESIFISEER) MET
INKPEN. GEEN POTLODE OF "TIPPEX" TOEGELAAT NIE. VRAESTEL BESTAAN UIT 4 BLADSYE EN 16
VRAE. AFRIKAANSE VRAE WORD GEVOLG DEUR ENGELSE VRAE. SAKREKENAARS TOEGELAAT. /
ANSWER ALL QUESTIONS AND SHOW ALL CALCULATIONS (UNLESS SPECIFIED OTHERWISE) WITH AN
INK PEN. NO PENCILS OR "TIPPEX" ALLOWED. PAPER CONSISTS OF 4 PAGES AND 16 QUESTIONS.
ENGLISH QUESTIONS FOLLOW AFRIKAANS QUESTIONS. POCKET CALCULATORS ALLOWED.
1. Verduidelik kortliks watter benadering gevolg kan word om ‘n oplossing wat twee katione bevat m.b.v. EDTA
te analiseer. / Explain briefly what approach may be followed to analyze two different cations in the same
solution during the use of EDTA as analytical reagent.
(2)
A masking agent like CN- may be used ; OR else selectively adjust the pH .
2. Bepaal die oorgangsgebied vir Eriochroomswart-T indikator in die titrasie van Mg2+ by pH 10.00, gegee dat die
tweede dissosiasiekonstante vir die indikator is: / Determine the transition ranges for Eriochrome Black T
indicator in titration of Mg2+ at pH 10.0, given that the second acid dissociation constant for the indicator is
HIn2- + H2O ↔ In3- + H3O+
K2 = 2.8 x 10-12,
en die vormingskonstante vir MgIn- / and the formation constant for MgIn- is
Mg2+ + In3- ↔ MgIn-
Kf = 1.0 x 107.
(6)
Multiplication of K2 for the indicator by Kf for MgIn- gives an expression that contains this ratio:
[MgIn-][H3O+]/{[HIn2-][Mg2+]} = 2.8 x 10-12 x 1.0 x 107 = 2.8 x 10-5 
Which rearranges to:
[Mg2+] = [MgIn-]/[HIn2-] x [H3O+]/2.8 x 10-5 
The substitution of 1.0 x 10-10 M for [H3O+]; and 10 & 0.10 for the ratio  yields the range of [Mg2+] over
which colour change occurs:
[Mg2+] = 3.6 x 10-5 M to 3.6 x 10-7 M 
pMg = 5.4 ± 1.0 
3. Kalamien, wat gebruik word om velirritasies mee te behandel, is ‘n mengsel van sink- en ysteroksiede. ‘n
0.1022 g monster droë kalamien is opgelos in suur en verdun tot 250.0 mL. KF is by ‘n 10.00 mL monster van
die verdunde oplossing gevoeg om die yster te maskeer; na geskikte aanpassing v/d pH het Zn2+ 38.71 mL van
‘n 0.01294 M EDTA oplossing in ‘n titrasie gebruik. ‘n Tweede 50.00 mL monster is na wense gebuffer en
getitreer met 2.40 mL van ‘n 0.002727 M ZnY2- oplossing:
/ Calamine, which is used for relief of skin
irritations, is a mixture of zinc and iron oxides. A 1.022 g sample of dried calamine was dissolved in acid
and diluted to 250.0 mL. KF was added to a 10.00 mL aliquot of the diluted solution to mask the iron; after
suitable adjustment of the pH, Zn2+ consumed 38.71 mL of 0.01294 M EDTA. A second 50.00 mL aliquot was
suitably buffered and titrated with 2.40 mL of 0.002727 M ZnY2- solution:
Fe3+ + ZnY2- ↔ FeY- + Zn2+
Bereken die persentasies ZnO en Fe2O3 in die monster. /
Calculate the percentages of ZnO and Fe2O3 in the sample.
(6)
 0.01294 mmol EDTA
1 mmol ZnO 81.39 g ZnO 


 38.71 mL EDTA 

mL
mmol
EDTA
1000
mmol

  100%
10.00 mL
1.022 g sample 
250.0 mL
 99.7% ZnO
 0.002727 mmol ZnY 2
1 mmol Fe 2 O 3 159.69 g Fe 2 O 3 


 2.40 mL ZnY 2 

2
mL
2
mmol
ZnY
1000
mmol

  100%
50.00 mL
1.022 g sample 
250.0 mL
 0.256% Fe 2 O 3
4. Gee ‘n kort beskrywing van absorbansie verskynsels in / Give a short description of absorbance phenomena
in
4.1 anorganiese komplekse, en / inorganic complexes, and
(2)
4.2 ladingsoordrag komplekse / charge-transfer complexes.
(2)
4.1
The ions and complexes of elements in the first row transition series absorb broad bands of
visible radiation in at least one of their oxidation states and are, as a result, coloured.  Here, absorption
involves transitions between filled and unfilled d-orbitals with energies that depend on the ligands
bonded to metal ions. 
4.2
The charge-transfer complex consists of an e-donor group bonded to an e-acceptor. When this
product absorbs radiation, an e from the donor is transferred to an orbital that is largely associated with
the acceptor. Excited state = product of type of internal redox process.  (For quantitative purposes,
charge-transfer absorption is particularly important because molar absorptivities are unusually large (ε >
10 000), which leads to high sensitivity.)
2
5. Die molare absorptiwiteit vir ‘n kompleks wat gevorm word tussen bismut(III) en tiorea is 9.32 x 10 3 L cm-1
mol-1 by 470 nm. Bereken die reeks toelaatbare konsentrasies vir die kompleks indien die absorbansie nie
minder as 0.10 of groter as 0.90 mag wees vir metings in ‘n 1.00-cm sel. / The molar absorptivity for the
complex formed between bismuth(III) and thiourea is 9.32 x 103 L cm-1 mol-1 at 470 nm. Calculate the range
of permissible concentrations for the complex if the absorbance is to be not less than 0.10 nor greater than
0.90, when the measurements are made in 1.00-cm cells.
(3)
A = εbc
cmin = A / ε b = 0.10 / (9.32  103  1.00) = 1.1  10-5 M 
cmax = A / ε b = 0.90 / (9.32  103  1.00) = 9.7  10-5 M 
6. Verduidelik die verskil tussen ‘n enkelstraal en ‘n dubbelstraal UV/sigbare spektrofotometer? /
Explain the difference between a single-beam and a double-beam UV/visible spectrophotometer?
(2)
The double-beam instrument has 2 light paths / beams measuring the reference/blank solution
simultaneously with the sample solution, and do the blank subtraction automatically. 
7.
UV/sigbare absorpsie spektroskopie is een van die mees bruikbare tegnieke beskikbaar aan die chemikus vir
analise. Bespreek, met spesifieke verwysing na voorbeelde van anorganiese en organiese reagense, die
toepassing van hierdie tegniek op nie-absorberende verbindings. Teken ook 'n struktuur van 'n tipiese
chromofoor om u antwoord verder toe te lig. / UV/visible absorption spectroscopy is one of the most useful
tools available to the chemist for analysis. Discuss, with specific reference to examples of inorganic and
organic reagents, the application of this technique to nonabsorbing species. Draw also a structure of a
typical chromophore as further explanation to your answer.
(5)
8. Die enkelpunt standaardaddisiemetode word soms gebruik om ‘n onbekende konsentrasie, Cx,
spektrofotometries te bepaal. Tydens hierdie metode word identiese volumes, Vx, van die onbekende
oplossing, met konsentrasie Cx, gevoeg in twee volumetriese flesse met volume Vt, elk. In die een fles is ‘n
volume Vs, van die standaard oplossing met bekende konsentrasie Cs, gevoeg. Die kleurreagense word dan
3
bygevoeg en elke oplossing verdun tot by die merk. Lei ‘n vergelyking af vir Cx./ The single-point standard
addition method is sometimes used to spectrophotometrically determine the unknown concentration Cx of
a sample. In this method identical aliquots Vx of the unknown solution with concentration Cx are
transferred to two volumetric flasks each of volume Vt. To the one flask is added a volume Vs of a standard
solution having a known concentration Cs. The colouring reagents are then added and each solution diluted
to the mark. Derive an equation for Cx.
(5)
MEMO: For the unknown alone, we can write Beer’s law in the form
Ax = εbcxVx / Vt 
where Vt is the total volume of solution.
For the solution after standard addition:
As = εb (cxVx + csVs) / Vt 
Dividing the first equation by the second gives
Ax
c xV x

As c xV x  c sVs

AxcxVx + AxcsVs = AscxVx
This equation rearranges to
cx (AsVx – AxVx) = AxcsVs 
cx =
Ax c sVs

Vx ( As  Ax )
9. Yster(III) vorm ‘n kompleks met tiosianaat, met die formule Fe(SCN)2+. Die kompleks het ‘n absorpsie
maksimum van 580 nm. ‘n Boorgatwatermonster is ondersoek op die wyse soos hier onder in die tabel
aangedui. Bereken die yster konsentrasie in dele per miljoen in die boorgatwater. / Iron(III) forms a complex
with thiocyanate, with formula Fe(SCN)2+. The complex has an absorption maximum at 580 nm. A sample of
well water was assayed according to the scheme shown in the table here below. Calculate the
concentration of iron in parts per million in the well water.
(3)
Volumes (mL)
Monster
Sample
Oxidizing
Fe(II)
KSCN
Sample
Volume
reagent
2.75 ppm
0.050 M
H2O
Absorbance,
580 nm
1
50.00
5.00
5.00
20.00
20.00
0.549
2
50.00
5.00
0.00
20.00
25.00
0.231
cx =
Ax c sVs
0.231  2.75  5.00
=
50.0(0.549  0.231)
Vx ( As  Ax )

= 0.200 ppm Fe 
4
10.
'n Spektrofotometriese titrasie van 10.00 ml Pd(II)-oplossing met 2.44  10-4 M Nitroso R is in 1.00
cm kuvette uitgevoer. Die kompleks [Pd(Nitroso R)2] is gevorm. Bereken die Pd(II)-konsentrasie
vanaf onderstaande data (handig grafiekblaai in):/ A spectrophotometric titration of 10.00 ml Pd(II)
solution with 2.44  10-4 M Nitroso R reagent in 1.00 cm cuvettes was performed. The [Pd(Nitroso
R)2] complex was formed. Calculate the Pd(II) concentration from the data below (graph must be
handed in):
(6)
MEMO:
The data must be corrected for dilution so:
For 1.00 ml,
Vol Nitroso R (ml)
0
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
A500 ongekorr/uncorr
0
0.147
0.271
0.375
0.371
0.347
0.325
0.306
0.289
5
A500 corrected
0
0.162
0.325
0.488
0.519
0.521
0.520
0.520
0.520
Palladium(II) concentration:
11. Bereken die potensiaal van die volgende sel. Dui aan of die reaksie spontaan in die aangeduide rigting sal
plaasvind (LR). / Calculate the potential of the following cell. Indicate whether the reaction will proceed
spontaneously in the direction considered (LR).
(4)
PtTiO2+ (0.0790 M), Ti3+ (0.00918 M), H+ (1.47 x 10-2 M) VO2+(0.1340M), V3+(0.0784 M), H+ (0.0538 M) Pt
Eo (TiO2+ + 2H+ + e-  Ti3+ + H2O) = +0.099 V
Eo (VO2+ + 2H+ + e-  V3+ + H2O) = +0.359 V


0.0784
  0.223 V
E right  0.359  0.0592 log 
2 
 0.13400.0538 


0.00918
  0.063 V
Eleft  0.099  0.0592 log 
 0.07901.47  10 2 2 


Ecell  E right  Eleft  0.223  ( 0.063)  0.286 V
Because Ecell is positive, the reaction proceeds spontaneously in the direction considered.
12. 12.1 Bepaal 'n ewewigskonstante uitdrukking vir die volgende reaksie. Bereken die numeriese waarde van
Keq. / Generate an equilibrium constant expression for the following reaction. Calculate numerical
values for Keq.
(4)
2V(OH)4+ + H3SO3  SO42- + 2VO2+ + 5H2O
(0.100 M H+)
Eo` (V(OH)4+ + 2H+ + e-  VO2+ + 3H2O)
= +1.70 V
Eo` (SO42-+ 4H+ + 2e-  H2SO3 +H2O)
= +0.450 V

0.0592 
[ VO 2 ]2
0.0592  [ H 2SO 3 ] 



0.45
V 
1.7
V 
1.00
log 

0
.
172
log 
 2
2
 4 
 4 
2
2
 [ V(OH ) 4 ] [ H ] 
 [SO 4 ][ H ] 

(1.7- 0.45
V)  2
1.00
0.172
 log 
2
[ VO 2 ]2 [SO 4 ] 

42.23
.97
 [ V(OH )  ]2 [ H SO ]   log K eq  27
4
2
3 

0.0592
2
[ VO 2 ]2 [SO 4 ]
27
42
 K eq  91.7.4x1010
 9  10 27
 2
[ V(OH ) 4 ] [ H 2SO 3 ]
6
12.2 Bereken die electrode potensiaal van die sisteem by die ewewigspunt vir bg. Reaksie. / Calculate the
electrode potential of the system at the equivalence point for the above reaction.
(4)
At equivalence, [V(OH)4+] = 2[H2SO3] and [VO2+] = 2[SO42-]


[ VO 2 ]

Eeq  11.7
.00  0.0592 log 

 2 
[
V
(
OH
)
][
H
]
4


 [ H 2SO 3 ] 

2 Eeq  20.0.45
172   0.0592 log 
2
 4 
[
SO
][
H
]
4


3Eeq


[ VO 2 ][ H 2SO 3 ]


2

 [ V(OH ) ][SO ][ H  ]6 
4
4


 1.1.7
00 +2(0.45)
20.172   0.0592 log 


1
  1.2.6
V 1 V
3Eeq  1.1.7
00 +2(0.45)
20.172   0.0592 log 
344  0.355  9.2.245
89  10
6 
 0.100 
V 1
92.245
.89  10
Eeq 
 00.748
.330 VV
3
V1.7 V
13. Skets 'n tipiese elektrodeopstelling (nie kombinasie nie) met byskrifte vir pH-metings./
Sketch a typical electrode system with captions (not combination electrode) for pH
measurements.
(5)
(5)
14. 14.1 Lei 'n vergelyking af (vanaf die basiese beginsels waarop die KMR verskynsel gebaseer is) vir die Larmorfrekwensie (i) van kerne met spin-kwantum getal = ½ / Derive an equation (from the basic principles
on which the NMR phenomenon is based) for the Larmor frequency (i) of nuclei with spin quantum
number = ½.
(5)
7
14.1 The relative energy (E) of each orientation is dependent on its magnetic quantum number (m I), the
gyro magnetic ratio (), which is a constant for a particular isotope (e.g. 1H  = 26.7519), the magnetic field
strength (H0) and Planck's constant (h).
E = -(mI)(H0h/2) 
Nuclei with I = ½ therefore have two possible orientations with two energy levels, corresponding to mIvalues of +½ and -½. An orientation parallel to the field direction of H0 is considered to be the low energy
state (-state) with mI = +½. Conversely an anti-parallel orientation is the high energy state (-state) with
mI = -½.
Subtraction of the relative energies of the two states gives the energy difference (E) between the states.
This is the energy required to promote a magnetic dipole moment from a parallel to an anti-parallel
orientation in an applied magnetic field (H0).
E = (-(-½)H0h/2) - (-(+½)H0h/2) = H0h/2 
According to quantum mechanics, transition from the low energy state to the high energy state may be
achieved by electromagnetic radiating of the nuclei with the correct frequency (i).
E = hI 
Since this energy (E) corresponds to the energy difference between the levels, the frequency (i) can be
related to the magnetic field strength (H0).
E = H0h/2 = hI 
thus i = H0/2 
14.2 Die multiplisiteit van spin-spin koppeling is baie afhanklik van die getal naburige kerne. Drie algemene
This frequency (i) is referred to as the resonant frequency or Larmor frequency.
reels is van toepassing in die bepaling van sein multiplisiteite. Gee hierdie reëls. / The multiplicity of
spin-spin coupling is strictly dependant on the number of neighboring nuclei. Three general rules
may be applied in determining signal multiplicities. Give these rules.
(3)
 A proton with a set of N equivalent protons as neighbours will split into (N+1) lines. 
 A proton with a set of N unequivalent protons as neighbours will split into 2N lines. 
 Equivalent protons do not exhibit spin-spin coupling with each other (no splitting). 
15.
Is 'n toename in elektrondigtheid om 'n spesifieke kern verantwoordelik vir / Does an increase in electron
density around a specific nucleus give rise to,
15.1 'n laer, of hoër effektiewe magneetveld (H0) wat deur die kern ondervind
higher effective magnetic field (H0) experienced by the nucleus?
word? / a lower, or
(2)
15.2 seine wat na hoër, of laer ppm waardes skuif? / signals that shift to higher, or lower ppm values?
(2)
15.1
Lower H0 
15.2
Lower ppm values 
8
16. Teken die meegaande molekulêre struktuur oor in u antwoordblad en ken die 1H KMR pieke toe deur van die
gegewe simbole gebruik te maak. Dui ook spin-multiplisiteite en integrale van elk aan. / Redraw the
accompanying molecular structure on your answer sheet and assign the 1H NMR peaks by means of the
given symbols. Also indicate spin multiplicities and integrals of each.
(9)
O
OH
f
a
OH
O
d
c
a
b
12
10
e
8
6
PPM
4
2
0
Mark half points:
O
Position √ Coupling √ Integration √
f
1.01
a
a (2 x -OH) – singulet – int.2
b (1 x Ph) – triplet – int.1
c (2 x Ph) – doublet – int.2
d (2 x CH) – doublet – int.2
e (1 x CH) – multiplet – int.1
f (6 x CH) – doublet – int.6
e
OH
11.0
c
b
8.82
8.33
2.22
OH
1.01
2.51
f
a
d
9
a
8.33
c
11.0
O