HOOF- & QWAQWA KAMPUS / MAIN & QWAQWA CAMPUS
CEM 314 MEMORANDUM
DIE UNIVERSITEIT VAN DIE VRYSTAAT
UNIVERSITY OF THE FREE STATE
TOETS/TEST: 10 MEI/10 MAY 2008
MEMORANDUM
ASSESSORE / ASSESSORS:
DR. K.G. VON ESCHWEGE
MNR. /MR. T.A. TSOTETSI
MODERATOR / MODERATOR:
PROF W. PURCELL
TYD / TIME: 2 URE / HOURS
PUNTE / MARKS: 76
BEANTWOORD ALLE VRAE EN TOON ALLE BEREKENINGE (TENSY ANDERS
GESPESIFISEER) MET INKPEN. GEEN POTLODE OF "TIPPEX" TOEGELAAT NIE.
VRAESTEL BESTAAN UIT 4 BLADSYE EN 9 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 9 QUESTIONS.
ENGLISH QUESTIONS FOLLOW AFRIKAANS
QUESTIONS. POCKET CALCULATORS ALLOWED.
1.
Limiting law – only dilute solutions 
Chemical deviations – for example association, dissociation, etc

Instrumental deviations polychromatic radiation; Stray light
Mismatched cells
2.
Give a short description of absorbance phenomena in
2.1
inorganic complexes, and
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. 
2.2
charge-transfer complexes.
[2]
The charge-transfer complex consists of an e-donor group bonded to an eacceptor. 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]
1
3.
At 470 nm, the wavelength of maximum absorption of FeR has a molar absorptivity
of 10400 L cm-1 mol-1. Calculate
3.1
the absorbance of a 4 × 10-5 M solution in a 1 cm cell, and
A = bc = 10400  1.00  4  10-5 = 0.416
3.2

the % transmittance of the solution.
[2]
T = antilog(-A) = antilog(-0.416) = 0.384 (%T = 38.4)
4.
[2]

What is 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.  
5.
Comment on the molar absorptivities (A, P & T) of the following titration curve:
A
Volume Titrant
[3]
P > T > 0,   A = 0 
6.
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 )
2

7.
Volumes, ml
Sample
Sample
Oxidizing
Fe(II)
KSCN
Volume
reagent
2.75 ppm
0.050 M
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
MEMO: cx =
8.
H2O
Ax c sVs
0.231  2.75  5.00
=
50.0(0.549  0.231)
Vx ( As  Ax )
 =
0.200 ppm Fe 
In a continuous variation method’s application to the reaction between Hg2+ and
complexing reagent R, the following absorbance measurements using 1.0 cm cuvettes
were made:
Reagent Volume, mL
8.1
Solution
1.25 x 10-4 M Hg
1.25 x 10-4 M R
A390
0
10.00
0.00
0.00
1
9.00
1.00
0.30
2
8.00
2.00
0.58
3
7.00
3.00
0.85
4
6.00
4.00
0.80
5
5.00
5.00
0.68
6
4.00
6.00
0.55
7
3.00
7.00
0.42
8
2.00
8.00
0.25
9
1.00
9.00
0.12
10
0.00
10.00
0.00
Determine graphically (hand in your graph) and through subsequent
calculations the formula of this complex.
3
[5]
MEMO:
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1.2

From the plot it is clear that the lines intersect at VM / (VM + VL) = 0.33, thus the
Cd2+ to R ratio is 1 to 2, i.e. ML2.

Calculate the molar extinction coefficient, ε, of the complex from the data of
8.2
solution number 2. Assume that in the linear portions of the plot the metal is
completely complexed.
[3]
MEMO:
[HgR2+] = 2/10 x 1.25 x 10-4 = 2.5 x 10-5 M 
ε=
9.
Absorbance
0.58

 2.3 x 104 M -1 cm -1 
2
[HgR ]
2.5 x 10-5
9.1
MEMO:
 = (I - 0) / Instrument Frequency (MHz)

I = Larmor frequency of the sample 
0 = Larmor frequency of the internal standard 
9.2
MEMO:
 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).

4
9.3
9.4
Atoommassa /
Atoomgetal /
Spin Kwantum Getal /
Atomic Mass
Atomic Number
Spin Quantum Number
Uneven
Even or Uneven
1/2, 3/2, 5/2 
even 
even 
0
even
uneven 
1, 2, 3, 4, …
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.
[5]
d
MEMO:

(a)
singlet, 1
(b)
doublet, 2
(c)
doublet, 2


(d)
quartet, 2

(e)
triplet, 3
b
e
c
a

10.
Effectively shielding 
Effectively deshielding 
Induced magnetic field 
Circulation of π-electrons 
5
11.
Higher pH generally gives better endpoint, indicator behaviour etc.


As with higher pH certain cations also give better endpoints, etc.

p.470/1

[4]
12.
K2 =
[H3O  ] [In3  ]
 2.8 x 10-12
2
[HIn ]
Kf =
[CaIn  ]
 2.5 x 105
2
3[Ca ] [In ]
K2 X Kf =
[H3O  ] [In3  ]
[HIn 2 ]
X
[CaIn  ]
[Ca 2 ] [In3- ]
= 2.8 x 10-12 X 2.5 x 105 
[CaIn  ] [H3O  ]
x
 7.0 x 10- 7 
2
2
[Ca ] [HIn ]
Rearrange to
[Ca2+] =
[H3O ]
[CaIn  ]
x

[HIn 2  ] 7.0 x 10- 7
Substitution of [H3O+] = 10-10 (since pH=10), and 10 and 0.10 for the ratio yields the
range of [Ca2+] over which the colour change occurs:
6

[Ca2+] = 1.428 x 10-3 M to 1.428 x 10-5 M 
pCa = 3.85 ± 1.0 
13.
(i.e. 2.85-4.85)
An EDTA solution was prepared by dissolving 3.156 g of purified and dried
Na2H2Y2·H2O in sufficient water to give 1.000 L. Calculate the molar conentration,
given that the solute contained 0.3% excess moisture.
[3]
MEMO:
3.156 g reagent 
14.
99.7 g Na 2 H 2 Y  2 H 2 O
1 mole EDTA

100 g reagent
372.24 g Na 2 H 2 Y  2 H 2 O
 0.00845 M EDTA
1.000 L

A 1.509 g smaple of Pb/Cd alloy was dissolved in acid and dilute to exactly 250.0
mL in a volumetric flask. A 50.00 mL aliquot of the diluted solution was brought to a
pH of 10.0 with an NH4+/NH3 buffer; the subsequent titration involved both cations
and required 28.89 mL of 0.06950 M EDTA. A second 50.00 mL aliqout was
brougfht to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the
Cd2+; 11.56 mL of the EDTA solution were needed to titrate the Pb 2+. Calculate the
precentage of Pb and Cd in the sample.
[8]
MEMO:
n(Cd2+ + Pb2+) = 0.06950 M x 28.89 L / 1000 = 2.00786 x 10-3 mol 
n(Pb2+) = 0.06950 M x 11.56 L / 1000 = 8.0342 x 10-4 mol

n(Cd2+) = 2.00786 x 10-3 mol - 8.0342 x 10-4 mol = 1.20444 x 10-3 mol 
m(sample) = 1.509 g x 50.00 mL / 250.0 mL = 0.3018 g 
m(Cd2+) = 1.20444 x 10-3 x 112.41 = 0.1354 g 
m(Pb2+) = 8.0342 x 10-4 x 207.2 = 0.1665 g 
%Cd2+ = 0.1354 g / 0.3018 g X 100% = 44.86% 
%Pb2+ = 0.1665 g / 0.3018 g x 100% = 55.16% 
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