Practical part – June 5 th , 2002
Task 8: ....../......../17
Task 9: ....../......../10
Task 10: ....../......../13
Sum: .........../40
Name:........................................ Nr.:.........

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002
Task 8 17 points
Qualitative and instrumental analysis a) Qualitative analysis
The test tubes A to F contain aqueous solutions of inorganic substances as well as a solid organic compound. The samples which are the salts contain the identical cation.
Identify the substances. To reach this goal you have additionally to the samples themselves the following reagents and apparatus at your hand:
1% AgNO
3
-solution, 2M NH
4
OH-solution, 1M H
2
SO
4
, MgO-rod, Co-glas, pH-paper.
Write the names and the formulae (only for the inorganic substances) as well as your reasoning for the identification – if possible using ionic equations – into the following table:
Sample Formulae Name Reasons
A
B
C
D
E
F
2

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002 b) Instrumental analysis
In the bottle G you will find an aqueous solution of a monobasic acid which you can find in the following pK
A
-table. Identify the acid. For this purpose you have a NaOH-solution at your hand which has the same concentration like the acid, and you also may use pHmeter.
Which acid is it? Give reasons for your result by describing the steps you undertook and also by writing down your measurement. acid PK
A
4,75 acetic acid (ethanoic acid) lactic acid
(2-hydroxypropanoic acid) cyclopropanecarbonic acid resorcine (1,3-benzenediol)
3,91
4,84
9,81 phenylethanoic acid chlorous acid (HClO nitric acid
2
) chloroacetic acid salicylic acid
(2-Hydroxybenzenecarbonic acid)
4,28
2,37
-1,32
2,85
2,97
3

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002
Task 9 10 points
Quantitative determination of bleaching agents in household cleaners
1.
General
Liquid bleaching cleaners on the basis of chlorine contain sodium hypochlorite (Na + OCl ).
In this task you will determine the concentration of this substance in a common available product.
2.
Procedure
Your 100 mL-measuring flask contains 10.00 mL of a common bleaching cleaner. Fill up to the mark with water and homogenize. Pipette 10.00 mL of this solution into a titration flask and add about 10 mL of water (one test tube full), 3 mL of pure acetic acid
(measuring pipette) as well as 10 mL (measuring pipette) of a KI-solution (10%).
Let the sample stand for five minutes, then titrate with a solution of Na
2
S
2
O
3
(c = 0,100 mol/L) until the colour of the solution becomes bright yellow. Add 1 – 2 mL of a starch solution and continue the titration until the solution becomes colourless.
3.
Results a) Write your data into the following table:
Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 initial burette reading final burette reading titration volume mean value of the titration volumes: n(OCl ) = %Cl
2
= b) Write a balanced equation for the redox reaction between ClO and I . c) Write a balanced equation for the titration reaction. d) Calculate the total amount of OCl in your sample and insert this value into the table. e) Chlorine reacts in basic solution to give Cl and ClO . Write a balanced equation for this reaction. What is the name of such a redox reaction? f) Calculate the content of bleaching agent in your sample in %(m/m)Cl
2
and insert this value into the table ( ρ (bleaching cleaner) = 1.00 g/mL)
4

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002
Task 10 13 points
Organic synthesis of phenacetin
4.
General
Phenacetin is the international generic term for 4-Ethoxyacetanilide or ethanoic acid-
(4-ethoxyanilide), a analgesic and antipyretic pharmaceutical preparation which on the other hand is no longer in use in many European countries because of its side effects.
The synthesis starts from 4-Ethoxyaniline (= phenetidine), which reacts with acetic anhydride in weakly basic surrounding.
5.
Procedure
In a test tube 2.18 g of sodium salt of ethanoic acid (= sodium acetate-trihydrate) are dissolved in 10 mL of water. In the 100 mL Erlenmeyer flask with the narrow neck you have 1.92 g p-phenetidin (4-ethoxyaniline). Add 41 mL diluted HCl and mix until the ethoxyaniline is dissolved. In order to decolourize the solution 1 g of activated carbon are added (test tube at your place), the solution is mixed thoroughly and filtered using a folded filter into the 100 mL Erlenmeyer flask with the wide neck. The solution should be colourless. Then you add 1.43 g of acetic acid anhydride and the solution of sodium acetate mixing the content properly. Within the next 10 minutes the flask is shaken from time to time then cooled down in an ice bath for another 10 minutes.
The precipitated crystals are filtered off by vacuum using a sintered glass frit, are then washed twice with 10 mL of water, and then dried by vacuum suction.
For recrystallization the crude product is put back into the wide necked Erlenmeyer flask which meanwhile was cleaned. 10 mL of ethanol are added and the mixture is heated in the hot water bath until everything has dissolved. Then you add to the hot solution about 20 mL of hot water (While sucking off a test tube of water may be heated). Then you cool down the solution slowly, at first under flowing water then in an ice bath.
The substance is filtered off using the meanwhile cleaned sintered glass frit (cleaning with acetone) and dried by using vacuum suction for 5 minutes.
Hand over the watch glass with the number and with the product (weigh the tara!) to the supervisor who will put it in a drying cabinet at 110°C for 30 minutes to dry the substance.
After drying yield and melting point are measured and the product again is delivered to the supervisor. a) Write a balanced equation for the synthesis. b) Calculate the theoretical yield. c) Write down your yield and calculate the percentage of the theory. d) Write down the melting point of your product.
5

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002
Task 8 17 points(55 b.p.; f = 0.30909)
Qualitative and instrumental analysis sample formula name reasons
A
5 b.p.
B
5 b.p.
C
5 b.p.
D
3 b.p.
E
5 b.p.
KI
1 b.p.
KIO
1 b.p.
K
1 b.p.
KMnO
1 b.p.
H
2
2
SO
O
3
2
3
4
1 b.p. potassium idodide
1 b.p. potassium iodate
1 b.p. potassium sulfite
1 b.p. potassium permanganate
1 b.p. hydrogen peroxide
1 b.p. starch
1 b.p.
I
K
-
+
: I
or: 2 I + H
2
O
2
+ 2 H +  I
IO
SO
3
: violet flame 1 b.p.
3
-
-
+ 5 I
2-
+ Ag +  AgI, not sol. in NH
3
2MnO
+ 2 H
4
-
3
+ 5H
O
2
+
O
2
 SO
2
+ 6 H
1 b.p.
2
+ 2 H
2
O
+ 3 H
2
O 1 b.p.
Solution basic  not HSO
+
3
1 b.p.
 2Mn 2+ +5O
2
+ 8H
2
O
1 b.p.
2MnO
4
-
+ 6 H
+ 5H
2
O
2
+  3 I
+ 6 H +
2
+ 3 H
 or: 2 I + H
2
O
2
+ 2 H +  I
2
2Mn
2
O 1 b.p.
2+ +5O
+ 2 H
2
O
2
+ 8H
2
O
1 b.p.
F
5 b.p.
- blue complex with iodine 1 b.p. b) Instrumental analysis mixing of the acid and the NaOH in volume proportion 2:1 gives a buffer solution.
According to the equation by Henderson-Hasselbalch the measured pH equals the pK
A
of the acid: pH = pK
A
= 3.91  lactic acid
8 b.p. (only in the case of correct reasoning)
6

28th Austrian Chemistry Olympiad
National competition
Practical part
June 5 th , 2002
Task 9 10 points(21 b.p.; f = 0.47619)
Quantitative determination of bleaching agents in household cleaners a) for the region  V
M
- V
SOLL
  0.10 mL 15 b.p. for the region  V
M
- V
SOLL
 > 0.80 mL for the region 0.10 mL <  V
M
- V
SOLL
  0.80 mL the b.p. are calculated according to the formula (sliding scale):
0 b.p. b.p. = 15 – 21.4286
 (  V
M
- V
SOLL
 - 0.1) b) ClO + 2 I + 2 H +  Cl + I
2
+ H
2
O 1 b.p. c) 2 S
2
O
3
2 + I
2
 2 I + S
4
O
6
2d) n(ClO ) = 0,5.V
M
mmol e) Cl
2
+ 2 OH -  Cl + OCl + H
2
O
disproportionation f) %Cl
2
= 0.709
 n(OCl )
1 b.p.
1 b.p.
1 b.p.
1 b.p.
1 b.p.
Task 10 a)
C
2
H
5
O NH
2 b) 179  0.014 = 2.51 g
+ c) master value: 1.70 g
Organic synthesis of phenacetin
O
13 points(27 b.p.; f = 0.48148)
2 b.p.
O
O b.p. = 7.06
 individual yield in g
C
2
H
5
O NH COCH
3
+ CH
3
COOH
1 b.p.
12 b.p. for the correct mass for the correct percentage of theory d) 136°C
133°C – 135°C: 6 b.p.
130°C – 132°C: 4 b.p.
127°C – 129°C: 2 b.p. for the correct measurement of the melting point
1 b.p.
1 b.p.
8 b.p.
2 b.p.
7