A. Title : Two Phase Component Equilibrium B. Data of Experiment
Start
: 9 March 2012, at 7.15 AM
Finish
: 9 March 2012, at ? AM
C. Objective
a. Describe two phase component equilibrium of liquid-liquid phase (phenolwater). b.
Determine equivalent point in two phase component equilibrium of liquidliquid phase (phenol-water).
c. Determine phase, component, component, and degree of of freedom a system two phase component equilibrium liquid-liquid phase (phenol-water). D. Basic Theory
E. Chemicals and Equipments Chemicals :
1. Phenol
: 42 mL
2. Aquadest
: 40 mL
Equipments
1. Beaker Glass
?
2. Spatula
: 2 piece : 1 piece
3. Graduated Cylinder 10 mL : 2 piece 4. Pipette
: 2 piece
5. Thermometer
: 2 piece
6. Tripod
: 1 piece
7. Sepiritus burners
: 1 piece
F. Procedure
1. Standardization of Na2S2O3 0.3005 g of KIO 3
- Pouring into Volumetric Flask 100 mL - Adding water until V= 100 mL - Shake well Standard Solution -
Pipette 10 mL Pouring into Erlenmeyer 100 mL Adding 2 mL KI Solution 20% Adding 12,5 mL HCl 4N Titrating with Na2S2O3 three times
Analyte (yellow) - Adding 3 drops Starch solution - Titrating with Na2S2O3 Colorless 2. Determine Percentage Cl 2 in “Soklin pemutih”
2 mL “Soklin Pemutih” - Measuring density - Pouring 2 mL soklin pemutih into conical flask 100 mL - Adding 75 mL Aquadest - Adding ±3,0002 grams KI - Adding 8 mL H2SO4 1:6 - Adding 3 drops Ammonium Molibdat 3% - Titrating with Na2S2O3
Analyte (yellow)
Standardization of Na2S2O3
KIO3 (s) 0,30005 gram
- Adding 5 mL starch solution - Titrating with Na2S2O3
Colorless
KIO3 + KI + HCl titrated with Na2S2O3
KIO3 + KI + HCl + starch solution titrated with Na2S2O3
Determine Percentage Cl2 in Soklin pemutih “
“soklin pemutih” + 3,0002 grams KI
”
“soklin pemutih” + KI + Ammonium molibat titrated with Na2S2O3
“soklin pemutih” + KI + Ammonium molibat + ammilum titrated again Na2S2O3
G.Experimental Data No.
Procedure of experiment
Experiment result
Hypothesis / reaction
Conclusion
1
KIO3 (s) 0,30005
Dissolve in 100 ml volumetric
Analyte
pipette 10 ml by volumetric flask
poured in conical
Adding 2 ml KI
20%
Adding 2,5 ml HCl 4N
Titrating with
Na2S2O3
Analyte (yellow)
Adding 3 drop starch solution
Titrating with
Na2S2O3
KIO3 (s) = colorless KI = yellow KIO3 + KI = yellow KIO3 + KI + HCl + Na2S2O3 = Blackish brown KIO3 + KI + HCl + Na2S2O3 + starch indicator = Blackish purple KIO3 + KI + HCl + starch solution + Na2S2O3 = colorless st 1 titration V 1 = 7,7 ml nd 2 titration V2 = 7,6 ml th 3 titration V3 = 7,6 ml
Reaction : + 2 IO3 + 12 H + 10e → I2 + 6 H2O 10 l → 5 l2 + 10e + + 2 IO3 + 12 H + 10 I → 6 I2 + 6 H2O
“soklin pemutih” = yellow light soklin pemutih + KI = colorless soklin + KI + Ammonium molibat + H2SO4 + Na2S2O3 = Blackish brown soklin + KI + Ammonium molibat + H2SO4 + Na2S2O3 + Starch indicator = blackish purple soklin + KI + Ammonium molibat + ammilum +Na2S2O3 = colorless
Reaction : + OCl + 2I + 2H → I2 + Cl + H2O
-
N Na2S2O3 = 0,1103 N
-
I2 + 2e → 2 I 222 S2O3 → S4O6 + 2e 2
2S2O3
+ I2 →
2S4O6
+ +2I
Colorless
2
Soklin pemutih
measure
with picnometer
pouring 2 ml
“soklin pemutih” into
conical flask
adding 75 ml aquadest
3,0002 grams
KI
adding 3 drops ammonium molibat 3%
titrate with Na2S2O3
Analyte (yellow)
adding 3 drop of amilum
titrating with Na2S2O3
Colorless (analyte)
m KI 1 = 3,0005 gr m KI 2 = 3,0004 gr m KI 3 = 3,0002 gr V1 Na2S2O3 = 21,3 mL V2 Na2S2O3 =21,0 mL
-
I2 + 2e- → 2I 2-
2-
2 S2O3 → S4O6 + 2e
-
+ I2 + 2 S2O3 -
+2I
2-
2-
→ S4O6
% Cl2 = 4,3890 %
V3 Na2S2O3 =21,3 mL
H. Analysis and discustion
Analysis
Standardization Na 2S 2O 3 solution Thiosulfate solution before used as standard solutions in the iodometric process should be standardized first by potassium iodate which is primary standard. KIO 3 salt can oxidizing iodide to iodine quantitatively in acid solution. Therefore used as standard solutions in the iodometric titration. The color of KIO 3 is colorless after adding KI by the color of solution become -
yellow. Addition of KI is use for get excess of I , because I 2 ilustrate the sample which is calculated. Then we adding HCl 4N, the analyte become blackish brown. It caused by the present of I 2. The function of the addition of HCl in the solution is to provide acidic conditions, because the solution consisting of potassium iodate and potassium iodida are in neutral or has a low acidity. This reaction is as follows:
-
+
2IO3 + 12H + 10e
-
-
2I -
+
2IO3 + 12H + 10 I
-
I2
+ 6H2O
x1
I2
+ 2e
-
x5
+
6 I2 + 6H2O
Indicators used in this standardization process is starch indicator. Addition of starch indicator is use when it approaches end point, it means that starch can not wrap the iod, because it will make starch difficult to titrate and back in the first compound. The process of titration should be done as soon as possible, this is due to the nature of the I2 easy to evaporate. In the end point of titration, iod which is bonded is also lose and it react with Na 2S2O3 solution, so the blackish purple color become colorless. Using this indicator to make clear the solution color which is occur in the end point of titration.. This reaction is as follows: I2 + 2e 2S2O3
2-
2I
-
2-
S4O6 + 2e
+
2-
2-
-
2S2O3 + I2 → S4O6 + 2I
From calculation, normality of sodium thiosulfate is 0.1103 N. The calculation is: Know : mass of IO3
-
= 0,3005 gr
V1 Na2S2O3
= 7,7 ml
V2 Na2S2O3
= 7,6 ml
V3 Na2S2O3
= 7,6 ml
Mr KIO3
= 214,0042
Asked : N Na2S2O3 Answer
-
mole eqivalent KIO3 = mole eqivalent Na2S2O3
= N Na2S2O3 .V -3
= N Na2S2O3 .7,7 x 10 L
N Na2S2O3
= 0,1094
-
mole eqivalent KIO3 = mole eqivalent Na2S2O3
= N Na2S2O3 . V -3
= N Na2S2O3 .7,6 x 10 L
N Na2S2O3
= 0,1108
-
mole eqivalent KIO3 = mole eqivalent Na2S2O3
= N Na2S2O3 . V
N Na2S2O3
N average
-3
= N Na2S2O3 .7,6 x 10 L = 0,1108
= =
= 0,1103 N Determine percentage Cl 2 in soklin pemutih The color of “Soklin Pemutih” is yellow light. After then we add 75 ml aquades, evidently the color of solution become colorless. We adding KI (s) in order that I
-
excess. So the color of solution become colorless too. It caused by they don ’t react each other, because in iodometric titration is only occur when it is in the strong acid condition. After then we add H 2SO4 solution 1:6 to make acid condition. The color of solution become to blackish brown. After then we adding Amonium molibdat 3% (colorless) as catalys. We titration by Na 2S2O3 solution until the color of analyte become yellow. We use starch indicator. After we adding starch indicator the color -
become blackish purple. It is because there is I . The reaction is: -
-
+
OCl + 2I + 2H
I2 + 2e 2-
2S2O3
I2
-
2I
S4O6 + 2e
I2 + 2S2O3
2-
-
+ Cl + H2O
2-
2-
-
S4O6 + 2I
From calculation, percentage Cl 2 in “Soklin Pemutih” is 3,7909 %. The calculation is: Known
: N Na2S2O3
= 0,1103
V Cl2
= 2 ml
Mr Cl2
= 70,906
V Na2S2O3
= 21,3 ml
V Na2S2O3
= 21,0 ml
V Na2S2O3
= 21,3 ml
m empty picnometer = 26,8294 gr m picnometer + “soklin pemutih” = 81, 4994 gr Asked
: % Cl2 in “Soklin Pemutih”
Answer
:
Dencity =
=
–
= 1,0934 gr/L
m sample = 1,0934 gr/L x 2 mL = 2,1868 gr
mole eqivalent Cl2 = mole eqivalent Na 2S2O3
N Cl2 x V Cl2
= N Na2S2O3 x V Na2S2O3
-3
-3
N Cl2 x 2x10 L = 0,1103 x 21,30x10 L N Cl2 M Cl2 =
=
= 1,1747
M
m Cl2 = M Cl2 x Mr Cl2 x V Cl2 -3
= M x 70,906 x 2x10 L = 0,0833 gr % Cl2 =
N Cl2 x V Cl2
= N Na2S2O3 x V Na2S2O3
-3
-3
N Cl2 x 2x10 L = 0,1103 x 21,00x10 L N Cl2 M Cl2 =
=
= 1,1582
M
m Cl2 = M Cl2 x Mr Cl2 x V Cl2 -3
= M x 70,906 x 2x10 L = 0,0821 gr % Cl2 =
N Cl2 x V Cl2
= N Na2S2O3 x V Na2S2O3
-3
-3
N Cl2 x 2x10 L = 0,1103 x 21,30x10 L N Cl2 M Cl2 =
=
= 1,1747
M
m Cl2 = M Cl2 x Mr Cl2 x V Cl2 -3
= M x 70,906 x 2x10 L = 0,0833 gr % Cl2 =
% Cl2 average =
Discussion
%
From the result of experiment the percentage of Cl 2 not appropriate with the table composition in “Soklin Pemutih”. The percentage of Cl2 in “Soklin Pemutih” is 5,25% but in the my experiment is 3,7909%. To determine end point
a titration must be done carefully and
thoroughly, the excess Na 2S2O3 solution when the end point has been reached will make analyte become colorless should be pale yellow and vice versa if the Na2S2O3 solution is still less so yellow color desirable not appropriate because the color is less light, so that it will affect the results of calculations to determine the normality of Na2S2O3. Titration end point is not much different from the equivalent point, but because of the limitations of sense sight make the end point titration is not exactly with equivalent point. I. Conclusion
1. The Normality of Na2S2O3 solution is 0,1103 N. 2. For the Iodometri application, obtained the percentage Cl 2 in “Soklin Pemutih” solution is 3,7909%. J. Question Answer Standardization A. 1. Write reaction that occurs in permanganometry titration, if reductor is ferrous
ions! Each mole of ferrous ions equal to how the equivalence? Answer 2+
Fe
-
+
MnO4 + 8H + 5e 2+
-
+
5Fe + MnO4 + 8H
3+
Fe
+e
2+
Mn + 4H2O
3+
5Fe
2+
+ Mn
+ 4H2O
Each 1 mole KIO 3 = 5 eqivalent 2. Why in the permanganometry titration no need add by indicator again? Answer -
Because MnO 4 purple color can function as indicator (auto indicator) B. 1. What is difference iodometric and iodimetri titration?
Answer
Iodimetry is the direct method, using standered solutions of iodine to titrate against another reagent.
Iodometry is an indirect method or procedure in which the titration(using sodium thiosulphate ) of the iodine liberated in the reaction takes place.
2. How reaction between KIO3 + KI + HCl? Each 1 mole KIO 3 equal to how the equivalence? Answer
-
+
2IO3 + 12H + 10e I2 + 6H2O -
2I
-
+
-
2IO3 + 12H + 10I
I2 + 2e
6I2 + 6H2O
Each 1 mole KIO3 = 5 eqivalent Aplication
1. Explain some lack of starch is used as an indicator! Answer : (1) The insolubility of starch in cold water ; (ii) The instability of starch dispersions in water, in consequence of which a stock solution soon deposits a flocculent precipitate of retrograded starch ; (iii) That starch gives with iodine a water-insoluble complex, the formation of which precludes the addition of the indicator early in the titration ; (iv) The “drift” of end-point which is particularly marked when the solutions used are dilute. 2. Why on iodometric titration starch indicator is added at the time of approaching the equivalence point? Answer : 1. Amilum-I2 complex dissociates very slowly as a result many I 2 to be absorbed by amilum if starch is added at the beginning titration. 2. Usually Iodometric titration in strong acid medium so that it will avoid the occurrence of hydrolysis of amilum. 3. Why adding Na2S2O3 solution use boiling aquadest?
0
Answer : Because it use for make CO 2 lose, and the temperature must be 70 0
0
90 C, if the temperature more than 90 C oxalate acid is straggling. K. References
Day, R. A, and Underwood. A.L. 2002. Analisis Kimia Kuantitatif. Edisi ke-6. Jakarta: Erlangga. Day,R.A.,Underwood,A.L.(1991).Quantitative Analysis (Sixth ed).New York: Prentice Hall. Poedjiastoeti, Sri. dkk. 20011. Panduan Praktikum Dasar Dasar Kimia Analitik. Surabaya: Jurusan Kimia FMIPA Universitas Negeri Surabaya. 2011.http://www.nature.com/nature/journal/v159/n4050/abs/159810b0.html. Accesed on Friday, 23 December 2011 at 08.00 AM.
ATTACHMENT Standardization of Na 2S2O3 Reaction
KIO3 Solution
KIO3 Solution + KI Solution
KIO3 Solution + KI Solution 20 % + HCl 4 N
Picture
KIO3 Solution + KI Solution 20 % + HCl 4 N + Na2S2O3
KIO3 Solution + KI Solution 20 % + HCl 4 N + Na2S2O3 + Strach indicator
Titration again with Na2S2O3
Determine Percentage Cl 2 in Soklin pemutih “
Reaction
“ Soklin pemutih” solution
”
Picture
“ Soklin pemutih” solution + KI(s) + H2SO4 Solution 1:6 + amonium molibdat 3%
“ Soklin pemutih” solution + KI(s) + H2SO4 Solution 1:6 + amonium molibdat 3% + Na2S2O3
“ Soklin pemutih” solution + KI(s) + H2SO4 Solution 1:6 + amonium molibdat 3% + Na2S2O3 + Starch indicator
Titration again with Na2S2O3
