IODOMETRY TITRATION

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 (phenol-

water).

b. Determine equivalent point in two phase component equilibrium of liquid-

liquid phase (phenol-water).

c. Determine phase, component, and degree 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 piece

2. Spatula : 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 KIO3

2. Determine Percentage Cl2 in “Soklin pemutih”

Standardization of Na2S2O3

KIO3 (s) 0,30005 gram

Standard Solution

Analyte (yellow)

Colorless

- Adding 3 drops Starch solution- Titrating with Na2S2O3

- 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

- Pouring into Volumetric Flask 100 mL- Adding water until V= 100 mL- Shake well

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)

- 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

KIO3 (s) 0,30005 gram

Analyte

Analyte (yellow)

Colorless

Dissolve in 100 ml volumetric flask

pipette 10 ml by volumetric flaskpoured in conicalAdding 2 ml KI 20%Adding 2,5 ml HCl 4N Titrating with Na2S2O3

Adding 3 drop starch solutionTitrating with Na2S2O3

Soklin pemutih

Analyte (yellow)

Colorless (analyte)

measure with picnometerpouring 2 ml “soklin pemutih” into conical flaskadding 75 ml aquadest 3,0002 grams KI adding 3 drops ammonium molibat 3%titrate with Na2S2O3

adding 3 drop of amilumtitrating with Na2S2O3

1 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

1st titration V 1 = 7,7 ml 2nd titration

V2 = 7,6 ml 3th titration

V3 = 7,6 ml

Reaction :2 IO3

- + 12 H+ + 10e → I2 + 6 H2O10 l- → 5 l2 + 10e-

+2 IO3

- + 12 H+ + 10 I- → 6 I2 + 6 H2O

I2 + 2e- → 2 I-

2 S2O32- → S4O6

2- + 2e-

+2 S2O3

2- + I2 → S4O62- + 2 I-

N Na2S2O3 = 0,1103 N

2 “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

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

Reaction :OCl- + 2I- + 2H+ → I2 + Cl- + H2O

I2 + 2e- → 2I-

2 S2O32- → S4O6

2- + 2e-

+I2 + 2 S2O3

2- → S4O62- +

2 I-

% Cl2 = 4,3890 %

V3 Na2S2O3 =21,3 mL

H. Analysis and discustion

Analysis

Standardization Na2S2O3 solution

Thiosulfate solution before used as standard solutions in the iodometric process

should be standardized first by potassium iodate which is primary standard. KIO3 salt

can oxidizing iodide to iodine quantitatively in acid solution. Therefore used as

standard solutions in the iodometric titration.

The color of KIO3 is colorless after adding KI by the color of solution become

yellow. Addition of KI is use for get excess of I- , because I2 ilustrate the sample

which is calculated. Then we adding HCl 4N, the analyte become blackish brown. It

caused by the present of I2. 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- I2 + 6H2O x 1

2I- I2 + 2e- x 5 +

2IO3- + 12H+ + 10 I- 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 Na2S2O3 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 2I-

2S2O32- S4O6

2- + 2e +

2S2O32- + I2 → S4O6

2- + 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

mMrn

. VxV

= N Na2S2O3 .V

0,3005 gr214,0042

6x 0,1 L

X 0,01 L= N Na2S2O3 .7,7 x 10-3 L

N Na2S2O3 = 0,1094

mole eqivalent KIO3- = mole eqivalent Na2S2O3

mMrn

. VxV

= N Na2S2O3 . V

0,3005 gr214,0042

6x 0,1 L

X 0,01 L = N Na2S2O3 .7,6 x 10-3 L

N Na2S2O3 = 0,1108

mole eqivalent KIO3- = mole eqivalent Na2S2O3

mMrn

. VxV

= N Na2S2O3 . V

0,3005 gr214,0042

6x 0,1 L

X 0,01 L = N Na2S2O3 .7,6 x 10-3 L

N Na2S2O3 = 0,1108

N average = 0,1094 N+0,1108 N+0,1108 N

3

= 0,331

3

= 0,1103 N

Determine percentage Cl2 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 H2SO4 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 Na2S2O3 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 + Cl- + H2O

I2 + 2e 2I-

2S2O32- S4O6

2- + 2e

I2 + 2S2O32- S4O6

2- + 2I-

From calculation, percentage Cl2 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 = mV

= 81 , 4994 gr – 26 , 8294 gr

50 ml = 1,0934 gr/L

m sample = 1,0934 gr/L x 2 mL = 2,1868 gr

mole eqivalent Cl2 = mole eqivalent Na2S2O3

N Cl2 x V Cl2 = N Na2S2O3 x V Na2S2O3

N Cl2 x 2x10-3 L = 0,1103 x 21,30x10-3 L

N Cl2 = 1,1747

M Cl2 = N Cl2

n =

1,17472

=0,5873 M

m Cl2 = M Cl2 x Mr Cl2 x V Cl2

= 0,5873 M x 70,906 x 2x10-3 L

= 0,0833 gr

% Cl2 = mCl2

m sample=0,0833 gr

2,1868 grx100 %=3,8092%

N Cl2 x V Cl2 = N Na2S2O3 x V Na2S2O3

N Cl2 x 2x10-3 L = 0,1103 x 21,00x10-3 L

N Cl2 = 1,1582

M Cl2 = N Cl2

n =

1,15822

=0,5791M

m Cl2 = M Cl2 x Mr Cl2 x V Cl2

= 0,5791M x 70,906 x 2x10-3 L

= 0,0821 gr

% Cl2 = mCl2

m sample=0,0821 gr

2,1868 grx100 %=3,7543 %

N Cl2 x V Cl2 = N Na2S2O3 x V Na2S2O3

N Cl2 x 2x10-3 L = 0,1103 x 21,30x10-3 L

N Cl2 = 1,1747

M Cl2 = N Cl2

n =

1,17472

=0,5873 M

m Cl2 = M Cl2 x Mr Cl2 x V Cl2

= 0,5873 M x 70,906 x 2x10-3 L

= 0,0833 gr

% Cl2 = mCl2

m sample=0,0833 gr

2,1868 grx100 %=3,8092%

% Cl2 average = 3,8092%+3,7543 %+3,8092 %

3=3,7909%

Discussion

From the result of experiment the percentage of Cl2 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 Na2S2O3 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 Cl2 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

Fe2+ Fe3+ + e

MnO4- + 8H+ + 5e Mn2+ + 4H2O

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Each 1 mole KIO3 = 5 eqivalent

2. Why in the permanganometry titration no need add by indicator again?

Answer

Because MnO4- 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 KIO3 equal to how the

equivalence?

Answer

2IO3- + 12H+ + 10e I2 + 6H2O

2I- I2 + 2e

2IO3- + 12H+ + 10I- 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 I2 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?

Answer : Because it use for make CO2 lose, and the temperature must be 700-

900C, if the temperature more than 900C 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 Na2S2O3

Reaction Picture

KIO3 Solution

KIO3 Solution + KI Solution

KIO3 Solution + KI Solution 20 % + HCl 4 N

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 Cl2 in “Soklin pemutih”

Reaction Picture

“ Soklin pemutih” solution

“ 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