coef distribution iod

Fractional Distillation Page 1

I. EXPERIMENT TITLE : Distribution Coefficient of Iod

II. EXPERIMENT DATE : May 4th , 2015 at 7.30 p.m

III. END OF THE EXPERIMENT : May 4th , 2015 at 10.30p.m

IV. EXPERIMENT PURPOSE :

- Extraction Iodium into organic solvent

- Calculate the distribution coefficient of Iod

V. BASIC THEORY

A. Extraction

Crystallization, purification, and isolation (may only be restricted to a solid)

are insufficient ways to separate mixtures of compounds. Extraction is the recovery

of a substance from a mixture by bringing it into contact with a solvent, which

dissolves the desired material. Partitioning is the separation between two distinct

phases (immiscible liquids) and also called fractional separation.

Like recrystallization and distillation, extraction is a separation technique

frequently employed in the laboratory to isolate one or more components from a

mixture. Unlike recrystallization and distillation, it does not yield a pure product;

thus, the former techniques may be required to purify a product isolated by

extraction. In the technical sense extraction is based on the principle of the

equilibrium distribution of a substance (solute) between two immiscible phases, one

of which is usually a solvent. The solvent need not be a pure liquid but may be a

mixture of several solvents or a solution of some chemical reagent that will react

with one or more components of the mixture being extracted to form a new

substance soluble in the solution. The material being extracted may be a liquid, a

solid, or a mixture of these. Extraction is a very general, highly versatile technique

that is of great value not only in the laboratory but also in everyday life. Extraction

is a convenient method for separating an organic substance from a mixture, such as

an aqueous reaction mixture or a steam distillate. The extraction solvent is usually a

volatile organic liquid that can be removed by evaporation after the desired

component has been extracted.

The extraction technique is based on the fact that if a substance is insoluble

to some extent in two immiscible liquids, it can be transferred from one liquid to the

other by shaking it together with the two liquids. For example, acetanilide is partly

soluble in both water and ethyl ether. If a solution of acetanilide in water is shaken

with a portion of ethyl ether (which is immiscible with water), some of the


acetanilide will be transferred to the ether layer. The ether layer, being less dense

than water, separates out above the water layer and can be removed and replaced

with another portion of ether. When this in turn is shaken with the aqueous solution,

more acetanilide passes into the new ether layer. This new layer can be removed and

combined with the first. By repeating this process enough times, virtually all of the

acetanilide can be transferred from the water to the ether.

As we stated above, the substance being extracted may be a solid.

Extractions of this type will not be conducted here, but they are probably already a

part of your own experience. The brewing of tea from tea leaves (or the tea bag that

combines extraction and filtration) and of coffee from the ground bean are excellent

examples of the extraction of a solid mixture with a hot solvent (water).

B. Coefficient Distribution (KD)

In the laboratory one of the more important applications of the extraction

process has been its use to remove an organic compound from a solution when

distillation is not feasible. Extraction is accomplished by shaking the solution in a

separatory funnel with a second solvent that is immiscible with the one in which the

compound is dissolved, but dissolves the compound more readily. Two liquid layers

are formed, and the layer that has most of the desired product in it can be separated

from the other. Sometimes not the entire product is extracted in a single operation

and the process must be repeated once or twice more to assure a clean separation. It

has been found that when two immiscible solvents are shaken together, the solute

distributes itself between them in a ratio roughly proportional to its solubility in

each. The ratio of the concentration of the solute in each solvent at equilibrium is a

constant called the distribution ratio(d) or partition coefficient (K ).

The larger the value of Kd, the more solute will be transferred to the ether

with each extraction, and the fewer portions of ether will be required for essentially

complete removal of the solute.

[ ]

[ ] =

=

,where o and aq refer to the organic (ether) and aqueous layers, respectively,

and Wo and Waq are the weights in grams of material dissolved in each respective

layer. Co = concentration of organic solution and Caq = concentration of aqueous

solution.


C. Separatory Funnel

A separatory funnel is used to separate immiscible liquids. When two

immiscible liquids are placed in a separatory funnel, two layers are seen. The denser

solvent will be the bottom layer. Most halogenated solvents are denser than water,

most non-halogenated solvents are less dense than water. If you are not sure which

layer is which, add a drop of water and see which layer it joins.

The following is an example in which iodine (I2) in being extracted from an

aqueous solution into dichloromethane (methylene chloride). Iodine, while slightly

soluble in aqueous solutions, is more soluble in dichloromethane. Therefore, when

aqueous solution containing iodine is placed in contact with dichloromethane, the

iodine migrates into (is extracted by) the dichloromethane layer. Iodine in an aqueous

solution appears brown. Dichloromethane is a colorless liquid. Since dichloromethane

is a volatile solvent, this extraction should be done in a hood. The performing is

shown below:

The aqueous iodine solution (the solution

to be extracted) and dichloromethane

poured.

A portion of the dichloromethane is

added to the sep funnel and the funnel

shaken gently.


Most of the iodine has been extracted into

the dichloromethane, the aqueous layer

becomes lighter brown. The iodine in the

dichloromethane layer makes it appear

purplish.

Remove the stopper and drain the lower

layer (the dichloromethane layer in this

case) through the stopcock into a beaker

or flask. The upper layer should remain in

the sep funnel.

The dichloromethane now contains

almost all the iodine, while only a small

amount of I2 remains in the aqueous

layer.

D. Titration

Iodine in water solutions is usually colored strong enough so that its presence can be

detected visually. However, close to the end point, when the iodine concentration is

very low, its yellowish color is very pale and can be easily overlooked. Thus for the

end point detection starch solutions are used. Iodine gets adsorbed on the starch

molecule surface and product of adsorption has strong, blue color. Exact mechanism

behind adsorption and color change is not known, see for example this explanation of

starch as an indicator usage. In the presence of small amounts of iodine adsorption

and desorption are fast and reversible. However, when the concentration of iodine is

high, it gets bonded with starch relatively strong, and desorption becomes slow, which

makes detection of the end point relatively difficult. Luckily high concentrations of

iodine are easily visible, so if we are using thiosulfate to titrate solution that initially


contains high iodine concentration, we can titrate till the solution gets pale and add

starch close to the end point. In the case of titration with iodine solution we can add

starch at the very beginning, as high iodine concentrations are not possible before we

are long past the end point. At the elevated temperatures adsorption of the iodine on

the starch surface decreases, so titrations should be done in the cold.

VI. TOOLS AND MATERIALS

A. Tools

Measuring glass

Volumetric flask

Erlenmeyer flask

Burette

Clamp

Separated funnel

Beaker glass

Pipette

B. Materials

Iodium 0,1 M

Chloroform p.a

Starch solution

H2SO4(aq)

Aquades

Na2S2O3 0,01M

(c) Adding starch forms the deep purple starchI3

complex. (d) As the titration continues, the end point

is a sharp transition from a purple to a colorless

solution. The change in color from (c) to (d) typically

takes 12 drops of titrant.


VII. FLOW CHART

I.

II.

10 mL Iodium 0,01 M

100 mL Iodium

-Put into volumetric flask

Solute with water until

Volume was 100 mL

Organic layers Water layers

10 mL Iodium

Forming two layers

-Move into separated funnel

-Added 5mL CH3Cl(aq)

-Shake a few minutes

Separated


III.

KD Iod

-Determine initial

concentration of Iodium

- Repeat three times

Water layers

Blue color disappear

-Move into Erlenmeyer

-Added 4 mL of H2SO4 2M

-Titrate with Na2S2O3 0,001M


VIII. DATA OF EXPERIMENT

No. Procedure Data of Experiment Hypothesis/Rea

ctions Conclusion

Before After

1.

I2 =

yellowish

brown

Water =

colorless

CH3Cl =

colorless

I2+water = yellowish

brown

I2+CH3Cl two

layer = orangish

brown

Bottom layer = purple

I2 and CH3Cl is

non polar

solution, non-

polar and non-

polar, like

dissolve like

Reaction

I2+H2O

3I2(aq) +

2CH3Cl(aq)

2CH3I3(aq)

+3Cl2(aq)

Iodine can

extract in

organic solvent,

namely

chloroform.

That proven by

two layer

(water layer-

upper layer, and

organic layer-

bottom layer )

2.

10 mL Iodium 0,01 M

100 mL Iodium

-Put into volumetric flask

Solute with water until Volume was 100 mL

Organic layers Water layers

10 mL Iodium

Forming two layers

-Move into separated funnel

-Added 5mL CH3Cl(aq)

-Shake a few minutes

-Repeat 2-5 times

Separated


3.

Water layer

= orangish

brown

H2SO4 =

colorless

Na2S2O3 =

colorless

Starch

solution =

colorless

water layer + H2SO4 +

starch solution

blackish purple

water layer + H2SO4

+ starch solution +

Na2S2O3 colorless

Volume Na2S2O3

needed =

V1 = 3,7 mL

V2= 3,8 mL

V3 = 3,5 mL

Kd first titration =

2,86

KD first titration =

2,86

KD second titration =

3,14

KD third titration =

3,14

KD average =

I2+ 2e - 2I-

2S2O32-

S4O6+ 4e-

2I2 + 2S2O32-

4I

- + S4O6

Coefficient

distribution

Iodine (KD) is 3

KD Iod

-Determine initial

concentration of Iodium

- Repeat three times

Water layers

Blue color disappear

-Move into Erlenmeyer

-Added 4 mL of H2SO4 2M

-Titrate with Na2S2O3 0,001M


IX. EXPLANATION

This experiment is have purpose to extract iodine in organic solvent. First, is to

determine mole of iodine as of standard solutions. In determining mole iodine standard

solution , the first step that done is take 10 mL of iodine solution 0,1 M reddish brown ,

then the solution diluted with aquadest use volumetric flask 100 mL and have result

reddish brown color solution.The reaction that occur is

I2(aq) + H2O(l) no reaction occur

After a solution of iodine obtained, the results of dilution is taken as many as 10

mL using volumetric pippette and then in put into erlenmeyer .After that added 2 mL

H2SO4 2M that will make change color become brownish orange. The role of H2SO4

added is because in iodometric titrations are carried out in strongly acid media, a situation

that promotes the reaction between oxidizing agents and iodide, and also H2SO4 is a

strong acid and will deliver H+ which will help to proceed the reaction much faster or as a

catalyst. Then added 1 mL starch solution 2 % , and undergo a change colors become

purpleish black. Starch solution is added as an indicators for titration, when starch is

mixed with iodine in water, an intensely colored starch/iodine complex is formed. Many

of the details of the reaction are still unknown. The iodine/starch complex has energy

level spacings that are just so for absorbing visible light- giving the complex its intense

blue color. The complex is very useful for indicating redox titrations, when there is

excess oxidizing agent, the complex is blue; when there is excess reducing agent, the I5-

breaks up into iodine and iodide and the color disappears. After that a solution is titrate

using Na2S2O3 0,01 M . The reaction that occur is

I2 + 2e- 2I-

2 S2O3 2-

S4O62-

+ 2e-

I2 + 2 S2O3 2-

2I- + S4O6 2-

Based on the theory, the end point of I2 is changing color became colorless. From

theory we know that the color is shown in the picture below (left picture) and our

experiment (right picture). We can assume that based on the theory our end point is right.


In this titration, obtained changing color solution become colorless at the volume

of titrant Na2S2O3 is 9 mL. By knowing the volume of Na2S2O3 that needed , it can use to

find out a mole from I2 by calculation below:

Based on the calculation mole Iodine as standard is 0,045 mmol. Which will be

used to acquire mmol I2 organic in phase, to determine coefficient distribution of Iodine.

Second steps is to extract iodine, by using a water solvents and organic solvents

that in this experiment using chloroform . Using chloroform to extract because, iodine is

a non-polar molecule this has a weak interaction with the hydrogen bonded water

molecules. The energy associated with the iodine/ water interactions is not enough to

compensate for the lost energy of the water/water interactions. This ultimately means that

not much iodine will dissolve in water. If a solvent with weaker solvent/solvent

interactions than water were introduced to this system, iodine would find it easier to

disrupt these interactions and insert itself (dissolve) between the solvent molecules,

cyclohexane or chloroform are such solvents. They have no hydrogen bonding and are

Theory

Color I2solution +

H2SO4 + Starch

Color at end

Point

Color I2solution +

H2SO4 + Starch

Color at end

Point

Our Experiment


only very slightly polar. Energywise, it is more favorable for the iodine to dissolve in the

non-polar solvent than in the water, so it exists preferentially (but not totally, an

equilibrium will exist) in the non-polar solvent. The release of energy as the iodine

switches solvents is partially responsible for the initial build up of pressure in the

separatory funnel, so two layers are produced when the two solvents are added together.

In short Iodine, is slightly soluble in aqueous solutions, is more soluble in

dichloromethane. Therefore, when aqueous solution containing iodine is placed in contact

with dichloromethane, the iodine migrates into (is extracted by) the dichloromethane

layer. The first step that done is take 10 mL of I2 solution 0,1 M brownish orange .Then

diluted with aquadest on volumetric flask 100 mL . After diluted , take solution of the

results of dilution with using volumetric pipette 10 mL and poured into separating funnel.

The reaction occur is

I2(aq) + H2O(l) no reaction occur

After that added chloroform solution 5 mL as an organic solvent and shake 2-5

times until two layer separated perfectly. That was obtained two layers, upper layer is

water layer brownish orange and bottom layer is organic layer purple. The reaction that

occur is

3I2(aq) + 2CHCl3(aq) 2CHCl3(aq) + 3Cl2(g)

After that separate organic layer with take out into beaker glass, it do repeatedly

until all the organic layer are strictly separated entirely. Based on the theory, when

aqueous solution containing iodine is placed in contact with dichloromethane, the iodine

migrates into (is extracted by) the dichloromethane layer. Iodine in an aqueous solution

appears brown. From theory we know that the color is shown in the picture below (left

picture) and our experiment (right picture). We can assume that based on the theory our

result is right


The water layer moved into erlenmeyer .After moved into erlenmeyer added 2 mL

H2SO4 2M. The role of H2SO4 added is because in iodometric titrations are carried out in

strongly acid media, a situation that promotes the reaction between oxidizing agents and

iodide, and also H2SO4 is a strong acid and will deliver H+ which will help to proceed the

reaction much faster or as a catalyst. And then added 1 mL starch solution 2 % that make

changing color to blackish blue. Starch solution is added as an indicators for titration,

when starch is mixed with iodine in water, an intensely colored starch/iodine complex is

formed. Many of the details of the reaction are still unknown. The iodine/starch complex

has energy level spacings that are just so for absorbing visible light- giving the complex

its intense blue color. The complex is very useful for indicating redox titrations, when

there is excess oxidizing agent, the complex is blue; when there is excess reducing agent,

the I5-

breaks up into iodine and iodide and the color disappears. Titrate solution using

Na2S2O3 0,01M. The reaction that occur is

I2 + 2e- 2I-

2 S2O3 2-

S4O62-

+ 2e-

I2 + 2 S2O3 2-

2I- + S4O6 2-

Extraction iodine do 3 times, which also continue with the titration repeated 3

times to get valid results. Based on the theory, the end point of I2 is changing color

Color of layer that

already wholly

separated

Theory

Color of layer

when Iodine

extracted with

Color of layer that

already wholly

separated

Color of layer

when Iodine

extracted with

Our Experiment


became colorless. From theory we know that the color is shown in the picture below (left

picture) and our experiment (right picture). We can assume that based on the theory our

end point is right.

In first titration, volume of Na2S2O3 needed when solution change from purpleish

black to colorless is 3,7 mL. With known volume of Na2S2O3 in titrations, we can

calculate coefficient distribution iodine, the calculation below:

Theory

Color

I2solution +

H2SO4 +

Color at end

Point

Color

I2solution +

H2SO4 +

Color at end

Point

Our Experiment First

Color

I2solution +

H2SO4 +

Color at end

Point

Our Experiment Second Titration

Color

I2solution +

H2SO4 +

Color at end

Point

Our Experiment Third


( )

( ) ( )

( )

( )

[ ]( )

[ ]( )

[ ]( )

[ ]( )

Coefficient distribution (KD) iodine in first titration based on the calculation is

2,86.

In second titration, volume of Na2S2O3 needed when solution change from

purpleish black to colorless is 3,8 mL. With known volume of Na2S2O3 in titrations, we

can calculate coefficient distribution iodine, the calculation below:

( )

( ) ( )

( )

( )

[ ]( )

[ ]( )

[ ]( )

[ ]( )

Coefficient distribution (KD) iodine in second titration based on the calculation is

3.

In third titration, volume of Na2S2O3 needed when solution change from purpleish

black to colorless is 3,5 mL. With known volume of Na2S2O3 in titrations, we can

calculate coefficient distribution iodine, the calculation below:


( )

( ) ( )

( )

( )

[ ]( )

[ ]( )

[ ]( )

[ ]( )

Coefficient distribution (KD) iodine in third titration based on the calculation is 3,14.

Based on the calculation that have been gained above the value of KD on each titration

the value of KD iodine average is 3.

X. DISCUSSION

From the results that we obtained, the value of coefficient distribution of iodine equal

to 3 it is deviated from the theory. While value of coefficient distribution of iodine from

theory is 10. This may due to some of the errors or mistakes have been occurred during the

experiment. First and foremost is, iodine solution that we use in experiment is maybe has

contaminant so it can affect the end point of titration, so the volume of Na2S2O3 that needed

for determine mole of I2 is decrease and make whole calculation to determine coefficient

distribution of iodine became not closely to coefficient distribution of iodine in theory.

Second, is mistake occur when we do the titration, the reading of meniscus in burrette that

contain Na2S2O3 that needed to achieve end point in titration is deviated, so the volume that

needed may not be accurate. Beside it, human error may occur when measure the volume of

materials that used in experiment, so the volume that needed may not be accurate and all this

careless mistake may contribute to the inaccuracy of the result of value of coefficient

distribution of iodine.


XI. CONCLUSION

From this experiment we conclude that; first Iodine can extract in organic

solvent, namely chloroform, that proven by two layer of organic layer(bottom) and

water layer (upper)that formed. Second, the coefficient distribution (KD) Iodine is 3.

XII. QUESTION AND ANSWER

A. Question

1. What is the difference between KD and D ?

2. What if value of KD is same with D?

3. How to find the value of the relationship between KD and D for a

weak acid HB ? A weak acid HB that experienced dimerization in an organic

solvents ?

4. How to find the value of the relationship between KD and D for a

weak base that ionize in water solvent and not react in organic solvent?

5. Prove that with multiple extraction will have increases percent result,

than just one extraction?

Answer

1. KD is the distribution coefficient or partition coefficient a special type

of equilibrium constant which is related to the relative solubilities of the solute

in the two solvents. Often one solvent is water and the other is an organic

solvent. D is the distribution ratio, Distribution ratio(D) to account for the total

concentration of species in two phases.

The difference between D and KD is we use D when we calculate the total

concentration of solute on organic solvent and water solvent, but in KD we just

use the ratio concentration of solute into organic solvent and water solvent.

2. Its mean that the only reaction affecting extraction efficiency is the

solutes partitioning between the two phases.

We can assume that the solute is initially present in the aqueous phase and that

we are extracting it into the organic phase. A conservation of mass requires

that the moles of solute initially present in the aqueous phase equal the


combined moles of solute in the aqueous phase and the organic phase after the

extraction.

3. The equilibrium reactions affecting the extraction of the weak acid,

HA, by an organic phase in which ionic species are not soluble. In this case

the partition coefficient and the distribution ratio are

Although the weak acid is soluble in both phases, its conjugate weak base, A,

is soluble only in the aqueous phase. The Ka reaction, which is called a

secondary equilibrium reaction, affects the extraction efficiency because it

controls the relative abundance of HA in solution.

4. Because the weak base exists in two forms, only one of which extracts

into the organic phase, the partition coefficient, KD, and the distribution ratio,

D, are not identical.

Using the Kb expression for the weak base,

For the concentration of HB+ and substitute back into the equation for D,

obtaining


5. Multiple extraction is always more efficient than simple one simple

extraction, as explained when a solute is being extracted from an aqueous

phase using an organic solvent, a better recovery will be obtained by using two

equal volumes of solvent than the recovery that would be obtained using all

the solvent in one large volume. This is all related to partition coefficients.

The solubility of a material in two phases. If the material has a partition

coefficient lets say of 4:1 . A single extraction will yield a ratio of 4:1,

however if you do multiple extractions with a lower volume, each one will

yield 4:1. So in total you will achieve a greater yield with multiple extractions,

using smaller quantities.


REFERENCES

Franek, Joseph.2000. Extractions.(online)

(https://www.chem.umn.edu/services/lecturedemo/info/Extraction.html ) (access in

13/5/2015 1:55AM)

Harper college tutor.(noyear). Separatory Funnel. (online)

(http://www.harpercollege.edu/tm-

ps/chm/100/dgodambe/thedisk/labtech/sepfun2.htm ) (access in 13/5/2015 2:10 AM)

Harvey ,L David.2011. LiquidLiquid Extractions.(online)

(http://chemwiki.ucdavis.edu/Analytical_Chemistry/Analytical_Chemistry_2.0/07%3

A_Collecting_and_Preparing_Samples/7G%3A_LiquidLiquid_Extractions ) (access

in 13/5/2015 1:50AM)

Matthew. 2011. Extraction and Determination of a Distribution Coefficient

(Kd) . (online)

(http://www.mendelset.com/articles/685/extraction_and_determination_distribution_c

oefficient_kd ) (access in 13/5/2015 2:10 AM)

Pahlavan. (noyear) . Determination of Distribution Coefficient. (online)

(http://swc2.hccs.edu/pahlavan/2423L6.pdf ) (access in 13/5/2015 2:13 AM)

TimPenyusun .2015. Panduan Praktikum Kimia Analitik II: DDPK. Surabaya:

Jurusan Kimia FMIPA Unesa

Underwood,A.L dan Day.R.A.Jr.2002. Analisis Kimia Kuantitatif Edisi

Keenam. Jakarta: Erlangga

Surabaya, May 18th

, 2015

Mengetahui,

Dosen/Asisten/Pembimbing, Praktikan,

(.) ()


ATTACHMENT

Iodium solution 0,1M

has yellowish brown

color

Making 100mL iodium

solution, the color was

yellowish brown

Put the 10 mL of iodium

solution into separated

funnel

Chloroform solution was

colorless

The solution formed two

layers after added by

chloroform

Water Phase has

orangish brown Color

of solution

Organic Phase has

purple Color of

solution


Shaking the solution to separate the

organic phase and water phase

Turn the tap to separate

the organic phase

The organic phase was

purple solution

The orangish brown

water phase in separated

funnel

Shaking again to separated organic

phase with water phase certainly

Separated again The water phase that

left in separated funnel

Pouring to Erlenmeyer

flask, repeat until 3 times


Water phase was adding

with H2SO4

Water phase was adding

with Starch solution

The color of solution

become blackish purple

Na2S2O3(aq) 0,01M,

colorless solution

Titrate with Na2S2O3(aq)

0,01M

Initial scale

was 0 mL

Final scale

was 3,7 mL

The solution after titrate

become colorless solution

Second water phase was

added with H2SO4 too.

And adding with Starch

solution



0,01M


become colorless

solution

Initial scale

was 5 mL

Final scale

was 8,8mL

Initial scale

was 9 mL

Final scale was

12,5 mL

Third water phase was added

with H2SO4 and Starch solution


0,01M

Three of Erlenmeyer which is after titrated


Iodium solution that added

with H2SO4 and Starch

solution

The color of solution

become blackish purple


0,01M

Initial scale was

13 mL

Final scale was 22

mL


become colorless

solution


CALCULATION

Reaction :

I2+ 2e - 2I- x 2

2S2O32-

S4O6+ 4e-

x 1

2I2 + 2S2O32-

4I- + S4O6

- Comparison solution / I2 standard

N Na2S2O3 = 0,01 N

V Na2S2O3 = 9 mL

mmolek I2 = mmolek S2O32-

mmolek I2 = VS2O32-

x N S2O32-

mmolek I2 = 9 mL x 0,01 N

I2 = 0,09 mmolek x

I2 standard = 0,045 mmol

- First titration

VNa2SO3 = 3,7 mL


mmolek I2 = V S2O3-x Na2SO3

2-

mmolek I2 = 3,7 mmolek x 0,01 N

I2 = 3,7 mL x

I2 (a ) = 0,0185 mmol

Mmol I2(0) = mmol I2 standard mmol I2(a)

= 0,045 0,0185

= 0,0265 mmol

[I2](a) =

= 0,00185

[I2](0) =

= 0,0530

KD = [ ]( )

[ ]( )


- Second Titration

VNa2SO3 = 3,8 mL



2-

mmolek I2 = 3,8 mL x 0,01 N

= 3,8 mL x

I2 (a ) = 0,0180 mmol


= 0,045 0,0180

= 0,027 mmol

[I2](a) =

= 0,00180

[I2](0) =

= 0,0054

KD = [ ]( )

[ ]( )

- Third Titration

VNa2SO3 = 3,5 mL



2-

mmolek I2 = 3,5 mmolek x 0,01 N

I2 = 0,035 mmolek x

I2 (a ) = 0,0175 mmol


= 0,045 0,0175

= 0,0275 mmol

[I2](a) =

= 0,00175

[I2](0) =

= 0,0055

KD = [ ]( )

[ ]( )

KD average =

= 3

coef distribution iod

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