A LABORATORY WORK THAT WILL HELP ANYONE TO DETERMINE EQUILIBRIUM CONSTANT OF TRI-IODIDE ION. THE EQUILIBRIUM CONSTANT IS NOT DIRECTLY DETERMINED, THIS PIECE WILL GIVE A WAY TO HOW IT IS DONE.

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AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY, KUMASI
KNUST
COLLEGE OF SCIENCE
DEPARTMENT OF CHEMISTRY
ANALYTICAL LABORATORY REPORT
TITLE: DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION (I3-)
NAME: AUGUSTINE ADU
DATE: 23RD SEPTEMBER, 2014
E-mail :adu.aug@gmail.com

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AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION (I3-)
AIMS AND OBJECTIVES:
 To determine the equilibrium constant for the formation of tri-iodide ion
 To determine the partition coefficient of iodine between water and N- Hexane
 To demonstrate solvent extraction
INTRODUCTION
Solvent extraction is a very common laboratory separation procedure used when isolating or purifying a product. It is extremely useful for very rapid and clean separation of both organic and inorganic substances. It involves the distribution of a solute between two immiscible liquid phases. A separation funnel is used for this process. The density of the phases must differ appreciably so that they can separate easily after mixing. The organic solvent used for extraction must meet the following criteria:
 Should readily dissolve substances to be extracted.
 Should not react with the substance to be extracted.
 Should not react with or be miscible with water.
 Should have a low boiling point so it can be easily evaporated.
 Should be less viscous
Elemental iodine dissolves easily in carbon tetrachloride but its solubility in water can vastly be increased by the addition of potassium iodide. The molecular iodine reacts reversibly with the negative ion, thus creating the triiodide anion, I3−, which dissolves well in water. This is also the formulation of some types of medicinal (antiseptic) iodine, although tincture of iodine classically dissolves the element in alcohol. The deep blue colour of starch-iodine complexes is produced only by the free element.
In a chemical reaction, the whole of the reactants do not get converted into the products. After some time, there will come a point when a fixed amount of reactants will exist in harmony with a fixed amount of products, the amounts of neither changing anymore. This is called chemical equilibrium. The equilibrium constant for a reaction is indeed a constant, independent of the activities of the various species involved, though it does depend on temperature as observed by the Van’t Hoff equation. Adding a catalyst will affect both the forward reaction and the reverse reaction in the same way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached. In the fields of organic and medicinal chemistry, a partition or distribution coefficient (KD) is the ratio of concentrations of a compound in the two phases of a mixture of two immiscible solvents at equilibrium. It is concentration equilibrium constant, representing the equilibrium constant for the process of a compound, which begins in the mobile phase and is dissolved into the solution (stationary) phase. The distribution equilibrium in the simplest case involves the same molecular species in each phase. A(aq) = A(org)
Hence these coefficients are a measure of differential solubility of the compound between these two solvents. Normally one of the solvents chosen is water while the second is hydrophobic (an organic solvent). Hence both the partition and distribution coefficient are measures of how hydrophilic or hydrophobic a chemical substance is. Partition coefficients are useful in estimating distribution of drugs within the body.
Iodine (I2) is sparingly soluble in pure water but dissolves it readily in aqueous solutions of potassium iodide (KI) owing to the formation of complex tri-iodide (I3 -) ions which exist in equilibrium with free iodine molecules and iodide ions.
I2 + I- ⇌ I3 –

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AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
A complex ion, the tri-iodide (I3−) ion, is formed from the reaction between I− and I2. In the two-phase system two equilibria occur simultaneously, each competing for I2
The concentration of all species at equilibrium can be determined titrimetrically, and the equilibrium constant for the formation of the complex ion can be calculated. Titration of the aqueous layer gives the total iodine present in the aqueous layer, that is,
[I2(total)] = [I2(aq)] + [I3−(aq)]
but [I2(aq)] = [I2(organic)] / K′
[I3−(aq)] = [I2(total)] – [I2(aq)]
[I−(aq)] = [I−(aq)] initial – [I3−(aq)]
The equilibrium constant (K) for the formation of the complex ion is then calculated with Eq (1). Actually the equilibrium constant should be written in terms of the activities and the activity coefficients.
[I (aq)] 3−γI−
K = [I (aq)] [I (aq)] 2−γ γ3 I2 I− Equ 2
However, I− and I3−, with the same charge and in the same ionic environment, have very nearly the same activity coefficient, which then tend to cancel out in Eq (2). Molecular iodine (I2) in both water and dichloromethane behaves quite ideally, so its activity coefficient is very nearly unity.
NOTE: N-HEXANE WAS USED INSTEAD OF CARBON TETRECHLORIDE.
CHEMICALS AND APPARATUS
i. Beakers
ii. Separating funnel
iii. Solid iodine
iv. Clamp
v. 0.1M potassium iodide
vi. 0.05M sodium thiosulphate
vii. Starch indicator
viii. Distilled water
ix. Wash bottle
x. Hexane (organic solvent)
xi. Burette
xii. Volumetric flask

4. pg. 4
AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
PROCEDURE
Weigh accurately 0.5g of iodine. Add it to 50cm3 of n-hexane in a separating funnel. 50cm3 of water is added to the separating funnel and well shaken. Leave it to stand for some time for the phase to separate.
Run off the lower water layer into a container and the hexane layer is also run into a different container.
Take 25cm3 of the hexane layer into a conical flask. Add 25cm3 of 0.1M potassium iodide (KI) solution and well shaken.
Titrate the mixture against 0.05M sodium thiosulphate solution using starch indicator.
Towards the end point of the titration the flask stoppered and well shaken after each addition of sodium thiosulphate solution. The end point is reach when the n-hexane layer just becomes colourless.
Another 0.5g iodine is weighed and added to 50cm3 of n-hexane in a separating funnel.
Add 50cm3 of 0.10M potassium iodide (KI) solution and shake well. Leave it to stand for some time to allow the phases to separate.
Run off the lower KI layer into a container and then the n-hexane into another container.
25cm3 of the hexane layer into a conical flask and add 25cm3 of 0.10M KI solution.
Titrate as before against 0.05M sodium thiosulphate solution.
EXPERIMENTAL DATA
1. Water + hexane against sodium thiosulphate
Burette readinga / cm3
I
Final rading
33.40
Initial reading
0.00
Titre value
33.40
2. KI + hexane against sodium thiosulphate
Burette readinga / cm3
I
Final rading
14.80
Initial reading
0.00
Titre value
14.80

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AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
The partition coefficient between water and hexane
Iodine reacts with thiosulphate in the ff. Way
I3- + 2 S2O32- ⇌ 3 I- + S4O62-
I2 + 2 S2O32- ⇌ 2 I- + S4O62-
I2 + I- ⇌ I3-
Mass of iodine = 0.5g, molar mass of iodine = 126.9 x 2 = 253.8g/mol
∴푖푛푖푡푖푎푙 푚표푙푒푠 표푓 퐼 2 is given by 0.5푔 253.8푔/푚표푙 = 1.97x10-3moles
푛[푆2푂32−] 푛[퐼2] = 21 ⟹ 푛[퐼2]= 12 푛[푆2푂32−]
푛[푆2푂32−]=퐶표푛푐[푆2푂32−]×푉표푙[푆2푂32−] =0.05푀×0.0334푑푚3=1.67×10−3푚표푙
푛[퐼2]= 12× 1.67×10−3푚
Number of moles of iodine in organic phase =8.35×10−4푚표푙
Concentration of iodine in organic phase = =8.35×10−4푚표푙 0.05푑푚3 = 0.0167 mol/dm3
Number of moles in aqueous phase = initial moles of iodine – moles of iodine in organic phase
=1.97x10-3moles - 8.35×10−4푚표푙푒푠 = 1.135 푥 10−3moles
Concentration in aqueous phase = 1.135 푥 10−3 moles 0.05푑푚3 = 0.0227M
푝푎푟푡푖푡푖표푛 푐표푒푓푓푖푐푖푒푛푡,퐾퐷 = 푐표푛푐 푖푛 표푟푔푎푛푖푐 푝ℎ푎푠푒 푐표푛푐 푖푛 푎푞. 푝ℎ푎푠푒 = 0.0167 푀 0.0227푀 = 0.736
The formation constant for the tri-iodide ion formation
I2 + I- ⇌ I3-............... (I)
The formation constant is given by Ka
Ka of I3- = [퐼3−] [퐼2][퐼−]
n(I2) = n(I3-)
m(I2) = 0.5g, M(I2) = 126.9 x 2 = 253.8gmol-1
∴푛(퐼2)= 0.5푔 253.8푔푚표푙−1=1.97×10−3
At equilibrium, n(I3-) produced is the amount of I3- that reacted with the thiosulphate.
I3- + 2 S2O32- ⇌ 3 I- + S4O62- 푛(퐼3−)= 12 푛(푆2푂3 2−)= 12 0.05 ×14.8×10−3=3.7×10−4푚표푙푒푠 ∴푛(퐼3−) 푎푡 푒푞푢푖푙푖푏푟푖푢푚=3.7×10−4푚표푙푒푠
[퐼3 –]푎푡 푒푞푢푖푙푖푏푟푖푢푚 = 3.7×10−4푚표푙푒푠 50×10−3=7.4×10−3푀
Initial conc. of I- is 0.1M
Mol of 퐼−= conc ×푣표푙=0.1 ×0.05
= 5.0×10-3푚표푙

6. pg. 6
AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
푛(퐼−) 푎푡 푒푞푢푖푙푖푏푟푖푢푚= 5×10−3푚표푙푒푠−3.7×10−4푚표푙푒푠
=4.63 ×10−3푚표푙푒푠 [퐼−]푎푡 푒푞푢푖푙푖푏푟푖푢푚= 푛 푎푡 푒푞푢. 푣 푎푡 푒푞푢. = 4.63 ×10−350×10−3
=0.0926푀 푛(퐼2) 푎푡 푒푞푢푖푙푖푏푟푖푢푚=1.97×10−3푚표푙푒푠−3.7×10−4푚표푙푒푠
=1.6×10−3푚표푙푒푠 [퐼2]푎푡 푒푞푢푖푙푖푏푟푖푢푚= 푛 푎푡 푒푞푢. 푣 푎푡 푒푞푢. = 1.6×10−450×10−3
=0.0320푀
퐾= [퐼3−] [퐼2][퐼−] = 7.4×10−3(0.0320)(0.0926) =2.50
DISCUSSION.
The reaction between the thiosulphate and the iodine was a redox reaction with the I2 been reduced and the thiosulphate being oxidized. In the dissolution of the solid iodine in the hexane, it was noticed that the iodine did not completely dissolve but there was some remains in the solid state. The concentration of the iodine in the organic phase was found to be smaller than that in the aqueous phase.
The partition coefficient which was observed to be 0.736, which clearly shows that the I2 molecules clearly tend to quantitative distribution in the hexane; this implies that a large amount of the solid iodine was extracted into the organic phase, the hexane.
In the second test, the iodide was reacted directly with the solid iodine in the hexane phase. This was done in order to bring about extraction of most of the solid iodine into the KI solution phase. The hexane layer that was drawn from the beaker was found to contain just some small amount of the iodine. This was observed from the small volume of the thiosulphate that was obtained in the titration. This implies that the amount of the iodine that reacted with the second addition of the potassium iodide was very small showing that most of the initial iodine present was extracted into the KI solution phase. Hence, the total tri-iodide ion concentration was the addition of the iodide concentration in the KI solution and the one that reacted with the thiosulphate solution.
PRECAUTIONS
 There was vigorous shaking during the separation analysis in order to ensure higher extraction.
 The separating solutions were allowed to stand without any interference to ensure higher efficiency.
 During the draining, the funnel was opened bit by bit in order to ensure that pressure did not build up in the funnel.
 In the second test, the titration was done in drop wise in order to avoid exceeding the endpoint since just a small amount of the iodine was left in the hexane phase.

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AUGUSTINE ADU-KNUST
E-mail:adu.aug@gmail.com
 A little of the aqueous layer was left at the tip of the funnel in order to prevent some of the organic phase from entering the aqueous phase that was collected.
ERROR ANALYSIS:
 Decomposition of Iodine molecule
 Indeterminate error
 PH and temperature as a factor
CONCLUSION
The partition coefficient was 0.736
The formation constant for the tri-iodide ion was found to be 2.50
REFERENCES
1. S. P. Mushtakova, G. V. Gerasimova, and T. M. Varlamova “Distribution of Iodine in the
Systems Iodine–Water–Chloroform (Dichloromethane) and Iodine–Potassium Iodide– Water–Chloroform (Dichloromethane)”, Journal of Analytical Chemistry 2009, 64, 125128.
2. General Chemistry Laboratories, University of Alberta; http://www.chem.ualberta.ca/~ngee/Expt.N_05.html
3. G. Jones, and B.B. Kaplan, “THE IODIDE, IODINE, TRI-IODIDE EQUILIBRIUM AND THE FREE ENERGY OF FORMATION OF SILVER IODIDE”, J. Amer. Chem.,
1928, 50, 1845-1864.
4. H.M. Dawson, “On the nature of polyiodides and their dissociation in aqueous solution”, J. Chem. Soc. 1901, 79, 238-247.
5. http://en.wikipedia.org/wiki/Chemical_equilibrium
6. http://www.answers.com/topic/partition-coefficient