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TITLE: Equilibrium distribution of solute between immiscible solvents. AIM: To study the equilibrium distribution of a solute between two immiscible solvents. THEORY OF THE EXPERIMENT A solution containing solvent A is well mixed with a second solvent, which A is soluble but does not form homogeneous mixture when mixed with solvent A. at equilibrium, some of the solute is contained in each of the non-homogeneous mixture producing solvents and the solute concentration in each of the two solutions are related by heterogeneous equilibrium distribution constant (KD).

At constant temperature and pressure, the change in state for the distribution of a solute between two non-homogeneous mixture producing solvents is written as: A(phase 1) = A(phase 2) The equilibrium constant for this change in state is given by: K = C2/C1 ; where C is the concentration of the solute molecules distributed between two solvents, and K is the equilibrium distribution constant that changes its value as temperature changes.

The distribution constant(partition ratio KD) is the equilibrium constant for the distribution of an analyte in two non-homogeneous mixture producing solvents. In chromatography, for a particular solvent, it is equal to the ratio of its molar concentration in the stationary phase to its molar concentration in the mobile phase, also approximating the ratio of the solubility of the solvent in each phase. PROCEDURE Benzene water system 50 ml of 0.50 mol.dm-3, 0.25 mol.dm-3, 0.125 mol.dm-3 and 0.0625 mol.dm-3 of acetic acid were measured into separate separating funnels. Then 50 ml of benzene was added to each funnel. Separating funnels were stopped and shake well for about 20 seconds. Opened the tap to relieve the pressure inside and was allowed to separate for 5 minutes. This process was repeated three time it was left for 20 minutes for equilibration.

After equilibrium was established, two portions of the benzene were withdrawn for titration in each of the funnels. Two 10 ml portions of extracted aqueous layer

(benzene) from the funnel with 0.50 ml concentration of acetic acid was titrated followed six 20 ml portions of benzene from the other funnels were withdrawn and titrated. This was followed by the titration of the remaining Organic layer(acetic acid).

Two mixtures of 75 ml pure acetic acid and benzene were prepared added into two separate separating funnel. One mixture was washed with 25 ml of water and the other with 75 ml of water, 25 ml of water was added into this funnel and was shaked, relieved the pressure and allowed the layers to separate. This process was repeated three times. The procedure was performed to the other separating funnel that washed with 75 ml of water. The aqueous layer was extracted and titrated in three portions for both funnels. Phenolphthalein indicator was used in all titrations. Results: Table 1: Titration of two portion of Benzene(aqueous layer) in each funnel, two 10 ml portions for the Benzene dissolved with 0.50 M Acetic Acid and two 20 ml portions for the Benzene dissolved with 0.25 M, 0.125 M and 0.0626 M Acetic acid making six portions of 20 ml Benzene. Concentrati on of Organic Layers ( mol.dm-3) 0.50 0.125 0.25 0.0625

Volume of Initial Final Volume the solvent Volume of Volume of Change (aqueous Titration Titration Layer). (ml) (ml) (ml) 10 ml 0 91.2 91.2 10 ml 0 56.2 56.2 20 ml 0 1.2 1.2 20 ml 0 2.9 2.9 20 ml 0 13.5 13.5 20 ml 13.5 28.4 14.9 20 ml 0 6.4 6.4 20 ml 6.4 12.8 6.4

Colour Change.

Light Pink Dark Pink Light Pink Light Pink Dark Pink Dark Pink Dark Pink Light Pink

Table 2: Titration of the remaining Organic layers from separate separating funnels Concentration Initial Volume Final Volume Volume Colour of Organic of Titration of Titration Change Change. Layers (ml) (ml) (ml) ( mol.dm-3) 0.50 0 5.9 5.9 Dark Pink micelles-like at the bottom and colourless solvent 0.125 0 0.5 0.5 Dark Pink 0.25 0 3.6 3.6 Dark Pink micelles-like at the bottom and colourless solvent 0.0625 0 1.5 1.5 Dark Pink

Table 3: Titration of pure acetic acid acid mixed with benzene Volume Final Mixtures of Portions of Initial Volume of Volume of Change acetic acid Layers (ml) Titration Titration and benzene (ml) (ml) 75 ml 20 ml 0 1.0 1.0 mixture 20 ml 1.0 2.5 1.5 washed with 25 ml 20 ml 2.5 5.2 2.7 of water. 75 ml 20 ml 0 12.3 12.3 mixture 20 ml 12.3 24.1 11.8 washed with 75 ml 20 ml 0 14.5 14.5 of water.

Colour Change.

Light Pink Light Pink Dark Pink Dark Pink Dark Pink Light Pink

DISCUSSION For the titration of two portions of Benzene(aqueous layer), the two 10 ml portions for the Benzene dissolved with 0.50 M Acetic Acid, the results showed very large values of change in volume or volume used to reach the end point of that titration, this is because benzene was affected by the concentration of the organic layer(acetic acid) and it was made concentrated. From the results of the first titration one could observe that the more concentrated the acetic acid which the portions were extracted from, the higher the volume is being used to titrate the portions(aqueous layer) and vice versa.

In the second titration (Table 2),the titration of different concentrations of organic layer (Acetic acid). The results also shows that the volume used to neutralize the analyte is widely dependent on the concentration of organic layer portions as discussed above. The micelles-like molecules were the traces of benzene left on the organic layer.

For the last titration of pure acetic acid acid mixed with benzene (Table 3).No difference was expected in end-point volume of the three portions for both the three portions of mixture washed with 75 ml of water and three portions of mixture washed with 25 ml of water, this is because all three portions contains the same volume and same chemical properties. But the difference in end-point volume between washed with 75 ml water and 25 ml water portions are expected to be different. In simple terms all three portions of 75 ml washed with water solution should be having same end-point but different from those of 25 ml washed with water solution. All color changes for this experiment were suppose to be light pink

CONCLUSION From this experiment one could have observed that the end-point is dependent on the concentration of the analyte as the titrant concentration was held unchanged, also depends on the volume of the analyte and the type of indicator that is being used. Errors that occurred in color change could be due to misusing of the indicator and over titration.

REFERENCE https://www.scribd.com/doc/54974436/Phase-Diagram-of-a-Three-ComponentLiquid-System https://www.scribd.com/document/373940936/Distribution-of-a-SoluteBetween-Immiscible-Solvents Rice, N. M.; Irving, H. M. N. H.; Leonard, M. A. (1993), "Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993)", Pure and Applied Chemistry, https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Map %3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/UNIT_4%3A_EQUILI BRIUM_IN_CHEMICAL_REACTIONS/14%3A_Chemical_Equilibrium/14.8%3 A_Distribution_of_a_Single_Species_between_Immiscible_Phases %3A_Extraction_and_Separation_Processes

1. (a) density = 1.64 g/cm3 mass = 0.0124 g Molar mass = 204.22 g/mol

n

=

m M

0.0124 g 204.22 g/mol = 5 x 10-4 mol =

m V V = m/d =0.1024 g/1.64g.cm-3 =6.24 x 10-2 cm3

d=

n V =5 x 10-4 mol/ 6.24 x 10-2 cm3 The concentration of KHP is 8.01 x 10-3 mol.cm-3

C=

(b) NaOH mass= 10.003 g Molar mass= 40 g/mol density= 2.13 g/cm3 m V m V= d d=

= 10.003 g/ 2.13 g/cm3 =4.7 cm3

C=

1 n m = x V V M =10.003g /40.01 g/mol x 1/4.7 cm3

The concentration of 0.01mol dm-3 = 5.32 x 10-2 mol/cm3 (iii) NaOH m= 1.001 g M=40g/mol d=2.13 g/cm3

d=

m V

v=

m d

c=

n V

1.001 g 2.13 g /cm = 0.46 cm3 =

= m/M x 1/V

=1.001 g/40 g/mol x 1/0.46 cm3 = 5.44 x10 -2 mol/cm3 (c) (i)ACETIC ACID m= ? M= 60.25 g/mol d= 1.05 g/cm3 v= 125 cm3=0.5 mol /dm3 m V m=d Xv =1.05 g/cm3 x 125 cm3 =131.25 g d=

c=

n V

= m/M x 1/V =131.25 g/60.05 g/mol x 1/125cm3 =1.75 x 10-2 mol/cm3

(ii) ACETIC ACID M= 60.20 g/mol d= 1.05 g/cm3 v= 62.5 ml = 0.25 mol/dm3 m=? c=? m d= V m=d x v =1.05 g/cm3 x 62.5 cm3 =65.63 g c=

n V

= m/M x 1/V =131.25 g/60.05 g/mol x 1/125cm3 =1.75 x 10-2 mol/cm3

SURNAME INITIALS STUDENT# COURSE CODE PRACTICAL# TITLE distribution of between immiscible solvents. DUE DATE

:CWAYI :H.Q :201614438 :PAC 224 :01 :Equilibrium solute

:22/08/2018...